[{"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, we were able to show that any lambda that satisfies this equation for some non-zero vectors v, then the determinant of lambda times the identity matrix minus a must be equal to 0. Or we could rewrite this as saying lambda is an eigenvalue of a if and only if. I'll write it as if. If and only if the determinant of lambda times the identity matrix minus a is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2. Let's do an R2."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "If and only if the determinant of lambda times the identity matrix minus a is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2. Let's do an R2. Let's say that a is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of a. So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's do an R2. Let's say that a is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of a. So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1 minus a, 1, 2, 4, 3, is going to be equal to 0. Well, what does this equal to? This right here is the determinant of lambda times this is just lambda times all of these terms."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1 minus a, 1, 2, 4, 3, is going to be equal to 0. Well, what does this equal to? This right here is the determinant of lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda. Lambda times 0 is 0. Lambda times 0 is 0."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "This right here is the determinant of lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda. Lambda times 0 is 0. Lambda times 0 is 0. Lambda times 1 is lambda. And from that we'll subtract a. So you get 1, 2, 4, 3."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Lambda times 0 is 0. Lambda times 1 is lambda. And from that we'll subtract a. So you get 1, 2, 4, 3. And this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of, this first term is going to be lambda minus 1."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So you get 1, 2, 4, 3. And this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of, this first term is going to be lambda minus 1. The second term is 0 minus 2. So it's just minus 2. The third term is 0 minus 4."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is the determinant of, this first term is going to be lambda minus 1. The second term is 0 minus 2. So it's just minus 2. The third term is 0 minus 4. So it's just minus 4. And then the fourth term is lambda minus 3. Lambda minus 3, just like that."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "The third term is 0 minus 4. So it's just minus 4. And then the fourth term is lambda minus 3. Lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well, everything became a negative. We negated everything."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well, everything became a negative. We negated everything. And the terms around the diagonal, we had a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix?"}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "We negated everything. And the terms around the diagonal, we had a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well, the determinant of this is just this times that minus this times that. So it's lambda minus 1 times lambda minus 3 minus these two guys multiplied by each other. So minus 2 times minus 4 is plus 8."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So what's the determinant of this 2 by 2 matrix? Well, the determinant of this is just this times that minus this times that. So it's lambda minus 1 times lambda minus 3 minus these two guys multiplied by each other. So minus 2 times minus 4 is plus 8. Minus 8. This is the determinant of this matrix right here, or this matrix right here, which simplify to that matrix. And that has got to be equal to 0."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So minus 2 times minus 4 is plus 8. Minus 8. This is the determinant of this matrix right here, or this matrix right here, which simplify to that matrix. And that has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible. And its determinant has to be equal to 0."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And that has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible. And its determinant has to be equal to 0. So now we have an interesting polynomial equation right here. We can multiply it out. We get what?"}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And its determinant has to be equal to 0. So now we have an interesting polynomial equation right here. We can multiply it out. We get what? Let's multiply it out. We get lambda squared minus 3 lambda plus 3 minus 8 is equal to 0, or lambda squared minus 4 lambda minus 5 is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "We get what? Let's multiply it out. We get lambda squared minus 3 lambda plus 3 minus 8 is equal to 0, or lambda squared minus 4 lambda minus 5 is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology. Polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology. Polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product as minus 5."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product as minus 5. When you add them, you get minus 4. It's minus 5 and plus 1. So you get lambda minus 5 times lambda plus 1 is equal to 0, right?"}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, two numbers and you take the product as minus 5. When you add them, you get minus 4. It's minus 5 and plus 1. So you get lambda minus 5 times lambda plus 1 is equal to 0, right? Minus 5 times 1 is minus 5. And then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So you get lambda minus 5 times lambda plus 1 is equal to 0, right? Minus 5 times 1 is minus 5. And then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors."}, {"video_title": "Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors. We know that this equation can be satisfied with lambdas equaling 5 or minus 1. So we know the eigenvalues, but we're yet to determine the actual eigenvectors. So that's what we're going to do in the next video."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So in the last video, I said, look, in standard coordinates, if you have some vector x in your domain and you apply some transformation, then let's say that A is the transformation matrix with respect to the standard basis, then you're just going to have this mapping. You take x, you multiply it by A, you're going to get the transformation of x. Now in the last video and a couple of videos before that, or actually the one right before that, we said, well look, you can do the same mapping but just in an alternate coordinate system. You could do it with respect to some coordinate system or in some coordinate system with respect to some basis B. And that should be the same thing. It should just be a different transformation matrix. And in the last video, we actually figured out what that different transformation matrix is."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "You could do it with respect to some coordinate system or in some coordinate system with respect to some basis B. And that should be the same thing. It should just be a different transformation matrix. And in the last video, we actually figured out what that different transformation matrix is. We had a change of basis. So let's say we had this basis right here. Let me actually copy and paste everything so that we understand what we did."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And in the last video, we actually figured out what that different transformation matrix is. We had a change of basis. So let's say we had this basis right here. Let me actually copy and paste everything so that we understand what we did. So this was the example. Let me copy it, paste it up here, put all of our takeaways from the last video up here. Let me paste it right here."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Let me actually copy and paste everything so that we understand what we did. So this was the example. Let me copy it, paste it up here, put all of our takeaways from the last video up here. Let me paste it right here. So in the last video, we said, OK, this is my basis right there, and then we said, let me copy and paste. Let me copy and paste. That was my alternate basis."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Let me paste it right here. So in the last video, we said, OK, this is my basis right there, and then we said, let me copy and paste. Let me copy and paste. That was my alternate basis. And then I have my change of basis matrix and its inverse. Those will be useful to deal with. So let me copy and paste that."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "That was my alternate basis. And then I have my change of basis matrix and its inverse. Those will be useful to deal with. So let me copy and paste that. OK, copy. And then I'm going to paste it. Edit, paste."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So let me copy and paste that. OK, copy. And then I'm going to paste it. Edit, paste. Maybe I'll just write it over there. Not maybe the best order. Maybe I should have written that first, but I think we get the idea."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Edit, paste. Maybe I'll just write it over there. Not maybe the best order. Maybe I should have written that first, but I think we get the idea. And then we want to figure out, we want to write what our transformation matrix is with respect to the standard basis. And I wrote that right over here. This was all from the last problem, if you're wondering where I got all this stuff."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I should have written that first, but I think we get the idea. And then we want to figure out, we want to write what our transformation matrix is with respect to the standard basis. And I wrote that right over here. This was all from the last problem, if you're wondering where I got all this stuff. Let me copy and then paste that. Edit, paste. So I'll paste that."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "This was all from the last problem, if you're wondering where I got all this stuff. Let me copy and then paste that. Edit, paste. So I'll paste that. And then the whole point of the last video is we figured out what the transformation matrix is with respect to this basis, right here. So D, which was the big takeaway from the last video, was equal to this right here. Let me copy and paste that."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So I'll paste that. And then the whole point of the last video is we figured out what the transformation matrix is with respect to this basis, right here. So D, which was the big takeaway from the last video, was equal to this right here. Let me copy and paste that. And now we have all of our takeaways in one place. Edit, paste. What I want to do in this video is to verify that D actually works."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Let me copy and paste that. And now we have all of our takeaways in one place. Edit, paste. What I want to do in this video is to verify that D actually works. That I could start with some vector x that I can, let me write it up here. Let's take some example vector. So this transformation, its entire domain is R2."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "What I want to do in this video is to verify that D actually works. That I could start with some vector x that I can, let me write it up here. Let's take some example vector. So this transformation, its entire domain is R2. So let's start with some vector x. Let's say that x is equal to 1 minus 1. Now, we could just apply the transformation in the traditional way and get the transformation of x."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So this transformation, its entire domain is R2. So let's start with some vector x. Let's say that x is equal to 1 minus 1. Now, we could just apply the transformation in the traditional way and get the transformation of x. So let's just do that. The transformation of x is just this matrix times x. And so what is that going to equal?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Now, we could just apply the transformation in the traditional way and get the transformation of x. So let's just do that. The transformation of x is just this matrix times x. And so what is that going to equal? Let me see, maybe I can just do it right here in this corner to save space. So it's going to be this matrix times x. So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And so what is that going to equal? Let me see, maybe I can just do it right here in this corner to save space. So it's going to be this matrix times x. So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? Minus 2 times minus 1 is just 2. So it's going to be 3 plus 2. So it's going to be equal to 5."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? Minus 2 times minus 1 is just 2. So it's going to be 3 plus 2. So it's going to be equal to 5. And then the second term right here is going to be 2 times 1 plus minus 2 times minus 1. Well, that's just positive 2. So it's going to be 2 plus 2."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be equal to 5. And then the second term right here is going to be 2 times 1 plus minus 2 times minus 1. Well, that's just positive 2. So it's going to be 2 plus 2. So that is 4. So that's just the transformation of x. Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be 2 plus 2. So that is 4. So that's just the transformation of x. Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates? So what is that vector x represented in coordinates with respect to this basis right here? Well, you saw before, I wrote them out here. Maybe it'll be useful to do it right here."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates? So what is that vector x represented in coordinates with respect to this basis right here? Well, you saw before, I wrote them out here. Maybe it'll be useful to do it right here. The coordinate, I'll copy this. Actually, let me copy both of these. These will both be useful."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Maybe it'll be useful to do it right here. The coordinate, I'll copy this. Actually, let me copy both of these. These will both be useful. Edit, copy. As you can see, if you want to go from x to the x in an alternate basis, or the alternate coordinate representations of x, you essentially multiply x times c inverse. But that's why I'm copying and pasting it."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "These will both be useful. Edit, copy. As you can see, if you want to go from x to the x in an alternate basis, or the alternate coordinate representations of x, you essentially multiply x times c inverse. But that's why I'm copying and pasting it. Let me copy and then let me put it up here so that we can apply these. So then paste it right there. So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "But that's why I'm copying and pasting it. Let me copy and then let me put it up here so that we can apply these. So then paste it right there. So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse. c inverse is this thing right here. So if I take x and I multiply it times c inverse, I'll get this version of x. So let's do that."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse. c inverse is this thing right here. So if I take x and I multiply it times c inverse, I'll get this version of x. So let's do that. So this times that, let me just put the minus 1 third out front, so it's going to be equal to minus 1 third times, let's see if we can do this one in our head as well. So it's going to be 1 times 1 plus minus 2 times minus 1, which is just positive 2. So it's going to be 1 plus 2."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So this times that, let me just put the minus 1 third out front, so it's going to be equal to minus 1 third times, let's see if we can do this one in our head as well. So it's going to be 1 times 1 plus minus 2 times minus 1, which is just positive 2. So it's going to be 1 plus 2. So it's going to be equal to 3. And then it's minus 2 times 1, which is minus 2, plus 1 times minus 1, which is just minus 1. So it's minus 2 minus 1."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be 1 plus 2. So it's going to be equal to 3. And then it's minus 2 times 1, which is minus 2, plus 1 times minus 1, which is just minus 1. So it's minus 2 minus 1. It's minus 3. So if we have minus 1 third times this, the b coordinate representation of our vector x is going to be equal to minus 1 and then 1, just like that. Which is actually interesting for this example."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So it's minus 2 minus 1. It's minus 3. So if we have minus 1 third times this, the b coordinate representation of our vector x is going to be equal to minus 1 and then 1, just like that. Which is actually interesting for this example. It just kind of swapped the first entry and the second entry. Now let's see what happens when we apply d to x. So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Which is actually interesting for this example. It just kind of swapped the first entry and the second entry. Now let's see what happens when we apply d to x. So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates. So let's see what happens. So if we apply d to x, let me scroll over a little bit, just so that we get a little bit more real estate. So if we apply d to x, what do we get?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates. So let's see what happens. So if we apply d to x, let me scroll over a little bit, just so that we get a little bit more real estate. So if we apply d to x, what do we get? And so this is going to be the transformation, or this should be the transformation of x in b coordinates. So what is this going to be equal to? We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So if we apply d to x, what do we get? And so this is going to be the transformation, or this should be the transformation of x in b coordinates. So what is this going to be equal to? We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. So it's just minus 1 times minus 1, which is 1. And then we're going to get 0 times minus 1 plus 2 times 1. So 2 times 1 is just 2."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. So it's just minus 1 times minus 1, which is 1. And then we're going to get 0 times minus 1 plus 2 times 1. So 2 times 1 is just 2. Now, for everything to work together, and assuming I haven't made any careless mistakes, this thing, this vector right here, should be the same as this vector if I change my basis. So if I go from the standard basis to the basis b, and when you go in that direction, you just multiply this guy times c inverse. So I'm just using this formula right here."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So 2 times 1 is just 2. Now, for everything to work together, and assuming I haven't made any careless mistakes, this thing, this vector right here, should be the same as this vector if I change my basis. So if I go from the standard basis to the basis b, and when you go in that direction, you just multiply this guy times c inverse. So I'm just using this formula right here. If I'm in the standard basis, I multiply by c inverse, I'm going to get the b basis. So let's see what I get. So this guy, I'm going to multiply him times c inverse."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just using this formula right here. If I'm in the standard basis, I multiply by c inverse, I'm going to get the b basis. So let's see what I get. So this guy, I'm going to multiply him times c inverse. Let me do it up here just to get some extra space. Do it right here. So I'm going to multiply the vector 5, 4."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So this guy, I'm going to multiply him times c inverse. Let me do it up here just to get some extra space. Do it right here. So I'm going to multiply the vector 5, 4. I'm going to multiply that by c inverse. We're going to have minus 1 third times 1 minus 2 minus 2 1, just like that. So this is going to be equal to, I'll just write the minus 1 third out front."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to multiply the vector 5, 4. I'm going to multiply that by c inverse. We're going to have minus 1 third times 1 minus 2 minus 2 1, just like that. So this is going to be equal to, I'll just write the minus 1 third out front. And we have 1 times 5, which is 5. Let me just write this right. 5 plus minus 2 times 4."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to, I'll just write the minus 1 third out front. And we have 1 times 5, which is 5. Let me just write this right. 5 plus minus 2 times 4. So 5 minus 8. And then we have minus 2 times 5, which is minus 10. And then we have plus 1 times 4."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "5 plus minus 2 times 4. So 5 minus 8. And then we have minus 2 times 5, which is minus 10. And then we have plus 1 times 4. Minus 2 times 5, which is minus 10 plus 1 times 4. So this is equal to minus 1 third times minus 3. And this is what?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And then we have plus 1 times 4. Minus 2 times 5, which is minus 10 plus 1 times 4. So this is equal to minus 1 third times minus 3. And this is what? This is minus 6. If you multiply the minus 1 third times, all the negatives cancel out and you get 1 and 2. Which is exactly what we needed to get."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And this is what? This is minus 6. If you multiply the minus 1 third times, all the negatives cancel out and you get 1 and 2. Which is exactly what we needed to get. When you take this guy and you change its basis to basis b, or you change its coordinate system to the coordinate system with respect to b, you multiply it by c inverse, you get that right there. So this literally is the b coordinate representation of the transformation of x. We just did it by multiplying it by c inverse."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Which is exactly what we needed to get. When you take this guy and you change its basis to basis b, or you change its coordinate system to the coordinate system with respect to b, you multiply it by c inverse, you get that right there. So this literally is the b coordinate representation of the transformation of x. We just did it by multiplying it by c inverse. Which is exactly what we got when we took the b coordinate version of x and we applied that matrix that we found, that transformation matrix with respect to the b coordinates, and you multiply it times this guy right here, we got the same answer. So it didn't matter whether we went this way around this little cycle or we went this way, we got the same answer. This isn't a proof, but it shows us that what we did in the last video at least works for this case."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "We just did it by multiplying it by c inverse. Which is exactly what we got when we took the b coordinate version of x and we applied that matrix that we found, that transformation matrix with respect to the b coordinates, and you multiply it times this guy right here, we got the same answer. So it didn't matter whether we went this way around this little cycle or we went this way, we got the same answer. This isn't a proof, but it shows us that what we did in the last video at least works for this case. And I literally did pick a random x here, and you can verify it if you like for other things. Now, you should hopefully be reasonably convinced that we can do this, that you can change your basis and find a transformation matrix. We've shown how to do it."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "This isn't a proof, but it shows us that what we did in the last video at least works for this case. And I literally did pick a random x here, and you can verify it if you like for other things. Now, you should hopefully be reasonably convinced that we can do this, that you can change your basis and find a transformation matrix. We've shown how to do it. But the obvious question is, why do you do it? Why? Why do you do it?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "We've shown how to do it. But the obvious question is, why do you do it? Why? Why do you do it? And someone actually wrote a comment on the last video, which I think is actually, it kind of captures the art of why you do it. I'm not looking at the comment right now, but if I remember correctly, they said their linear algebra teacher said that linear algebra is the art of choosing the right basis. Let me write that down."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Why do you do it? And someone actually wrote a comment on the last video, which I think is actually, it kind of captures the art of why you do it. I'm not looking at the comment right now, but if I remember correctly, they said their linear algebra teacher said that linear algebra is the art of choosing the right basis. Let me write that down. The art of choosing the right basis. Or you could imagine the right coordinate system. And why is there a right coordinate system?"}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. The art of choosing the right basis. Or you could imagine the right coordinate system. And why is there a right coordinate system? Maybe I'll put little quotes inside the quotation. What does it mean to have the right coordinate system? Well, if you look at the original transformation matrix with respect to the standard basis, it's fine."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And why is there a right coordinate system? Maybe I'll put little quotes inside the quotation. What does it mean to have the right coordinate system? Well, if you look at the original transformation matrix with respect to the standard basis, it's fine. It's got this 2 by 2. But if you performed matrix operations with this, you've got to do some math. And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Well, if you look at the original transformation matrix with respect to the standard basis, it's fine. It's got this 2 by 2. But if you performed matrix operations with this, you've got to do some math. And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is. But when you transfer your bases, when you go to a new basis, when you went to this basis right here, all of a sudden you find that the transformation matrix is much simpler. It's a diagonal matrix. When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is. But when you transfer your bases, when you go to a new basis, when you went to this basis right here, all of a sudden you find that the transformation matrix is much simpler. It's a diagonal matrix. When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms. When you multiply this guy times some vector, we did it here, when you multiplied this guy times this vector, you literally just scaled the first term times minus 1, and you scaled the second term by 2. So it's a much simpler operation. And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms. When you multiply this guy times some vector, we did it here, when you multiplied this guy times this vector, you literally just scaled the first term times minus 1, and you scaled the second term by 2. So it's a much simpler operation. And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates. You know, that's a lot more work than just what you save here. But imagine if you had to apply d multiple times. Imagine if you had to apply d times d times d times d to x."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates. You know, that's a lot more work than just what you save here. But imagine if you had to apply d multiple times. Imagine if you had to apply d times d times d times d to x. Or, you know, let me say it this way. Imagine if you had to apply a times a times a. Like, you have to apply a 100 times to some vector up here."}, {"video_title": "Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3", "Sentence": "Imagine if you had to apply d times d times d times d to x. Or, you know, let me say it this way. Imagine if you had to apply a times a times a. Like, you have to apply a 100 times to some vector up here. Then applying a 100 times to some vector, it would be much more computationally intensive than applying d 100 times to this vector, even though you had a little bit of overhead from converting in this direction and then converting back. So in a lot of problems, especially in computer science frankly, or some other applications you might be doing, you want to pick the right basis. The problems, or at least the computation for many problems, get a lot simpler if you pick the right coordinate system."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have some matrix, let's call it matrix x. Matrix x is equal to, I'll just start with a 3 by 3 case because I think the 2 by 2 case is a bit trivial. Actually, why don't I just start with a 2 by 2 case? Let's say matrix x is a, b, and then it has x1, x2. I could have called these c and d, but you'll see why I call them x1 and x2 in a second. And now let's say I have another matrix. Let's say matrix y is identical to matrix x, except for this row. So matrix y is a, b, y1, and y2."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I could have called these c and d, but you'll see why I call them x1 and x2 in a second. And now let's say I have another matrix. Let's say matrix y is identical to matrix x, except for this row. So matrix y is a, b, y1, and y2. And let's say we have a third matrix, z. Let's say we have a third matrix, z, that's identical to the first two matrices on the first row. So a, b."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So matrix y is a, b, y1, and y2. And let's say we have a third matrix, z. Let's say we have a third matrix, z, that's identical to the first two matrices on the first row. So a, b. But on the second row, it's actually the sum of the two rows of x and y. So this entry is going to be x1 plus y1. And this entry right here is y, sorry, is x2 plus y2."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So a, b. But on the second row, it's actually the sum of the two rows of x and y. So this entry is going to be x1 plus y1. And this entry right here is y, sorry, is x2 plus y2. Just like that. I want to be very clear. z is not x plus y."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this entry right here is y, sorry, is x2 plus y2. Just like that. I want to be very clear. z is not x plus y. All of the terms of z are not the sum of all the terms of x and y. I'm only focusing on one particular row. And this is just a general theme that you'll see over and over again, and we saw it in the last video, and I guess you'll see it here, is that the determinants, or finding the determinants, of matrices aren't linear on matrix operations, but they are linear on operations that you do just to one row. So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "z is not x plus y. All of the terms of z are not the sum of all the terms of x and y. I'm only focusing on one particular row. And this is just a general theme that you'll see over and over again, and we saw it in the last video, and I guess you'll see it here, is that the determinants, or finding the determinants, of matrices aren't linear on matrix operations, but they are linear on operations that you do just to one row. So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys. So let's explore how the determinants of these guys relate. So the determinant, let me do it in x's color. The determinant of x, I'll write it like that, is equal to ax2 minus bx1."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys. So let's explore how the determinants of these guys relate. So the determinant, let me do it in x's color. The determinant of x, I'll write it like that, is equal to ax2 minus bx1. We've seen that multiple times. The determinant of y is equal to ay2 minus by1. And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The determinant of x, I'll write it like that, is equal to ax2 minus bx1. We've seen that multiple times. The determinant of y is equal to ay2 minus by1. And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1. And if we just rearrange things, this is equal to a, let me write it this way, this is equal to ax2 minus bx1, so that's that term and that term. We switch colors. So that's those two guys."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1. And if we just rearrange things, this is equal to a, let me write it this way, this is equal to ax2 minus bx1, so that's that term and that term. We switch colors. So that's those two guys. And then plus ay2 minus by1. Now, what is this right here? That is the determinant of x."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's those two guys. And then plus ay2 minus by1. Now, what is this right here? That is the determinant of x. And this right here is the determinant of y. So there you have it. If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is the determinant of x. And this right here is the determinant of y. So there you have it. If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants. So this is a very special case. I want to keep reiterating, it only works in the case where this row and only this row is the sum of this row and this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case, and I think it'll make it a little bit more general."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants. So this is a very special case. I want to keep reiterating, it only works in the case where this row and only this row is the sum of this row and this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case, and I think it'll make it a little bit more general. And then we'll go to n by n. The n by n is actually, on some level, the easiest to do, but it's kind of abstract, so I like to save that for the end. So let's redefine all those guys into the 3 by 3 case. So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you the 3 by 3 case, and I think it'll make it a little bit more general. And then we'll go to n by n. The n by n is actually, on some level, the easiest to do, but it's kind of abstract, so I like to save that for the end. So let's redefine all those guys into the 3 by 3 case. So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant. a, b, c, d, e, f. Actually, let me do the middle row, because I don't want to make you think it always has to be the last row. So let's say it's x1, x2, x3, and then you have a d, e, f. And what's the determinant of x going to be? The determinant of x is going to be equal to, let's say we're going along this row right here."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant. a, b, c, d, e, f. Actually, let me do the middle row, because I don't want to make you think it always has to be the last row. So let's say it's x1, x2, x3, and then you have a d, e, f. And what's the determinant of x going to be? The determinant of x is going to be equal to, let's say we're going along this row right here. That's the row in question. It's going to be equal to, well, you remember your checkerboard pattern, so it's going to be, remember, plus, minus, plus, minus, plus. You remember all the rest, how it goes."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The determinant of x is going to be equal to, let's say we're going along this row right here. That's the row in question. It's going to be equal to, well, you remember your checkerboard pattern, so it's going to be, remember, plus, minus, plus, minus, plus. You remember all the rest, how it goes. So it's going to start with a minus x1 times the submatrix. You get rid of that column, that row. b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You remember all the rest, how it goes. So it's going to start with a minus x1 times the submatrix. You get rid of that column, that row. b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix. Get rid of that column, that row. a, c, d, f. And then finally, minus x3. You get rid of its row and column."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix. Get rid of that column, that row. a, c, d, f. And then finally, minus x3. You get rid of its row and column. You have a, b, d, e. Now, let me define another matrix, y, that is identical to matrix x, except for that row. So it's a, b, c, and then down here, d, e, f. But that middle row is different. It's y1, y2, and y3."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get rid of its row and column. You have a, b, d, e. Now, let me define another matrix, y, that is identical to matrix x, except for that row. So it's a, b, c, and then down here, d, e, f. But that middle row is different. It's y1, y2, and y3. What's the determinant of y going to be? Well, it's going to be identical to the determinant of x, because all the submatrices are going to be the same when you cross out this row and each of the columns. But the coefficients are going to be different."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's y1, y2, and y3. What's the determinant of y going to be? Well, it's going to be identical to the determinant of x, because all the submatrices are going to be the same when you cross out this row and each of the columns. But the coefficients are going to be different. Instead of an x1, you have a y1. So it's going to be equal to minus y1 times the determinant, b, c, e, f, plus y2 times the determinant of a, c, d, f, minus y3 times the determinant of a, b, d, e. I think you see where this is going. Now I'm going to create another matrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But the coefficients are going to be different. Instead of an x1, you have a y1. So it's going to be equal to minus y1 times the determinant, b, c, e, f, plus y2 times the determinant of a, c, d, f, minus y3 times the determinant of a, b, d, e. I think you see where this is going. Now I'm going to create another matrix. I'm going to create another matrix, z. I'm going to create another matrix, z, just like that. That is equal to, it's identical to these two guys on the first and third rows, a, b, c, d, e, f, just like that. But this row just happens to be the sum of this row and this row."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now I'm going to create another matrix. I'm going to create another matrix, z. I'm going to create another matrix, z, just like that. That is equal to, it's identical to these two guys on the first and third rows, a, b, c, d, e, f, just like that. But this row just happens to be the sum of this row and this row. And when we figured out this determinant, we went along that row. You can see that right there. So this row right here is going to be x1 plus y1."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this row just happens to be the sum of this row and this row. And when we figured out this determinant, we went along that row. You can see that right there. So this row right here is going to be x1 plus y1. That's its first term. x2 plus y2. And then you have x3 plus y3."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this row right here is going to be x1 plus y1. That's its first term. x2 plus y2. And then you have x3 plus y3. Now what's the determinant of z going to be? Well, we can go down this row right there. So it's going to be minus x1 plus y1 times its submatrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have x3 plus y3. Now what's the determinant of z going to be? Well, we can go down this row right there. So it's going to be minus x1 plus y1 times its submatrix. Get rid of that row, that column. You get b, c, e, f. You get b, c, e, f. I think you definitely see where this is going. Plus this coefficient, plus x2 plus y2 times its submatrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be minus x1 plus y1 times its submatrix. Get rid of that row, that column. You get b, c, e, f. You get b, c, e, f. I think you definitely see where this is going. Plus this coefficient, plus x2 plus y2 times its submatrix. Get rid of that row, that column. a, c, d, f. And then you have minus this guy right here. x3 plus y3 times its submatrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Plus this coefficient, plus x2 plus y2 times its submatrix. Get rid of that row, that column. a, c, d, f. And then you have minus this guy right here. x3 plus y3 times its submatrix. Get rid of that column and row. a, b, d, e. Now what do you have right here? This is the determinant of z."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "x3 plus y3 times its submatrix. Get rid of that column and row. a, b, d, e. Now what do you have right here? This is the determinant of z. This is, let me do it in a, this right here is the determinant of z. This thing right here. I think you can see immediately that if you were to add this to this, you would get this right here, right?"}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the determinant of z. This is, let me do it in a, this right here is the determinant of z. This thing right here. I think you can see immediately that if you were to add this to this, you would get this right here, right? Because you would have this coefficient and this coefficient on that. If you added them up, you would get minus x1 plus y1. This guy and this guy add up to this guy."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think you can see immediately that if you were to add this to this, you would get this right here, right? Because you would have this coefficient and this coefficient on that. If you added them up, you would get minus x1 plus y1. This guy and this guy add up to this guy. And then if I were to do this guy and this guy add up to that guy. And let me do another one. And finally, that term plus that term add up to that term."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This guy and this guy add up to this guy. And then if I were to do this guy and this guy add up to that guy. And let me do another one. And finally, that term plus that term add up to that term. So you immediately see that the determinant, or hopefully you immediately see that the determinant of x plus the determinant of y is equal to the determinant of z. So we did it for the 2 by 2 case. We just did it for the 3 by 3 case."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And finally, that term plus that term add up to that term. So you immediately see that the determinant, or hopefully you immediately see that the determinant of x plus the determinant of y is equal to the determinant of z. So we did it for the 2 by 2 case. We just did it for the 3 by 3 case. Might as well do it for the n by n case so we know that it works. But the argument is identical to this 3 by 3 case. And it's good to keep in your mind, because 3 by 3 is easy to visualize."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just did it for the 3 by 3 case. Might as well do it for the n by n case so we know that it works. But the argument is identical to this 3 by 3 case. And it's good to keep in your mind, because 3 by 3 is easy to visualize. n by n is sometimes a little bit abstract. So let me define, let me redefine my matrices again. I'm just going to do the same thing over again."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's good to keep in your mind, because 3 by 3 is easy to visualize. n by n is sometimes a little bit abstract. So let me define, let me redefine my matrices again. I'm just going to do the same thing over again. So I'm going to have a matrix x, but it's an n by n matrix. So let me write it this way. Let's say it is a11, a12, all the way to a1n."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just going to do the same thing over again. So I'm going to have a matrix x, but it's an n by n matrix. So let me write it this way. Let's say it is a11, a12, all the way to a1n. And there's some row here, let's say that there's some row here on row i. Let's call this row i right here. And here it is, has the terms x1, x2, all the way to xn."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it is a11, a12, all the way to a1n. And there's some row here, let's say that there's some row here on row i. Let's call this row i right here. And here it is, has the terms x1, x2, all the way to xn. But then everything else is just the regular a's. So then you have a, let me make this as a21, all the way to a2n, and then if you went all the way down here, you would have a n1, and you'd go all the way to a nn. So essentially you could imagine our standard matrix where everything is defined in a, but I replaced row i with certain numbers that are maybe a little different."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And here it is, has the terms x1, x2, all the way to xn. But then everything else is just the regular a's. So then you have a, let me make this as a21, all the way to a2n, and then if you went all the way down here, you would have a n1, and you'd go all the way to a nn. So essentially you could imagine our standard matrix where everything is defined in a, but I replaced row i with certain numbers that are maybe a little different. And I think you'll see where I'm going. Now let me define my other matrix. Let me define matrix y."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So essentially you could imagine our standard matrix where everything is defined in a, but I replaced row i with certain numbers that are maybe a little different. And I think you'll see where I'm going. Now let me define my other matrix. Let me define matrix y. Let me define matrix y to be essentially the same thing. This is a11, it's the same a11. This is a12, all the way to a1n."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me define matrix y. Let me define matrix y to be essentially the same thing. This is a11, it's the same a11. This is a12, all the way to a1n. This is a21, we could go all the way to a2n. And then on row i, the same row, this is n by n, this is the same n by n. If this was 10 by 10, this is 10 by 10. If this is row 7, then this is row 7."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a12, all the way to a1n. This is a21, we could go all the way to a2n. And then on row i, the same row, this is n by n, this is the same n by n. If this was 10 by 10, this is 10 by 10. If this is row 7, then this is row 7. It has different terms. It's identical to matrix x, except for row i. In row i it is y1, y2, all the way to yn."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If this is row 7, then this is row 7. It has different terms. It's identical to matrix x, except for row i. In row i it is y1, y2, all the way to yn. And if you keep going down, of course you have a n1, all the way to a nn. Fair enough. Now let's say we have a third matrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In row i it is y1, y2, all the way to yn. And if you keep going down, of course you have a n1, all the way to a nn. Fair enough. Now let's say we have a third matrix. Let's have a third matrix. Let me draw it right here. So you have z. z is equal to, I think you could imagine where this is going."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's say we have a third matrix. Let's have a third matrix. Let me draw it right here. So you have z. z is equal to, I think you could imagine where this is going. z is identical to these two guys, except for row i. So let me write that out. So z looks like this."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you have z. z is equal to, I think you could imagine where this is going. z is identical to these two guys, except for row i. So let me write that out. So z looks like this. You have a11, a12, a12, all the way to a1n. And then you go down. And then row i happens to be the sum of the row i of matrix x and matrix y."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So z looks like this. You have a11, a12, a12, all the way to a1n. And then you go down. And then row i happens to be the sum of the row i of matrix x and matrix y. So it is x1 plus y1, x2 plus y2, all the way to xn plus yn. And then you keep going down. Everything else is identical."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then row i happens to be the sum of the row i of matrix x and matrix y. So it is x1 plus y1, x2 plus y2, all the way to xn plus yn. And then you keep going down. Everything else is identical. a n1, all the way to a nn. So all of these matrices are identical, except for row x has a different row i than matrix y does. And row z is identical everywhere, except its row i is the sum of this row i and that row i."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Everything else is identical. a n1, all the way to a nn. So all of these matrices are identical, except for row x has a different row i than matrix y does. And row z is identical everywhere, except its row i is the sum of this row i and that row i. So it's a very particular case. But we can figure out their determinants. So what are the determinants?"}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And row z is identical everywhere, except its row i is the sum of this row i and that row i. So it's a very particular case. But we can figure out their determinants. So what are the determinants? The determinant of x, the determinant of matrix x. Hopefully you're maybe a bit comfortable with writing sigma notation. We did this in the last matrix. We can go down this row."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what are the determinants? The determinant of x, the determinant of matrix x. Hopefully you're maybe a bit comfortable with writing sigma notation. We did this in the last matrix. We can go down this row. We can go down this row right there. And for each of these guys, we can say that this, so the determinant is going to be equal to the sum. Let's say we start from j is equal to 1. j is going to be the column."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can go down this row. We can go down this row right there. And for each of these guys, we can say that this, so the determinant is going to be equal to the sum. Let's say we start from j is equal to 1. j is going to be the column. So we're going to take the sum of each of these terms. From j is equal to 1 to n. And then remember our checkerboard pattern. So we don't know if this is a positive or negative."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we start from j is equal to 1. j is going to be the column. So we're going to take the sum of each of these terms. From j is equal to 1 to n. And then remember our checkerboard pattern. So we don't know if this is a positive or negative. But we can figure it out by taking negative 1 to the i plus j. Remember, this is the i-th row that we're talking about. Times xj."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we don't know if this is a positive or negative. But we can figure it out by taking negative 1 to the i plus j. Remember, this is the i-th row that we're talking about. Times xj. xj is the coefficient. x sub j. Times the sub matrix for x sub j."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times xj. xj is the coefficient. x sub j. Times the sub matrix for x sub j. So if you get rid of this guy's row and this guy's column, what is it going to be? We could say that that's the same thing as the sub matrix. If we called this guy, let me write it this way."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times the sub matrix for x sub j. So if you get rid of this guy's row and this guy's column, what is it going to be? We could say that that's the same thing as the sub matrix. If we called this guy, let me write it this way. If we got rid of this guy's row and this guy's column, if we had just our traditional matrix where this wasn't replaced, if we just had an ai1 here, ai2, its sub matrix would be the same thing. Because we're crossing out this row and this column. So it would be all of these guys and all of these guys down here."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we called this guy, let me write it this way. If we got rid of this guy's row and this guy's column, if we had just our traditional matrix where this wasn't replaced, if we just had an ai1 here, ai2, its sub matrix would be the same thing. Because we're crossing out this row and this column. So it would be all of these guys and all of these guys down here. So it would be the sub matrix. This is a n minus 1 by n minus 1 matrix. It would be the sub matrix for aij."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it would be all of these guys and all of these guys down here. So it would be the sub matrix. This is a n minus 1 by n minus 1 matrix. It would be the sub matrix for aij. And then that's for the first term. Sorry, the determinant. Don't want to lose the determinant there."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It would be the sub matrix for aij. And then that's for the first term. Sorry, the determinant. Don't want to lose the determinant there. Times the determinant of the sub matrix aij. And so that's for the first term. And then you're going to add it to the second term."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Don't want to lose the determinant there. Times the determinant of the sub matrix aij. And so that's for the first term. And then you're going to add it to the second term. And then you're just going to keep doing that. And that's what this sigma notation is. That's the determinant of x."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you're going to add it to the second term. And then you're just going to keep doing that. And that's what this sigma notation is. That's the determinant of x. Now, what's the determinant of y? The determinant of y is equal to the sum. We could do the same thing."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the determinant of x. Now, what's the determinant of y? The determinant of y is equal to the sum. We could do the same thing. j is equal to 1 to n of negative 1 to the i plus j. Each of them, we're going to go along this row right here. The ith row."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could do the same thing. j is equal to 1 to n of negative 1 to the i plus j. Each of them, we're going to go along this row right here. The ith row. So we're going to have y sub j. We're going to start with y sub 1, then plus y sub 2. Times the determinant of its sub matrix."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The ith row. So we're going to have y sub j. We're going to start with y sub 1, then plus y sub 2. Times the determinant of its sub matrix. Which is the same as the determinant of this sub matrix. When you get rid of that row and that column for each of these guys, everything else on the matrix is the same. So aij."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times the determinant of its sub matrix. Which is the same as the determinant of this sub matrix. When you get rid of that row and that column for each of these guys, everything else on the matrix is the same. So aij. The matrix of aij. Now, what is the determinant of z? I'm pretty sure you know exactly where this is going."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So aij. The matrix of aij. Now, what is the determinant of z? I'm pretty sure you know exactly where this is going. The determinant, this should be a capital Y right there. The determinant of z is equal to the sum from j is equal to 1 to n of negative 1 to the i plus j. We're going along this row."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm pretty sure you know exactly where this is going. The determinant, this should be a capital Y right there. The determinant of z is equal to the sum from j is equal to 1 to n of negative 1 to the i plus j. We're going along this row. But now the coefficients are xj. That's what we're indexing along. xj plus yj."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're going along this row. But now the coefficients are xj. That's what we're indexing along. xj plus yj. And then times its sub matrix, which is the same as these sub matrices. So aij. aij."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "xj plus yj. And then times its sub matrix, which is the same as these sub matrices. So aij. aij. Which you might immediately see is the sum of these two things. If I, for every j, I just summed these two things up, you're having two coefficients. You could have this coefficient and that coefficient on your aij term."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "aij. Which you might immediately see is the sum of these two things. If I, for every j, I just summed these two things up, you're having two coefficients. You could have this coefficient and that coefficient on your aij term. And then when you add them up, you can factor this guy out and you'll get this right here. You will get this right here. So you get the determinant of x plus the determinant of y is equal to the determinant of z."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could have this coefficient and that coefficient on your aij term. And then when you add them up, you can factor this guy out and you'll get this right here. You will get this right here. So you get the determinant of x plus the determinant of y is equal to the determinant of z. So hopefully that shows you the general case. But I want to make it very clear. This is just for a very particular scenario where three matrices are identical except for on one row."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you get the determinant of x plus the determinant of y is equal to the determinant of z. So hopefully that shows you the general case. But I want to make it very clear. This is just for a very particular scenario where three matrices are identical except for on one row. And one of the matrices on that special row just happens to be the sum of the other two matrices for that special row. And everything else is identical. That's the only time where the determinant of z, not the only time, but that's the only time we can make the general statement where the determinant of z is equal to the determinant of x plus the determinant of y."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is just for a very particular scenario where three matrices are identical except for on one row. And one of the matrices on that special row just happens to be the sum of the other two matrices for that special row. And everything else is identical. That's the only time where the determinant of z, not the only time, but that's the only time we can make the general statement where the determinant of z is equal to the determinant of x plus the determinant of y. It's not the case, let me write what is not the case. So not the case that if, let's say, z is equal to x plus y, it is not the case that the determinant of z is necessarily equal to the determinant of x plus the determinant of y. You cannot assume this."}, {"video_title": "Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the only time where the determinant of z, not the only time, but that's the only time we can make the general statement where the determinant of z is equal to the determinant of x plus the determinant of y. It's not the case, let me write what is not the case. So not the case that if, let's say, z is equal to x plus y, it is not the case that the determinant of z is necessarily equal to the determinant of x plus the determinant of y. You cannot assume this. Determinant operations are not linear on matrix addition. They're linear only on particular rows getting added. Anyway, hopefully you found that vaguely clear."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And that says, well, that means any value lambda that satisfies this equation for v is not a non-zero vector. We just did a little bit of vector algebra up here to come up with that. You can review that video if you like. And then we determined, look, the only way that this is going to have a non-zero solution is if this matrix has a non-trivial null space and only non-invertible matrices have a non-trivial null space, or only matrices that have a determinant of 0 have non-trivial null spaces. So you do that, you got your characteristic polynomial, and we were able to solve it. And we got our eigenvalues where lambda is equal to 3 and lambda is equal to minus 3. So now let's do what I consider the more interesting part, is actually find out the eigenvectors, or the eigenspaces."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then we determined, look, the only way that this is going to have a non-zero solution is if this matrix has a non-trivial null space and only non-invertible matrices have a non-trivial null space, or only matrices that have a determinant of 0 have non-trivial null spaces. So you do that, you got your characteristic polynomial, and we were able to solve it. And we got our eigenvalues where lambda is equal to 3 and lambda is equal to minus 3. So now let's do what I consider the more interesting part, is actually find out the eigenvectors, or the eigenspaces. So we can go back to this equation. For any eigenvalue, this must be true, but this is easier to work with. And so this matrix right here times your eigenvector must be equal to 0 for any given eigenvalue."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So now let's do what I consider the more interesting part, is actually find out the eigenvectors, or the eigenspaces. So we can go back to this equation. For any eigenvalue, this must be true, but this is easier to work with. And so this matrix right here times your eigenvector must be equal to 0 for any given eigenvalue. This matrix right here, I just copied and pasted from above. I marked it up with the rule of Cyrus, but you could ignore those lines. It's just this matrix right here for any lambda."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And so this matrix right here times your eigenvector must be equal to 0 for any given eigenvalue. This matrix right here, I just copied and pasted from above. I marked it up with the rule of Cyrus, but you could ignore those lines. It's just this matrix right here for any lambda. Lambda times the identity matrix minus a ends up being this. So let's take this matrix for each of our lambdas and then solve for our eigenvectors, or our eigenspaces. So let's take the case of lambda is equal to 3 first."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "It's just this matrix right here for any lambda. Lambda times the identity matrix minus a ends up being this. So let's take this matrix for each of our lambdas and then solve for our eigenvectors, or our eigenspaces. So let's take the case of lambda is equal to 3 first. So if lambda is equal to 3, this matrix becomes lambda plus 1 is 4, lambda minus 2 is 1, lambda minus 2 is 1. And then all of the other terms stay the same. Minus 2, minus 2, minus 2, 1, minus 2, and 1."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's take the case of lambda is equal to 3 first. So if lambda is equal to 3, this matrix becomes lambda plus 1 is 4, lambda minus 2 is 1, lambda minus 2 is 1. And then all of the other terms stay the same. Minus 2, minus 2, minus 2, 1, minus 2, and 1. And then this times that vector v, or our eigenvector v, is equal to 0. Or we could say that the eigenspace for the eigenvalue 3 is the null space of this matrix, which is not this matrix. It's lambda times identity minus a."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2, minus 2, minus 2, 1, minus 2, and 1. And then this times that vector v, or our eigenvector v, is equal to 0. Or we could say that the eigenspace for the eigenvalue 3 is the null space of this matrix, which is not this matrix. It's lambda times identity minus a. So the null space of this matrix is the eigenspace. So all of the values that satisfy this make up the eigenvectors of the eigenspace of lambda is equal to 3. So let's just solve for this."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "It's lambda times identity minus a. So the null space of this matrix is the eigenspace. So all of the values that satisfy this make up the eigenvectors of the eigenspace of lambda is equal to 3. So let's just solve for this. So the null space of this guy, we could just put it in reduced row echelon form. The null space of this guy is the same thing as the null space of this guy in reduced row echelon form. So let's put it in reduced row echelon form."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's just solve for this. So the null space of this guy, we could just put it in reduced row echelon form. The null space of this guy is the same thing as the null space of this guy in reduced row echelon form. So let's put it in reduced row echelon form. So the first thing I want to do, let me just do it down here, so let me, I'll keep my first row the same for now. 4, minus 2, minus 2. And let me replace my second row with my second row times 2 plus my first row."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's put it in reduced row echelon form. So the first thing I want to do, let me just do it down here, so let me, I'll keep my first row the same for now. 4, minus 2, minus 2. And let me replace my second row with my second row times 2 plus my first row. So minus 2 times 2 plus 1 is 0. 1 times 2 plus minus 2 is 0. 1 times 2 plus minus 2 is 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And let me replace my second row with my second row times 2 plus my first row. So minus 2 times 2 plus 1 is 0. 1 times 2 plus minus 2 is 0. 1 times 2 plus minus 2 is 0. This row is the same as this row, so I'm going to do the same thing. Minus 2 times 2 plus 4 is 0. 1 times 2 plus 2 is 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 times 2 plus minus 2 is 0. This row is the same as this row, so I'm going to do the same thing. Minus 2 times 2 plus 4 is 0. 1 times 2 plus 2 is 0. And then 1 times 2 plus minus 2 is 0. So the solutions to this equation are the same as the solutions to this equation. Let me write it like this."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 times 2 plus 2 is 0. And then 1 times 2 plus minus 2 is 0. So the solutions to this equation are the same as the solutions to this equation. Let me write it like this. Instead of just writing the vector v, let me write it out. So v1, v2, v3 are going to be equal to the 0 vector, 0, 0. Just rewriting it slightly different."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it like this. Instead of just writing the vector v, let me write it out. So v1, v2, v3 are going to be equal to the 0 vector, 0, 0. Just rewriting it slightly different. And so these two rows, or these two equations, give us no information. The only one is this row up here, which tells us that 4 times v1 minus 2 times v2, actually this wasn't complete reduced row echelon form, but close enough. It's easy for us to work with."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Just rewriting it slightly different. And so these two rows, or these two equations, give us no information. The only one is this row up here, which tells us that 4 times v1 minus 2 times v2, actually this wasn't complete reduced row echelon form, but close enough. It's easy for us to work with. 4 times v1 minus 2 times v2 minus 2 times v3 is equal to 0, and let's just divide by 4. I could have just divided by 4 here, which might have made me skip a step. If you divide by 4, you get v1 minus 1 half v2 minus 1 half v3 is equal to 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "It's easy for us to work with. 4 times v1 minus 2 times v2 minus 2 times v3 is equal to 0, and let's just divide by 4. I could have just divided by 4 here, which might have made me skip a step. If you divide by 4, you get v1 minus 1 half v2 minus 1 half v3 is equal to 0. Or v1 is equal to 1 half v2 plus 1 half v3. I just added these guys to both sides of the equation. Or we could say, if we say that v2 is equal to, I don't know, I'm just going to put some random number a, and v3 is equal to b, then we can say, and then v1 would be equal to 1 half a plus 1 half b."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "If you divide by 4, you get v1 minus 1 half v2 minus 1 half v3 is equal to 0. Or v1 is equal to 1 half v2 plus 1 half v3. I just added these guys to both sides of the equation. Or we could say, if we say that v2 is equal to, I don't know, I'm just going to put some random number a, and v3 is equal to b, then we can say, and then v1 would be equal to 1 half a plus 1 half b. We can say that the eigenspace for lambda is equal to 3. Is the set of all vectors, v1, v2, v3, that are equal to a times v2 is a, right? So v2 is equal to a times 1. v3 has no a in it, so it's a times 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Or we could say, if we say that v2 is equal to, I don't know, I'm just going to put some random number a, and v3 is equal to b, then we can say, and then v1 would be equal to 1 half a plus 1 half b. We can say that the eigenspace for lambda is equal to 3. Is the set of all vectors, v1, v2, v3, that are equal to a times v2 is a, right? So v2 is equal to a times 1. v3 has no a in it, so it's a times 0. I'll save v1. Plus b times, v2 is just a, right? v2 has no b in it, so it's 0. v3 is 1 times, so 0 times a plus 1 times b."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So v2 is equal to a times 1. v3 has no a in it, so it's a times 0. I'll save v1. Plus b times, v2 is just a, right? v2 has no b in it, so it's 0. v3 is 1 times, so 0 times a plus 1 times b. And then v1 is 1 half a plus 1 half b. 1 half and 1 half. For any a and b, I could, you know, such that a and b are members of the reals."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "v2 has no b in it, so it's 0. v3 is 1 times, so 0 times a plus 1 times b. And then v1 is 1 half a plus 1 half b. 1 half and 1 half. For any a and b, I could, you know, such that a and b are members of the reals. Just to be a little bit formal about it. So that's our eigen, any vector that satisfies this is an eigenvector, and they're the eigenvectors that correspond to the eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "For any a and b, I could, you know, such that a and b are members of the reals. Just to be a little bit formal about it. So that's our eigen, any vector that satisfies this is an eigenvector, and they're the eigenvectors that correspond to the eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Or you could say, let me write it this way. The eigenspace for lambda is equal to 3 is equal to the span, all of the potential linear combinations of this guy and that guy. So 1 half, 1 0, and 1 half, 0, 1."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Or you could say, let me write it this way. The eigenspace for lambda is equal to 3 is equal to the span, all of the potential linear combinations of this guy and that guy. So 1 half, 1 0, and 1 half, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So 1 half, 1 0, and 1 half, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3, I'll do it up here. I think I'll have enough space. Lambda is equal to minus 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3, I'll do it up here. I think I'll have enough space. Lambda is equal to minus 3. This matrix becomes, I'll do the diagonals, minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. Minus 3 minus 2 is minus 5."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Lambda is equal to minus 3. This matrix becomes, I'll do the diagonals, minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. Minus 3 minus 2 is minus 5. And then all the other things don't change. Minus 2, minus 2, 1, minus 2, minus 2, and 1. And then that times vectors and the eigenspace that corresponds to lambda is equal to minus 3 is going to be equal to 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Minus 3 minus 2 is minus 5. And then all the other things don't change. Minus 2, minus 2, 1, minus 2, minus 2, and 1. And then that times vectors and the eigenspace that corresponds to lambda is equal to minus 3 is going to be equal to 0. I'm just applying this equation right here, which we just derived from that one over there. So we're looking at the eigenspace that corresponds to lambda is equal to minus 3 is the null space of this matrix right here, or all the vectors that satisfy this equation. So the null space of this is the same thing as the null space of this in reduced row echelon form."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then that times vectors and the eigenspace that corresponds to lambda is equal to minus 3 is going to be equal to 0. I'm just applying this equation right here, which we just derived from that one over there. So we're looking at the eigenspace that corresponds to lambda is equal to minus 3 is the null space of this matrix right here, or all the vectors that satisfy this equation. So the null space of this is the same thing as the null space of this in reduced row echelon form. So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do, just because I think I'm going to run out of space."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So the null space of this is the same thing as the null space of this in reduced row echelon form. So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do, just because I think I'm going to run out of space. So minus 2, minus 2, minus 2. And then, actually, let me just do it this way. I will skip some steps."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to write a little bit smaller than I normally do, just because I think I'm going to run out of space. So minus 2, minus 2, minus 2. And then, actually, let me just do it this way. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0. Minus 5 plus minus, or let me say it this way. Let me replace it with the first row minus the second row, so minus 2 minus minus 2 is 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0. Minus 5 plus minus, or let me say it this way. Let me replace it with the first row minus the second row, so minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And let me do the last row."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let me replace it with the first row minus the second row, so minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And let me do the last row. I'll just do it in a different color for fun. And I'll do the same thing. I'll do this row minus this row."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And let me do the last row. I'll just do it in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "I'll do this row minus this row. So minus 2 minus minus 2 is 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5. So that's minus 2 plus 5. So that is 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5. So that's minus 2 plus 5. So that is 3. Now let me replace, and I'll do it in two steps. So this is 1, 1, 1. I'll just keep it like that."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So that is 3. Now let me replace, and I'll do it in two steps. So this is 1, 1, 1. I'll just keep it like that. And actually, well, yeah, let me just keep it like that. And then let me replace my third row with my third row plus my second row. It'll just zero out."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "I'll just keep it like that. And actually, well, yeah, let me just keep it like that. And then let me replace my third row with my third row plus my second row. It'll just zero out. You just add these terms. These all just become 0. That guy got zeroed out."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "It'll just zero out. You just add these terms. These all just become 0. That guy got zeroed out. And let me take my second row and divide it by 3. So this becomes 0, 1, minus 1. And then I'm almost there."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "That guy got zeroed out. And let me take my second row and divide it by 3. So this becomes 0, 1, minus 1. And then I'm almost there. I'll do it in orange. So let me replace my first row with my first row minus my second row. So this becomes 1, 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then I'm almost there. I'll do it in orange. So let me replace my first row with my first row minus my second row. So this becomes 1, 0. And then 1 minus minus 1 is 2. 1 minus minus 1 is 2. And then the second row is 0, 1, minus 1."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this becomes 1, 0. And then 1 minus minus 1 is 2. 1 minus minus 1 is 2. And then the second row is 0, 1, minus 1. And then the last row is 0, 0, 0. So any v that satisfies this equation will also satisfy this guy. If this guy's null space, it's going to be the null space of that guy in reduced row echelon form."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then the second row is 0, 1, minus 1. And then the last row is 0, 0, 0. So any v that satisfies this equation will also satisfy this guy. If this guy's null space, it's going to be the null space of that guy in reduced row echelon form. So v1, v2, v3 is equal to 0, 0, 0. Let me move this because I've officially run out of space. So let me move this lower down where I have some free real estate."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "If this guy's null space, it's going to be the null space of that guy in reduced row echelon form. So v1, v2, v3 is equal to 0, 0, 0. Let me move this because I've officially run out of space. So let me move this lower down where I have some free real estate. Let me move it down here. This corresponds to lambda is equal to minus 3. This was lambda is equal to minus 3, just to make us, you know, it's not related to this stuff right here."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let me move this lower down where I have some free real estate. Let me move it down here. This corresponds to lambda is equal to minus 3. This was lambda is equal to minus 3, just to make us, you know, it's not related to this stuff right here. So what are all of the v1's, v2's, and v3's that satisfy this? So if we say that v3 is equal to t. If v3 is equal to t, then what do we have here? This tells us that v2 minus v3 is equal to 0."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "This was lambda is equal to minus 3, just to make us, you know, it's not related to this stuff right here. So what are all of the v1's, v2's, and v3's that satisfy this? So if we say that v3 is equal to t. If v3 is equal to t, then what do we have here? This tells us that v2 minus v3 is equal to 0. So it tells us that v2 minus v3, right? v2, 0 times v1 plus v2 minus v3 is equal to 0. Or that v2 is equal to v3, which is equal to t. That's what that second equation tells us."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "This tells us that v2 minus v3 is equal to 0. So it tells us that v2 minus v3, right? v2, 0 times v1 plus v2 minus v3 is equal to 0. Or that v2 is equal to v3, which is equal to t. That's what that second equation tells us. And then the third equation tells us, or the top equation tells us v1 times 1. So v1 plus 0 times v2 plus 2 times v3 is equal to 0. Or v1 is equal to minus 2v3, which is equal to minus 2 times t. So the eigenspace that corresponds to lambda is equal to minus 3 is equal to the set of all the vectors v1, v2, and v3, where, well, it's equal to t times v3 is just t, right?"}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Or that v2 is equal to v3, which is equal to t. That's what that second equation tells us. And then the third equation tells us, or the top equation tells us v1 times 1. So v1 plus 0 times v2 plus 2 times v3 is equal to 0. Or v1 is equal to minus 2v3, which is equal to minus 2 times t. So the eigenspace that corresponds to lambda is equal to minus 3 is equal to the set of all the vectors v1, v2, and v3, where, well, it's equal to t times v3 is just t, right? v3 was just t. v2 also just ends up being t. So 1 times t. And v1 is minus 2 times t. For t is any real number. Or another way to say it is that the eigenspace for lambda is equal to minus 3 is equal to the span of the vector minus 2, 1, and 1. Just like that."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Or v1 is equal to minus 2v3, which is equal to minus 2 times t. So the eigenspace that corresponds to lambda is equal to minus 3 is equal to the set of all the vectors v1, v2, and v3, where, well, it's equal to t times v3 is just t, right? v3 was just t. v2 also just ends up being t. So 1 times t. And v1 is minus 2 times t. For t is any real number. Or another way to say it is that the eigenspace for lambda is equal to minus 3 is equal to the span of the vector minus 2, 1, and 1. Just like that. It looks interesting, because if you take this guy and dot it with either of these guys, I think you get 0. Is that definitely the case? If you take minus 2 times 1 half, you get a minus 1 there."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. It looks interesting, because if you take this guy and dot it with either of these guys, I think you get 0. Is that definitely the case? If you take minus 2 times 1 half, you get a minus 1 there. And then you have a plus 1. That's 0. And then minus 2 times 1 half."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "If you take minus 2 times 1 half, you get a minus 1 there. And then you have a plus 1. That's 0. And then minus 2 times 1 half. Yeah. You dot it with either of these guys, you get 0. So this line is orthogonal to that plane."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then minus 2 times 1 half. Yeah. You dot it with either of these guys, you get 0. So this line is orthogonal to that plane. Very interesting. So let's just graph it, just so we have a good visualization of what we're doing. So we had that 3 by 3 matrix A."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this line is orthogonal to that plane. Very interesting. So let's just graph it, just so we have a good visualization of what we're doing. So we had that 3 by 3 matrix A. It represents some transformation in R3. And it has two eigenvalues. And each of those have a corresponding eigenspace."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we had that 3 by 3 matrix A. It represents some transformation in R3. And it has two eigenvalues. And each of those have a corresponding eigenspace. So the eigenspace that corresponds to the eigenvalue 3 is a plane in R3. So this is the eigenspace for lambda is equal to 3. And it's the span of these two vectors right there."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And each of those have a corresponding eigenspace. So the eigenspace that corresponds to the eigenvalue 3 is a plane in R3. So this is the eigenspace for lambda is equal to 3. And it's the span of these two vectors right there. So if I draw, maybe they're like that, just like that. And then the eigenspace for lambda is equal to minus 3 is a line. It's a line that's perpendicular to this plane."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "And it's the span of these two vectors right there. So if I draw, maybe they're like that, just like that. And then the eigenspace for lambda is equal to minus 3 is a line. It's a line that's perpendicular to this plane. It's a line like that. It's the span of this guy. Maybe if I draw that vector, that vector might look something like this."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "It's a line that's perpendicular to this plane. It's a line like that. It's the span of this guy. Maybe if I draw that vector, that vector might look something like this. It's the span of that guy. So what this tells us, this right here is the eigenspace for lambda is equal to minus 3. So what that tells us, just to make sure we are interpreting our eigenvalues and eigenspaces correctly, is look, you give me any eigenvector in this."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Maybe if I draw that vector, that vector might look something like this. It's the span of that guy. So what this tells us, this right here is the eigenspace for lambda is equal to minus 3. So what that tells us, just to make sure we are interpreting our eigenvalues and eigenspaces correctly, is look, you give me any eigenvector in this. You give me any vector right here. Let's say that is vector x. If I apply the transformation, if I multiply it by a, I'm going to have 3 times that, because it's in the eigenspace for lambda is equal to 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "So what that tells us, just to make sure we are interpreting our eigenvalues and eigenspaces correctly, is look, you give me any eigenvector in this. You give me any vector right here. Let's say that is vector x. If I apply the transformation, if I multiply it by a, I'm going to have 3 times that, because it's in the eigenspace for lambda is equal to 3. So if I were to apply a times x, a times x would be just 3 times that. So that would be a times x. That's what it tells me."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "If I apply the transformation, if I multiply it by a, I'm going to have 3 times that, because it's in the eigenspace for lambda is equal to 3. So if I were to apply a times x, a times x would be just 3 times that. So that would be a times x. That's what it tells me. This would be true for any of these guys. If this was x, and you took a times x, it's going to be 3 times as long. Now these guys over here, if you have some vector in this eigenspace that corresponds to lambda is equal to 3, and you apply the transformation, let's say that this is x right there."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "That's what it tells me. This would be true for any of these guys. If this was x, and you took a times x, it's going to be 3 times as long. Now these guys over here, if you have some vector in this eigenspace that corresponds to lambda is equal to 3, and you apply the transformation, let's say that this is x right there. If you took the transformation of x, it's going to make it 3 times longer and in the opposite direction. It's still going to be on this line, so it's going to go down like this, and that would be a times x. It would be the same, it would be 3 times this length, but in the opposite direction, because it corresponds to lambda is equal to minus 3."}, {"video_title": "Eigenvectors and eigenspaces for a 3x3 matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now these guys over here, if you have some vector in this eigenspace that corresponds to lambda is equal to 3, and you apply the transformation, let's say that this is x right there. If you took the transformation of x, it's going to make it 3 times longer and in the opposite direction. It's still going to be on this line, so it's going to go down like this, and that would be a times x. It would be the same, it would be 3 times this length, but in the opposite direction, because it corresponds to lambda is equal to minus 3. So anyway, we've, I think, made a great achievement. We've not only figured out the eigenvalues for a 3 by 3 matrix, we now have figured out all of the eigenvectors, which are, you know, there's an infinite number, but they represent two eigenspaces that correspond to those two eigenvalues, or minus 3 and 3. See you in the next video."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's just some set or some subset of Rn where if I take any two members of that subset, so let's say I take the members a and b, they're both members of my subspace. By the fact that this is a subspace, we then know that the addition of these two vectors, or a plus b, is also in my subspace, and this is our closure under addition. And by the fact that it's a subspace, we also know that if we multiply any member of our subspace by a scalar, so the fact that those guys are members of our subspace, we also know that if I pick one of them, let's say a, and I multiply a by some scalar, that this is also going to be a member of our subspace. And we sometimes call this closure under scalar multiplication. And then a somewhat redundant statement is that V must contain the zero vector. And that's true of all subspaces. V, let me write it this way, the zero vector is a member of V, and it would be the zero vector with n components here, because V is a subspace of Rn."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we sometimes call this closure under scalar multiplication. And then a somewhat redundant statement is that V must contain the zero vector. And that's true of all subspaces. V, let me write it this way, the zero vector is a member of V, and it would be the zero vector with n components here, because V is a subspace of Rn. And why I say that's redundant, because if I say that any multiple of these vectors is also in V, I could just set the scalar to be equal to 0. So this statement kind of takes this statement into account, but in a lot of textbooks, it'll always write, oh, and the zero vector has to be a member of V, although that's kind of redundant with the closure under scalar multiplication. Fair enough."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "V, let me write it this way, the zero vector is a member of V, and it would be the zero vector with n components here, because V is a subspace of Rn. And why I say that's redundant, because if I say that any multiple of these vectors is also in V, I could just set the scalar to be equal to 0. So this statement kind of takes this statement into account, but in a lot of textbooks, it'll always write, oh, and the zero vector has to be a member of V, although that's kind of redundant with the closure under scalar multiplication. Fair enough. Now, let's say that I also have some transformation T. It is a mapping, a function from Rn to Rm. What I want to understand in this video is I have a subspace right here, V. I want to understand whether the transformation of the subspace, and what did we call that, we call that the image of our subspace or our subset, either way, the image of V under T. In the last video, just to kind of help you visualize it, how did that work, or we had some subset of Rn that looked like this, it was a triangle that looked something like that, and that was in Rn, this was actually in R2. It was a triangle that looked something like that, and we figured out its image under T. So we went from R2 to R2, and we had our transformation, and it ended up looking something like this, if I remember it properly."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now, let's say that I also have some transformation T. It is a mapping, a function from Rn to Rm. What I want to understand in this video is I have a subspace right here, V. I want to understand whether the transformation of the subspace, and what did we call that, we call that the image of our subspace or our subset, either way, the image of V under T. In the last video, just to kind of help you visualize it, how did that work, or we had some subset of Rn that looked like this, it was a triangle that looked something like that, and that was in Rn, this was actually in R2. It was a triangle that looked something like that, and we figured out its image under T. So we went from R2 to R2, and we had our transformation, and it ended up looking something like this, if I remember it properly. Actually, I don't remember it fully, but it was like a triangle that was skewed like this, rotated. So it was, actually, I think it was more like, I think that's right, it was rotated a bit clockwise like that, and it was skewed. But the exact particulars of that last video aren't what matter."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It was a triangle that looked something like that, and we figured out its image under T. So we went from R2 to R2, and we had our transformation, and it ended up looking something like this, if I remember it properly. Actually, I don't remember it fully, but it was like a triangle that was skewed like this, rotated. So it was, actually, I think it was more like, I think that's right, it was rotated a bit clockwise like that, and it was skewed. But the exact particulars of that last video aren't what matter. What matters is that you are able to visualize what an image under transformation means. It means you take some subset of R2, all of the vectors that define this triangle right here, that some subset of R2. You transform all of them, and then you get some subset in your co-domain."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But the exact particulars of that last video aren't what matter. What matters is that you are able to visualize what an image under transformation means. It means you take some subset of R2, all of the vectors that define this triangle right here, that some subset of R2. You transform all of them, and then you get some subset in your co-domain. And this is the, you could call this the image, call this the transformation of that triangle, or if we call this S, it's equal to the transformation of S. Or you could say it's the image of, you could call it the set S, but maybe it helps you to visualize, call it the image of this triangle under T. Or maybe even a neater way, if you're thinking about it, is this triangle, that skewed rotated triangle, this one is the image of this kind of right triangle under T. And I think that might make a little bit of visual sense to you. And just as a bit of a reminder, in that last video, these triangles, these weren't subspaces. And just as you could take scalar multiples of some of the vectors that are subsets, that are members of this triangle, and you'll find that they're not going to be in that triangle."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You transform all of them, and then you get some subset in your co-domain. And this is the, you could call this the image, call this the transformation of that triangle, or if we call this S, it's equal to the transformation of S. Or you could say it's the image of, you could call it the set S, but maybe it helps you to visualize, call it the image of this triangle under T. Or maybe even a neater way, if you're thinking about it, is this triangle, that skewed rotated triangle, this one is the image of this kind of right triangle under T. And I think that might make a little bit of visual sense to you. And just as a bit of a reminder, in that last video, these triangles, these weren't subspaces. And just as you could take scalar multiples of some of the vectors that are subsets, that are members of this triangle, and you'll find that they're not going to be in that triangle. So this wasn't a subspace. This was just a subset of R2. All subsets are not subspaces, but all subspaces are definitely subsets, although something can be a subset of itself."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And just as you could take scalar multiples of some of the vectors that are subsets, that are members of this triangle, and you'll find that they're not going to be in that triangle. So this wasn't a subspace. This was just a subset of R2. All subsets are not subspaces, but all subspaces are definitely subsets, although something can be a subset of itself. I don't want to wander off too much. But this just helps you visualize what we mean by an image. It means all of the vectors that are mapped to from the members of your subset."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All subsets are not subspaces, but all subspaces are definitely subsets, although something can be a subset of itself. I don't want to wander off too much. But this just helps you visualize what we mean by an image. It means all of the vectors that are mapped to from the members of your subset. So I want to know whether the image of V under T is a subspace. So in order for it to be a subspace, if I take the transformation, let me find two members of T. Well, clearly, if I take the transformation of any members of V, I'm getting members of the image. So I can write this."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It means all of the vectors that are mapped to from the members of your subset. So I want to know whether the image of V under T is a subspace. So in order for it to be a subspace, if I take the transformation, let me find two members of T. Well, clearly, if I take the transformation of any members of V, I'm getting members of the image. So I can write this. Clearly, the transformation of A and the transformations of B, these are both members of our images of V under T. These are both members of that right there. So my question to you is, what is the transformation of A plus the transformation of B? And the way I've written this, these are two arbitrary members of our image of V under T. Or maybe I should call it T of capital V. These are two arbitrary members."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I can write this. Clearly, the transformation of A and the transformations of B, these are both members of our images of V under T. These are both members of that right there. So my question to you is, what is the transformation of A plus the transformation of B? And the way I've written this, these are two arbitrary members of our image of V under T. Or maybe I should call it T of capital V. These are two arbitrary members. So what is this equal to? Well, we know from our properties, our definition of linear transformations, this is equal to the sum of the transformations of two vectors is equal to the transformation of the sum of their vectors. Now, is the transformation of A plus B a member of TV?"}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the way I've written this, these are two arbitrary members of our image of V under T. Or maybe I should call it T of capital V. These are two arbitrary members. So what is this equal to? Well, we know from our properties, our definition of linear transformations, this is equal to the sum of the transformations of two vectors is equal to the transformation of the sum of their vectors. Now, is the transformation of A plus B a member of TV? Is it a member of our image? Well, A plus B is a member of V, and the image contains the transformation of all of the members of V. So the image contains the transformation of this guy. This guy, A plus B, is a member of V. So you're taking the transformation of a member of V, which by definition is in your image of V under T. So this is definitely true."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, is the transformation of A plus B a member of TV? Is it a member of our image? Well, A plus B is a member of V, and the image contains the transformation of all of the members of V. So the image contains the transformation of this guy. This guy, A plus B, is a member of V. So you're taking the transformation of a member of V, which by definition is in your image of V under T. So this is definitely true. Now, let's ask the next question. If I take a scalar multiple of some member of my image of V under T, or my T of capital V right there, what is this equal to? By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This guy, A plus B, is a member of V. So you're taking the transformation of a member of V, which by definition is in your image of V under T. So this is definitely true. Now, let's ask the next question. If I take a scalar multiple of some member of my image of V under T, or my T of capital V right there, what is this equal to? By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector. Now, is this going to be a member of our image of V under T? Well, we know that CA is definitely in V. That's from the definition of a subspace. This is definitely in V. And so if this is in V, the transformation of this has to be in V's image under T. So this is also a member of V. And obviously, you can set this equal to 0."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector. Now, is this going to be a member of our image of V under T? Well, we know that CA is definitely in V. That's from the definition of a subspace. This is definitely in V. And so if this is in V, the transformation of this has to be in V's image under T. So this is also a member of V. And obviously, you can set this equal to 0. The 0 vector is a member of V, so any transformation of, if you just put a 0 here, you'll get the 0 vector. So the 0 vector is definitely, I don't care what this is, if you multiply it times 0, you're going to get the 0 vector. So the 0 vector is definitely also a member of TV."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is definitely in V. And so if this is in V, the transformation of this has to be in V's image under T. So this is also a member of V. And obviously, you can set this equal to 0. The 0 vector is a member of V, so any transformation of, if you just put a 0 here, you'll get the 0 vector. So the 0 vector is definitely, I don't care what this is, if you multiply it times 0, you're going to get the 0 vector. So the 0 vector is definitely also a member of TV. So we come on the result that the image of V under T is a subspace, which is a useful result, which we'll be able to use later on. But this, I guess, might naturally lead to the question, what if we go, everything we've been dealing with so far have been subsets, in the case of this triangle, or subspaces, in the case of V. But what if I were to take the image of Rn under T? This is the image of Rn under T. Let's think about what this means."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the 0 vector is definitely also a member of TV. So we come on the result that the image of V under T is a subspace, which is a useful result, which we'll be able to use later on. But this, I guess, might naturally lead to the question, what if we go, everything we've been dealing with so far have been subsets, in the case of this triangle, or subspaces, in the case of V. But what if I were to take the image of Rn under T? This is the image of Rn under T. Let's think about what this means. This means, what do we get when we take any member of Rn, what is the set of all of the vectors, when we take the transformation of all of the members of Rn? Let me write this. This is equal to the set of the transformation of all of the x's where each x is a member of Rn."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the image of Rn under T. Let's think about what this means. This means, what do we get when we take any member of Rn, what is the set of all of the vectors, when we take the transformation of all of the members of Rn? Let me write this. This is equal to the set of the transformation of all of the x's where each x is a member of Rn. So you take each of the members of Rn and transform them, and you create this new set. This is the image of Rn under T. Well, there's a couple of ways you can think of this. Remember when we defined, let's see, T is a mapping from Rn to Rm."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to the set of the transformation of all of the x's where each x is a member of Rn. So you take each of the members of Rn and transform them, and you create this new set. This is the image of Rn under T. Well, there's a couple of ways you can think of this. Remember when we defined, let's see, T is a mapping from Rn to Rm. We defined this as the domain, all of the possible inputs for our transformation, and we defined this the codomain. Remember I told you that the codomain is essentially part of the definition of the function or of the transformation, and it's the space that we map to. It's not necessarily all of the things that we're mapping to."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember when we defined, let's see, T is a mapping from Rn to Rm. We defined this as the domain, all of the possible inputs for our transformation, and we defined this the codomain. Remember I told you that the codomain is essentially part of the definition of the function or of the transformation, and it's the space that we map to. It's not necessarily all of the things that we're mapping to. For example, the image of Rn under transformation, maybe it's all of Rm, or maybe it's some subset of Rn. The way you can think about it, and I touched on this in that first video, and at least the linear algebra books I looked at, they didn't specify this, but you can kind of view this as the range of T. These are the actual members of Rm that T maps to. That if you take the image of Rn under T, you're actually finding, so let's say that Rm looks like that."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's not necessarily all of the things that we're mapping to. For example, the image of Rn under transformation, maybe it's all of Rm, or maybe it's some subset of Rn. The way you can think about it, and I touched on this in that first video, and at least the linear algebra books I looked at, they didn't specify this, but you can kind of view this as the range of T. These are the actual members of Rm that T maps to. That if you take the image of Rn under T, you're actually finding, so let's say that Rm looks like that. Obviously it'll go in every direction. And let's say that when you take, let me draw Rn right here, say this is Rn, and we know that T is a mapping from Rn to Rm, but let's say when you take every element of Rn and you map them into Rm, let's say you get some subset of Rm, let's say you get something that looks like this. So let me see if I can draw this nicely."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That if you take the image of Rn under T, you're actually finding, so let's say that Rm looks like that. Obviously it'll go in every direction. And let's say that when you take, let me draw Rn right here, say this is Rn, and we know that T is a mapping from Rn to Rm, but let's say when you take every element of Rn and you map them into Rm, let's say you get some subset of Rm, let's say you get something that looks like this. So let me see if I can draw this nicely. So you literally map every point here, and it goes to one of these guys, or one of these guys can be represented as a mapping from one of these members right here. So if you map all of them, you get this subset right here. This subset is T, the image of Rn under T, and in the terminology that you don't normally see in linear algebra a lot, you can also kind of consider it its range."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me see if I can draw this nicely. So you literally map every point here, and it goes to one of these guys, or one of these guys can be represented as a mapping from one of these members right here. So if you map all of them, you get this subset right here. This subset is T, the image of Rn under T, and in the terminology that you don't normally see in linear algebra a lot, you can also kind of consider it its range. The range of T. Now, this has a special name. This is called, and I don't want you to get confused, this is called the image of T. It might be a little confusing, image of T. So this is sometimes written as just Im of T. Now, you're a little confused here. You're like, before when we were talking about subsets, we would call this the image of our subset under T, and that is the correct terminology when you're dealing with a subset."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This subset is T, the image of Rn under T, and in the terminology that you don't normally see in linear algebra a lot, you can also kind of consider it its range. The range of T. Now, this has a special name. This is called, and I don't want you to get confused, this is called the image of T. It might be a little confusing, image of T. So this is sometimes written as just Im of T. Now, you're a little confused here. You're like, before when we were talking about subsets, we would call this the image of our subset under T, and that is the correct terminology when you're dealing with a subset. But when you take all of a sudden the entire n-dimensional space, and you're finding that image, we call that the image of the actual transformation. So we can also call this set right here, we can call this the image of T. And now what is the image of T? Well, we know that we can write any, and this is literally any."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're like, before when we were talking about subsets, we would call this the image of our subset under T, and that is the correct terminology when you're dealing with a subset. But when you take all of a sudden the entire n-dimensional space, and you're finding that image, we call that the image of the actual transformation. So we can also call this set right here, we can call this the image of T. And now what is the image of T? Well, we know that we can write any, and this is literally any. So T is going from Rn to Rm. We can write T of x, we can write any linear transformation like this as being equal to some matrix, some m by n matrix, m by n, times a vector. And these vectors obviously are going to be members of Rn."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we know that we can write any, and this is literally any. So T is going from Rn to Rm. We can write T of x, we can write any linear transformation like this as being equal to some matrix, some m by n matrix, m by n, times a vector. And these vectors obviously are going to be members of Rn. Rn. Times some Rn. And what is this?"}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And these vectors obviously are going to be members of Rn. Rn. Times some Rn. And what is this? So what is the image? Let me write it in a bunch of different ways. What is the image of Rn under T?"}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is this? So what is the image? Let me write it in a bunch of different ways. What is the image of Rn under T? So we could write that as T of Rn, which is the same thing as the image of T. Notice we're not saying under anything else, because now we're saying the image of the actual transformation, which we could also write as the image of T. Well, what are these equal to? These are equal to the sets, this is equal to the set of all the transformations of x. Well, all the transformations of x are going to be Ax, where x is a member of Rn."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is the image of Rn under T? So we could write that as T of Rn, which is the same thing as the image of T. Notice we're not saying under anything else, because now we're saying the image of the actual transformation, which we could also write as the image of T. Well, what are these equal to? These are equal to the sets, this is equal to the set of all the transformations of x. Well, all the transformations of x are going to be Ax, where x is a member of Rn. So x is going to be an n-tuple where each element just has to be a real number. So what is this? So if we write A, let me write my matrix A, it's just a bunch of column vectors, A1, A2, it's going to have n of these, because it has n columns."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, all the transformations of x are going to be Ax, where x is a member of Rn. So x is going to be an n-tuple where each element just has to be a real number. So what is this? So if we write A, let me write my matrix A, it's just a bunch of column vectors, A1, A2, it's going to have n of these, because it has n columns. And so A times any x is going to be, so if I multiply that times any x that's a member of Rn, so I multiply x1, x2, all the way to xn, we've seen this multiple, multiple times. This is equal to x1, the scalar x1, times A1 plus x2 times A2, all the way to plus xn times An. And we're saying we want the set of all of these sums of these column vectors where x can take on any vector in Rn, which means that the elements of x can take on any real scalar values."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we write A, let me write my matrix A, it's just a bunch of column vectors, A1, A2, it's going to have n of these, because it has n columns. And so A times any x is going to be, so if I multiply that times any x that's a member of Rn, so I multiply x1, x2, all the way to xn, we've seen this multiple, multiple times. This is equal to x1, the scalar x1, times A1 plus x2 times A2, all the way to plus xn times An. And we're saying we want the set of all of these sums of these column vectors where x can take on any vector in Rn, which means that the elements of x can take on any real scalar values. So the set of all of these is essentially all of the linear combinations of the columns of A, right? Because I can set these guys to be equal to any value. So what is that equal to?"}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we're saying we want the set of all of these sums of these column vectors where x can take on any vector in Rn, which means that the elements of x can take on any real scalar values. So the set of all of these is essentially all of the linear combinations of the columns of A, right? Because I can set these guys to be equal to any value. So what is that equal to? That is equal to, and we touched on this, or we actually talked about this when we introduced the idea, this is equal to the column span, sorry, the column space of A, or we just denoted it sometimes as C of A. So that's a pretty neat result. If you take, well, it's almost obvious, I mean, it's just I'm playing with words a little bit, but any linear transformation can be represented as a matrix vector product."}, {"video_title": "im(T) Image of a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is that equal to? That is equal to, and we touched on this, or we actually talked about this when we introduced the idea, this is equal to the column span, sorry, the column space of A, or we just denoted it sometimes as C of A. So that's a pretty neat result. If you take, well, it's almost obvious, I mean, it's just I'm playing with words a little bit, but any linear transformation can be represented as a matrix vector product. And so the image of any linear transformation, which means the subset of its codomain, when you map all of the elements of its domain into its codomain, this is the image of your transformation, this is equivalent to the column space of the matrix that your transformation could be represented as. And the column space, of course, is the span of all the column vectors of your matrix. This is just all of the linear combinations, or the span of all of your column vectors, which we do right here."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We saw this several videos ago. And the definition of our linear transformation, or the composition of our linear transformation, so the composition of s with t applied to some vector x in our set x, our domain, is equal to s of t of x. This was our definition. And then we went on and we said, look, if s of x can be represented as the matrix multiplication A x, the matrix vector product, and if t of x can be represented, or the transformation t can be represented as the product of the matrix B with x, we saw that this thing right here, which is, of course, if we just write it this way, this is equal to A times t times x, which is just Bx. We saw in multiple videos now that this is equivalent to, by our definition of matrix products, this is equal to the matrix AB, right? When you take the product of two matrices, you just get another matrix, the product AB times x. So you take, essentially, the first linear transformation in your composition, its matrix, which was A, and then you take the product with the second one."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we went on and we said, look, if s of x can be represented as the matrix multiplication A x, the matrix vector product, and if t of x can be represented, or the transformation t can be represented as the product of the matrix B with x, we saw that this thing right here, which is, of course, if we just write it this way, this is equal to A times t times x, which is just Bx. We saw in multiple videos now that this is equivalent to, by our definition of matrix products, this is equal to the matrix AB, right? When you take the product of two matrices, you just get another matrix, the product AB times x. So you take, essentially, the first linear transformation in your composition, its matrix, which was A, and then you take the product with the second one. Fair enough. All of this is review so far. So let's take three linear transformations."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you take, essentially, the first linear transformation in your composition, its matrix, which was A, and then you take the product with the second one. Fair enough. All of this is review so far. So let's take three linear transformations. Let's say that I have the linear transformation H, and when I apply that to a vector x, it's equivalent to multiplying my vector x by the matrix A. Let's say I have the linear transformation G. When I apply that to a vector x, it's equivalent to multiplying that vector. There should be a new concept called a vectrix."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's take three linear transformations. Let's say that I have the linear transformation H, and when I apply that to a vector x, it's equivalent to multiplying my vector x by the matrix A. Let's say I have the linear transformation G. When I apply that to a vector x, it's equivalent to multiplying that vector. There should be a new concept called a vectrix. It's equivalent to multiplying that vector times the matrix B, and then I have a final linear transformation F. Applied, when it's applied to some vector x, it's equivalent to multiplying that vector x times the matrix C. Now, what I'm curious about is what happens when I take the composition of H with G, and then I take the composition of that with F, these are all linear transformations, and then I apply that to some vector x, which is necessarily going to be in the domain of this guy. I haven't actually drawn out their domain and codomain definitions, but I think you get the idea. So let's explore what this is a little bit."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "There should be a new concept called a vectrix. It's equivalent to multiplying that vector times the matrix B, and then I have a final linear transformation F. Applied, when it's applied to some vector x, it's equivalent to multiplying that vector x times the matrix C. Now, what I'm curious about is what happens when I take the composition of H with G, and then I take the composition of that with F, these are all linear transformations, and then I apply that to some vector x, which is necessarily going to be in the domain of this guy. I haven't actually drawn out their domain and codomain definitions, but I think you get the idea. So let's explore what this is a little bit. Well, by the definition of what composition even means, we can just apply that to this right here. So we could just imagine this as being our S. If we imagine this was our S, and then this is our T right there, then what is this going to be equal to? If we just do a straight up pattern match right there, this is going to be equal to S, the transformation S, applied to the transformation F applied to x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's explore what this is a little bit. Well, by the definition of what composition even means, we can just apply that to this right here. So we could just imagine this as being our S. If we imagine this was our S, and then this is our T right there, then what is this going to be equal to? If we just do a straight up pattern match right there, this is going to be equal to S, the transformation S, applied to the transformation F applied to x. So S is H of G. So it is H, or I shouldn't say H of G, the composition of H with G, that is our S. And then I apply that to F applied to x. F is our T. I apply that to F applied to x, just like that. Now, what is this equal to? Now we can imagine that this is our x, if we just pattern match according to this definition, that this is this guy right here, that this is our T, and that this is our S. And so if we just pattern match here, this is equal to what?"}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we just do a straight up pattern match right there, this is going to be equal to S, the transformation S, applied to the transformation F applied to x. So S is H of G. So it is H, or I shouldn't say H of G, the composition of H with G, that is our S. And then I apply that to F applied to x. F is our T. I apply that to F applied to x, just like that. Now, what is this equal to? Now we can imagine that this is our x, if we just pattern match according to this definition, that this is this guy right here, that this is our T, and that this is our S. And so if we just pattern match here, this is equal to what? This is just straight from the definition of a composition. So it's equal to S of, S is our H, so H of T, which in this case is G, G applied to x. But instead of an x, we have a whole, this vector here, which was the transformation F applied to x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now we can imagine that this is our x, if we just pattern match according to this definition, that this is this guy right here, that this is our T, and that this is our S. And so if we just pattern match here, this is equal to what? This is just straight from the definition of a composition. So it's equal to S of, S is our H, so H of T, which in this case is G, G applied to x. But instead of an x, we have a whole, this vector here, which was the transformation F applied to x. So G of F of x. That's what this is equal to. The composition of H with G, or the composition of F with the composition of H and G, all of that applied to x is equal to H of G of F of x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But instead of an x, we have a whole, this vector here, which was the transformation F applied to x. So G of F of x. That's what this is equal to. The composition of H with G, or the composition of F with the composition of H and G, all of that applied to x is equal to H of G of F of x. Now what is this equal to? What is this equal to? Well, this is equal to, I'll do it right here, this is equal to H, the transformation H applied to, what is this term right here?"}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The composition of H with G, or the composition of F with the composition of H and G, all of that applied to x is equal to H of G of F of x. Now what is this equal to? What is this equal to? Well, this is equal to, I'll do it right here, this is equal to H, the transformation H applied to, what is this term right here? I'll do it in pink. What is this? That is the composition of G and F applied to x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is equal to, I'll do it right here, this is equal to H, the transformation H applied to, what is this term right here? I'll do it in pink. What is this? That is the composition of G and F applied to x. You can just replace S with G and F with T and you'll get that right there. So this is just equal to the composition of G with F applied to x. That's all that is."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is the composition of G and F applied to x. You can just replace S with G and F with T and you'll get that right there. So this is just equal to the composition of G with F applied to x. That's all that is. Now, what is this equal to right there? And it's probably confusing to see two parentheses in different colors, but I think you get the idea. What is this equal to?"}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's all that is. Now, what is this equal to right there? And it's probably confusing to see two parentheses in different colors, but I think you get the idea. What is this equal to? Well, just go back to your definition of the composition. I just want to make it very clear what we're doing. This is, if you imagine this being your T and then this being your S, this is just the composition of S with T applied to x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is this equal to? Well, just go back to your definition of the composition. I just want to make it very clear what we're doing. This is, if you imagine this being your T and then this being your S, this is just the composition of S with T applied to x. So this is just equal to, I'll write it like this way, this is equal to, I just shouldn't write S's, this is the composition of H with the composition of G and F. And then all of that applied to x. Now why did I do all of this? Well, one, to show you that the composition is associative."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is, if you imagine this being your T and then this being your S, this is just the composition of S with T applied to x. So this is just equal to, I'll write it like this way, this is equal to, I just shouldn't write S's, this is the composition of H with the composition of G and F. And then all of that applied to x. Now why did I do all of this? Well, one, to show you that the composition is associative. I went all the way here and then I went all the way back. And essentially, it doesn't matter where you put the parentheses. The composition of the composition of H with G with F is equivalent to the composition of H with the composition of G and F. These two things are equivalent."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, one, to show you that the composition is associative. I went all the way here and then I went all the way back. And essentially, it doesn't matter where you put the parentheses. The composition of the composition of H with G with F is equivalent to the composition of H with the composition of G and F. These two things are equivalent. And essentially, you can just rewrite them. The parentheses are essentially unnecessary. You can write this as the composition of H with G with F, all of that applied to x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The composition of the composition of H with G with F is equivalent to the composition of H with the composition of G and F. These two things are equivalent. And essentially, you can just rewrite them. The parentheses are essentially unnecessary. You can write this as the composition of H with G with F, all of that applied to x. Now, I took the time to say that I can represent H as a matrix, that each of these linear transformations I can represent as matrix multiplications. Why did I do that? Well, we saw before that any composition, when you take the composition of S with T, the matrix version of this transformation, of this composition, is going to be equal to the product, by our definition of matrix-matrix products, the product of the S's transformation matrix and T's transformation matrix."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You can write this as the composition of H with G with F, all of that applied to x. Now, I took the time to say that I can represent H as a matrix, that each of these linear transformations I can represent as matrix multiplications. Why did I do that? Well, we saw before that any composition, when you take the composition of S with T, the matrix version of this transformation, of this composition, is going to be equal to the product, by our definition of matrix-matrix products, the product of the S's transformation matrix and T's transformation matrix. So what are these going to be equal to? So this one right here, if you think of this transformation right here, this statement right here, it's matrix version of it. So let me write that, the matrix version of the composition of H with G, and then the composition of that with F applied to x, is going to be equal to, and we've seen this before, the product of these matrices."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we saw before that any composition, when you take the composition of S with T, the matrix version of this transformation, of this composition, is going to be equal to the product, by our definition of matrix-matrix products, the product of the S's transformation matrix and T's transformation matrix. So what are these going to be equal to? So this one right here, if you think of this transformation right here, this statement right here, it's matrix version of it. So let me write that, the matrix version of the composition of H with G, and then the composition of that with F applied to x, is going to be equal to, and we've seen this before, the product of these matrices. So this composition, its matrix is going to be AB. H and G, their matrices are A and B. So it's going to be AB and I'll do it in parentheses."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that, the matrix version of the composition of H with G, and then the composition of that with F applied to x, is going to be equal to, and we've seen this before, the product of these matrices. So this composition, its matrix is going to be AB. H and G, their matrices are A and B. So it's going to be AB and I'll do it in parentheses. And then you take that matrix, and you take the product, so this guy's matrix representation is AB, and this guy's matrix representation is C. So the matrix representation of this whole thing is this guy taking the product of AB and then taking the product of that with C. So AB and then C. And then if you look at this guy right here, and of course all of that times a vector x. All of that times some vector x right there. That's the vector x."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be AB and I'll do it in parentheses. And then you take that matrix, and you take the product, so this guy's matrix representation is AB, and this guy's matrix representation is C. So the matrix representation of this whole thing is this guy taking the product of AB and then taking the product of that with C. So AB and then C. And then if you look at this guy right here, and of course all of that times a vector x. All of that times some vector x right there. That's the vector x. Now let's look at this one right here. If we take the composition of H with the composition of G and F and apply all of that to some vector x, what is that equivalent to? Well, this composition right here, the matrix version of it I guess we can say, is going to be the product BC."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the vector x. Now let's look at this one right here. If we take the composition of H with the composition of G and F and apply all of that to some vector x, what is that equivalent to? Well, this composition right here, the matrix version of it I guess we can say, is going to be the product BC. And we're going to apply that to x. So we're going to have the product BC. And then we're going to take the product of that with this guy's matrix representation, which is A."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this composition right here, the matrix version of it I guess we can say, is going to be the product BC. And we're going to apply that to x. So we're going to have the product BC. And then we're going to take the product of that with this guy's matrix representation, which is A. And we've shown this before. We never showed it with 3, but it extends. I mean, I kind of showed it extends."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to take the product of that with this guy's matrix representation, which is A. And we've shown this before. We never showed it with 3, but it extends. I mean, I kind of showed it extends. You can just keep applying the definition. You can keep applying this property right here, and so it'll just naturally extend. Because every time we're just taking the composition of 2 things, even though it looks like we're taking the composition of 3, we're taking the composition of 2 things first here, and then we get its matrix representation."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I mean, I kind of showed it extends. You can just keep applying the definition. You can keep applying this property right here, and so it'll just naturally extend. Because every time we're just taking the composition of 2 things, even though it looks like we're taking the composition of 3, we're taking the composition of 2 things first here, and then we get its matrix representation. And then we take the composition of that with this other thing. So the matrix representation of the entire composition is going to be this matrix times this matrix, which I did here. Similarly, here we take first the matrix, the composition of these 2 linear transformations, and their matrix representation will be that right there."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because every time we're just taking the composition of 2 things, even though it looks like we're taking the composition of 3, we're taking the composition of 2 things first here, and then we get its matrix representation. And then we take the composition of that with this other thing. So the matrix representation of the entire composition is going to be this matrix times this matrix, which I did here. Similarly, here we take first the matrix, the composition of these 2 linear transformations, and their matrix representation will be that right there. And then we take the linear, we take the composition of that with that. So the entire matrix representation is going to be this guy's matrix times this guy's matrix. So A times BC."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Similarly, here we take first the matrix, the composition of these 2 linear transformations, and their matrix representation will be that right there. And then we take the linear, we take the composition of that with that. So the entire matrix representation is going to be this guy's matrix times this guy's matrix. So A times BC. And of course, all of that applied to the vector x. Now, in this video, I've showed you that these 2 things are equivalent. If anything, the parentheses are completely unnecessary."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A times BC. And of course, all of that applied to the vector x. Now, in this video, I've showed you that these 2 things are equivalent. If anything, the parentheses are completely unnecessary. And I showed you that there. They both essentially come boiled down to h of g of f of x, so these 2 things are equivalent. So we could say, essentially, that these 2 things over here are equivalent."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If anything, the parentheses are completely unnecessary. And I showed you that there. They both essentially come boiled down to h of g of f of x, so these 2 things are equivalent. So we could say, essentially, that these 2 things over here are equivalent. Or that AB, the product AB, and then taking the product of that matrix with the matrix C, is equivalent to taking the product A with the matrix BC, which is just another product matrix. Or another way of saying it is that these parentheses don't matter, that all of these is just equal to ABC. Or, I mean, this is just a statement, that matrix products exhibit the associative property."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say, essentially, that these 2 things over here are equivalent. Or that AB, the product AB, and then taking the product of that matrix with the matrix C, is equivalent to taking the product A with the matrix BC, which is just another product matrix. Or another way of saying it is that these parentheses don't matter, that all of these is just equal to ABC. Or, I mean, this is just a statement, that matrix products exhibit the associative property. It doesn't matter where you put the parentheses. And sometimes it's confusing the word associative. It just means it doesn't matter where you put the parentheses."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or, I mean, this is just a statement, that matrix products exhibit the associative property. It doesn't matter where you put the parentheses. And sometimes it's confusing the word associative. It just means it doesn't matter where you put the parentheses. Matrix products do not exhibit the commutative property. We saw that in the last video. We cannot, in general, make the statement that AB is equal to BA."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It just means it doesn't matter where you put the parentheses. Matrix products do not exhibit the commutative property. We saw that in the last video. We cannot, in general, make the statement that AB is equal to BA. We cannot do that. And in fact, in the last video, I think it was the last video, I showed you that if AB is defined, sometimes BA is not even defined. Or if BA is defined, sometimes AB isn't defined."}, {"video_title": "Matrix product associativity Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We cannot, in general, make the statement that AB is equal to BA. We cannot do that. And in fact, in the last video, I think it was the last video, I showed you that if AB is defined, sometimes BA is not even defined. Or if BA is defined, sometimes AB isn't defined. So it's not commutative. It is associative, though. In the next video, I'll see if matrix products are actually distributive."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then my set that I'm mapping into, set Y, that's the co-domain. We know that T is a transformation that if you take any member of X and you transform it, you'll get, or you'll associate it with a member of set Y. You'll map it to a member of set Y. That's what the transformation or the function does. Now, if we have some subset of T, let's call A to be some subset of T. So let me draw A like that. This notation right here just means subset. Some subset of T. We've defined the notion of an image of T, of A like that, which is the image of A, of our subset A under T. We've defined this to be equal to the set, let me write it here, the set of all, where if we take each of the members of our subset, so if we take each of the members of our subset, it's the set of all of their transformations."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what the transformation or the function does. Now, if we have some subset of T, let's call A to be some subset of T. So let me draw A like that. This notation right here just means subset. Some subset of T. We've defined the notion of an image of T, of A like that, which is the image of A, of our subset A under T. We've defined this to be equal to the set, let me write it here, the set of all, where if we take each of the members of our subset, so if we take each of the members of our subset, it's the set of all of their transformations. And of course, these are going to be some subset of Y right there. So we essentially take each of the members of A, this was one of them, you find its transformation, it's that point. You take another member of A, this is all set A right here, take another member of A, find its transformation, maybe it's that point."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Some subset of T. We've defined the notion of an image of T, of A like that, which is the image of A, of our subset A under T. We've defined this to be equal to the set, let me write it here, the set of all, where if we take each of the members of our subset, so if we take each of the members of our subset, it's the set of all of their transformations. And of course, these are going to be some subset of Y right there. So we essentially take each of the members of A, this was one of them, you find its transformation, it's that point. You take another member of A, this is all set A right here, take another member of A, find its transformation, maybe it's that point. You keep doing that, find its transformation, maybe it's that point. And then the set of all of those transformations, maybe it's this blob right here, we call this the image of A under T. Now, what if we wanted to think about the opposite problem? What if we were to start with set Y, which is our codomain, so that's Y, and we were to have some subset of Y, let's call our subset of Y S. So S is a subset of our codomain Y, and I'm curious about what subset of X maps into S. I'm curious about this set, I'm curious about the set of all vectors that are members of my domain, such that their mapping, or the transformation of those vectors, ends up in my subset of S. So what I'm saying is, look, if I take my domain, there must be some subset of vectors right here, where if I take any member of this set, it will map into these guys."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You take another member of A, this is all set A right here, take another member of A, find its transformation, maybe it's that point. You keep doing that, find its transformation, maybe it's that point. And then the set of all of those transformations, maybe it's this blob right here, we call this the image of A under T. Now, what if we wanted to think about the opposite problem? What if we were to start with set Y, which is our codomain, so that's Y, and we were to have some subset of Y, let's call our subset of Y S. So S is a subset of our codomain Y, and I'm curious about what subset of X maps into S. I'm curious about this set, I'm curious about the set of all vectors that are members of my domain, such that their mapping, or the transformation of those vectors, ends up in my subset of S. So what I'm saying is, look, if I take my domain, there must be some subset of vectors right here, where if I take any member of this set, it will map into these guys. And that's what I'm defining right here, this is equal to that guy. So I'm literally saying, what are all of the members of X, where those members of X all map into S. Now, I want to make a very subtle nuance here, or point out something here. I'm not saying that every point in S necessarily gets mapped to."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What if we were to start with set Y, which is our codomain, so that's Y, and we were to have some subset of Y, let's call our subset of Y S. So S is a subset of our codomain Y, and I'm curious about what subset of X maps into S. I'm curious about this set, I'm curious about the set of all vectors that are members of my domain, such that their mapping, or the transformation of those vectors, ends up in my subset of S. So what I'm saying is, look, if I take my domain, there must be some subset of vectors right here, where if I take any member of this set, it will map into these guys. And that's what I'm defining right here, this is equal to that guy. So I'm literally saying, what are all of the members of X, where those members of X all map into S. Now, I want to make a very subtle nuance here, or point out something here. I'm not saying that every point in S necessarily gets mapped to. For example, maybe there's some element in S right here, right there, that no element in X ever gets mapped to from our transformation T. That's okay. All I'm saying is, is that everything in this set maps to something within S right here. And what we call this, we call this set right here, we call this, the notation is the inverse T of S, but this is equal to the preimage of S under T. So this is S, this is the preimage of S under T, and that makes sense."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm not saying that every point in S necessarily gets mapped to. For example, maybe there's some element in S right here, right there, that no element in X ever gets mapped to from our transformation T. That's okay. All I'm saying is, is that everything in this set maps to something within S right here. And what we call this, we call this set right here, we call this, the notation is the inverse T of S, but this is equal to the preimage of S under T. So this is S, this is the preimage of S under T, and that makes sense. The image, we go from a subset of our domain to a subset of our codomain. Preimage, we go from a subset of our codomain, and we say what subset of our domain maps into that subset of our codomain. Now, let me ask you an interesting question, and this is kind of for bonus points."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what we call this, we call this set right here, we call this, the notation is the inverse T of S, but this is equal to the preimage of S under T. So this is S, this is the preimage of S under T, and that makes sense. The image, we go from a subset of our domain to a subset of our codomain. Preimage, we go from a subset of our codomain, and we say what subset of our domain maps into that subset of our codomain. Now, let me ask you an interesting question, and this is kind of for bonus points. What is the image of our preimage under S? So if we take this guy, this is essentially the image of this guy right here. This part right here is the preimage of S right here."}, {"video_title": "Preimage of a set Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let me ask you an interesting question, and this is kind of for bonus points. What is the image of our preimage under S? So if we take this guy, this is essentially the image of this guy right here. This part right here is the preimage of S right here. Now if we take the image of this, we're saying what, if we take every member of this, what vectors do they map into? All of them are going to be within S, so they're going to map within S, but they don't necessarily map to everything within S. So this is going to be some subset of S. So this right here is going to be some subset of our original S. It's not necessarily equal to S, but it's a subset of it, and so this is, I think, the motivation for where the notation comes from. We can construct a subset of S by taking the image of the preimage of S. You can kind of view the image and the preimage kind of cancelling out, and that's why the inverse notation was probably introduced."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have some function f, and it's a mapping from the set x to y. So if I were to draw the set x right there, that's my set x. And then if I were to draw the set y just like that, we know and I've done this several videos ago, that a function just associates any member of our set x. So I have some member of my set x there. If I apply the function to it, or if we're dealing with vectors, we could imagine instead of using the word function, we would use the word transformation, but it's the same thing. We would associate with this element, or this member of x, a member of y. So that's why we call it a mapping."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I have some member of my set x there. If I apply the function to it, or if we're dealing with vectors, we could imagine instead of using the word function, we would use the word transformation, but it's the same thing. We would associate with this element, or this member of x, a member of y. So that's why we call it a mapping. It says, hey, that guy, when I apply this function, let me do it in a different color, this little member right there is associated, this little member of x is associated with this member of y. And if this is right here, this is a capital X. So let's say we call this a, and let's call that b."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's why we call it a mapping. It says, hey, that guy, when I apply this function, let me do it in a different color, this little member right there is associated, this little member of x is associated with this member of y. And if this is right here, this is a capital X. So let's say we call this a, and let's call that b. And we would say that the function where a is a member of x and b is a member of y, we would say that f of a is equal to b. This is all a review of everything that we've learned already about functions. Now I'm going to define a couple of interesting functions."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we call this a, and let's call that b. And we would say that the function where a is a member of x and b is a member of y, we would say that f of a is equal to b. This is all a review of everything that we've learned already about functions. Now I'm going to define a couple of interesting functions. The first one, and I guess it's really just one function, I said it's a couple, but I'll call it the identity function. And this is a function, I'll just call it a big capital I, and its identity function operates on some set. So let's say this is the identity function on set x, and it's a mapping from x to x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now I'm going to define a couple of interesting functions. The first one, and I guess it's really just one function, I said it's a couple, but I'll call it the identity function. And this is a function, I'll just call it a big capital I, and its identity function operates on some set. So let's say this is the identity function on set x, and it's a mapping from x to x. And what's interesting about the identity function is that if you give it some a that is a member of x, so let's say you give it that a, the identity function applied to that member of x, so the identity function of a, is going to be equal to a. So it literally just maps things back to itself. So the identity function, if I were to draw it on this diagram right here, would look like this."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say this is the identity function on set x, and it's a mapping from x to x. And what's interesting about the identity function is that if you give it some a that is a member of x, so let's say you give it that a, the identity function applied to that member of x, so the identity function of a, is going to be equal to a. So it literally just maps things back to itself. So the identity function, if I were to draw it on this diagram right here, would look like this. It would look like, let me pick a nice suitable color, it would look like this. It would just kind of be a circle. It just points back at the point that you started off with, it associates all points with themselves."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the identity function, if I were to draw it on this diagram right here, would look like this. It would look like, let me pick a nice suitable color, it would look like this. It would just kind of be a circle. It just points back at the point that you started off with, it associates all points with themselves. That's the identity function on x. If it, especially as it applies to the point a, if you apply it to some other point in x, it would just refer back to itself. Now that's the identity function on x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It just points back at the point that you started off with, it associates all points with themselves. That's the identity function on x. If it, especially as it applies to the point a, if you apply it to some other point in x, it would just refer back to itself. Now that's the identity function on x. You could also have an identity function on y. So let's say that b is a member of y, so I drew b right there, then the y identity function, so this would be the identity function on y, applied to b, would just refer back to itself. And so that would be equal to b."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now that's the identity function on x. You could also have an identity function on y. So let's say that b is a member of y, so I drew b right there, then the y identity function, so this would be the identity function on y, applied to b, would just refer back to itself. And so that would be equal to b. This is the identity function on y. And so you might say, hey Sal, these are kind of silly functions, but we'll use them. They're actually at least a useful notation to use as we kind of progress through our explorations of linear algebra."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so that would be equal to b. This is the identity function on y. And so you might say, hey Sal, these are kind of silly functions, but we'll use them. They're actually at least a useful notation to use as we kind of progress through our explorations of linear algebra. But I'm going to make a new definition. I'm going to say that a function, let me pick a nice color, pink, I'm going to say that a function, or let me say f, since we already established it right over here, I'm going to say that f is invertible. Invertible."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They're actually at least a useful notation to use as we kind of progress through our explorations of linear algebra. But I'm going to make a new definition. I'm going to say that a function, let me pick a nice color, pink, I'm going to say that a function, or let me say f, since we already established it right over here, I'm going to say that f is invertible. Invertible. Introducing some new terminology. f is invertible if and only if the following is true. So if and only if the following is true."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Invertible. Introducing some new terminology. f is invertible if and only if the following is true. So if and only if the following is true. I could either write it with this two-way arrows like that, or I could write it as if, with two f's. That means that if this is true, then this is true, and only if this is true. So this implies that, and that implies this."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if and only if the following is true. I could either write it with this two-way arrows like that, or I could write it as if, with two f's. That means that if this is true, then this is true, and only if this is true. So this implies that, and that implies this. So f is invertible. I'm kind of making a definition right here. If and only if there exists a function, we'll call it, I'll call it nothing just yet."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this implies that, and that implies this. So f is invertible. I'm kind of making a definition right here. If and only if there exists a function, we'll call it, I'll call it nothing just yet. I'll call it something in a second. And I'll write it as this f with this negative 1 superscript on it. So f is invertible if and only if there exists a function, f inverse, well, I guess I just called it something, f inverse such that, let me do that in purple, such that if I apply f, remember f is just a mapping from x to y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If and only if there exists a function, we'll call it, I'll call it nothing just yet. I'll call it something in a second. And I'll write it as this f with this negative 1 superscript on it. So f is invertible if and only if there exists a function, f inverse, well, I guess I just called it something, f inverse such that, let me do that in purple, such that if I apply f, remember f is just a mapping from x to y. So this function f inverse is going to be a mapping from y to x. So I'm saying that f is invertible if there exists a function, f inverse, that's a mapping from y to x such that if I apply, if I take the composition of f inverse with f, this is equal to the identity function over x. So let's think about what's happening."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So f is invertible if and only if there exists a function, f inverse, well, I guess I just called it something, f inverse such that, let me do that in purple, such that if I apply f, remember f is just a mapping from x to y. So this function f inverse is going to be a mapping from y to x. So I'm saying that f is invertible if there exists a function, f inverse, that's a mapping from y to x such that if I apply, if I take the composition of f inverse with f, this is equal to the identity function over x. So let's think about what's happening. This is just part of it, actually. Let me just complete the whole definition. This is true, this has to be true, and f, the composition of f with the identity function, has to be, sorry, the composition of f with the inverse function has to be equal to the identity function over y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's think about what's happening. This is just part of it, actually. Let me just complete the whole definition. This is true, this has to be true, and f, the composition of f with the identity function, has to be, sorry, the composition of f with the inverse function has to be equal to the identity function over y. So let's think about what's this saying. There's some function, f, we'll just call it, well, I'll call it right now, this is called the inverse of f. This is the inverse of f. And it's a mapping from y to x. So f was a mapping from, let me draw it up here."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is true, this has to be true, and f, the composition of f with the identity function, has to be, sorry, the composition of f with the inverse function has to be equal to the identity function over y. So let's think about what's this saying. There's some function, f, we'll just call it, well, I'll call it right now, this is called the inverse of f. This is the inverse of f. And it's a mapping from y to x. So f was a mapping from, let me draw it up here. So f is a mapping from x to y. We showed that. This is the mapping of f right there."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So f was a mapping from, let me draw it up here. So f is a mapping from x to y. We showed that. This is the mapping of f right there. It goes in that direction. We're saying there has to be some other function, f inverse, that's a mapping from y to x. So let's write it here."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the mapping of f right there. It goes in that direction. We're saying there has to be some other function, f inverse, that's a mapping from y to x. So let's write it here. So f inverse is a mapping from y to x. So f inverse, if you give me some value in set y, I go to set x. So this guy's domain is this guy's codomain."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's write it here. So f inverse is a mapping from y to x. So f inverse, if you give me some value in set y, I go to set x. So this guy's domain is this guy's codomain. And this guy's codomain is this guy's domain. Fair enough. But let's see what it's saying."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this guy's domain is this guy's codomain. And this guy's codomain is this guy's domain. Fair enough. But let's see what it's saying. It's saying that the composition of f inverse with f has to be equal to the identity matrix. So essentially it's saying, if I apply f to some value in x, if you think about what's this composition doing? This guy's going from x to y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But let's see what it's saying. It's saying that the composition of f inverse with f has to be equal to the identity matrix. So essentially it's saying, if I apply f to some value in x, if you think about what's this composition doing? This guy's going from x to y. And then this guy goes from y to x. So let's think about what's happening here. f is going from x to y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This guy's going from x to y. And then this guy goes from y to x. So let's think about what's happening here. f is going from x to y. And then f inverse is going from y to x. So this composition is going to be a mapping from x to x, which the identity matrix, or the identity function, needs to do. It needs to go from x to x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "f is going from x to y. And then f inverse is going from y to x. So this composition is going to be a mapping from x to x, which the identity matrix, or the identity function, needs to do. It needs to go from x to x. And they're saying this equals identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It needs to go from x to x. And they're saying this equals identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item. These two statements are equivalent. And so by definition, this thing is going to be your original thing. Or another way of writing this is that f inverse applied to f of a is going to be equal to a."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item. These two statements are equivalent. And so by definition, this thing is going to be your original thing. Or another way of writing this is that f inverse applied to f of a is going to be equal to a. That's what this first statement tells us. And if you think of it visually, it's saying you start with an a, you apply f to it, and you get this value right here, that is f of a. I'm saying it equals b, or I said it equals b earlier on. But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way of writing this is that f inverse applied to f of a is going to be equal to a. That's what this first statement tells us. And if you think of it visually, it's saying you start with an a, you apply f to it, and you get this value right here, that is f of a. I'm saying it equals b, or I said it equals b earlier on. But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this. By definition, it needs to go back to original a. It has to be equivalent to just doing this little closed loop right here when I introduce you to the identity function. Now that's what this statement is telling us right here."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this. By definition, it needs to go back to original a. It has to be equivalent to just doing this little closed loop right here when I introduce you to the identity function. Now that's what this statement is telling us right here. The second statement is saying, look, if I apply f to f inverse, I'm getting the identity function on y. So if I start at some point in y right there, and I apply f inverse first, maybe I go right here. Maybe this point, let's call that lowercase y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now that's what this statement is telling us right here. The second statement is saying, look, if I apply f to f inverse, I'm getting the identity function on y. So if I start at some point in y right there, and I apply f inverse first, maybe I go right here. Maybe this point, let's call that lowercase y. So this would be f inverse of lowercase y. And then if I were to apply f to that, I know this chart is getting very confusing. If I apply f to this right here, I need to go right back to my original y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe this point, let's call that lowercase y. So this would be f inverse of lowercase y. And then if I were to apply f to that, I know this chart is getting very confusing. If I apply f to this right here, I need to go right back to my original y. So when I apply f to f inverse of y, this has to be equivalent of just doing the identity function on y. So that's what the second statement is saying. Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I apply f to this right here, I need to go right back to my original y. So when I apply f to f inverse of y, this has to be equivalent of just doing the identity function on y. So that's what the second statement is saying. Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y. And you've been exposed to the idea of an inverse before. We're just doing it a little bit more precisely because we're going to start dealing with these notions, with transformations and matrices in the very near future. So it's good to be exposed to it in this kind of more precise form."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y. And you've been exposed to the idea of an inverse before. We're just doing it a little bit more precisely because we're going to start dealing with these notions, with transformations and matrices in the very near future. So it's good to be exposed to it in this kind of more precise form. Now the first thing you might ask is, hey, let's say that I have a function f. Let's say I have a function f. And there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question, is, is f inverse unique?"}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's good to be exposed to it in this kind of more precise form. Now the first thing you might ask is, hey, let's say that I have a function f. Let's say I have a function f. And there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question, is, is f inverse unique? And actually, probably the obvious question is, how do you know when something's invertible? And we're going to talk a lot about that in the very near future. But let's say we know that f is invertible."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The obvious question, or maybe it's not an obvious question, is, is f inverse unique? And actually, probably the obvious question is, how do you know when something's invertible? And we're going to talk a lot about that in the very near future. But let's say we know that f is invertible. How do we know, or do we know, whether f inverse is unique? And to answer that question, let's assume it's not unique. So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But let's say we know that f is invertible. How do we know, or do we know, whether f inverse is unique? And to answer that question, let's assume it's not unique. So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them. So let's say g is a mapping. Remember, f is a mapping from x to y. Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them. So let's say g is a mapping. Remember, f is a mapping from x to y. Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. So I'm saying that g is an inverse of f. This assumption implies these two things."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. So I'm saying that g is an inverse of f. This assumption implies these two things. Now let's say that h is another inverse. Let's say that h is another inverse of f. h is another inverse. By definition, by what I just called an inverse, h has to satisfy two requirements."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm saying that g is an inverse of f. This assumption implies these two things. Now let's say that h is another inverse. Let's say that h is another inverse of f. h is another inverse. By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from y to x. And then if I take the composition of h with f, I have to get the identity matrix on the set x. Now that wasn't just part of the definition."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from y to x. And then if I take the composition of h with f, I have to get the identity matrix on the set x. Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The inverse, the composition of the inverse with the function, has to become the identity matrix on x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The inverse, the composition of the inverse with the function, has to become the identity matrix on x. And then the composition of the function with the inverse has to be the identity function on y. So let's write that. So g is an inverse of f. It implies this."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The inverse, the composition of the inverse with the function, has to become the identity matrix on x. And then the composition of the function with the inverse has to be the identity function on y. So let's write that. So g is an inverse of f. It implies this. And it also implies that the composition of f with g is equal to the identity matrix. The identity function on y. And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So g is an inverse of f. It implies this. And it also implies that the composition of f with g is equal to the identity matrix. The identity function on y. And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. And just as a reminder, let me redraw what I drew at the beginning, just so we know what we're doing. So if this is the set x right here, let me do it in different color. Let's say this right here is the set y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. And just as a reminder, let me redraw what I drew at the beginning, just so we know what we're doing. So if this is the set x right here, let me do it in different color. Let's say this right here is the set y. We know that f is a mapping from x to y. What we're seeing is, what we're trying to determine is, is f's inverse unique? So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say this right here is the set y. We know that f is a mapping from x to y. What we're seeing is, what we're trying to determine is, is f's inverse unique? So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. So f does that. If you take g, you're going to go back to the same point. So it's equivalent."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. So f does that. If you take g, you're going to go back to the same point. So it's equivalent. So taking the composition of g with f, that means doing f first, then g. This is equivalent of just taking the identity function in x. So just taking an x and going back to an x. It's equivalent to that."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's equivalent. So taking the composition of g with f, that means doing f first, then g. This is equivalent of just taking the identity function in x. So just taking an x and going back to an x. It's equivalent to that. So this is g right here. And the same thing is true with h. h should also be, if I start with some element in x and go into y and then apply h like that, it should also be equivalent to the identity transformation. That's what this statement and this statement are saying."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equivalent to that. So this is g right here. And the same thing is true with h. h should also be, if I start with some element in x and go into y and then apply h like that, it should also be equivalent to the identity transformation. That's what this statement and this statement are saying. Now this statement is saying that if I start with some entry in y here and I apply g, which is the inverse of f, I'm going to go here. So g will take me there. And that when I apply f then to that, I'm going to go back to that same element of y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what this statement and this statement are saying. Now this statement is saying that if I start with some entry in y here and I apply g, which is the inverse of f, I'm going to go here. So g will take me there. And that when I apply f then to that, I'm going to go back to that same element of y. And that's equivalent to just doing the identity function on y. So that's the same thing as the identity function of y. And I could do the same thing here with h. Just take a point here, apply h, then apply f back."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that when I apply f then to that, I'm going to go back to that same element of y. And that's equivalent to just doing the identity function on y. So that's the same thing as the identity function of y. And I could do the same thing here with h. Just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I could do the same thing here with h. Just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions, g and h? So let's start with g. Remember, g is just a mapping from y to x. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's go back to the question of whether g is unique. Can we have two different inverse functions, g and h? So let's start with g. Remember, g is just a mapping from y to x. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly. So let's say this is x and this is y. Remember, g is a mapping from y to x. So g will take us there as a mapping from y to x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly. So let's say this is x and this is y. Remember, g is a mapping from y to x. So g will take us there as a mapping from y to x. And I'm saying that this g is equivalent to the identity mapping or the identity function in composition with this, because all this is saying is you apply g and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So g will take us there as a mapping from y to x. And I'm saying that this g is equivalent to the identity mapping or the identity function in composition with this, because all this is saying is you apply g and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well, by definition, if h is another inverse of f, this is true."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well, by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f and the composition of that with g. And you might want to put parentheses here. Actually, I'll do it very lightly. You might want to put parentheses there."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f and the composition of that with g. And you might want to put parentheses here. Actually, I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions or of transformations is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually, I'll do that."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions or of transformations is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually, I'll do that. I'll put the parentheses there at first, just so you can understand that this is the same thing as that right there, but we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to? The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Actually, I'll do that. I'll put the parentheses there at first, just so you can understand that this is the same thing as that right there, but we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to? The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember, h is a function, is a mapping from y to x."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember, h is a function, is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h takes some element in y and gives me some element in x. Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, h is a function, is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h takes some element in y and gives me some element in x. Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. That's the same thing as just applying h to the function to begin with. So just going through this little exercise, we've shown, even though we've started off saying, hey, I have these two different inverses, we've just shown that g must be equal to h. So any function has a unique inverse. You can't set up two different inverses."}, {"video_title": "Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. That's the same thing as just applying h to the function to begin with. So just going through this little exercise, we've shown, even though we've started off saying, hey, I have these two different inverses, we've just shown that g must be equal to h. So any function has a unique inverse. You can't set up two different inverses. And if you do, you'll find that they're always going to be equal to each other. So so far we know what an inverse is and we don't know what causes someone to be able to have an inverse or not. But we know if they have an inverse, how to think about it, and we also know that that inverse is unique."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "What I want to know is, what does this imply about this equation right here? The equation f of x is equal to y. I want to know that for every y that's a member of our codomain, so for every y, let me write this down, for every y that's a member of my codomain, is there a unique, I'll write it in caps, unique solution x that's a member of our domain such that, and I could write such that, well, I'll just write it out. I was going to write it the mathy way, but I think it's nicer to write it in the actual word sometimes. Such that f of x is equal to y. So let me just draw everything out a little bit. We have our set x right here. This is x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Such that f of x is equal to y. So let me just draw everything out a little bit. We have our set x right here. This is x. We have our codomain y here. We know that f, if you take some point here, let's call that a, it's a member of x. And you apply the function f to it, it'll map you to some element in set y."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "This is x. We have our codomain y here. We know that f, if you take some point here, let's call that a, it's a member of x. And you apply the function f to it, it'll map you to some element in set y. So that's f of a right there. This is so far what this tells us. Now, I want to look at this equation here."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And you apply the function f to it, it'll map you to some element in set y. So that's f of a right there. This is so far what this tells us. Now, I want to look at this equation here. And I want to know that if I can pick any y in this set, or any lowercase y in this set y, so let's say I pick something here, let's say that's b. I want to know is there a unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess you have to think, is there a solution?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Now, I want to look at this equation here. And I want to know that if I can pick any y in this set, or any lowercase y in this set y, so let's say I pick something here, let's say that's b. I want to know is there a unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess you have to think, is there a solution? So is there a solution is saying, look, is there some x here that if I apply the transformation f to it, that I get there? And I also want to know, is it unique? For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So one, I guess you have to think, is there a solution? So is there a solution is saying, look, is there some x here that if I apply the transformation f to it, that I get there? And I also want to know, is it unique? For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution. If there's some other guy in x that if I apply the transformation, I also go to b. This would make it non-unique. Yeah, not unique."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution. If there's some other guy in x that if I apply the transformation, I also go to b. This would make it non-unique. Yeah, not unique. So what I want to concern ourselves with in this video is somehow is invertibility related to the idea of a unique solution to this for any y in our codomain. So let's just work through our definitions of invertibility and see if we can get anywhere constructive. So by definition, f is invertible implies that there exists this little backward looking, three looking thing."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Yeah, not unique. So what I want to concern ourselves with in this video is somehow is invertibility related to the idea of a unique solution to this for any y in our codomain. So let's just work through our definitions of invertibility and see if we can get anywhere constructive. So by definition, f is invertible implies that there exists this little backward looking, three looking thing. This means there exists. I think it's nice to be exposed sometimes to the mathy notations, let me just write there, there exists. That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So by definition, f is invertible implies that there exists this little backward looking, three looking thing. This means there exists. I think it's nice to be exposed sometimes to the mathy notations, let me just write there, there exists. That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out. Such that the composition of f inverse with f is equal to the identity on x. So essentially it's saying, look, if I apply f to something in x, and then I apply f inverse to that, I'm going to get back to that point, which is essentially equivalent, or it isn't just essentially equivalent, it is equivalent to just applying the identity function. So that's ix."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out. Such that the composition of f inverse with f is equal to the identity on x. So essentially it's saying, look, if I apply f to something in x, and then I apply f inverse to that, I'm going to get back to that point, which is essentially equivalent, or it isn't just essentially equivalent, it is equivalent to just applying the identity function. So that's ix. So you just get what you put into it. Such that this inverse function, the composition of the inverse with the function is equal to the identity function, and that the composition of the function with the inverse function is equal to the identity function on y. So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So that's ix. So you just get what you put into it. Such that this inverse function, the composition of the inverse with the function is equal to the identity function, and that the composition of the function with the inverse function is equal to the identity function on y. So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function. So this is what invertibility tells me. This is how I defined invertibility in the last video. Now we're concerned with this equation up here."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function. So this is what invertibility tells me. This is how I defined invertibility in the last video. Now we're concerned with this equation up here. We're concerned with the equation, I'll write it in pink, f of x is equal to y. And we want to know for any y, or any lowercase cursive y in our big set y, is there a unique x solution to this? So what we can do is, we know that f is invertible."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Now we're concerned with this equation up here. We're concerned with the equation, I'll write it in pink, f of x is equal to y. And we want to know for any y, or any lowercase cursive y in our big set y, is there a unique x solution to this? So what we can do is, we know that f is invertible. I told you that from the get go. So given that f is invertible, we know that there is this f inverse function. We know that there's this f inverse function, and I can apply that f inverse function."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So what we can do is, we know that f is invertible. I told you that from the get go. So given that f is invertible, we know that there is this f inverse function. We know that there's this f inverse function, and I can apply that f inverse function. It's a mapping from y to x. So I can apply it to any element in y. So for any y, let's say that this is my y right there."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "We know that there's this f inverse function, and I can apply that f inverse function. It's a mapping from y to x. So I can apply it to any element in y. So for any y, let's say that this is my y right there. So I can apply my f inverse to that y, and I'm going to go over here. And of course, y is equal to f of x. These are the exact same points."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So for any y, let's say that this is my y right there. So I can apply my f inverse to that y, and I'm going to go over here. And of course, y is equal to f of x. These are the exact same points. So let's apply our f inverse function to this. So if I apply the f inverse function to both sides of the equation, both sides of this right here is an element in y, and this is the same element in y, right? They're the same element."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "These are the exact same points. So let's apply our f inverse function to this. So if I apply the f inverse function to both sides of the equation, both sides of this right here is an element in y, and this is the same element in y, right? They're the same element. Now, if I apply the mapping, the inverse mapping, to both of that, that's going to take me to some element in x. So let's do that. So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "They're the same element. Now, if I apply the mapping, the inverse mapping, to both of that, that's going to take me to some element in x. So let's do that. So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x. And what is this going to be equal to? Well, on the right-hand side, we could just write the f inverse of y. That's going to be some element over here."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x. And what is this going to be equal to? Well, on the right-hand side, we could just write the f inverse of y. That's going to be some element over here. But what does the left-hand side of this equation translate to? The definition of this inverse function is that when you take the composition with f, you're going to end up with the identity function. This is going to be equivalent to, let me write it this way."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be some element over here. But what does the left-hand side of this equation translate to? The definition of this inverse function is that when you take the composition with f, you're going to end up with the identity function. This is going to be equivalent to, let me write it this way. This is equal to the composition of f inverse with f of x, which is equivalent to the identity function being applied to x. And then the identity function being applied to x is what? That's just x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equivalent to, let me write it this way. This is equal to the composition of f inverse with f of x, which is equivalent to the identity function being applied to x. And then the identity function being applied to x is what? That's just x. This thing right here just reduces to x. So we started with the idea that f is invertible. We used the definition of invertibility, that there exists this inverse function right there."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "That's just x. This thing right here just reduces to x. So we started with the idea that f is invertible. We used the definition of invertibility, that there exists this inverse function right there. And then we essentially applied the inverse function to both sides of this equation. Say, look, you give me any y, any lowercase cursive y in this set y, and I will find you a unique x. This is the only x that satisfies this equation."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "We used the definition of invertibility, that there exists this inverse function right there. And then we essentially applied the inverse function to both sides of this equation. Say, look, you give me any y, any lowercase cursive y in this set y, and I will find you a unique x. This is the only x that satisfies this equation. This is the only x. Remember, if, and how do I know it's the only x? Because this is the only possible inverse function."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "This is the only x that satisfies this equation. This is the only x. Remember, if, and how do I know it's the only x? Because this is the only possible inverse function. This is the only possible inverse function. Only one inverse function for which this is true. I proved that to you in the last video."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Because this is the only possible inverse function. This is the only possible inverse function. Only one inverse function for which this is true. I proved that to you in the last video. That if f is invertible, it only has one unique inverse function. We tried before to have maybe two inverse functions, but we saw that they have to be the same thing. So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "I proved that to you in the last video. That if f is invertible, it only has one unique inverse function. We tried before to have maybe two inverse functions, but we saw that they have to be the same thing. So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution. And because it's only one inverse function, and functions only map to one value in this case, then we know this is a unique solution. So let's write this down. So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution. And because it's only one inverse function, and functions only map to one value in this case, then we know this is a unique solution. So let's write this down. So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution. And that unique solution, if you really care about it, is going to be the inverse function applied to y. It might seem like a bit of a no-brainer, but you can see you have to be a little bit precise about it in order to get to the point you want. But let's see if the opposite is true."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution. And that unique solution, if you really care about it, is going to be the inverse function applied to y. It might seem like a bit of a no-brainer, but you can see you have to be a little bit precise about it in order to get to the point you want. But let's see if the opposite is true. Let's see if we start from the assumption that for all y that is a member of our set y, that the solution, that the equation f of x is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "But let's see if the opposite is true. Let's see if we start from the assumption that for all y that is a member of our set y, that the solution, that the equation f of x is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. So let's think about it the first way. So we're saying that for any y, so let me draw my sets again, so this is my set x and this is my set y right there. Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "If given this, we can prove invertibility. So let's think about it the first way. So we're saying that for any y, so let me draw my sets again, so this is my set x and this is my set y right there. Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution. Let's call that unique solution, well, we could call it whatever, but a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that look, you pick a point in y, I can find you some point in x such that f of x is equal to y. And not only can I find that for you, that that is a unique solution."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution. Let's call that unique solution, well, we could call it whatever, but a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that look, you pick a point in y, I can find you some point in x such that f of x is equal to y. And not only can I find that for you, that that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And not only can I find that for you, that that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x. And s of, let's say, s of y, where of course y is a member of our set capital Y, s of y is equal to the unique solution in x to f of x is equal to y. Now you're saying, hey Sal, that looks a little convoluted, but think about it, this is a completely valid function definition, right? We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "It's a mapping from y to x. And s of, let's say, s of y, where of course y is a member of our set capital Y, s of y is equal to the unique solution in x to f of x is equal to y. Now you're saying, hey Sal, that looks a little convoluted, but think about it, this is a completely valid function definition, right? We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation. Well, OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So why don't I just define a function that says, look, I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation. Well, OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So why don't I just define a function that says, look, I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. And of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value, because any value y, any lowercase value y in this set, has a unique solution to f of x is equal to y. So this can only equal one value, so it's well defined."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "That's how I'm defining this function right here. And of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value, because any value y, any lowercase value y in this set, has a unique solution to f of x is equal to y. So this can only equal one value, so it's well defined. So let's apply, let's take some element here. Let's take some element, let me do a good color. Let's say this is b."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So this can only equal one value, so it's well defined. So let's apply, let's take some element here. Let's take some element, let me do a good color. Let's say this is b. And b is a member of y. So let's just map it using our new function right here. So let's take it and map it."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Let's say this is b. And b is a member of y. So let's just map it using our new function right here. So let's take it and map it. And this is s of b right here, which is a member of x. Now, we know that s of b is a unique solution by definition. I know it seems a little circular, but it's not."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So let's take it and map it. And this is s of b right here, which is a member of x. Now, we know that s of b is a unique solution by definition. I know it seems a little circular, but it's not. We know that s of b is a solution. So we know that s of b is a unique solution to f of x is equal to b. Well, if this is the case, we just got this because this is what this function does."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "I know it seems a little circular, but it's not. We know that s of b is a solution. So we know that s of b is a unique solution to f of x is equal to b. Well, if this is the case, we just got this because this is what this function does. It maps every y to the unique solution to this equation. Because we said that every y has a unique solution. So if this is the case, then what happens if I take f of s of b?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Well, if this is the case, we just got this because this is what this function does. It maps every y to the unique solution to this equation. Because we said that every y has a unique solution. So if this is the case, then what happens if I take f of s of b? Well, I just said this is the unique solution to this. So if I put this guy in here, what am I going to get? I'm going to get b."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So if this is the case, then what happens if I take f of s of b? Well, I just said this is the unique solution to this. So if I put this guy in here, what am I going to get? I'm going to get b. Or another way of saying this is that the composition of f with s applied to b is equal to b. Or another way to say it is that when you take the composition of f with s, this is the same thing. Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to get b. Or another way of saying this is that the composition of f with s applied to b is equal to b. Or another way to say it is that when you take the composition of f with s, this is the same thing. Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b. That's what's happening here. So this is the same thing as the identity function on y being applied to b. So it's equal to b."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b. That's what's happening here. So this is the same thing as the identity function on y being applied to b. So it's equal to b. So we can say that the composition, we can say that there exists, and we know that this function exists, or that we can always construct this. So we already know that this exists. This existed by me constructing it, but I've hopefully shown you that this is well-defined."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to b. So we can say that the composition, we can say that there exists, and we know that this function exists, or that we can always construct this. So we already know that this exists. This existed by me constructing it, but I've hopefully shown you that this is well-defined. From our assumption that this always has a unique solution in x for any y here, I can define this in a fairly reasonable way. So it definitely exists. And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "This existed by me constructing it, but I've hopefully shown you that this is well-defined. From our assumption that this always has a unique solution in x for any y here, I can define this in a fairly reasonable way. So it definitely exists. And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. Now, let's do another little experiment. Let's take a particular, let me draw our sets again. Let me take some, this is our set x, and let me take some member of set x, call it a."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. Now, let's do another little experiment. Let's take a particular, let me draw our sets again. Let me take some, this is our set x, and let me take some member of set x, call it a. Let me take my set y right there. And so we can apply the function to a, and we'll get a member of set y. Let's call that right there, let's call that f of a right there."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Let me take some, this is our set x, and let me take some member of set x, call it a. Let me take my set y right there. And so we can apply the function to a, and we'll get a member of set y. Let's call that right there, let's call that f of a right there. Now, if I apply my magic function here, that always I can get you any member of set y, and I'll give you the unique solution in x to this equation. So let me apply that to this. Let me apply s to this."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that right there, let's call that f of a right there. Now, if I apply my magic function here, that always I can get you any member of set y, and I'll give you the unique solution in x to this equation. So let me apply that to this. Let me apply s to this. So if I apply s to this, it'll give me the unique solution. Let me write this down. So if I apply s to this, I'm going to apply s to this."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Let me apply s to this. So if I apply s to this, it'll give me the unique solution. Let me write this down. So if I apply s to this, I'm going to apply s to this. And maybe I shouldn't point it back at that. I don't want to imply that it necessarily points back at that. So let me apply s to that."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So if I apply s to this, I'm going to apply s to this. And maybe I shouldn't point it back at that. I don't want to imply that it necessarily points back at that. So let me apply s to that. s to this. So what is this going to point to? What is that point going to be right there?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So let me apply s to that. s to this. So what is this going to point to? What is that point going to be right there? So that's going to be s of this point, which is f of a, which we know is the unique solution. So this is equal to the unique solution to the equation f of x is equal to this y right here. Or this y right here is just called f of a, right?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "What is that point going to be right there? So that's going to be s of this point, which is f of a, which we know is the unique solution. So this is equal to the unique solution to the equation f of x is equal to this y right here. Or this y right here is just called f of a, right? Remember, the mapping s just maps you from any member of a to the unique solution to the equation f of x is equal to that, so this is the mapping from f of a to the unique. So this s of f of a is going to be a mapping to the, or this right here is going to be the unique solution to the equation f of x is equal to this member of y. And what's this member of y called?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Or this y right here is just called f of a, right? Remember, the mapping s just maps you from any member of a to the unique solution to the equation f of x is equal to that, so this is the mapping from f of a to the unique. So this s of f of a is going to be a mapping to the, or this right here is going to be the unique solution to the equation f of x is equal to this member of y. And what's this member of y called? It's called f of a. Well, you could say this in a very convoluted way, but if I were to just, before you learned any linear algebra, if I said, look, if I have the equation f of x is equal to f of a, what is the unique solution to this equation? What does x equal?"}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And what's this member of y called? It's called f of a. Well, you could say this in a very convoluted way, but if I were to just, before you learned any linear algebra, if I said, look, if I have the equation f of x is equal to f of a, what is the unique solution to this equation? What does x equal? Well, x would have to be equal to a. So the unique solution to the equation f of x is equal to f of a is equal to a. And we know that there's only one solution to that."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "What does x equal? Well, x would have to be equal to a. So the unique solution to the equation f of x is equal to f of a is equal to a. And we know that there's only one solution to that. Because that was one of our starting assumptions. So this thing is equal to a. Or we could write s of f of a is equal to a."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And we know that there's only one solution to that. Because that was one of our starting assumptions. So this thing is equal to a. Or we could write s of f of a is equal to a. Or that the composition of s with f is equal, or applied to a is equal to a, or that the composition of s with f is just the identity function on the set x. This is a mapping right here from x to x. So we could write that the composition of s with f is the identity on x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Or we could write s of f of a is equal to a. Or that the composition of s with f is equal, or applied to a is equal to a, or that the composition of s with f is just the identity function on the set x. This is a mapping right here from x to x. So we could write that the composition of s with f is the identity on x. So what have we done so far? We started with the idea that you pick any y in our set capital Y here, and we're going to have a unique solution x such that this is true. Such that f of x is equal to y."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So we could write that the composition of s with f is the identity on x. So what have we done so far? We started with the idea that you pick any y in our set capital Y here, and we're going to have a unique solution x such that this is true. Such that f of x is equal to y. That's what the assumption we started off with. We constructed this function s that immediately maps any member here with its unique solution to this equation. Fair enough."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Such that f of x is equal to y. That's what the assumption we started off with. We constructed this function s that immediately maps any member here with its unique solution to this equation. Fair enough. Now from that, we say, OK, this definitely exists. Not only does it exist, but we figured out that the composition of f with our constructed function is equal to the identity on the set y. And then we also learned that s, the composition of s with f, is the identity function on x."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now from that, we say, OK, this definitely exists. Not only does it exist, but we figured out that the composition of f with our constructed function is equal to the identity on the set y. And then we also learned that s, the composition of s with f, is the identity function on x. Let me write this. So we learned this, and we also learned that the composition of f with s is equal to the identity on y. And s clearly exists because I constructed it, and we know it's well-defined because every y, for every y here, there is a solution to this."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And then we also learned that s, the composition of s with f, is the identity function on x. Let me write this. So we learned this, and we also learned that the composition of f with s is equal to the identity on y. And s clearly exists because I constructed it, and we know it's well-defined because every y, for every y here, there is a solution to this. So given that I was able to find a function that these two things are true, this is by definition what it means to be invertible. Remember, so this means that f is invertible. Remember, f being invertible, in order for f to be invertible, that means that there must exist some function from, so if f is a mapping from x to y, invertibility means that there must be some function, f inverse, that is a mapping from y to x such that, so I could write there exists a function, such that the inverse function composed with our function should be equal to the identity on x, and the inverse, and the function, and the composition of the function with the inverse function should be the identity on y."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "And s clearly exists because I constructed it, and we know it's well-defined because every y, for every y here, there is a solution to this. So given that I was able to find a function that these two things are true, this is by definition what it means to be invertible. Remember, so this means that f is invertible. Remember, f being invertible, in order for f to be invertible, that means that there must exist some function from, so if f is a mapping from x to y, invertibility means that there must be some function, f inverse, that is a mapping from y to x such that, so I could write there exists a function, such that the inverse function composed with our function should be equal to the identity on x, and the inverse, and the function, and the composition of the function with the inverse function should be the identity on y. Well, we found a function. It exists, and that function is s, where both of these things are true. We can say that s is equal to f inverse."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "Remember, f being invertible, in order for f to be invertible, that means that there must exist some function from, so if f is a mapping from x to y, invertibility means that there must be some function, f inverse, that is a mapping from y to x such that, so I could write there exists a function, such that the inverse function composed with our function should be equal to the identity on x, and the inverse, and the function, and the composition of the function with the inverse function should be the identity on y. Well, we found a function. It exists, and that function is s, where both of these things are true. We can say that s is equal to f inverse. So f is definitely invertible. So hopefully you found this satisfying. This proof is very subtle and very nuanced because we kind of keep bouncing between our sets x and y, but what we've shown is that if f is, in the beginning part of this video, we showed that if f is invertible, then for any y there is a unique solution to the equation f of x equals y."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "We can say that s is equal to f inverse. So f is definitely invertible. So hopefully you found this satisfying. This proof is very subtle and very nuanced because we kind of keep bouncing between our sets x and y, but what we've shown is that if f is, in the beginning part of this video, we showed that if f is invertible, then for any y there is a unique solution to the equation f of x equals y. And in the second part of the video, we showed that the other way is true. That if, let me put it this way, that if for all y, a member of capital Y, there is a unique solution to f of x is equal to x, then f is invertible. So the fact that both of these assumptions imply each other we can write our final conclusion of the video."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "This proof is very subtle and very nuanced because we kind of keep bouncing between our sets x and y, but what we've shown is that if f is, in the beginning part of this video, we showed that if f is invertible, then for any y there is a unique solution to the equation f of x equals y. And in the second part of the video, we showed that the other way is true. That if, let me put it this way, that if for all y, a member of capital Y, there is a unique solution to f of x is equal to x, then f is invertible. So the fact that both of these assumptions imply each other we can write our final conclusion of the video. That f being invertible, if f, which is a mapping from x to y is invertible, this is true if and only if, and we could write that either as a two-way arrow or we could write if for if and only if. So both of these statements imply each other. If and only if for all y, for every y that is a member of our set Y, there exists a unique, I could actually write that like that."}, {"video_title": "Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3", "Sentence": "So the fact that both of these assumptions imply each other we can write our final conclusion of the video. That f being invertible, if f, which is a mapping from x to y is invertible, this is true if and only if, and we could write that either as a two-way arrow or we could write if for if and only if. So both of these statements imply each other. If and only if for all y, for every y that is a member of our set Y, there exists a unique, I could actually write that like that. That means there exists a unique x for the, or let me write it this way. There exists a unique solution to the equation f of x is equal to y. So that was our big takeaway in this video."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I have this matrix here, this matrix A, and I guess a good place to start is let's just figure out its column space and its null space. And the column space is actually super easy to figure out. It's just the span of the column vectors of A. So we can right from the get-go write that the column space of our matrix A is equal to the span of the vectors 1, 2, 3, 1, 1, 1, 4, and 1, 1, 4, 1, 4, 1, and 1, 3, 2. I'm done. That was pretty straightforward. A lot easier than finding null spaces."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we can right from the get-go write that the column space of our matrix A is equal to the span of the vectors 1, 2, 3, 1, 1, 1, 4, and 1, 1, 4, 1, 4, 1, and 1, 3, 2. I'm done. That was pretty straightforward. A lot easier than finding null spaces. Now, this may or may not be satisfying to you, and there's a lot of open questions. Is this a basis for the space? For example, is this a linear independent set of vectors?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A lot easier than finding null spaces. Now, this may or may not be satisfying to you, and there's a lot of open questions. Is this a basis for the space? For example, is this a linear independent set of vectors? How can we visualize the space? And I haven't answered any of those yet. But if someone just says, hey, what's the column space of A?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For example, is this a linear independent set of vectors? How can we visualize the space? And I haven't answered any of those yet. But if someone just says, hey, what's the column space of A? This is the column space of A. And now we can answer some of those other questions. If this is a linearly independent set of vectors, then these vectors would be a basis for the column space of A."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if someone just says, hey, what's the column space of A? This is the column space of A. And now we can answer some of those other questions. If this is a linearly independent set of vectors, then these vectors would be a basis for the column space of A. We don't know that yet. We don't know whether these are linearly independent. But we can figure out if they're linearly independent by looking at the null space of A."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If this is a linearly independent set of vectors, then these vectors would be a basis for the column space of A. We don't know that yet. We don't know whether these are linearly independent. But we can figure out if they're linearly independent by looking at the null space of A. Remember, these are linearly independent if the null space of A only contains the zero vector. So let's figure out what the null space of A is. And remember, we can do a little shortcut here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But we can figure out if they're linearly independent by looking at the null space of A. Remember, these are linearly independent if the null space of A only contains the zero vector. So let's figure out what the null space of A is. And remember, we can do a little shortcut here. The null space of A is equal to the null space of the reduced row echelon form of A. And I showed you that when we first calculated the null space of a vector. Because when you perform these, essentially if you want to solve for the null space of A, you create an augmented matrix."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And remember, we can do a little shortcut here. The null space of A is equal to the null space of the reduced row echelon form of A. And I showed you that when we first calculated the null space of a vector. Because when you perform these, essentially if you want to solve for the null space of A, you create an augmented matrix. And you put the augmented matrix in reduced row echelon form. But the zeros never change. So essentially you're just taking A and putting it in reduced row echelon form."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because when you perform these, essentially if you want to solve for the null space of A, you create an augmented matrix. And you put the augmented matrix in reduced row echelon form. But the zeros never change. So essentially you're just taking A and putting it in reduced row echelon form. So let's do that. So I'll keep row 1 the same. 1, 1, 1, 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So essentially you're just taking A and putting it in reduced row echelon form. So let's do that. So I'll keep row 1 the same. 1, 1, 1, 1. And then let me replace row 2 with row 2 minus row 1. Row 2 minus row 1. So what do I get?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1, 1, 1, 1. And then let me replace row 2 with row 2 minus row 1. Row 2 minus row 1. So what do I get? 2. No, actually I want to zero this out here. So row 2 minus 2 times row 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what do I get? 2. No, actually I want to zero this out here. So row 2 minus 2 times row 1. Actually, even better, because I eventually want to get a 1 here. So let me do 2 times row 1 minus row 2. So let me say 2 times row 1 and I'm going to minus row 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So row 2 minus 2 times row 1. Actually, even better, because I eventually want to get a 1 here. So let me do 2 times row 1 minus row 2. So let me say 2 times row 1 and I'm going to minus row 2. So 2 times 1 minus 2 is 0, which is exactly what I wanted there. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me say 2 times row 1 and I'm going to minus row 2. So 2 times 1 minus 2 is 0, which is exactly what I wanted there. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. Alright. Now let me see if I can zero out this guy here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. Alright. Now let me see if I can zero out this guy here. So what can I do? Let me take, and I can do any combination, anything that essentially zeros this guy out. But I want to minimize my number of negative numbers."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let me see if I can zero out this guy here. So what can I do? Let me take, and I can do any combination, anything that essentially zeros this guy out. But I want to minimize my number of negative numbers. So let me take this third row minus 3 times this first row. So I'm going to take minus 3 times that first row and add it to this third row. So 3 minus 3 times 1 is 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But I want to minimize my number of negative numbers. So let me take this third row minus 3 times this first row. So I'm going to take minus 3 times that first row and add it to this third row. So 3 minus 3 times 1 is 0. 4 minus 3 times 1. These are just going to be a bunch of 3's. 4 minus 3 times 1 is 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 3 minus 3 times 1 is 0. 4 minus 3 times 1. These are just going to be a bunch of 3's. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And 2 minus 3 times 1 is minus 1. Now if we want to get this into reduced row echelon form, we need to target that one there and that one there."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And 2 minus 3 times 1 is minus 1. Now if we want to get this into reduced row echelon form, we need to target that one there and that one there. And what can we do? So let's keep my middle row the same. My middle row is not going to change."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now if we want to get this into reduced row echelon form, we need to target that one there and that one there. And what can we do? So let's keep my middle row the same. My middle row is not going to change. 0, 1, minus 2, minus 1. And to get rid of this one up here, I can just replace my first row with my first row minus my second row. Because then this won't change."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My middle row is not going to change. 0, 1, minus 2, minus 1. And to get rid of this one up here, I can just replace my first row with my first row minus my second row. Because then this won't change. I have 1 minus 0 is 1. 1 minus 1 is 0. That's what we wanted."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because then this won't change. I have 1 minus 0 is 1. 1 minus 1 is 0. That's what we wanted. 1 minus minus 2 is 3. Right? That's 1 plus 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's what we wanted. 1 minus minus 2 is 3. Right? That's 1 plus 2. 1 minus minus 1, that's 1 plus 1. That is 2. Fair enough."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's 1 plus 2. 1 minus minus 1, that's 1 plus 1. That is 2. Fair enough. Now let me do my third row. The third row, let me just add, or let me just subtract, let me replace my third row with my third row subtracted from my first row. They're obviously the same thing."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now let me do my third row. The third row, let me just add, or let me just subtract, let me replace my third row with my third row subtracted from my first row. They're obviously the same thing. So if I subtract this row from, if I subtract the third row from the second row, I'm just going to get a bunch of 0's. 0 minus 0 is 0. 1 minus 1 is 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're obviously the same thing. So if I subtract this row from, if I subtract the third row from the second row, I'm just going to get a bunch of 0's. 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0. And minus 1 minus minus 1, that's minus 1 plus 1. That's equal to 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 1 is 0. Minus 2 minus minus 2 is 0. And minus 1 minus minus 1, that's minus 1 plus 1. That's equal to 0. And just like that, we have it now in reduced row echelon form. So this right here is the reduced row echelon form of A. That straightforward."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's equal to 0. And just like that, we have it now in reduced row echelon form. So this right here is the reduced row echelon form of A. That straightforward. Now, the whole reason why we even went through this exercise is we wanted to figure out the null space of A, and we already know that the null space of A is equal to the null space of the reduced row echelon form of A. So if this is the reduced row echelon form of A, let's figure out its null space. So the null space is the set of all vectors in R4, because we have 4 columns here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That straightforward. Now, the whole reason why we even went through this exercise is we wanted to figure out the null space of A, and we already know that the null space of A is equal to the null space of the reduced row echelon form of A. So if this is the reduced row echelon form of A, let's figure out its null space. So the null space is the set of all vectors in R4, because we have 4 columns here. 1, 2, 3, 4. The null space is the set of all vectors that satisfy this equation. We're going to have 3 0's right here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the null space is the set of all vectors in R4, because we have 4 columns here. 1, 2, 3, 4. The null space is the set of all vectors that satisfy this equation. We're going to have 3 0's right here. That's the 0 vector in R3, because we have 3 rows right there. And you can figure it out. This times this has to equal that 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We're going to have 3 0's right here. That's the 0 vector in R3, because we have 3 rows right there. And you can figure it out. This times this has to equal that 0. That dotted with that, essentially, is going to equal that 0. That dotted with that is equal to that 0. I say essentially because I didn't define a row vector dot a column vector."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This times this has to equal that 0. That dotted with that, essentially, is going to equal that 0. That dotted with that is equal to that 0. I say essentially because I didn't define a row vector dot a column vector. I've only defined column vectors dotted with other column vectors, but we've covered that in a previous video, where we just say this is a transpose of a column vector. So let's just take this and write a system of equations with this. So we get 1 times x1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I say essentially because I didn't define a row vector dot a column vector. I've only defined column vectors dotted with other column vectors, but we've covered that in a previous video, where we just say this is a transpose of a column vector. So let's just take this and write a system of equations with this. So we get 1 times x1. So this times this is going to be equal to that 0. 1 times x1, that is x1, plus 0 times x2, plus 3 times x3, plus 2 times x4 is equal to that 0. Then in yellow right here, I have 0 times x1 plus 1 times x2 minus 2 times x3 minus x4 is equal to 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we get 1 times x1. So this times this is going to be equal to that 0. 1 times x1, that is x1, plus 0 times x2, plus 3 times x3, plus 2 times x4 is equal to that 0. Then in yellow right here, I have 0 times x1 plus 1 times x2 minus 2 times x3 minus x4 is equal to 0. And then this gives me no information. 0's times all of this is equal to 0. So it just turns into 0 equals 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then in yellow right here, I have 0 times x1 plus 1 times x2 minus 2 times x3 minus x4 is equal to 0. And then this gives me no information. 0's times all of this is equal to 0. So it just turns into 0 equals 0. So let's see if we can solve for our pivot entries or our pivot variables. What are our pivot entries? The left of the pivot entry, that's a pivot entry."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it just turns into 0 equals 0. So let's see if we can solve for our pivot entries or our pivot variables. What are our pivot entries? The left of the pivot entry, that's a pivot entry. That's what reduced row echelon form is all about. Getting these entries that are 1 and they're the only non-zero term in their respective columns, and that every pivot entry is to the right of a pivot entry above it. And then the columns that don't have pivot entries, these columns represent the free variables."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The left of the pivot entry, that's a pivot entry. That's what reduced row echelon form is all about. Getting these entries that are 1 and they're the only non-zero term in their respective columns, and that every pivot entry is to the right of a pivot entry above it. And then the columns that don't have pivot entries, these columns represent the free variables. So this column has no pivot entry, and so when you take the dot product, you get this column turned into this column in our system of equations. So we know that x3 is a free variable. We can set it equal to anything."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then the columns that don't have pivot entries, these columns represent the free variables. So this column has no pivot entry, and so when you take the dot product, you get this column turned into this column in our system of equations. So we know that x3 is a free variable. We can set it equal to anything. Likewise, x4 is a free variable. x1 and x2 are pivot variables because their corresponding columns in our reduced row echelon form have pivot entries in them. So let's see if we can simplify this into a form we know, and we've seen this before."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can set it equal to anything. Likewise, x4 is a free variable. x1 and x2 are pivot variables because their corresponding columns in our reduced row echelon form have pivot entries in them. So let's see if we can simplify this into a form we know, and we've seen this before. So if I solve for x1, this 0 I can ignore, that 0 I can ignore. I could say that x1 is equal to minus 3 x3 minus 2 x4. I just subtracted these two from both sides of the equation, and I can say that x2 is equal to 2 x3 plus x4."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can simplify this into a form we know, and we've seen this before. So if I solve for x1, this 0 I can ignore, that 0 I can ignore. I could say that x1 is equal to minus 3 x3 minus 2 x4. I just subtracted these two from both sides of the equation, and I can say that x2 is equal to 2 x3 plus x4. And if we want to write our solution set now, so if I wanted to find the null space of A, which is the same thing as the null space of the reduced row echelon form of A, is equal to all of the vectors x1, x2, x3, x4, that are equal to minus 3 x3 minus 2 x4. So just to be clear, these are free variables because I can set these to be anything. And these are pivot variables because I can't just set them to anything."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just subtracted these two from both sides of the equation, and I can say that x2 is equal to 2 x3 plus x4. And if we want to write our solution set now, so if I wanted to find the null space of A, which is the same thing as the null space of the reduced row echelon form of A, is equal to all of the vectors x1, x2, x3, x4, that are equal to minus 3 x3 minus 2 x4. So just to be clear, these are free variables because I can set these to be anything. And these are pivot variables because I can't just set them to anything. When I determine what my x3's and my x4's are, they determine what my x1's and my x2's have to be. So these are pivot variables, these are free variables. I can make this guy pi, and I can make this guy minus 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And these are pivot variables because I can't just set them to anything. When I determine what my x3's and my x4's are, they determine what my x1's and my x2's have to be. So these are pivot variables, these are free variables. I can make this guy pi, and I can make this guy minus 2. We can set them to anything. So x1 is equal to, let's see, let me write it this way. They're equal to x3, let me do it in a different color, do x3 like this."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can make this guy pi, and I can make this guy minus 2. We can set them to anything. So x1 is equal to, let's see, let me write it this way. They're equal to x3, let me do it in a different color, do x3 like this. So it's equal to x3 times some vector plus x4 times some other vector. So any solution set in my null space is going to be a linear combination of these two vectors. And we can figure out what these two vectors are just from these two constraints right here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're equal to x3, let me do it in a different color, do x3 like this. So it's equal to x3 times some vector plus x4 times some other vector. So any solution set in my null space is going to be a linear combination of these two vectors. And we can figure out what these two vectors are just from these two constraints right here. So let me do it in a neutral color. x1 is equal to minus 3 times x3, so minus 3 times x3, minus 2 times x4. Straightforward enough."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we can figure out what these two vectors are just from these two constraints right here. So let me do it in a neutral color. x1 is equal to minus 3 times x3, so minus 3 times x3, minus 2 times x4. Straightforward enough. x2 is equal to 2 times x3 plus x4. What's x3 equal to? Well x3 is equal to itself."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Straightforward enough. x2 is equal to 2 times x3 plus x4. What's x3 equal to? Well x3 is equal to itself. Whatever we said x3 equal to, that's going to be x3. So x3 is going to be 1 times x3 plus 0 times x4. It's not going to have any x4 in it."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well x3 is equal to itself. Whatever we said x3 equal to, that's going to be x3. So x3 is going to be 1 times x3 plus 0 times x4. It's not going to have any x4 in it. x3 is going to be kind of an independent variable where it's going to be free. We can set it to whatever it is. So we set it and then that's going to be our x3 in our solution set."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's not going to have any x4 in it. x3 is going to be kind of an independent variable where it's going to be free. We can set it to whatever it is. So we set it and then that's going to be our x3 in our solution set. x4 is not going to have any x3 in it. It's just going to be 1 times x4. And so our null space is essentially all of the linear combinations of these two vectors."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we set it and then that's going to be our x3 in our solution set. x4 is not going to have any x3 in it. It's just going to be 1 times x4. And so our null space is essentially all of the linear combinations of these two vectors. This can be any real number. This is just any real number and x4 is just any member of the real space. So all of these, the set of all of the valid solutions to Ax is equal to 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so our null space is essentially all of the linear combinations of these two vectors. This can be any real number. This is just any real number and x4 is just any member of the real space. So all of these, the set of all of the valid solutions to Ax is equal to 0. Where did I write that? Did I even write that down? No, I haven't even written that anywhere."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So all of these, the set of all of the valid solutions to Ax is equal to 0. Where did I write that? Did I even write that down? No, I haven't even written that anywhere. The set of all Ax is equal to 0 where this is my x, it equals all the linear combinations of this vector and that vector right there. And we know what all of the linear combinations mean. It means my null space is equal to the span of these two guys."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "No, I haven't even written that anywhere. The set of all Ax is equal to 0 where this is my x, it equals all the linear combinations of this vector and that vector right there. And we know what all of the linear combinations mean. It means my null space is equal to the span of these two guys. The span of minus 3, 2, 1, 0 and minus 2, 1, 0, 1. Now, let me ask you a question. Are the columns in A, are they a linearly independent set?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It means my null space is equal to the span of these two guys. The span of minus 3, 2, 1, 0 and minus 2, 1, 0, 1. Now, let me ask you a question. Are the columns in A, are they a linearly independent set? Let me write that down. So if we write these vectors right there, these are the vectors, the column vectors of A. So let me write that down."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Are the columns in A, are they a linearly independent set? Let me write that down. So if we write these vectors right there, these are the vectors, the column vectors of A. So let me write that down. So are the column vectors of A, what were they? 1, 2, 3, 1, 1, 4, 1, 4, 1 and 1, 3, 2. So this is just the column vectors of A. I could just write A as just this bunch of columns."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that down. So are the column vectors of A, what were they? 1, 2, 3, 1, 1, 4, 1, 4, 1 and 1, 3, 2. So this is just the column vectors of A. I could just write A as just this bunch of columns. But my question is, is this a linearly independent set? And here you might immediately start thinking, well, when we said that something is linearly independent, so linearly independence, let me just write it like this. Linear independence implies that there's only one solution."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is just the column vectors of A. I could just write A as just this bunch of columns. But my question is, is this a linearly independent set? And here you might immediately start thinking, well, when we said that something is linearly independent, so linearly independence, let me just write it like this. Linear independence implies that there's only one solution. We saw this, I think, two videos ago that there's only one solution. One solution to Ax is equal to 0, and that is the 0 solution, that x is equal to the 0 vector. Or another way to say that is that the null space of my matrix A is equal to just the 0 vector."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Linear independence implies that there's only one solution. We saw this, I think, two videos ago that there's only one solution. One solution to Ax is equal to 0, and that is the 0 solution, that x is equal to the 0 vector. Or another way to say that is that the null space of my matrix A is equal to just the 0 vector. That's what linear independence implies, and it goes both ways. If my null space is just the 0 vector, then I know it's linearly independent. If my null space includes other vectors, then I am not linearly independent."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say that is that the null space of my matrix A is equal to just the 0 vector. That's what linear independence implies, and it goes both ways. If my null space is just the 0 vector, then I know it's linearly independent. If my null space includes other vectors, then I am not linearly independent. Now, my null space of A, what does it include? Is it just the 0 vector? Well, no, it includes every linear combination of these guys."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If my null space includes other vectors, then I am not linearly independent. Now, my null space of A, what does it include? Is it just the 0 vector? Well, no, it includes every linear combination of these guys. It includes actually an infinite number of vectors. It's not just one solution. Obviously, 0 vector is contained here."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, no, it includes every linear combination of these guys. It includes actually an infinite number of vectors. It's not just one solution. Obviously, 0 vector is contained here. If you just multiply both of these, if you pick 0 for that and that, it's contained. But you can get a whole set of vectors. Because the span of this guy, so you can kind of, you know, the null span of A, the null space, sorry, the null space of A does not just contain the 0 vector."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Obviously, 0 vector is contained here. If you just multiply both of these, if you pick 0 for that and that, it's contained. But you can get a whole set of vectors. Because the span of this guy, so you can kind of, you know, the null span of A, the null space, sorry, the null space of A does not just contain the 0 vector. So it has more than just 0. So what does that mean? Well, that means that there's more than one solution to this, and that means that this is a linearly dependent set."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because the span of this guy, so you can kind of, you know, the null span of A, the null space, sorry, the null space of A does not just contain the 0 vector. So it has more than just 0. So what does that mean? Well, that means that there's more than one solution to this, and that means that this is a linearly dependent set. And what does that mean? Well, at the very beginning of the video, I said, what's the column space of A? And we said, oh, the column space of A is just the span of the column vectors, right?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, that means that there's more than one solution to this, and that means that this is a linearly dependent set. And what does that mean? Well, at the very beginning of the video, I said, what's the column space of A? And we said, oh, the column space of A is just the span of the column vectors, right? I just wrote it out like that. And I said, well, it's not clear whether this is a valid basis for the column space of A. And what's a basis?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we said, oh, the column space of A is just the span of the column vectors, right? I just wrote it out like that. And I said, well, it's not clear whether this is a valid basis for the column space of A. And what's a basis? A basis is a set of vectors that span a subspace, and they are also linearly independent. And we just showed that these guys are not linearly independent. So that means that they are not a basis for the column space of A."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what's a basis? A basis is a set of vectors that span a subspace, and they are also linearly independent. And we just showed that these guys are not linearly independent. So that means that they are not a basis for the column space of A. They do span the column space of A, by definition, really, but they're not a basis. They're just linearly independent for them to be a basis. So let's see if we can figure out what a basis for this column space would be."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that means that they are not a basis for the column space of A. They do span the column space of A, by definition, really, but they're not a basis. They're just linearly independent for them to be a basis. So let's see if we can figure out what a basis for this column space would be. And to do that, we just have to get rid of some redundant vectors. If I can show you that this guy can be represented by some combination of these two guys, then I can get rid of that guy. He's not adding any new information."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can figure out what a basis for this column space would be. And to do that, we just have to get rid of some redundant vectors. If I can show you that this guy can be represented by some combination of these two guys, then I can get rid of that guy. He's not adding any new information. Same with that guy. Who knows? So let's see if we can figure this piece of the puzzle out."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "He's not adding any new information. Same with that guy. Who knows? So let's see if we can figure this piece of the puzzle out. We know already that x1 times 1, 2, 3, plus x2 times 1, 1, 4, plus x3 times 1, 4, 1, plus x4 times 1, 3, 2. We know that this is equal to 0. Now, if we're able to solve for x4 in terms of, let's say I can solve them in terms of, let me just think that I can solve for the vectors that are associated with my free variables using the other vectors."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can figure this piece of the puzzle out. We know already that x1 times 1, 2, 3, plus x2 times 1, 1, 4, plus x3 times 1, 4, 1, plus x4 times 1, 3, 2. We know that this is equal to 0. Now, if we're able to solve for x4 in terms of, let's say I can solve them in terms of, let me just think that I can solve for the vectors that are associated with my free variables using the other vectors. Let me see if I can do that, and you'll see it's actually pretty straightforward. So let's say I want to figure out, I want to solve for x4. So if I subtract this from both sides of this equation, I get what?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, if we're able to solve for x4 in terms of, let's say I can solve them in terms of, let me just think that I can solve for the vectors that are associated with my free variables using the other vectors. Let me see if I can do that, and you'll see it's actually pretty straightforward. So let's say I want to figure out, I want to solve for x4. So if I subtract this from both sides of this equation, I get what? I get minus, let me put it this way, let me set x3 equal to 0. It was a free variable. I can do that."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I subtract this from both sides of this equation, I get what? I get minus, let me put it this way, let me set x3 equal to 0. It was a free variable. I can do that. So if I set x3 is equal to 0, then what do I get here? I get, if I set x3 equal to 0, this guy disappears. And if I subtract this from both sides of this equation, I get x1 times 1, 2, 3, plus x2 times 1, 1, 4 is equal to, I'm just setting x3 equal to 0."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can do that. So if I set x3 is equal to 0, then what do I get here? I get, if I set x3 equal to 0, this guy disappears. And if I subtract this from both sides of this equation, I get x1 times 1, 2, 3, plus x2 times 1, 1, 4 is equal to, I'm just setting x3 equal to 0. That was a free variable. So I'm setting x3 equal to 0, so this whole thing disappears. So that is equal to minus x4 times 1, 3, 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if I subtract this from both sides of this equation, I get x1 times 1, 2, 3, plus x2 times 1, 1, 4 is equal to, I'm just setting x3 equal to 0. That was a free variable. So I'm setting x3 equal to 0, so this whole thing disappears. So that is equal to minus x4 times 1, 3, 2. Now I set x3 equal to 0, let me set x1, sorry, let me set x4 to be equal to minus 1. x4 is equal to minus 1. If x4 is equal to minus 1, what is minus x4? Well then this thing will just be equal to 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that is equal to minus x4 times 1, 3, 2. Now I set x3 equal to 0, let me set x1, sorry, let me set x4 to be equal to minus 1. x4 is equal to minus 1. If x4 is equal to minus 1, what is minus x4? Well then this thing will just be equal to 1. And I'll have x1 times 1, 2, 3 plus x2 times 1, 1, 4 will equal this fourth vector right here. And can I always find things like this? Well sure, I can actually find the particular ones."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well then this thing will just be equal to 1. And I'll have x1 times 1, 2, 3 plus x2 times 1, 1, 4 will equal this fourth vector right here. And can I always find things like this? Well sure, I can actually find the particular ones. If x3 is equal to 0 and x4 is minus 1, let me copy and paste this that I have up here. Let me copy and paste this. Sometimes it doesn't."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well sure, I can actually find the particular ones. If x3 is equal to 0 and x4 is minus 1, let me copy and paste this that I have up here. Let me copy and paste this. Sometimes it doesn't. Edit, copy, edit, paste. There you go. Let me scroll down a little bit."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Sometimes it doesn't. Edit, copy, edit, paste. There you go. Let me scroll down a little bit. This is what we got when we figured out our null space right there. So, if I'm setting, remember these are the free variables. If I set x3 equal to 0 and x4 is equal to minus 1, what is x1?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll down a little bit. This is what we got when we figured out our null space right there. So, if I'm setting, remember these are the free variables. If I set x3 equal to 0 and x4 is equal to minus 1, what is x1? Then this will imply that x1 is equal to minus 3 times x3, that's just 0, minus 2 times x4. If x4 is minus 1, minus 2 times minus 1, x1 will equal 2. And then what will x2 be equal to?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I set x3 equal to 0 and x4 is equal to minus 1, what is x1? Then this will imply that x1 is equal to minus 3 times x3, that's just 0, minus 2 times x4. If x4 is minus 1, minus 2 times minus 1, x1 will equal 2. And then what will x2 be equal to? x2 is equal to 2 times x3, which is 0, plus x4. So it's equal to minus 1. I just showed you that if I set this equal to 2 and this equal to minus 1, I have a linear combination of this vector and this vector that can add up to this fourth vector."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then what will x2 be equal to? x2 is equal to 2 times x3, which is 0, plus x4. So it's equal to minus 1. I just showed you that if I set this equal to 2 and this equal to minus 1, I have a linear combination of this vector and this vector that can add up to this fourth vector. You can even verify it. 2 times 1 minus 1 is equal to 1. 2 times 2 minus 1 is equal to 3."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just showed you that if I set this equal to 2 and this equal to minus 1, I have a linear combination of this vector and this vector that can add up to this fourth vector. You can even verify it. 2 times 1 minus 1 is equal to 1. 2 times 2 minus 1 is equal to 3. 2 times 3 is 6. Minus 4 is equal to 2. So it checks out."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times 2 minus 1 is equal to 3. 2 times 3 is 6. Minus 4 is equal to 2. So it checks out. I just showed you using really our definitions, really looking at what were our free variables versus our pivot variables, we were able to show you, kind of just very simply solve for this fourth vector in terms of these first two. We know, if we go back to the set, that this fourth vector is really unnecessary, really not adding anything to, I guess, the span of the set of vectors because this guy can be written as a combination of this guy and this guy. Now let's see if this guy, this third guy, we can do the same exercise."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it checks out. I just showed you using really our definitions, really looking at what were our free variables versus our pivot variables, we were able to show you, kind of just very simply solve for this fourth vector in terms of these first two. We know, if we go back to the set, that this fourth vector is really unnecessary, really not adding anything to, I guess, the span of the set of vectors because this guy can be written as a combination of this guy and this guy. Now let's see if this guy, this third guy, we can do the same exercise. This is also dictated by a free variable. So let's see if I can write him as a combination of these first two. Well, we'll do the exact same thing."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's see if this guy, this third guy, we can do the same exercise. This is also dictated by a free variable. So let's see if I can write him as a combination of these first two. Well, we'll do the exact same thing. Instead of setting x3 equal to 0 and x4 equal to minus 1, let's set, let us set x4 is equal to 0 because I want to cross that out. And let me set x3 is equal to minus 1. If x3 is equal to minus 1, so this equals minus 1, what is our, what is this equation reduced to?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we'll do the exact same thing. Instead of setting x3 equal to 0 and x4 equal to minus 1, let's set, let us set x4 is equal to 0 because I want to cross that out. And let me set x3 is equal to minus 1. If x3 is equal to minus 1, so this equals minus 1, what is our, what is this equation reduced to? We get x1 times 1, 2, 3 plus x2 times 1, 1, 4 is equal to, if this is minus 1 times 1, 4, 1 and then we add it to both sides of this equation, we get plus 1 times 1, 4, 1. And once again, we can just solve for our x1's and x2. If x4 is 0 and x3 is minus 1, then x1, x4 is 0, so x3 is just minus 3 times x3."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If x3 is equal to minus 1, so this equals minus 1, what is our, what is this equation reduced to? We get x1 times 1, 2, 3 plus x2 times 1, 1, 4 is equal to, if this is minus 1 times 1, 4, 1 and then we add it to both sides of this equation, we get plus 1 times 1, 4, 1. And once again, we can just solve for our x1's and x2. If x4 is 0 and x3 is minus 1, then x1, x4 is 0, so x3 is just minus 3 times x3. So x1 would be equal to 3, right? Minus 3 times minus 1. And what would x2 be equal to?"}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If x4 is 0 and x3 is minus 1, then x1, x4 is 0, so x3 is just minus 3 times x3. So x1 would be equal to 3, right? Minus 3 times minus 1. And what would x2 be equal to? x4 is 0. We can ignore that. x2 would be equal to minus 2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what would x2 be equal to? x4 is 0. We can ignore that. x2 would be equal to minus 2. So this would be 3 and then this would be minus 2. Let's see if it works out. 3 times 1 minus 2 is 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x2 would be equal to minus 2. So this would be 3 and then this would be minus 2. Let's see if it works out. 3 times 1 minus 2 is 1. 3 times 2 minus 2 is 4. 3 times 3 minus 8 is 1. It checks out."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "3 times 1 minus 2 is 1. 3 times 2 minus 2 is 4. 3 times 3 minus 8 is 1. It checks out. So I'm able to write this vector that was associated with a free variable as a linear combination of these two. So we can get rid of him from our set. So now I've shown that this guy can be written as a linear combination of these two."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It checks out. So I'm able to write this vector that was associated with a free variable as a linear combination of these two. So we can get rid of him from our set. So now I've shown that this guy can be written as a linear combination of these two. So the span of all of those guys should be equal to the span. So let me write it this way. The column space of A, I can now rewrite."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now I've shown that this guy can be written as a linear combination of these two. So the span of all of those guys should be equal to the span. So let me write it this way. The column space of A, I can now rewrite. Before it was the span of all of those vectors. It was the span of all of the column vectors, v1, v2, v3, and v4. Now I just showed you that v3 and v4 can be rewritten in terms of v1 and v2."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The column space of A, I can now rewrite. Before it was the span of all of those vectors. It was the span of all of the column vectors, v1, v2, v3, and v4. Now I just showed you that v3 and v4 can be rewritten in terms of v1 and v2. So they're redundant. So that is equal to the span of v1 and v2, which are just those two vectors. One vector 1, 2, 3 and vector 1, 1, 4."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I just showed you that v3 and v4 can be rewritten in terms of v1 and v2. So they're redundant. So that is equal to the span of v1 and v2, which are just those two vectors. One vector 1, 2, 3 and vector 1, 1, 4. Now, are any of these guys redundant? Can I express one of them as a linear combination of the other? And essentially when I'm talking about the linear combination of only one other vector, it's just multiplying it by a scalar."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "One vector 1, 2, 3 and vector 1, 1, 4. Now, are any of these guys redundant? Can I express one of them as a linear combination of the other? And essentially when I'm talking about the linear combination of only one other vector, it's just multiplying it by a scalar. Well, let's think about that. There's multiple ways you can show this, but the easiest way is, well, look. To go from this entry to that entry, I'm just multiplying by 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And essentially when I'm talking about the linear combination of only one other vector, it's just multiplying it by a scalar. Well, let's think about that. There's multiple ways you can show this, but the easiest way is, well, look. To go from this entry to that entry, I'm just multiplying by 1. But if I multiply this whole vector times 1, then I'm going to get a 2 here and I'm going to get a 3 here. So it won't work. Let me put it this way."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "To go from this entry to that entry, I'm just multiplying by 1. But if I multiply this whole vector times 1, then I'm going to get a 2 here and I'm going to get a 3 here. So it won't work. Let me put it this way. If I want to represent this guy as a scalar multiple of that guy, so any scalar multiple of 1, 2, 3 is going to be equal to 1c, 2c, 3c. And so we're saying this guy has to be represented somehow like that. If we say that this guy is somehow a scalar, it somehow can be represented by that guy."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me put it this way. If I want to represent this guy as a scalar multiple of that guy, so any scalar multiple of 1, 2, 3 is going to be equal to 1c, 2c, 3c. And so we're saying this guy has to be represented somehow like that. If we say that this guy is somehow a scalar, it somehow can be represented by that guy. So that would have to be equal to 1, 1, 4. When you look at this top entry, it implies that c would have to be equal to 1. When you just, c is equal to 1."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we say that this guy is somehow a scalar, it somehow can be represented by that guy. So that would have to be equal to 1, 1, 4. When you look at this top entry, it implies that c would have to be equal to 1. When you just, c is equal to 1. But when you look at this second entry, you think that c would have to be equal to 1 half. So you get a contradiction over here. c would have to be equal to 4 thirds."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "When you just, c is equal to 1. But when you look at this second entry, you think that c would have to be equal to 1 half. So you get a contradiction over here. c would have to be equal to 4 thirds. So there's no c where this will work. There's no multiple of c, and you can work that both ways. So there's no way that you can represent one of these guys as a linear combination of the other."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "c would have to be equal to 4 thirds. So there's no c where this will work. There's no multiple of c, and you can work that both ways. So there's no way that you can represent one of these guys as a linear combination of the other. And you can actually prove other ways, maybe more formally, that this is linearly independent. But given that this is linearly independent, I think you're satisfied with that. We can then say that the set of vectors 1, 2, 3, and 1, 1, 4, this is a basis for the column span of A."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So there's no way that you can represent one of these guys as a linear combination of the other. And you can actually prove other ways, maybe more formally, that this is linearly independent. But given that this is linearly independent, I think you're satisfied with that. We can then say that the set of vectors 1, 2, 3, and 1, 1, 4, this is a basis for the column span of A. Now, I'm going to let you go in this video because I think I've gone well over time. But what I'm going to do in the next few videos is now that I've established that this is a basis for the column span of A, we can attempt to visualize it because we can say that the column span of A is equal to the span of these two vectors. And we can think about what the span of those two vectors are."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can then say that the set of vectors 1, 2, 3, and 1, 1, 4, this is a basis for the column span of A. Now, I'm going to let you go in this video because I think I've gone well over time. But what I'm going to do in the next few videos is now that I've established that this is a basis for the column span of A, we can attempt to visualize it because we can say that the column span of A is equal to the span of these two vectors. And we can think about what the span of those two vectors are. We're going to see that it's a plane in R3. Span of 1, 1, 4. And just as a quick reminder, I said it a couple of times."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we can think about what the span of those two vectors are. We're going to see that it's a plane in R3. Span of 1, 1, 4. And just as a quick reminder, I said it a couple of times. I said it's a basis. All I'm saying is that these guys, they both span the column space of A. When I had four vectors, they also span the column space of A."}, {"video_title": "Null space and column space basis Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And just as a quick reminder, I said it a couple of times. I said it's a basis. All I'm saying is that these guys, they both span the column space of A. When I had four vectors, they also span the column space of A. But what makes them a basis is that these guys are linearly independent. There's no extra information or redundant vectors that can be represented by other vectors within the basis. They're linearly independent."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "We used these exact examples where if this was matrix A, I just took it and I put it in reduced row echelon form, and I figured out which of these columns in my reduced row echelon form of A are pivot columns. It turned out to be the first one, the second one, and the fourth one. Then the method is, you say, look, the corresponding columns in A, so the first one, the second one, and the fourth one, form my basis for my column space. Since they form the basis, and if you want to know the dimension of your basis of your column space, which is also called the rank, you just say, well, there's three in there. It has a rank of 1, 2, 3. In this video, I want to discuss a little bit about why this worked. Why were we able to just take the corresponding columns?"}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "Since they form the basis, and if you want to know the dimension of your basis of your column space, which is also called the rank, you just say, well, there's three in there. It has a rank of 1, 2, 3. In this video, I want to discuss a little bit about why this worked. Why were we able to just take the corresponding columns? Why did linear independence of these three guys imply linear independence of these three guys? Why was the fact that I can represent these guys, this guy right here is a linear combination of these three, or this guy is a linear combination of these three, why does that imply that I can construct this guy as a linear combination of my basis vectors? The first thing that wasn't too much of a stretch of the imagination in the last video was the idea that these pivot vectors are linearly independent, so R1, R2, and R4."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "Why were we able to just take the corresponding columns? Why did linear independence of these three guys imply linear independence of these three guys? Why was the fact that I can represent these guys, this guy right here is a linear combination of these three, or this guy is a linear combination of these three, why does that imply that I can construct this guy as a linear combination of my basis vectors? The first thing that wasn't too much of a stretch of the imagination in the last video was the idea that these pivot vectors are linearly independent, so R1, R2, and R4. Everything I'm doing, I'm kind of applying to this special case just so that it's easier to understand, but it should be generalizable. In fact, it definitely is generalizable that all of the pivot columns in reduced row echelon form are linearly independent. That's because the very nature of reduced row echelon form is that you are the only pivot column that has a 1 in that respective row."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "The first thing that wasn't too much of a stretch of the imagination in the last video was the idea that these pivot vectors are linearly independent, so R1, R2, and R4. Everything I'm doing, I'm kind of applying to this special case just so that it's easier to understand, but it should be generalizable. In fact, it definitely is generalizable that all of the pivot columns in reduced row echelon form are linearly independent. That's because the very nature of reduced row echelon form is that you are the only pivot column that has a 1 in that respective row. The only way to construct it is with that vector. You can't construct it with the other pivot columns because they're all going to have 0 in that row. When I say it's linearly independent, I'm just saying the set of pivot columns."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "That's because the very nature of reduced row echelon form is that you are the only pivot column that has a 1 in that respective row. The only way to construct it is with that vector. You can't construct it with the other pivot columns because they're all going to have 0 in that row. When I say it's linearly independent, I'm just saying the set of pivot columns. Let me say this in general. The set of pivot columns for any reduced row echelon form matrix is linearly independent. It's just a very straightforward argument because each column is going to have 1 in a very unique place."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "When I say it's linearly independent, I'm just saying the set of pivot columns. Let me say this in general. The set of pivot columns for any reduced row echelon form matrix is linearly independent. It's just a very straightforward argument because each column is going to have 1 in a very unique place. All of the other pivot columns are going to have a 0 in that same place, so you can't take any linear combination to get to that 1 because 0 times anything minus or plus 0 times anything can never be equal to 1. I think you can accept that. That means that the solution to C1 times R1 plus C2 times R2 plus C4 times R4, the solution to this equation, because these guys are linearly independent, we know that this only has one solution and that's C1, C2, and C4 is equal to 0."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "It's just a very straightforward argument because each column is going to have 1 in a very unique place. All of the other pivot columns are going to have a 0 in that same place, so you can't take any linear combination to get to that 1 because 0 times anything minus or plus 0 times anything can never be equal to 1. I think you can accept that. That means that the solution to C1 times R1 plus C2 times R2 plus C4 times R4, the solution to this equation, because these guys are linearly independent, we know that this only has one solution and that's C1, C2, and C4 is equal to 0. That's the only solution to that. Another way we could say it is if we write R times some vector x, I'll just write it times this particular x where I write it as C1, C2, 0, C4, and 0 is equal to 0. This will be some special member of your null space."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "That means that the solution to C1 times R1 plus C2 times R2 plus C4 times R4, the solution to this equation, because these guys are linearly independent, we know that this only has one solution and that's C1, C2, and C4 is equal to 0. That's the only solution to that. Another way we could say it is if we write R times some vector x, I'll just write it times this particular x where I write it as C1, C2, 0, C4, and 0 is equal to 0. This will be some special member of your null space. It's a particular solution to the equation. This is equal to 1, 2, 3, 4, 0 because we have 4 rows here. Now, if we just expand this out, if we just multiply 1 times C1 plus 0 times C2 minus 1 times 0 plus 4 times 0, you'll get... Actually, let me think."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "This will be some special member of your null space. It's a particular solution to the equation. This is equal to 1, 2, 3, 4, 0 because we have 4 rows here. Now, if we just expand this out, if we just multiply 1 times C1 plus 0 times C2 minus 1 times 0 plus 4 times 0, you'll get... Actually, let me think. A better way to explain it. This multiplication right here can be written as, and we've seen this multiple times, C1 times R1 plus C2 times R2 plus 0 times R3, so we could just ignore that term, plus C4 times R4 plus 0 times R5. That's R5 right there."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "Now, if we just expand this out, if we just multiply 1 times C1 plus 0 times C2 minus 1 times 0 plus 4 times 0, you'll get... Actually, let me think. A better way to explain it. This multiplication right here can be written as, and we've seen this multiple times, C1 times R1 plus C2 times R2 plus 0 times R3, so we could just ignore that term, plus C4 times R4 plus 0 times R5. That's R5 right there. All of that equal to 0. The only solution to this, because we know that these 3 columns are linearly independent, or the set of just those 3 columns, those 3 pivot columns are linearly independent, the only solution here is all of these equal to 0. That's exactly what I said right up here."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "That's R5 right there. All of that equal to 0. The only solution to this, because we know that these 3 columns are linearly independent, or the set of just those 3 columns, those 3 pivot columns are linearly independent, the only solution here is all of these equal to 0. That's exactly what I said right up here. The only solution here, where if these 2 are 0, is that these guys also all have to equal 0, if I already constrain these 2. Now, the one thing that we've done over and over again, we know that the solution set of this equation, the solution set of Rx is equal to 0, is the same as the solution set of Ax is equal to 0. How do we know that, or what do I mean?"}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "That's exactly what I said right up here. The only solution here, where if these 2 are 0, is that these guys also all have to equal 0, if I already constrain these 2. Now, the one thing that we've done over and over again, we know that the solution set of this equation, the solution set of Rx is equal to 0, is the same as the solution set of Ax is equal to 0. How do we know that, or what do I mean? The solution set of this is just the null space. The solution set is just the null space of R. It's all of the x that satisfy this equation, and we know that that is equal to the null space of A, where R is just A in reduced row echelon form. The null space of A is all of the x's that satisfy this equation."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "How do we know that, or what do I mean? The solution set of this is just the null space. The solution set is just the null space of R. It's all of the x that satisfy this equation, and we know that that is equal to the null space of A, where R is just A in reduced row echelon form. The null space of A is all of the x's that satisfy this equation. Now, the only version of this that satisfied this equation was when C1, C2, and C4 are equal to 0. That tells us that the only version of this, C1, C2, 0, C4, 0, that satisfies this equation, or this equation, is when C1, C2, and C4 is equal to 0. Another way of saying that, if this is vector A1, A2, A4 right here, if you multiply this out, you get C1 times A1 plus C2 times A2, and then 0 times A3 plus C4 times A4 is equal to 0."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "The null space of A is all of the x's that satisfy this equation. Now, the only version of this that satisfied this equation was when C1, C2, and C4 are equal to 0. That tells us that the only version of this, C1, C2, 0, C4, 0, that satisfies this equation, or this equation, is when C1, C2, and C4 is equal to 0. Another way of saying that, if this is vector A1, A2, A4 right here, if you multiply this out, you get C1 times A1 plus C2 times A2, and then 0 times A3 plus C4 times A4 is equal to 0. Now, these guys are going to be linearly independent if and only if the only solution to this equation is they all equal to 0. Well, we know that the only solution to this is that they all equal to 0 because anything that's a solution to this is a solution to this. The only solution to this was, if I go ahead and I constrain these two terms to being equal to 0, the only solution to this is all of these C's have to be 0."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "Another way of saying that, if this is vector A1, A2, A4 right here, if you multiply this out, you get C1 times A1 plus C2 times A2, and then 0 times A3 plus C4 times A4 is equal to 0. Now, these guys are going to be linearly independent if and only if the only solution to this equation is they all equal to 0. Well, we know that the only solution to this is that they all equal to 0 because anything that's a solution to this is a solution to this. The only solution to this was, if I go ahead and I constrain these two terms to being equal to 0, the only solution to this is all of these C's have to be 0. Likewise, if I constrain these to be 0, the only solution to this is that C1, C2, and C4 have to be 0. Those guys have to be 0, which imply that these three vectors, A1, A2, and A4, so that implies that the set A1, A2, and A4 are linearly independent. We're halfway there."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "The only solution to this was, if I go ahead and I constrain these two terms to being equal to 0, the only solution to this is all of these C's have to be 0. Likewise, if I constrain these to be 0, the only solution to this is that C1, C2, and C4 have to be 0. Those guys have to be 0, which imply that these three vectors, A1, A2, and A4, so that implies that the set A1, A2, and A4 are linearly independent. We're halfway there. We've shown that, look, because the pivot columns here are linearly independent, we can show that they have the same solution set. The null space of the reduced row echelon form is the same as the null space of our original matrix. We were able to show that the only solution to C1 times this plus C2 times this plus C4 times this is when all the constants are 0, which shows that these three vectors, or a set of those three vectors, are definitely linearly independent."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "We're halfway there. We've shown that, look, because the pivot columns here are linearly independent, we can show that they have the same solution set. The null space of the reduced row echelon form is the same as the null space of our original matrix. We were able to show that the only solution to C1 times this plus C2 times this plus C4 times this is when all the constants are 0, which shows that these three vectors, or a set of those three vectors, are definitely linearly independent. The next thing to prove that they are a basis is to show that, look, all of the other column vectors can be represented as multiples of these three guys. I realize just for the sake of clarity, or maybe not boring you too much, I'll do that in the next video. In this one, we saw that, look, if the pivot columns are linearly independent, they always are."}, {"video_title": "Showing relation between basis cols and pivot cols Linear Algebra Khan Academy.mp3", "Sentence": "We were able to show that the only solution to C1 times this plus C2 times this plus C4 times this is when all the constants are 0, which shows that these three vectors, or a set of those three vectors, are definitely linearly independent. The next thing to prove that they are a basis is to show that, look, all of the other column vectors can be represented as multiples of these three guys. I realize just for the sake of clarity, or maybe not boring you too much, I'll do that in the next video. In this one, we saw that, look, if the pivot columns are linearly independent, they always are. All pivot columns, by definition, are linearly independent, or the set of pivot columns are always linearly independent when you take away the non-pivot columns. Then the corresponding columns in your original vector are also linearly independent. The next one will show that these three guys also span your column space."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the function just maps each of the specific entries of x to an entry in y. When I say map, it really just creates an association. If we think of these in maybe even less abstract terms, on some levels it's more abstract, you could view x as a basket of bananas and y as a basket of apples. And for every banana, you're associating it with one of the apples. That would be the definition, the mapping of going from each of those bananas to each of those apples would be a function. I don't know if that helps you or not. But I just want to kind of broaden your already preconceived notion of what a function is."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And for every banana, you're associating it with one of the apples. That would be the definition, the mapping of going from each of those bananas to each of those apples would be a function. I don't know if that helps you or not. But I just want to kind of broaden your already preconceived notion of what a function is. I mean, everything that you've probably seen before probably took a form that looks something like that, where you said, oh, a function is just give me some number and I'll give you another number, or I'll do something to that number. Well, it can be much more general than that. It's association between any member of one set and some other members of another set."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But I just want to kind of broaden your already preconceived notion of what a function is. I mean, everything that you've probably seen before probably took a form that looks something like that, where you said, oh, a function is just give me some number and I'll give you another number, or I'll do something to that number. Well, it can be much more general than that. It's association between any member of one set and some other members of another set. Now, we know that vectors are members of sets. Vectors. In particular, if we say that some vector x is a member of some set, let me just say it's a member of Rn, because that's what we deal with, all that means is that this is just a particular representation of an n-tuple."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's association between any member of one set and some other members of another set. Now, we know that vectors are members of sets. Vectors. In particular, if we say that some vector x is a member of some set, let me just say it's a member of Rn, because that's what we deal with, all that means is that this is just a particular representation of an n-tuple. Remember what Rn was. Rn we defined way back, I think, maybe at the beginning of the linear algebra playlist. We defined it as the set of all n-tuples, x1, x2, xn, where your x1s, x2s, all the way to xn's are a member of the real numbers."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In particular, if we say that some vector x is a member of some set, let me just say it's a member of Rn, because that's what we deal with, all that means is that this is just a particular representation of an n-tuple. Remember what Rn was. Rn we defined way back, I think, maybe at the beginning of the linear algebra playlist. We defined it as the set of all n-tuples, x1, x2, xn, where your x1s, x2s, all the way to xn's are a member of the real numbers. So your Rn is most definitely a set. This could be Rn, and obviously the use of the letter n is arbitrary. It could be Rm, it could be Rs."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We defined it as the set of all n-tuples, x1, x2, xn, where your x1s, x2s, all the way to xn's are a member of the real numbers. So your Rn is most definitely a set. This could be Rn, and obviously the use of the letter n is arbitrary. It could be Rm, it could be Rs. Rn is just kind of a placeholder for how many tuples we have. This could be R5, it could be 5 tuples. When we say that a vector x is a member of Rn, we're just saying that it's another way of writing one of these n-tuples."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It could be Rm, it could be Rs. Rn is just kind of a placeholder for how many tuples we have. This could be R5, it could be 5 tuples. When we say that a vector x is a member of Rn, we're just saying that it's another way of writing one of these n-tuples. All of our vectors so far were our column vectors. That's the only type that we've defined so far. We say it's this ordered list where each of the members are a member of R. It's an ordered list of n components, x1, x2, all the way to xn, where each of those guys, where each of those x1s, x2s, all the way to xn's, are a member of the real numbers."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When we say that a vector x is a member of Rn, we're just saying that it's another way of writing one of these n-tuples. All of our vectors so far were our column vectors. That's the only type that we've defined so far. We say it's this ordered list where each of the members are a member of R. It's an ordered list of n components, x1, x2, all the way to xn, where each of those guys, where each of those x1s, x2s, all the way to xn's, are a member of the real numbers. That's by definition what we mean when we say that x is a member of Rn. So if x is a member of Rn, let me draw two sets right here. Let's say that this set right here is Rn."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We say it's this ordered list where each of the members are a member of R. It's an ordered list of n components, x1, x2, all the way to xn, where each of those guys, where each of those x1s, x2s, all the way to xn's, are a member of the real numbers. That's by definition what we mean when we say that x is a member of Rn. So if x is a member of Rn, let me draw two sets right here. Let's say that this set right here is Rn. Then let me just change, just to be general, let me create another set right there and call that set right there Rm. Just a different number. It could be the same as n, it could be different."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that this set right here is Rn. Then let me just change, just to be general, let me create another set right there and call that set right there Rm. Just a different number. It could be the same as n, it could be different. This is m-tuples, that's n-tuples. The vectors we've defined, that vectors can be members of Rn. So you could have some vector here, and then if you associate with that vector in Rm, if you associate it with some vector in Rm, if you associate it with, let's call that vector y, if you make this association, that too is a function."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It could be the same as n, it could be different. This is m-tuples, that's n-tuples. The vectors we've defined, that vectors can be members of Rn. So you could have some vector here, and then if you associate with that vector in Rm, if you associate it with some vector in Rm, if you associate it with, let's call that vector y, if you make this association, that too is a function. That might have already been obvious to you. This would be a function that's mapping from Rn to Rm. Actually, I just want to make one little special note here."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you could have some vector here, and then if you associate with that vector in Rm, if you associate it with some vector in Rm, if you associate it with, let's call that vector y, if you make this association, that too is a function. That might have already been obvious to you. This would be a function that's mapping from Rn to Rm. Actually, I just want to make one little special note here. When I just drew the arrow like this, this shows that I'm mapping between two sets. I'm taking elements of this set and I'm associating with them elements of that set. In the last video, you probably saw this."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Actually, I just want to make one little special note here. When I just drew the arrow like this, this shows that I'm mapping between two sets. I'm taking elements of this set and I'm associating with them elements of that set. In the last video, you probably saw this. I want to do the side note because I realize it might have been confusing. I introduced you to another way of writing a function like this, where I said f could be defined as a mapping for any given x to x squared. I just want to make a note on the notation."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, you probably saw this. I want to do the side note because I realize it might have been confusing. I introduced you to another way of writing a function like this, where I said f could be defined as a mapping for any given x to x squared. I just want to make a note on the notation. When I just have a regular arrow, I'm going between sets. When I have this little vertical line at the base of the arrow, that's kind of the function definition. It tells me for any x you give me in the first set, in the second set I'm going to associate this x with, in this case, x squared."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I just want to make a note on the notation. When I just have a regular arrow, I'm going between sets. When I have this little vertical line at the base of the arrow, that's kind of the function definition. It tells me for any x you give me in the first set, in the second set I'm going to associate this x with, in this case, x squared. I just wanted to make that side note. The whole direction I was going in is that vectors are valid elements of sets. Functions are just mappings between elements of sets."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It tells me for any x you give me in the first set, in the second set I'm going to associate this x with, in this case, x squared. I just wanted to make that side note. The whole direction I was going in is that vectors are valid elements of sets. Functions are just mappings between elements of sets. You could have functions of vectors. I even touched on that a little bit in the last video when I talked about vector-valued functions. If your codomain is a subset of our m, where m is greater than 1, then we say your function is vector-valued."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Functions are just mappings between elements of sets. You could have functions of vectors. I even touched on that a little bit in the last video when I talked about vector-valued functions. If your codomain is a subset of our m, where m is greater than 1, then we say your function is vector-valued. It's not just mapping into the real numbers. It's mapping into some m-tuple of real numbers. If you mapped to two-dimensional space, you're dealing with a vector-valued function."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If your codomain is a subset of our m, where m is greater than 1, then we say your function is vector-valued. It's not just mapping into the real numbers. It's mapping into some m-tuple of real numbers. If you mapped to two-dimensional space, you're dealing with a vector-valued function. I've been all abstract and whatnot, so let me actually deal with some vectors, and it might make everything a little bit more concrete. Let's say I define the function f as f of x1, x2, and x3 is equal to x1 plus 2x2, and the second coordinate is just 3x3. I actually haven't formally defined coordinates for you yet, but I think you understand that just from your basic algebra training."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you mapped to two-dimensional space, you're dealing with a vector-valued function. I've been all abstract and whatnot, so let me actually deal with some vectors, and it might make everything a little bit more concrete. Let's say I define the function f as f of x1, x2, and x3 is equal to x1 plus 2x2, and the second coordinate is just 3x3. I actually haven't formally defined coordinates for you yet, but I think you understand that just from your basic algebra training. Let's say that that's my function definition. Based on the notation that we've been introduced to, we could say that f is a mapping, its domain is R3, and it maps from R3, or it associates all values in R3 with some value in R2. This is a 2-tuple."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I actually haven't formally defined coordinates for you yet, but I think you understand that just from your basic algebra training. Let's say that that's my function definition. Based on the notation that we've been introduced to, we could say that f is a mapping, its domain is R3, and it maps from R3, or it associates all values in R3 with some value in R2. This is a 2-tuple. This is in R2. This is a 3-tuple. Or another way we could do this, if we just wanted to write it in vector notation, I could write that if you pass f to f, some vector, x1, x2, x3, I could say this will be equal to the vector, and now it's going to have a two-component vector."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a 2-tuple. This is in R2. This is a 3-tuple. Or another way we could do this, if we just wanted to write it in vector notation, I could write that if you pass f to f, some vector, x1, x2, x3, I could say this will be equal to the vector, and now it's going to have a two-component vector. It's going to be a vector in R2, where the first term is x1 plus 2x2, and the second term is 3x3. Let's play around with this a little bit. See what it does to the vectors."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way we could do this, if we just wanted to write it in vector notation, I could write that if you pass f to f, some vector, x1, x2, x3, I could say this will be equal to the vector, and now it's going to have a two-component vector. It's going to be a vector in R2, where the first term is x1 plus 2x2, and the second term is 3x3. Let's play around with this a little bit. See what it does to the vectors. What is f of the vector 1, 1, 1? Well, I get 1 plus 2 times 1 is I get the vector 3, and then my second term is just 3 times this one, so I get the vector 3, 3. Fair enough."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "See what it does to the vectors. What is f of the vector 1, 1, 1? Well, I get 1 plus 2 times 1 is I get the vector 3, and then my second term is just 3 times this one, so I get the vector 3, 3. Fair enough. Let me do another one. Just to really experiment with this mapping, if I take the f of the vector in R3, 2, 4, 1, what do I get? That equals 2 plus 2 times 4."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Let me do another one. Just to really experiment with this mapping, if I take the f of the vector in R3, 2, 4, 1, what do I get? That equals 2 plus 2 times 4. That goes to the vector 10. I get 2 plus 2 times 4, and then 3 times the third term right there, so the vector 10, 3. How can we visualize this?"}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That equals 2 plus 2 times 4. That goes to the vector 10. I get 2 plus 2 times 4, and then 3 times the third term right there, so the vector 10, 3. How can we visualize this? Well, three-dimensional vectors or vectors in R3 are not always the easiest to visualize, but I think we can attempt to visualize these two guys. Let's see. The first, let me do it a little bit better than that."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How can we visualize this? Well, three-dimensional vectors or vectors in R3 are not always the easiest to visualize, but I think we can attempt to visualize these two guys. Let's see. The first, let me do it a little bit better than that. Let's say that this is the x1 axis, that's the x2 axis, that's the x3 axis. This first vector right here, this yellow one, 1, 1, 1, it'll look like this, 1, 1, 1. If I were to go out here, then go out here, then go up 1, the point would be right there, and if I were to draw it in standard position, I'd start at the origin, and the vector looks something like that."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The first, let me do it a little bit better than that. Let's say that this is the x1 axis, that's the x2 axis, that's the x3 axis. This first vector right here, this yellow one, 1, 1, 1, it'll look like this, 1, 1, 1. If I were to go out here, then go out here, then go up 1, the point would be right there, and if I were to draw it in standard position, I'd start at the origin, and the vector looks something like that. Then the second guy, 2, 4, 1, it would look like this. It would be, we'd go 2 out here, we'd go 4 in this direction, 1, 2, 3, 4, and then we'd go 1 up, so it'd look something like this. 2, 4, 1."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I were to go out here, then go out here, then go up 1, the point would be right there, and if I were to draw it in standard position, I'd start at the origin, and the vector looks something like that. Then the second guy, 2, 4, 1, it would look like this. It would be, we'd go 2 out here, we'd go 4 in this direction, 1, 2, 3, 4, and then we'd go 1 up, so it'd look something like this. 2, 4, 1. I think you get the idea. I've drawn these two vectors, these two vectors that are essentially in my domain. Our domain is R3."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2, 4, 1. I think you get the idea. I've drawn these two vectors, these two vectors that are essentially in my domain. Our domain is R3. This is R3 right here. Let's see what our function maps these vectors to. It maps, what's our codomain?"}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our domain is R3. This is R3 right here. Let's see what our function maps these vectors to. It maps, what's our codomain? Our codomain is R2, so this is much easier to visualize for us. We just have to draw two axes, just have to draw two axes like this. Let's call this X1, and let's call this X2."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It maps, what's our codomain? Our codomain is R2, so this is much easier to visualize for us. We just have to draw two axes, just have to draw two axes like this. Let's call this X1, and let's call this X2. What does F of 1, 1, 1 of this yellow vector, it becomes 3, 3. If I do it in yellow, 1, 2, 3, 1, 2, 3. It gets, you mean this one."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's call this X1, and let's call this X2. What does F of 1, 1, 1 of this yellow vector, it becomes 3, 3. If I do it in yellow, 1, 2, 3, 1, 2, 3. It gets, you mean this one. If I draw it in standard position, the vector looks like this. We literally, our function went from, mapped from this vector in R3 to this vector in R2. That was what our function did."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It gets, you mean this one. If I draw it in standard position, the vector looks like this. We literally, our function went from, mapped from this vector in R3 to this vector in R2. That was what our function did. Likewise, if we take the other vector, we went from this 2, 4, 1 vector to this vector 10, 3. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. It's going to look something like this."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That was what our function did. Likewise, if we take the other vector, we went from this 2, 4, 1 vector to this vector 10, 3. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. It's going to look something like this. It's going to be 3 up, so it's going to look something like this. This vector right here by our function F got mapped, let me do a different color, got mapped to this vector. This vector right here in R3 got mapped to this vector in R2 by our function."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to look something like this. It's going to be 3 up, so it's going to look something like this. This vector right here by our function F got mapped, let me do a different color, got mapped to this vector. This vector right here in R3 got mapped to this vector in R2 by our function. This is just a switch of terminology. When we talk about functions of vectors, the term that we tend to use is the word transformation. It really is the exact same thing as a function."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This vector right here in R3 got mapped to this vector in R2 by our function. This is just a switch of terminology. When we talk about functions of vectors, the term that we tend to use is the word transformation. It really is the exact same thing as a function. I don't want to confuse you because if you've watched the differential equations playlist, you saw the idea of a Laplace transformation, which is really an operation that takes a function as an argument. In this case, and when we're dealing in the linear algebra world, a transformation is really just a function operating on vectors or the way we're going to use it, it's just a function operating on vectors. The general notation, instead of writing a lowercase f like that, for a function, people use an uppercase T to say it's a transformation."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It really is the exact same thing as a function. I don't want to confuse you because if you've watched the differential equations playlist, you saw the idea of a Laplace transformation, which is really an operation that takes a function as an argument. In this case, and when we're dealing in the linear algebra world, a transformation is really just a function operating on vectors or the way we're going to use it, it's just a function operating on vectors. The general notation, instead of writing a lowercase f like that, for a function, people use an uppercase T to say it's a transformation. It doesn't have to be an uppercase T, but that's the one that people use the most. This could be a G or an H, but people always use a lowercase f. The same way we could have written this, we could have called this a transformation. My sense of why in the linear algebra world they use this is because you kind of imagine that this vector is being changed into that vector or that this vector is being transformed into that vector."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The general notation, instead of writing a lowercase f like that, for a function, people use an uppercase T to say it's a transformation. It doesn't have to be an uppercase T, but that's the one that people use the most. This could be a G or an H, but people always use a lowercase f. The same way we could have written this, we could have called this a transformation. My sense of why in the linear algebra world they use this is because you kind of imagine that this vector is being changed into that vector or that this vector is being transformed into that vector. I think that's why they call it a transformation as opposed to a function. It actually makes a lot more sense when you start going into things like video game programming and a lot of what we're embarking on with our transformations is key to video game programming, but you're kind of transforming one image into another image if you're viewing at it from a different angle or whatever else you might want to do. We'll talk a lot more about that in the future."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "My sense of why in the linear algebra world they use this is because you kind of imagine that this vector is being changed into that vector or that this vector is being transformed into that vector. I think that's why they call it a transformation as opposed to a function. It actually makes a lot more sense when you start going into things like video game programming and a lot of what we're embarking on with our transformations is key to video game programming, but you're kind of transforming one image into another image if you're viewing at it from a different angle or whatever else you might want to do. We'll talk a lot more about that in the future. I just want to introduce you to this notation. These statements, I could have just as easily written my, I could have replaced all my f's with t's and I could have defined some transformation. I just want to make you comfortable with the notation."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We'll talk a lot more about that in the future. I just want to introduce you to this notation. These statements, I could have just as easily written my, I could have replaced all my f's with t's and I could have defined some transformation. I just want to make you comfortable with the notation. I could have defined it similarly a transformation from R3 to R2. I could have said that t of x1, x2, x3 is equal to the 2-tuple x1 plus 2x1, 3x3. I could have just as similarly put a t up here because I've defined it the same way."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I just want to make you comfortable with the notation. I could have defined it similarly a transformation from R3 to R2. I could have said that t of x1, x2, x3 is equal to the 2-tuple x1 plus 2x1, 3x3. I could have just as similarly put a t up here because I've defined it the same way. I could have said t of my vector 1, 1, 1 is equal to 3, 3. Now you might say, hey Sal, why are you going through all this trouble of replacing the t's with s? I'm just doing this so you don't get confused."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I could have just as similarly put a t up here because I've defined it the same way. I could have said t of my vector 1, 1, 1 is equal to 3, 3. Now you might say, hey Sal, why are you going through all this trouble of replacing the t's with s? I'm just doing this so you don't get confused. So that when you see, in your linear algebra book, when you see linear algebra problems where you see this big capital T and you're like, wow, I've never seen that before and they're using this fancy word called a transformation, this is completely identical to your notion of a function. It is a function. In the next video, I'm going to talk about linear transformations."}, {"video_title": "Vector transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just doing this so you don't get confused. So that when you see, in your linear algebra book, when you see linear algebra problems where you see this big capital T and you're like, wow, I've never seen that before and they're using this fancy word called a transformation, this is completely identical to your notion of a function. It is a function. In the next video, I'm going to talk about linear transformations. That's really just linear functions and I'll define that a little bit more precisely in the next video. But hopefully by watching this video, you at least have a sense that you can apply functions to vectors and in the linear algebra world, we tend to call those transformations and hopefully this example right here gives you at least a visual representation of why it's called a transformation. We're transforming from one vector to another."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at five miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving five miles per hour in. So this right over here, which is often referred to as a speed, this is a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We don't know what direction this thing is moving five miles per hour in. So this right over here, which is often referred to as a speed, this is a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving five miles per hour east. So let's say it's moving five miles per hour due east. So now this combined five miles per hour due east, this is a vector quantity, and now we wouldn't call it speed anymore, we would call it velocity."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving five miles per hour east. So let's say it's moving five miles per hour due east. So now this combined five miles per hour due east, this is a vector quantity, and now we wouldn't call it speed anymore, we would call it velocity. So velocity is a vector. We're specifying the magnitude, five miles per hour, and the direction east. But how can we actually visualize this?"}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now this combined five miles per hour due east, this is a vector quantity, and now we wouldn't call it speed anymore, we would call it velocity. So velocity is a vector. We're specifying the magnitude, five miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions, and what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three, and then even four, five, six, as many dimensions as we want. Our brains have trouble visualizing beyond three, but what's neat is we can mathematically deal with beyond three using linear algebra, and we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But how can we actually visualize this? So let's say we're operating in two dimensions, and what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three, and then even four, five, six, as many dimensions as we want. Our brains have trouble visualizing beyond three, but what's neat is we can mathematically deal with beyond three using linear algebra, and we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is five units long, we'll assume that each of our units, our unit here is miles per hour, and that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here, and I could make its length five. The length of the arrow specifies the magnitude, so one, two, three, four, five, and then the direction that the arrow is pointed in specifies its direction."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is five units long, we'll assume that each of our units, our unit here is miles per hour, and that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here, and I could make its length five. The length of the arrow specifies the magnitude, so one, two, three, four, five, and then the direction that the arrow is pointed in specifies its direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now what's interesting about vectors is that we only care about the magnitude and the direction."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The length of the arrow specifies the magnitude, so one, two, three, four, five, and then the direction that the arrow is pointed in specifies its direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now what's interesting about vectors is that we only care about the magnitude and the direction. We don't necessarily care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or it would be equivalent vector to this. This vector has the same length, so it has the same magnitude, has a length of five, and its direction is also due east, so these two vectors are equivalent."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what's interesting about vectors is that we only care about the magnitude and the direction. We don't necessarily care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or it would be equivalent vector to this. This vector has the same length, so it has the same magnitude, has a length of five, and its direction is also due east, so these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough, but how do we represent it with a little bit more mathematical notation so we don't have to draw it every time and we can start performing operations on it? Well, the typical way, one, if you want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it, but when you're doing it in your notebook, you would typically put a little arrow on top of it, and there's several ways that you could do it."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This vector has the same length, so it has the same magnitude, has a length of five, and its direction is also due east, so these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough, but how do we represent it with a little bit more mathematical notation so we don't have to draw it every time and we can start performing operations on it? Well, the typical way, one, if you want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it, but when you're doing it in your notebook, you would typically put a little arrow on top of it, and there's several ways that you could do it. You could literally say, hey, five miles per hour east, but that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions. So for example, this one only moves in the horizontal dimension, and so we'll put our horizontal dimension first, so you might call this vector five comma zero."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you're publishing a book, you can bold it, but when you're doing it in your notebook, you would typically put a little arrow on top of it, and there's several ways that you could do it. You could literally say, hey, five miles per hour east, but that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions. So for example, this one only moves in the horizontal dimension, and so we'll put our horizontal dimension first, so you might call this vector five comma zero. It's moving five, positive five, in the horizontal direction, and it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in a linear algebra context, it's more typical to write it as a column vector like this, five, zero."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So for example, this one only moves in the horizontal dimension, and so we'll put our horizontal dimension first, so you might call this vector five comma zero. It's moving five, positive five, in the horizontal direction, and it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in a linear algebra context, it's more typical to write it as a column vector like this, five, zero. This, once again, the first coordinate represents how much we're moving in the horizontal direction, and the second coordinate represents how much are we moving in the vertical direction. Now this one isn't that interesting. You could have other vectors."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You might also see notation, and actually in a linear algebra context, it's more typical to write it as a column vector like this, five, zero. This, once again, the first coordinate represents how much we're moving in the horizontal direction, and the second coordinate represents how much are we moving in the vertical direction. Now this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. It's moving three in the horizontal direction, and positive four, so one, two, three, four, in the vertical direction. So let me see."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could have other vectors. You could have a vector that looks like this. It's moving three in the horizontal direction, and positive four, so one, two, three, four, in the vertical direction. So let me see. It might look something like this. So this could be another vector right over here. Maybe we call this vector vector A, and once again, I want to specify that it is a vector, and you see here that if you were to break it down, in the horizontal direction, it is moving three, or it's shifting three in the horizontal direction, and it's shifting four, positive four, in the vertical direction, and we get that by literally thinking about how much we're moving up, and how much we're moving to the right when we start at the end of the arrow and go to the front of it."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me see. It might look something like this. So this could be another vector right over here. Maybe we call this vector vector A, and once again, I want to specify that it is a vector, and you see here that if you were to break it down, in the horizontal direction, it is moving three, or it's shifting three in the horizontal direction, and it's shifting four, positive four, in the vertical direction, and we get that by literally thinking about how much we're moving up, and how much we're moving to the right when we start at the end of the arrow and go to the front of it. So this vector might be specified as three, three, four. Three, four. And you could use the Pythagorean theorem to figure out the actual length of this vector, and you'll see, because this is a three, four, five triangle that this actually has a magnitude of five."}, {"video_title": "Vector intro for linear algebra Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe we call this vector vector A, and once again, I want to specify that it is a vector, and you see here that if you were to break it down, in the horizontal direction, it is moving three, or it's shifting three in the horizontal direction, and it's shifting four, positive four, in the vertical direction, and we get that by literally thinking about how much we're moving up, and how much we're moving to the right when we start at the end of the arrow and go to the front of it. So this vector might be specified as three, three, four. Three, four. And you could use the Pythagorean theorem to figure out the actual length of this vector, and you'll see, because this is a three, four, five triangle that this actually has a magnitude of five. And as we study more and more linear algebra, we're going to start extending these to multiple dimensions. Obviously, we can visualize up to three dimensions. In four dimensions, it becomes more abstract, and that's why this type of notation is useful, because it's very hard to draw a four, five, or 20-dimensional arrow like this."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's a very useful tool. And that just told us if I have two vectors, x and y, they're both members of Rn, and they're both non-zero vectors. And that was an assumption we had to make when we did the proof. Otherwise, there's a potential of dividing by one of their magnitudes, so that would have been a big no-no. But if we assume that they're both non-zero, then we can say that the absolute value of their dot products is going to be less than or equal to the products of their individual lengths. So that's the length of vector x, and we defined that a couple of videos ago. And then this is the length of vector y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Otherwise, there's a potential of dividing by one of their magnitudes, so that would have been a big no-no. But if we assume that they're both non-zero, then we can say that the absolute value of their dot products is going to be less than or equal to the products of their individual lengths. So that's the length of vector x, and we defined that a couple of videos ago. And then this is the length of vector y. And of course, this is just a regular number, and then each of these are just regular numbers. They're not vectors once you take a length. The length of a 50-dimensional vector could just be the number 3."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then this is the length of vector y. And of course, this is just a regular number, and then each of these are just regular numbers. They're not vectors once you take a length. The length of a 50-dimensional vector could just be the number 3. It's just a scalar value. So this is just scalar multiplication here. And we also learned that the only time that this inequality turns into an equality is a situation where x is equal to some scalar multiple of y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The length of a 50-dimensional vector could just be the number 3. It's just a scalar value. So this is just scalar multiplication here. And we also learned that the only time that this inequality turns into an equality is a situation where x is equal to some scalar multiple of y. And so in some textbooks, you'll say, and this has to be a non-zero scalar multiple, but that's a bit obvious. I told you that x and y are non-zero. So if this was zero, then x would be zero."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we also learned that the only time that this inequality turns into an equality is a situation where x is equal to some scalar multiple of y. And so in some textbooks, you'll say, and this has to be a non-zero scalar multiple, but that's a bit obvious. I told you that x and y are non-zero. So if this was zero, then x would be zero. And I just said that x is non-zero, but if you want to say that, you could say that c also is going to be non-zero. But that essentially just falls out of this information there. But if this is the case, and if and only if this is the case, then we can say that the absolute value of the dot product of the two vectors is equal to the product of their lengths."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if this was zero, then x would be zero. And I just said that x is non-zero, but if you want to say that, you could say that c also is going to be non-zero. But that essentially just falls out of this information there. But if this is the case, and if and only if this is the case, then we can say that the absolute value of the dot product of the two vectors is equal to the product of their lengths. Now, this is all just a review of what I did in the last video. Now, what else can we do that's useful with it? So let's just play around a little bit."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if this is the case, and if and only if this is the case, then we can say that the absolute value of the dot product of the two vectors is equal to the product of their lengths. Now, this is all just a review of what I did in the last video. Now, what else can we do that's useful with it? So let's just play around a little bit. But I can't claim to be experimenting. I know where this is going to go. Let's see what happens if I were to take the length of x plus y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's just play around a little bit. But I can't claim to be experimenting. I know where this is going to go. Let's see what happens if I were to take the length of x plus y. So I'm going to add the two vectors and then take the length of that vector squared. Well, we know from a couple of videos ago that the length squared can also be rewritten as the dot product of a vector with itself. This right here, x plus y, I know it looks like two vectors, but it's two vectors added to each other."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see what happens if I were to take the length of x plus y. So I'm going to add the two vectors and then take the length of that vector squared. Well, we know from a couple of videos ago that the length squared can also be rewritten as the dot product of a vector with itself. This right here, x plus y, I know it looks like two vectors, but it's two vectors added to each other. So it's really a vector. x plus y is a real vector. I could graph x plus y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This right here, x plus y, I know it looks like two vectors, but it's two vectors added to each other. So it's really a vector. x plus y is a real vector. I could graph x plus y. So the length of x plus y squared, I can rewrite it as the dot product of that vector with itself. So x plus y dot x plus y. And all of these are vectors."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could graph x plus y. So the length of x plus y squared, I can rewrite it as the dot product of that vector with itself. So x plus y dot x plus y. And all of these are vectors. These aren't just numbers. And this is the dot product. It's just not normal multiplication."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And all of these are vectors. These aren't just numbers. And this is the dot product. It's just not normal multiplication. But we saw two videos ago that the dot product has a distributive and the associative and the commutative properties, just like regular scalar multiplication. So you can kind of foil this out, if that's how you remember multiplying your binomials. Or I always think of it more as just doing the distributive property twice."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just not normal multiplication. But we saw two videos ago that the dot product has a distributive and the associative and the commutative properties, just like regular scalar multiplication. So you can kind of foil this out, if that's how you remember multiplying your binomials. Or I always think of it more as just doing the distributive property twice. So this can be rewritten as x dot x. Actually, let me write it as a distributive property, because that's sometimes not obvious to a lot of people. So let me write this x as a yellow x."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or I always think of it more as just doing the distributive property twice. So this can be rewritten as x dot x. Actually, let me write it as a distributive property, because that's sometimes not obvious to a lot of people. So let me write this x as a yellow x. And let me write this whole term as x plus y. So this right here can be rewritten as x dot, so this x dot this x plus y. And then it would be plus this y dot, I wanted to switch colors, plus this y dot x plus y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me write this x as a yellow x. And let me write this whole term as x plus y. So this right here can be rewritten as x dot, so this x dot this x plus y. And then it would be plus this y dot, I wanted to switch colors, plus this y dot x plus y. It's good to see that when you're multiplying these, you're just applying the distributive property. All I did is I distributed this term along each of these terms in this sum right here. So then I got this."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then it would be plus this y dot, I wanted to switch colors, plus this y dot x plus y. It's good to see that when you're multiplying these, you're just applying the distributive property. All I did is I distributed this term along each of these terms in this sum right here. So then I got this. And then I can distribute each of these into the sum. So then this becomes, I'll be careful with the colors, x dot x plus x dot y. Maybe this was a little bit overkill, but I think it's good to see that this isn't just some magic here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So then I got this. And then I can distribute each of these into the sum. So then this becomes, I'll be careful with the colors, x dot x plus x dot y. Maybe this was a little bit overkill, but I think it's good to see that this isn't just some magic here. And we're just using the exact properties that we proved of the dot product. So that's that right there. And then it's plus y dot x."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe this was a little bit overkill, but I think it's good to see that this isn't just some magic here. And we're just using the exact properties that we proved of the dot product. So that's that right there. And then it's plus y dot x. So y dot x plus this yellow y dot the blue y. So the magnitude or the length of our vector x plus y squared can be rewritten like this. And this, and I'll just switch back to one color."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then it's plus y dot x. So y dot x plus this yellow y dot the blue y. So the magnitude or the length of our vector x plus y squared can be rewritten like this. And this, and I'll just switch back to one color. So this equals that. And all of that, what does this equal? This is equal to x dot x."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this, and I'll just switch back to one color. So this equals that. And all of that, what does this equal? This is equal to x dot x. So this is equal to, what's x dot x? x dot x is just the magnitude. So let me write this is just equal to the magnitude of our vector x. I should stop using the word magnitude."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to x dot x. So this is equal to, what's x dot x? x dot x is just the magnitude. So let me write this is just equal to the magnitude of our vector x. I should stop using the word magnitude. The length of our vector x squared. And then I have two terms here. I have an x dot y and a y dot x."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me write this is just equal to the magnitude of our vector x. I should stop using the word magnitude. The length of our vector x squared. And then I have two terms here. I have an x dot y and a y dot x. But we know that x dot y and y dot x are really the same thing. We showed that order doesn't matter when you take the dot product, just like it doesn't matter with regular multiplication. So these are really the same terms."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have an x dot y and a y dot x. But we know that x dot y and y dot x are really the same thing. We showed that order doesn't matter when you take the dot product, just like it doesn't matter with regular multiplication. So these are really the same terms. So we could write plus 2 times x dot y. And then finally we have that last term sitting here. We have this y dot y. y dot y is the same thing as the length of our vector y squared."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So these are really the same terms. So we could write plus 2 times x dot y. And then finally we have that last term sitting here. We have this y dot y. y dot y is the same thing as the length of our vector y squared. Now, let's see if we can break out our Cauchy-Schwarz inequality. Or maybe Schwarz, I don't know if I'm pronouncing it right. But x dot y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have this y dot y. y dot y is the same thing as the length of our vector y squared. Now, let's see if we can break out our Cauchy-Schwarz inequality. Or maybe Schwarz, I don't know if I'm pronouncing it right. But x dot y. So x dot y, we have the absolute value of x dot y here. But we know that just x dot y is going to be, it has to be less than or equal to the absolute value of x dot y. Why is that?"}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But x dot y. So x dot y, we have the absolute value of x dot y here. But we know that just x dot y is going to be, it has to be less than or equal to the absolute value of x dot y. Why is that? Well, this could be negative. I could show you examples of dot products that are negative. In fact, if x has all positive terms and y has all negative terms, you're going to have a negative dot product."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Why is that? Well, this could be negative. I could show you examples of dot products that are negative. In fact, if x has all positive terms and y has all negative terms, you're going to have a negative dot product. So this could be negative. Or it could be positive. If it's positive, the absolute value, they're equal to each other."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In fact, if x has all positive terms and y has all negative terms, you're going to have a negative dot product. So this could be negative. Or it could be positive. If it's positive, the absolute value, they're equal to each other. If this is negative, then this absolute value is definitely going to be greater than it. So we can add to the Cauchy-Schwarz inequality. And this is a bit obvious."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If it's positive, the absolute value, they're equal to each other. If this is negative, then this absolute value is definitely going to be greater than it. So we can add to the Cauchy-Schwarz inequality. And this is a bit obvious. We could say, look, we could add a little x dot y is less than or equal to the absolute value of x dot y, which is less than or equal to the length of x times the length of y. So x dot y, the dot product of x with y is definitely less than its absolute value of that, which is definitely less than the lengths of those two multiplied. So if I rewrite this, this statement right here is definitely less than or equal to this exact statement."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is a bit obvious. We could say, look, we could add a little x dot y is less than or equal to the absolute value of x dot y, which is less than or equal to the length of x times the length of y. So x dot y, the dot product of x with y is definitely less than its absolute value of that, which is definitely less than the lengths of those two multiplied. So if I rewrite this, this statement right here is definitely less than or equal to this exact statement. But if I replace these with the lengths of the vectors, so that is definitely less than or equal to, I'm just rewriting this x squared, and I'll write the plus 2 there. But I want to make it very clear what I'm replacing here. And then I have the plus length of my vector y there squared."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I rewrite this, this statement right here is definitely less than or equal to this exact statement. But if I replace these with the lengths of the vectors, so that is definitely less than or equal to, I'm just rewriting this x squared, and I'll write the plus 2 there. But I want to make it very clear what I'm replacing here. And then I have the plus length of my vector y there squared. Now this, I'm saying, this is definitely less than the absolute value of x dot y, which is definitely less by the Cauchy-Schwarz inequality, definitely less than the product of the two lengths. So all I did here, so I'm just replacing this with the product of their two lengths. So I'm going to put x, the length of x times the length of y, right?"}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I have the plus length of my vector y there squared. Now this, I'm saying, this is definitely less than the absolute value of x dot y, which is definitely less by the Cauchy-Schwarz inequality, definitely less than the product of the two lengths. So all I did here, so I'm just replacing this with the product of their two lengths. So I'm going to put x, the length of x times the length of y, right? And since this is the same as that, this is the same as that, but this is definitely less than this. This whole term has to be less than this whole term. Now let me just remind you what we were doing."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to put x, the length of x times the length of y, right? And since this is the same as that, this is the same as that, but this is definitely less than this. This whole term has to be less than this whole term. Now let me just remind you what we were doing. I said that this thing that I wrote over here, this is the same as that. So this thing up here, which is the same as that, which is less than that, also. So we can write that the magnitude of x plus y squared, and not the magnitude, the length of the vector x plus y squared is less than this whole thing that I wrote out here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let me just remind you what we were doing. I said that this thing that I wrote over here, this is the same as that. So this thing up here, which is the same as that, which is less than that, also. So we can write that the magnitude of x plus y squared, and not the magnitude, the length of the vector x plus y squared is less than this whole thing that I wrote out here. Now, or less than or equal to. Now what is this thing? Remember, I mean this might look all fancy with my little double lines around everything, but these are just numbers."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we can write that the magnitude of x plus y squared, and not the magnitude, the length of the vector x plus y squared is less than this whole thing that I wrote out here. Now, or less than or equal to. Now what is this thing? Remember, I mean this might look all fancy with my little double lines around everything, but these are just numbers. This length of x squared, this is just a number. Each of these are numbers. And I can just say, hey, look, this looks like a perfect square to me."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, I mean this might look all fancy with my little double lines around everything, but these are just numbers. This length of x squared, this is just a number. Each of these are numbers. And I can just say, hey, look, this looks like a perfect square to me. This looks just like this term on the right-hand side is the exact same thing as the length of x plus the length of y plus the length of y, everything squared, right? If you just square this out, you'll get x squared plus 2 times the length of x times the length of y plus y squared, so our length of the vector x plus y squared is less than or equal to this quantity over here. And if we just take the square root of both sides of this, you get the length of our vector x plus y is less than or equal to the length of the vector x by itself plus the length of the vector y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I can just say, hey, look, this looks like a perfect square to me. This looks just like this term on the right-hand side is the exact same thing as the length of x plus the length of y plus the length of y, everything squared, right? If you just square this out, you'll get x squared plus 2 times the length of x times the length of y plus y squared, so our length of the vector x plus y squared is less than or equal to this quantity over here. And if we just take the square root of both sides of this, you get the length of our vector x plus y is less than or equal to the length of the vector x by itself plus the length of the vector y. And we call this the triangle inequality, which you might have remembered from geometry. Triangle inequality. Now, why is it called a triangle inequality?"}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if we just take the square root of both sides of this, you get the length of our vector x plus y is less than or equal to the length of the vector x by itself plus the length of the vector y. And we call this the triangle inequality, which you might have remembered from geometry. Triangle inequality. Now, why is it called a triangle inequality? Well, you can imagine each of these to be separate sides of a triangle. In fact, let's draw it. We can draw this in R2."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, why is it called a triangle inequality? Well, you can imagine each of these to be separate sides of a triangle. In fact, let's draw it. We can draw this in R2. Let me turn my graph paper on. Let me see where the graphs show up. If I turn my graph paper on right there, maybe I'll draw it here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can draw this in R2. Let me turn my graph paper on. Let me see where the graphs show up. If I turn my graph paper on right there, maybe I'll draw it here. So let's draw my vector x. So let's say that my vector x looks something like this. Let's say that's my vector x."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I turn my graph paper on right there, maybe I'll draw it here. So let's draw my vector x. So let's say that my vector x looks something like this. Let's say that's my vector x. It's the vector 2, 4. So that's my vector x. And let's say my vector y, well, I'm just going to do it head to tail because I'm going to add the two."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that's my vector x. It's the vector 2, 4. So that's my vector x. And let's say my vector y, well, I'm just going to do it head to tail because I'm going to add the two. So my vector y, I'm going to do it in non-standard position. Let's say my vector y looks something like this. I'm going to draw it properly."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And let's say my vector y, well, I'm just going to do it head to tail because I'm going to add the two. So my vector y, I'm going to do it in non-standard position. Let's say my vector y looks something like this. I'm going to draw it properly. So let's say that's my vector y. Now, what does x plus y look like? x plus y, and remember, I can't necessarily draw any two vectors on two-dimensional space like this."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to draw it properly. So let's say that's my vector y. Now, what does x plus y look like? x plus y, and remember, I can't necessarily draw any two vectors on two-dimensional space like this. I'm just assuming that these are in R2. But this is just to give you the idea. So then this is their sum, right?"}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x plus y, and remember, I can't necessarily draw any two vectors on two-dimensional space like this. I'm just assuming that these are in R2. But this is just to give you the idea. So then this is their sum, right? From the tail of x to the head of y. So this vector right here is the vector x plus y. And that's why it's called a triangle inequality."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So then this is their sum, right? From the tail of x to the head of y. So this vector right here is the vector x plus y. And that's why it's called a triangle inequality. It's just saying that, look, this thing is always going to be less than or equal to, or the length of this thing is always going to be less than or equal to the length of this thing plus the length of this thing. And that's kind of obvious when you just learn two-dimensional geometry. This is a much more efficient way of getting from this point to this point than going out here and then going out here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's why it's called a triangle inequality. It's just saying that, look, this thing is always going to be less than or equal to, or the length of this thing is always going to be less than or equal to the length of this thing plus the length of this thing. And that's kind of obvious when you just learn two-dimensional geometry. This is a much more efficient way of getting from this point to this point than going out here and then going out here. And then what's the limiting, or what's the case in which this length is equal to these lengths? Well, if you keep flattening this triangle out and you go to the extreme case where maybe the vector x looks like this, and if the vector y is just kind of going in the exact same direction, maybe it's going a little bit further, so this is vector x, this is vector y. Now x plus y will just be this whole vector."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a much more efficient way of getting from this point to this point than going out here and then going out here. And then what's the limiting, or what's the case in which this length is equal to these lengths? Well, if you keep flattening this triangle out and you go to the extreme case where maybe the vector x looks like this, and if the vector y is just kind of going in the exact same direction, maybe it's going a little bit further, so this is vector x, this is vector y. Now x plus y will just be this whole vector. Now that whole thing is x plus y. And this is the case now where the triangle inequality turns into an equality. That's why that little equal sign there."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now x plus y will just be this whole vector. Now that whole thing is x plus y. And this is the case now where the triangle inequality turns into an equality. That's why that little equal sign there. The extreme case where essentially x and y are collinear. And why does that work out? We can even go back to our math and understand that."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's why that little equal sign there. The extreme case where essentially x and y are collinear. And why does that work out? We can even go back to our math and understand that. Let me turn my graph off. We can go back to our math here. If I go back to this point, remember right here I made the statement, look, this thing is definitely less than this thing over here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can even go back to our math and understand that. Let me turn my graph off. We can go back to our math here. If I go back to this point, remember right here I made the statement, look, this thing is definitely less than this thing over here. But what if I made an assumption? What if I said that x is equal to some scalar times y? And actually, I have to be careful, because just some scalar times y, remember our Cauchy-Schwarz inequality said that look, the inequality turns into an equality if x is some non-zero scalar of y."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I go back to this point, remember right here I made the statement, look, this thing is definitely less than this thing over here. But what if I made an assumption? What if I said that x is equal to some scalar times y? And actually, I have to be careful, because just some scalar times y, remember our Cauchy-Schwarz inequality said that look, the inequality turns into an equality if x is some non-zero scalar of y. And then we can apply this. We can say that the absolute value of x dot y is the same as this over here. But I don't have the absolute value of x dot y here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And actually, I have to be careful, because just some scalar times y, remember our Cauchy-Schwarz inequality said that look, the inequality turns into an equality if x is some non-zero scalar of y. And then we can apply this. We can say that the absolute value of x dot y is the same as this over here. But I don't have the absolute value of x dot y here. I don't know that this is positive. I can say definitively that this is a positive quantity because I took the absolute value of it. No absolute value here."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But I don't have the absolute value of x dot y here. I don't know that this is positive. I can say definitively that this is a positive quantity because I took the absolute value of it. No absolute value here. So the only way that I can ensure that this is a positive quantity, that this is the same thing as the absolute value of x dot y, is to enforce, if I'm going to go down this road, is to enforce that this term right here, that c be positive. Because if c is positive, then x dot y, then that would be the same thing as c y dot y, which we know is just the same thing as c times the magnitude of y squared. And the only way that I can ensure that this expression right here is equal to the absolute value of x dot y, the only way I can assure this is if c is positive."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "No absolute value here. So the only way that I can ensure that this is a positive quantity, that this is the same thing as the absolute value of x dot y, is to enforce, if I'm going to go down this road, is to enforce that this term right here, that c be positive. Because if c is positive, then x dot y, then that would be the same thing as c y dot y, which we know is just the same thing as c times the magnitude of y squared. And the only way that I can ensure that this expression right here is equal to the absolute value of x dot y, the only way I can assure this is if c is positive. If c is negative, then this is going to be a negative number while this is a positive. So if I assume that this is positive, then I can say that x dot y is equal to the absolute value of x dot y. And since it's a scalar multiple, then I could say that that term is equal to, not just less than or equal to, the magnitude of x's and y's."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the only way that I can ensure that this expression right here is equal to the absolute value of x dot y, the only way I can assure this is if c is positive. If c is negative, then this is going to be a negative number while this is a positive. So if I assume that this is positive, then I can say that x dot y is equal to the absolute value of x dot y. And since it's a scalar multiple, then I could say that that term is equal to, not just less than or equal to, the magnitude of x's and y's. So hopefully I'm not confusing you. So all I'm saying is, if I can assume that x is some positive scalar multiple of y, that this wouldn't be a less than sign, then I could say that x dot y is the same thing as the absolute value of x dot y, since this is positive. And if it's the same thing as the absolute value of x dot y, and it's some scalar multiple of each other, then we could go down this other route."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And since it's a scalar multiple, then I could say that that term is equal to, not just less than or equal to, the magnitude of x's and y's. So hopefully I'm not confusing you. So all I'm saying is, if I can assume that x is some positive scalar multiple of y, that this wouldn't be a less than sign, then I could say that x dot y is the same thing as the absolute value of x dot y, since this is positive. And if it's the same thing as the absolute value of x dot y, and it's some scalar multiple of each other, then we could go down this other route. We could say that this thing here, I don't want to get too messy, we could say that this is equal to that. If this is equal to that, then this would have become an equal sign, not a less than or equal to sign. And then we would have had the limiting case."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if it's the same thing as the absolute value of x dot y, and it's some scalar multiple of each other, then we could go down this other route. We could say that this thing here, I don't want to get too messy, we could say that this is equal to that. If this is equal to that, then this would have become an equal sign, not a less than or equal to sign. And then we would have had the limiting case. But we could say that x plus y, we've done the same work, but we'd had an equal sign the whole way, would equal the length of x plus y. In the situation where x is equal to some positive scalar times y, so c is greater than 0. These two imply each other."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we would have had the limiting case. But we could say that x plus y, we've done the same work, but we'd had an equal sign the whole way, would equal the length of x plus y. In the situation where x is equal to some positive scalar times y, so c is greater than 0. These two imply each other. And we saw that geometrically. I lost my axes here, but we see that the only time that the length of x plus y is equal to the length of x plus the length of y is when they're collinear. Over here, this plus this is clearly, you can just visually look at it, longer than this right there."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These two imply each other. And we saw that geometrically. I lost my axes here, but we see that the only time that the length of x plus y is equal to the length of x plus the length of y is when they're collinear. Over here, this plus this is clearly, you can just visually look at it, longer than this right there. So you might be saying, Sal, once again, this linear algebra is a little bit silly. We learned the triangle inequality in eighth or ninth grade. Why did you go through all of this pain to redefine it?"}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Over here, this plus this is clearly, you can just visually look at it, longer than this right there. So you might be saying, Sal, once again, this linear algebra is a little bit silly. We learned the triangle inequality in eighth or ninth grade. Why did you go through all of this pain to redefine it? And this is the interesting thing. What I just drew here, and this is what you learned in ninth grade geometry, this is just an R2. This is just your Cartesian coordinates."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Why did you go through all of this pain to redefine it? And this is the interesting thing. What I just drew here, and this is what you learned in ninth grade geometry, this is just an R2. This is just your Cartesian coordinates. Or I don't want to use the word dimension too much, because we're going to define that formally, but this is kind of your two-dimensional space that's going on. But what's interesting or what's useful about linear algebra is we've just defined the triangle inequality for arbitrarily large vectors, or vectors that have an arbitrarily high number of components. Each of these, these don't have to be an R2."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is just your Cartesian coordinates. Or I don't want to use the word dimension too much, because we're going to define that formally, but this is kind of your two-dimensional space that's going on. But what's interesting or what's useful about linear algebra is we've just defined the triangle inequality for arbitrarily large vectors, or vectors that have an arbitrarily high number of components. Each of these, these don't have to be an R2. These could be, this is true if we're in R100, where every vector has 100 components to it. We've just defined some notion of the triangle inequality. We've abstracted well beyond just our two-dimensional Cartesian coordinates, well beyond even our three dimensions, to essentially n-dimensional space."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Each of these, these don't have to be an R2. These could be, this is true if we're in R100, where every vector has 100 components to it. We've just defined some notion of the triangle inequality. We've abstracted well beyond just our two-dimensional Cartesian coordinates, well beyond even our three dimensions, to essentially n-dimensional space. And I haven't defined dimensions yet, but I think you're starting to appreciate what they are. But anyway, hopefully you found that useful. And we can now take this result, and actually that result with this result, and define what the notion of an angle between two vectors are."}, {"video_title": "Vector triangle inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We've abstracted well beyond just our two-dimensional Cartesian coordinates, well beyond even our three dimensions, to essentially n-dimensional space. And I haven't defined dimensions yet, but I think you're starting to appreciate what they are. But anyway, hopefully you found that useful. And we can now take this result, and actually that result with this result, and define what the notion of an angle between two vectors are. And once again, on some level, you're like, wow, why do we have to define an angle? Isn't an angle just, isn't that just an angle? Well, yeah, we know what an angle is in two dimensions, but what does an angle mean when you abstract things to n dimensions, or when you're in Rn?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's say we know that the linear transformation T can be this transformation matrix, when you put it in reduced row echelon form, the transformation matrix, it is equal to an n by n identity matrix, or the n by n identity matrix. Well, this alone tells us a lot of things. First of all, if when you put this in reduced row echelon form, you get a square identity matrix, that tells us that the original matrix had to be n by n. And it also tells us that T is a mapping from Rn to Rn. And we saw in the last video, all of these are conditions, especially this one right here, for T to be invertible. So if we know that this is true, T is a linear transformation, it's the reduced row echelon form of its transformation matrix, is the identity matrix right there. We know that T is invertible. But let's remind ourselves what it even means to be invertible."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we saw in the last video, all of these are conditions, especially this one right here, for T to be invertible. So if we know that this is true, T is a linear transformation, it's the reduced row echelon form of its transformation matrix, is the identity matrix right there. We know that T is invertible. But let's remind ourselves what it even means to be invertible. To be invertible means that there exists some, we used the word function before, but now we're talking about transformations, they're really the same thing. So let's say there exists some transformation, let's call it T inverse, T inverse like that, or T to the minus 1, such that the composition of T inverse with T is equal to the identity transformation on your domain, and the composition of T with T inverse is equal to the identity transformation on your codomain, just like that. And just to remind you what this looks like, let's draw our domains and codomain."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But let's remind ourselves what it even means to be invertible. To be invertible means that there exists some, we used the word function before, but now we're talking about transformations, they're really the same thing. So let's say there exists some transformation, let's call it T inverse, T inverse like that, or T to the minus 1, such that the composition of T inverse with T is equal to the identity transformation on your domain, and the composition of T with T inverse is equal to the identity transformation on your codomain, just like that. And just to remind you what this looks like, let's draw our domains and codomain. Our domain is Rn, and our codomain is also Rn. So it's just like that. Codomain is also Rn."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And just to remind you what this looks like, let's draw our domains and codomain. Our domain is Rn, and our codomain is also Rn. So it's just like that. Codomain is also Rn. So if you take some vector in your domain, apply the transformation T, you're going to go into your codomain, so that is T. And then if you apply the T inverse after that, you're going to go back to that original x. So this says, look, you apply T and then you apply T inverse, you're just going to get back to where you started. It's equivalent to the identity transformation, just like that."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Codomain is also Rn. So if you take some vector in your domain, apply the transformation T, you're going to go into your codomain, so that is T. And then if you apply the T inverse after that, you're going to go back to that original x. So this says, look, you apply T and then you apply T inverse, you're just going to get back to where you started. It's equivalent to the identity transformation, just like that. This is saying, if you start here in your codomain, you apply your inverse first, you apply this inverse transformation first, then you apply your transformation, you're going to go back to the same point in your codomain. So it's equivalent to the identity transformation in your codomain. It just happens to be in this case that the domain and the codomain are the same set, Rn."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equivalent to the identity transformation, just like that. This is saying, if you start here in your codomain, you apply your inverse first, you apply this inverse transformation first, then you apply your transformation, you're going to go back to the same point in your codomain. So it's equivalent to the identity transformation in your codomain. It just happens to be in this case that the domain and the codomain are the same set, Rn. Now, we know what a transformation, what it means to be invertible. We know what the conditions are for invertibility. So this begs the next question."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It just happens to be in this case that the domain and the codomain are the same set, Rn. Now, we know what a transformation, what it means to be invertible. We know what the conditions are for invertibility. So this begs the next question. We know that this guy is a linear transformation. In fact, that's one of the conditions to be able to represent it as a matrix. Or any transformation that can be represented as a matrix vector product is a linear transformation."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this begs the next question. We know that this guy is a linear transformation. In fact, that's one of the conditions to be able to represent it as a matrix. Or any transformation that can be represented as a matrix vector product is a linear transformation. So this guy is a linear transformation. But the question is, is T inverse a linear transformation? Now, let's review what the two conditions are that we need to have to be a linear transformation."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or any transformation that can be represented as a matrix vector product is a linear transformation. So this guy is a linear transformation. But the question is, is T inverse a linear transformation? Now, let's review what the two conditions are that we need to have to be a linear transformation. So we know T is a linear transformation. So we know that if you apply the transformation T to two vectors, let's say x and y, if we apply it to the sum of those two vectors, it is equal to the transformation of the first vector plus the transformation of the second vector. That's one of the conditions, or one thing that we know is true for all linear transformations."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's review what the two conditions are that we need to have to be a linear transformation. So we know T is a linear transformation. So we know that if you apply the transformation T to two vectors, let's say x and y, if we apply it to the sum of those two vectors, it is equal to the transformation of the first vector plus the transformation of the second vector. That's one of the conditions, or one thing that we know is true for all linear transformations. And the second thing that we know is true for all linear transformations is if we take the transformation of some scaled version of a vector in our domain, it is equal to the scaling factor times the transformation of the vector itself. These are both conditions for linear transformations. So let's see if we can prove that both of these conditions hold for T inverse, for this guy right here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's one of the conditions, or one thing that we know is true for all linear transformations. And the second thing that we know is true for all linear transformations is if we take the transformation of some scaled version of a vector in our domain, it is equal to the scaling factor times the transformation of the vector itself. These are both conditions for linear transformations. So let's see if we can prove that both of these conditions hold for T inverse, for this guy right here. So to do this, let's do this little exercise right here. Let's apply T, let's take the composition of T with T inverse of two vectors, a plus b. Remember, T inverse is a mapping from your codomain to your domain, although they're both going to be Rn in this case, but T inverse maps from this set to that set."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can prove that both of these conditions hold for T inverse, for this guy right here. So to do this, let's do this little exercise right here. Let's apply T, let's take the composition of T with T inverse of two vectors, a plus b. Remember, T inverse is a mapping from your codomain to your domain, although they're both going to be Rn in this case, but T inverse maps from this set to that set. Let's write it up here. T inverse is a mapping from your codomain to your domain, although it looks identical just like that. OK, so what is this going to be equal to?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, T inverse is a mapping from your codomain to your domain, although they're both going to be Rn in this case, but T inverse maps from this set to that set. Let's write it up here. T inverse is a mapping from your codomain to your domain, although it looks identical just like that. OK, so what is this going to be equal to? Well, we just said, by definition, of your inverse transformation, this is going to be equal to the identity transformation on your codomain. So assuming that these guys are members of your codomain, in this case, Rn, this is just going to be equal to a plus b. This thing, the composition of T with its inverse, by definition, is just the identity transformation on your codomain."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "OK, so what is this going to be equal to? Well, we just said, by definition, of your inverse transformation, this is going to be equal to the identity transformation on your codomain. So assuming that these guys are members of your codomain, in this case, Rn, this is just going to be equal to a plus b. This thing, the composition of T with its inverse, by definition, is just the identity transformation on your codomain. So this is just whatever I put in here. If I put in an x here, this would be an x. If I put in an apple here, this would be an apple."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This thing, the composition of T with its inverse, by definition, is just the identity transformation on your codomain. So this is just whatever I put in here. If I put in an x here, this would be an x. If I put in an apple here, this would be an apple. It's going to be the identity transformation. Now what is this equal to? What is this equal to?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I put in an apple here, this would be an apple. It's going to be the identity transformation. Now what is this equal to? What is this equal to? Well, I could use the same argument to say that this right here is equal to the identity transformation applied to a. And I'm not writing the identity transformation. I'm writing this, but we know that this is equivalent to the identity transformation."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is this equal to? Well, I could use the same argument to say that this right here is equal to the identity transformation applied to a. And I'm not writing the identity transformation. I'm writing this, but we know that this is equivalent to the identity transformation. So we could say that equivalent to the composition of T with the inverse applied to a. And we could say that this is equivalent to the identity transformation, which we know is the same thing as the composition of T with T inverse applied to b. So we can rewrite this thing right here as being equal to the sum of these two things."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm writing this, but we know that this is equivalent to the identity transformation. So we could say that equivalent to the composition of T with the inverse applied to a. And we could say that this is equivalent to the identity transformation, which we know is the same thing as the composition of T with T inverse applied to b. So we can rewrite this thing right here as being equal to the sum of these two things. In fact, we don't even have to rewrite it. We can just write it equal to. This transformation is equal to this."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can rewrite this thing right here as being equal to the sum of these two things. In fact, we don't even have to rewrite it. We can just write it equal to. This transformation is equal to this. And maybe an easier way for you to, I guess, process it is, we could write this as T of the T inverse of a plus b is equal to T of the T inverse of a plus T of the T inverse of b. And this should, I don't know which one your brain processes easier, but either of these, when you take the composition of T with T inverse, you're just going to be left with an a plus b. You take the composition of T with T inverse, you're left with an a."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This transformation is equal to this. And maybe an easier way for you to, I guess, process it is, we could write this as T of the T inverse of a plus b is equal to T of the T inverse of a plus T of the T inverse of b. And this should, I don't know which one your brain processes easier, but either of these, when you take the composition of T with T inverse, you're just going to be left with an a plus b. You take the composition of T with T inverse, you're left with an a. You take the composition of T with T inverse, you're just left with a b there. So in either case, you get a plus b, the vector a plus, when you evaluate either side of this expression, you'll get the vector a plus the vector b. Now what can we do?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You take the composition of T with T inverse, you're left with an a. You take the composition of T with T inverse, you're just left with a b there. So in either case, you get a plus b, the vector a plus, when you evaluate either side of this expression, you'll get the vector a plus the vector b. Now what can we do? We know that T itself is a linear transformation. And since T is a linear transformation, we know that T applied to the sum of two vectors is equal to T applied to each of those vectors and summed up. Or we could take it the other way."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what can we do? We know that T itself is a linear transformation. And since T is a linear transformation, we know that T applied to the sum of two vectors is equal to T applied to each of those vectors and summed up. Or we could take it the other way. T applied to two separate vectors, so that we could call this one vector right here and this vector right here. So in this case, I have a T applied to one vector, and I'm summing it to a T applied to another vector. So it's this right here, which we know is equal to T applied to the sum of those two vectors."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or we could take it the other way. T applied to two separate vectors, so that we could call this one vector right here and this vector right here. So in this case, I have a T applied to one vector, and I'm summing it to a T applied to another vector. So it's this right here, which we know is equal to T applied to the sum of those two vectors. So this is T applied to the vector T inverse of a. Let me write it here. So this is going to be equal to T inverse of a plus T inverse of b."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's this right here, which we know is equal to T applied to the sum of those two vectors. So this is T applied to the vector T inverse of a. Let me write it here. So this is going to be equal to T inverse of a plus T inverse of b. It might look a little convoluted, but all I'm saying is, look, this looks just like this. If you say that x is equal to T inverse of a, and if you say that y is equal to T inverse of b. So if this looks just like that, it's going to be equal to the transformation T applied to the sum of those two vectors."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to T inverse of a plus T inverse of b. It might look a little convoluted, but all I'm saying is, look, this looks just like this. If you say that x is equal to T inverse of a, and if you say that y is equal to T inverse of b. So if this looks just like that, it's going to be equal to the transformation T applied to the sum of those two vectors. So it's going to be equal to the transformation T applied to the inverse of a, T inverse of a, plus T inverse of b. I just use the fact that T is linear to get here. Now what can I do? Let me simplify everything that I've written right here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this looks just like that, it's going to be equal to the transformation T applied to the sum of those two vectors. So it's going to be equal to the transformation T applied to the inverse of a, T inverse of a, plus T inverse of b. I just use the fact that T is linear to get here. Now what can I do? Let me simplify everything that I've written right here. So I now have, let me rewrite this. This thing up here, which is the same thing as this, T, the composition of T with T inverse applied to a plus b is equal to the composition, or actually not the composition, just T applied to two vectors. T inverse of a plus T inverse of vector b."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me simplify everything that I've written right here. So I now have, let me rewrite this. This thing up here, which is the same thing as this, T, the composition of T with T inverse applied to a plus b is equal to the composition, or actually not the composition, just T applied to two vectors. T inverse of a plus T inverse of vector b. That's what we've gotten so far. Now we're very close to proving that this condition is true for T inverse. If we can just get rid of these Ts."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "T inverse of a plus T inverse of vector b. That's what we've gotten so far. Now we're very close to proving that this condition is true for T inverse. If we can just get rid of these Ts. Well the best way to get rid of those Ts is to take the composition with T inverse on both sides, or take the T inverse transformation of both sides of this equation. So let's do that. So let's take T inverse of this side, T inverse of that side, should be equal to T inverse of this side."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we can just get rid of these Ts. Well the best way to get rid of those Ts is to take the composition with T inverse on both sides, or take the T inverse transformation of both sides of this equation. So let's do that. So let's take T inverse of this side, T inverse of that side, should be equal to T inverse of this side. Because these two things are the same thing, so if you put the same thing into a function, you should get the same value on both sides. So what is this thing on the left-hand side? What is this?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's take T inverse of this side, T inverse of that side, should be equal to T inverse of this side. Because these two things are the same thing, so if you put the same thing into a function, you should get the same value on both sides. So what is this thing on the left-hand side? What is this? This is the composition, let me write it this way, this is the composition of T inverse with T, that part, applied to this thing right here. I'm just changing the associativity of this. Applied to T inverse of the vector a plus the vector b."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is this? This is the composition, let me write it this way, this is the composition of T inverse with T, that part, applied to this thing right here. I'm just changing the associativity of this. Applied to T inverse of the vector a plus the vector b. That's what this left-hand side is. This part right here, T inverse of T of this, this first two steps, I'm just writing as the composition of T inverse with T, applied to this right here. That right there is the same thing as that right there."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Applied to T inverse of the vector a plus the vector b. That's what this left-hand side is. This part right here, T inverse of T of this, this first two steps, I'm just writing as the composition of T inverse with T, applied to this right here. That right there is the same thing as that right there. So that was another way to write that. And so that is going to be equal to the composition of T inverse with T, so I'll write it in the same color. The composition of T inverse with T, that's this part right here, which is very similar to that part right there, of this stuff right here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That right there is the same thing as that right there. So that was another way to write that. And so that is going to be equal to the composition of T inverse with T, so I'll write it in the same color. The composition of T inverse with T, that's this part right here, which is very similar to that part right there, of this stuff right here. Of T inverse of a plus T inverse of the vector b. Now, by definition of what T inverse is, what is this? This is the identity transformation on our domain."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The composition of T inverse with T, that's this part right here, which is very similar to that part right there, of this stuff right here. Of T inverse of a plus T inverse of the vector b. Now, by definition of what T inverse is, what is this? This is the identity transformation on our domain. This is the identity transformation on Rn. This is also the identity transformation on Rn. So if you apply the identity transformation to anything, you're just going to get anything."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the identity transformation on our domain. This is the identity transformation on Rn. This is also the identity transformation on Rn. So if you apply the identity transformation to anything, you're just going to get anything. So this is going to be equal to, and I'll do it on both sides of the equation, this whole expression on the left-hand side is just going to simplify to the T inverse of the vectors a plus the vector b. And the right-hand side is just going to simplify to this thing, is equal to, because this is just the identity transformation, so it's just equal to this one. T inverse of the vector a plus T inverse of the vector b."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you apply the identity transformation to anything, you're just going to get anything. So this is going to be equal to, and I'll do it on both sides of the equation, this whole expression on the left-hand side is just going to simplify to the T inverse of the vectors a plus the vector b. And the right-hand side is just going to simplify to this thing, is equal to, because this is just the identity transformation, so it's just equal to this one. T inverse of the vector a plus T inverse of the vector b. And just like that, T inverse has met its first condition for being a linear transformation. It's met its first condition. Now let's see if we can do the second condition."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "T inverse of the vector a plus T inverse of the vector b. And just like that, T inverse has met its first condition for being a linear transformation. It's met its first condition. Now let's see if we can do the second condition. Let's do the same type of thing. Let's take the composition of T with T inverse of, let's take the composition of that, on some vector, let's call it CA, just like that. Well, we know what this is equal to."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's see if we can do the second condition. Let's do the same type of thing. Let's take the composition of T with T inverse of, let's take the composition of that, on some vector, let's call it CA, just like that. Well, we know what this is equal to. This is equal to the identity transformation on Rn. So this is just going to be equal to CA. Now, what is A equal to?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we know what this is equal to. This is equal to the identity transformation on Rn. So this is just going to be equal to CA. Now, what is A equal to? What is A equal to? What is this thing right there equal to? I'll write it on the side right here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is A equal to? What is A equal to? What is this thing right there equal to? I'll write it on the side right here. Let me do it in an appropriate color. We could say that A, the vector A is equal to the transformation T with the composition of T with T inverse, T inverse applied to the vector A, because this is just the identity transformation. So we can rewrite this expression here as being equal to C times the composition of T with T inverse applied to my vector A."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll write it on the side right here. Let me do it in an appropriate color. We could say that A, the vector A is equal to the transformation T with the composition of T with T inverse, T inverse applied to the vector A, because this is just the identity transformation. So we can rewrite this expression here as being equal to C times the composition of T with T inverse applied to my vector A. And maybe it might be nice to rewrite it in this form instead of this composition form. So this left expression, we can just write it as saying T of the T inverse of C times the vector A. All I did is rewrite this left-hand side this way, is equal to this green thing right here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can rewrite this expression here as being equal to C times the composition of T with T inverse applied to my vector A. And maybe it might be nice to rewrite it in this form instead of this composition form. So this left expression, we can just write it as saying T of the T inverse of C times the vector A. All I did is rewrite this left-hand side this way, is equal to this green thing right here. Well, I'll rewrite similarly. This is equal to C times T, the transformation T applied to the transformation T inverse applied to A. This is by definition what composition means."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All I did is rewrite this left-hand side this way, is equal to this green thing right here. Well, I'll rewrite similarly. This is equal to C times T, the transformation T applied to the transformation T inverse applied to A. This is by definition what composition means. Now, T is a linear transformation, which means that if you take C times T times some vector, that is equivalent to T applied to C times that vector. This is one of the conditions of a linear transformation. So this is always going to be the case with T. So this is some vector that T is applying to."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is by definition what composition means. Now, T is a linear transformation, which means that if you take C times T times some vector, that is equivalent to T applied to C times that vector. This is one of the conditions of a linear transformation. So this is always going to be the case with T. So this is some vector that T is applying to. This is some scalar. So this thing, because we know that T is a linear transformation, we can rewrite as being equal to T applied to the scalar C times T inverse applied to A. And now what can we do?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is always going to be the case with T. So this is some vector that T is applying to. This is some scalar. So this thing, because we know that T is a linear transformation, we can rewrite as being equal to T applied to the scalar C times T inverse applied to A. And now what can we do? Well, let's apply the T inverse transformation to both sides of this. Let me rewrite it. So on this side, we get T of T inverse of CA is equal to T of C times T inverse times A."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And now what can we do? Well, let's apply the T inverse transformation to both sides of this. Let me rewrite it. So on this side, we get T of T inverse of CA is equal to T of C times T inverse times A. That's what we have so far. And wouldn't it be nice if we could get rid of these outer Ts? And the best way to do that is to take the T inverse transformation of both sides."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So on this side, we get T of T inverse of CA is equal to T of C times T inverse times A. That's what we have so far. And wouldn't it be nice if we could get rid of these outer Ts? And the best way to do that is to take the T inverse transformation of both sides. So let's do that. T inverse, let's take that of both sides of this equation. T inverse of both sides."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the best way to do that is to take the T inverse transformation of both sides. So let's do that. T inverse, let's take that of both sides of this equation. T inverse of both sides. And another way that this could be written, this is equivalent to the composition of T inverse with T applied to C times our vector A. This right here, I've just decided to keep writing it in this form. And I took these two guys out and I wrote them as a composition."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "T inverse of both sides. And another way that this could be written, this is equivalent to the composition of T inverse with T applied to C times our vector A. This right here, I've just decided to keep writing it in this form. And I took these two guys out and I wrote them as a composition. And this on the right-hand side, you can do something very similar. You could say that this is equal to the composition of T inverse with T times, or not times, let me be very careful, taking this composition, this transformation, and then taking that transformation on C times the inverse transformation applied to A. Let me be very clear what I did here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I took these two guys out and I wrote them as a composition. And this on the right-hand side, you can do something very similar. You could say that this is equal to the composition of T inverse with T times, or not times, let me be very careful, taking this composition, this transformation, and then taking that transformation on C times the inverse transformation applied to A. Let me be very clear what I did here. This thing right here is this thing right here. This thing right here is this thing right here. And I just rewrote this composition this way."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me be very clear what I did here. This thing right here is this thing right here. This thing right here is this thing right here. And I just rewrote this composition this way. I rewrote this composition this way. And the reason why I did this is because we know that this is just the identity transformation on Rn. And this is just the identity transformation on Rn."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I just rewrote this composition this way. I rewrote this composition this way. And the reason why I did this is because we know that this is just the identity transformation on Rn. And this is just the identity transformation on Rn. So the identity transformation applied to anything is just that anything. So this equation simplifies to the T inverse applied to C times some vector A is equal to this thing. C times T inverse times some vector A."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is just the identity transformation on Rn. So the identity transformation applied to anything is just that anything. So this equation simplifies to the T inverse applied to C times some vector A is equal to this thing. C times T inverse times some vector A. And just like that, we've met our second condition for being a linear transformation. We've met our second condition. The first condition was met up here."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "C times T inverse times some vector A. And just like that, we've met our second condition for being a linear transformation. We've met our second condition. The first condition was met up here. So now we know. And in both cases, we used the fact that T was a linear transformation to get to the result for T inverse. So now we know that if T is a linear transformation and T is invertible, then T inverse is also a linear transformation."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The first condition was met up here. So now we know. And in both cases, we used the fact that T was a linear transformation to get to the result for T inverse. So now we know that if T is a linear transformation and T is invertible, then T inverse is also a linear transformation. Which might seem like a little nice thing to know, but that's actually a big thing to know. Because now we know that T inverse can be represented as a matrix vector product. So that means that T inverse applied to some vector x could be represented as the product of some matrix times x."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So now we know that if T is a linear transformation and T is invertible, then T inverse is also a linear transformation. Which might seem like a little nice thing to know, but that's actually a big thing to know. Because now we know that T inverse can be represented as a matrix vector product. So that means that T inverse applied to some vector x could be represented as the product of some matrix times x. And what we're going to do is we're going to call that matrix. We're going to call that matrix the matrix A inverse. And I haven't defined this well, how do you construct this A inverse matrix."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that means that T inverse applied to some vector x could be represented as the product of some matrix times x. And what we're going to do is we're going to call that matrix. We're going to call that matrix the matrix A inverse. And I haven't defined this well, how do you construct this A inverse matrix. But we know that it exists. We know this exists now because T is a linear transformation. And we could take it even a step further."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I haven't defined this well, how do you construct this A inverse matrix. But we know that it exists. We know this exists now because T is a linear transformation. And we could take it even a step further. We know by the definition of invertibility that the composition of T inverse with T is equal to the identity transformation on Rn. Well, what is the composition? Let me put it this way."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we could take it even a step further. We know by the definition of invertibility that the composition of T inverse with T is equal to the identity transformation on Rn. Well, what is the composition? Let me put it this way. We know that T of x is equal to Ax. So if we write T inverse, the composition of T inverse with T applied to some vector x is going to be equal to first Ax, A being applied to x, is going to be equal to Ax, this A right here, Ax. And then you're going to apply A inverse x."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me put it this way. We know that T of x is equal to Ax. So if we write T inverse, the composition of T inverse with T applied to some vector x is going to be equal to first Ax, A being applied to x, is going to be equal to Ax, this A right here, Ax. And then you're going to apply A inverse x. You're going to apply this right here. And we got this, that this is equivalent to, when you take the composition, it's equivalent to the resulting transformation matrix of two composition transformations is equal to this matrix-matrix product. We got that a long time ago."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you're going to apply A inverse x. You're going to apply this right here. And we got this, that this is equivalent to, when you take the composition, it's equivalent to the resulting transformation matrix of two composition transformations is equal to this matrix-matrix product. We got that a long time ago. In fact, that was the motivation for how a matrix- matrix product was defined. But what's interesting here is this composition is equal to that, but it's also going to be equal to the identity transformation on Rn applied to that vector x, which is equal to the identity matrix. The identity matrix applied to x, right?"}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We got that a long time ago. In fact, that was the motivation for how a matrix- matrix product was defined. But what's interesting here is this composition is equal to that, but it's also going to be equal to the identity transformation on Rn applied to that vector x, which is equal to the identity matrix. The identity matrix applied to x, right? That is the n by n matrix. So when you multiply it by anything, you get that anything again. So we get a very interesting result."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The identity matrix applied to x, right? That is the n by n matrix. So when you multiply it by anything, you get that anything again. So we get a very interesting result. A inverse times A has to be equal to the identity matrix. A inverse, or the matrix transformation for T inverse, when you multiply that with the matrix transformation for T, you are going to get the identity matrix. And the argument actually holds both ways."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we get a very interesting result. A inverse times A has to be equal to the identity matrix. A inverse, or the matrix transformation for T inverse, when you multiply that with the matrix transformation for T, you are going to get the identity matrix. And the argument actually holds both ways. So we know this is true, but the other definition of an inverse or invertibility told us that the composition of T with T inverse is equal to the identity transformation in our codomain, which is also Rn, I Rn. So by the exact same argument, we know that when you go the other way, if you apply T inverse first, and then you apply T, so that's equivalent of saying you apply T inverse first, and then you apply T to some x vector, that's equivalent to multiplying that x vector by the identity matrix, the n by n identity matrix. Or you could say, you could switch the order."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the argument actually holds both ways. So we know this is true, but the other definition of an inverse or invertibility told us that the composition of T with T inverse is equal to the identity transformation in our codomain, which is also Rn, I Rn. So by the exact same argument, we know that when you go the other way, if you apply T inverse first, and then you apply T, so that's equivalent of saying you apply T inverse first, and then you apply T to some x vector, that's equivalent to multiplying that x vector by the identity matrix, the n by n identity matrix. Or you could say, you could switch the order. A times A inverse is also equal to the identity matrix. Which is neat, because we learned that matrix-matrix products, when you switch the order, they don't normally always equal each other. But in the case of an invertible matrix and its inverse, order doesn't matter."}, {"video_title": "Showing that inverses are linear Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or you could say, you could switch the order. A times A inverse is also equal to the identity matrix. Which is neat, because we learned that matrix-matrix products, when you switch the order, they don't normally always equal each other. But in the case of an invertible matrix and its inverse, order doesn't matter. You can take A inverse times A and get the identity matrix, or you could take A times A inverse and get the identity matrix. Now, we've gotten this far. The next step is to actually figure out how do you construct, we know that this thing exists, we know that the inverse is a linear transformation, that this matrix exists, we see this nice property that when you multiply it times the transformation matrix, you get the identity matrix."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "And this is, in general, terminology that you'll probably see in your mathematical careers. So let's say I have a function f. And it is a mapping from the set x to the set y. And we've drawn this diagram many times, but it never hurts to draw it again. So that is my set x or my domain. And then this is the set y over here. Or the codomain. Remember, the codomain is the set that you're mapping to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So that is my set x or my domain. And then this is the set y over here. Or the codomain. Remember, the codomain is the set that you're mapping to. You don't necessarily have to map to every element of the set or none of the elements of the set. This is just all of the elements, the set that you might map elements in your codomain to. So let's see."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Remember, the codomain is the set that you're mapping to. You don't necessarily have to map to every element of the set or none of the elements of the set. This is just all of the elements, the set that you might map elements in your codomain to. So let's see. If I have some element there, f will map it to some element in y, in my codomain. So the first idea or term I want to introduce you to is the idea of a function being surjective. And sometimes this is called onto."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So let's see. If I have some element there, f will map it to some element in y, in my codomain. So the first idea or term I want to introduce you to is the idea of a function being surjective. And sometimes this is called onto. And a function is surjective or onto if for every element in your codomain. So let me write it this way. If for every, let's say, y that is a member of my codomain, there exists, that's the little shorthand notation for exists, there exists at least one x."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "And sometimes this is called onto. And a function is surjective or onto if for every element in your codomain. So let me write it this way. If for every, let's say, y that is a member of my codomain, there exists, that's the little shorthand notation for exists, there exists at least one x. That's a member of x such that, and I can write such that like that. Actually, let me just write the word out. Such that f of x is equal to y."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "If for every, let's say, y that is a member of my codomain, there exists, that's the little shorthand notation for exists, there exists at least one x. That's a member of x such that, and I can write such that like that. Actually, let me just write the word out. Such that f of x is equal to y. So it's essentially saying, look, you can pick any y here and every y here is being mapped to by at least one of the x's over here. So for example, actually let me draw a simpler example. Instead of drawing these blurbs, let's say that I have a set y that literally looks like this."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Such that f of x is equal to y. So it's essentially saying, look, you can pick any y here and every y here is being mapped to by at least one of the x's over here. So for example, actually let me draw a simpler example. Instead of drawing these blurbs, let's say that I have a set y that literally looks like this. Let's say that a set y, I'll draw it very, and let's say it has four elements. It has the elements a, b, c, and d. This is my set y right there. And let's say my set x looks like that."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Instead of drawing these blurbs, let's say that I have a set y that literally looks like this. Let's say that a set y, I'll draw it very, and let's say it has four elements. It has the elements a, b, c, and d. This is my set y right there. And let's say my set x looks like that. And let's say it has the elements, I don't know, 1, 2, 3, and 4. Now, in order for my function f to be surjective or onto it means that every one of these guys have to be able to be mapped to. So what does that mean?"}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "And let's say my set x looks like that. And let's say it has the elements, I don't know, 1, 2, 3, and 4. Now, in order for my function f to be surjective or onto it means that every one of these guys have to be able to be mapped to. So what does that mean? So if every one of these guys, let me just draw some examples. Let's say that this guy maps to that. Let's say that this guy maps to that."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So what does that mean? So if every one of these guys, let me just draw some examples. Let's say that this guy maps to that. Let's say that this guy maps to that. Let's say that this guy maps to that. And let's say, let me draw a fifth one right here. Let's say that both of these guys right here map to d. So f of 4 is d and f of 5 is d. This is an example of a surjective function."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that this guy maps to that. Let's say that this guy maps to that. And let's say, let me draw a fifth one right here. Let's say that both of these guys right here map to d. So f of 4 is d and f of 5 is d. This is an example of a surjective function. So these are the mappings of f right here. This function right here is onto or surjective. Why is that?"}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that both of these guys right here map to d. So f of 4 is d and f of 5 is d. This is an example of a surjective function. So these are the mappings of f right here. This function right here is onto or surjective. Why is that? Because every element here is being mapped to. Now let me give you an example of a function that is not surjective. Let me add some more elements to y."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Why is that? Because every element here is being mapped to. Now let me give you an example of a function that is not surjective. Let me add some more elements to y. Let's say y has another element here called e. Now all of a sudden, this is not surjective. And why is that? Because there's some element in y that is not being mapped to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Let me add some more elements to y. Let's say y has another element here called e. Now all of a sudden, this is not surjective. And why is that? Because there's some element in y that is not being mapped to. So if I tell you that f is a surjective function, it means that if you map all of these values, everything here is being mapped to by at least one element here. So everything could be kind of a one-to-one mapping, and I'll define that a little bit better in the future. So it could just be like that and like that."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Because there's some element in y that is not being mapped to. So if I tell you that f is a surjective function, it means that if you map all of these values, everything here is being mapped to by at least one element here. So everything could be kind of a one-to-one mapping, and I'll define that a little bit better in the future. So it could just be like that and like that. And you could even have at least one. So you could even have two things in here mapping to one thing in here. But the main requirement is that everything here does get mapped to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So it could just be like that and like that. And you could even have at least one. So you could even have two things in here mapping to one thing in here. But the main requirement is that everything here does get mapped to. Another way to think about it is that if you take the image, so a surjective function, let me write this here. So a surjective function, or let me write it this way. So if I say that f is surjective or onto, these are equivalent terms, that means that the image of f, remember, the image was all of the values that f actually maps to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "But the main requirement is that everything here does get mapped to. Another way to think about it is that if you take the image, so a surjective function, let me write this here. So a surjective function, or let me write it this way. So if I say that f is surjective or onto, these are equivalent terms, that means that the image of f, remember, the image was all of the values that f actually maps to. So that means that the image of f is equal to y. Now we learned before that your image doesn't have to equal your codomain, but if you have a surjective or an onto function, your image is going to equal your codomain. Everything in your codomain gets mapped to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So if I say that f is surjective or onto, these are equivalent terms, that means that the image of f, remember, the image was all of the values that f actually maps to. So that means that the image of f is equal to y. Now we learned before that your image doesn't have to equal your codomain, but if you have a surjective or an onto function, your image is going to equal your codomain. Everything in your codomain gets mapped to. Actually, another word for image is range. You could also say that your range of f is equal to y. Remember, the difference, and I drew this distinction when we first talked about functions, the distinction between a codomain and a range, a codomain is a set that you can map to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Everything in your codomain gets mapped to. Actually, another word for image is range. You could also say that your range of f is equal to y. Remember, the difference, and I drew this distinction when we first talked about functions, the distinction between a codomain and a range, a codomain is a set that you can map to. You don't have to map to everything. The range is a subset of your codomain that you actually do map to. If you were to evaluate the function at all of these points, the point that you actually map to is your range."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Remember, the difference, and I drew this distinction when we first talked about functions, the distinction between a codomain and a range, a codomain is a set that you can map to. You don't have to map to everything. The range is a subset of your codomain that you actually do map to. If you were to evaluate the function at all of these points, the point that you actually map to is your range. And that's also called your image, and we use the word image is used more in a linear algebra context. But if your image or your range is equal to your codomain, if everything in your codomain does get mapped to, then you're dealing with a surjective function or an onto function. Now the next term I want to introduce you to is the idea of an injective function."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "If you were to evaluate the function at all of these points, the point that you actually map to is your range. And that's also called your image, and we use the word image is used more in a linear algebra context. But if your image or your range is equal to your codomain, if everything in your codomain does get mapped to, then you're dealing with a surjective function or an onto function. Now the next term I want to introduce you to is the idea of an injective function. Injective function. And this is sometimes called a one-to-one function. So let me draw my domain and codomain again."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Now the next term I want to introduce you to is the idea of an injective function. Injective function. And this is sometimes called a one-to-one function. So let me draw my domain and codomain again. So let's say that that is my domain, and this is my codomain. So this is x and this is y. If I say that f is injective, or one-to-one, that implies that for every value that is mapped to, so let me write it this way, for every value that is mapped to, so let's say, I'll say it a couple of different ways, there is at most one x that maps to it."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw my domain and codomain again. So let's say that that is my domain, and this is my codomain. So this is x and this is y. If I say that f is injective, or one-to-one, that implies that for every value that is mapped to, so let me write it this way, for every value that is mapped to, so let's say, I'll say it a couple of different ways, there is at most one x that maps to it. There is at most one x that maps to it. Or another way to say it is that for any y that's a member of y, let me write it this way, for any y that is a member of y, there is at most one, at most, let me write most in capital, at most one x such that f of x is equal to y. There might be no x's that map to it."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "If I say that f is injective, or one-to-one, that implies that for every value that is mapped to, so let me write it this way, for every value that is mapped to, so let's say, I'll say it a couple of different ways, there is at most one x that maps to it. There is at most one x that maps to it. Or another way to say it is that for any y that's a member of y, let me write it this way, for any y that is a member of y, there is at most one, at most, let me write most in capital, at most one x such that f of x is equal to y. There might be no x's that map to it. So for example, you could have a little member of y right here that just never gets mapped to. Everyone else in y gets mapped to, but that guy never gets mapped to. So this would be a case where we don't have a surjective function."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "There might be no x's that map to it. So for example, you could have a little member of y right here that just never gets mapped to. Everyone else in y gets mapped to, but that guy never gets mapped to. So this would be a case where we don't have a surjective function. This is not on to, because this guy is a member of the codomain, but he's not a member of the image or the range, he doesn't get mapped to. But this would still be an injective function as long as every x gets mapped to a unique y. Now how can a function not be injective or one to one?"}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So this would be a case where we don't have a surjective function. This is not on to, because this guy is a member of the codomain, but he's not a member of the image or the range, he doesn't get mapped to. But this would still be an injective function as long as every x gets mapped to a unique y. Now how can a function not be injective or one to one? And I think you get the idea when someone says one to one. Well if two x's here get mapped to the same y, or three get mapped to the same y, this would mean that we're not dealing with an injective or a one to one function. So that's all it means."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "Now how can a function not be injective or one to one? And I think you get the idea when someone says one to one. Well if two x's here get mapped to the same y, or three get mapped to the same y, this would mean that we're not dealing with an injective or a one to one function. So that's all it means. Let me draw another example here. So if I take, let's actually go back to this example right here. When I added this e here, we said this is not surjective anymore, because every one of these guys is not being mapped to."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "So that's all it means. Let me draw another example here. So if I take, let's actually go back to this example right here. When I added this e here, we said this is not surjective anymore, because every one of these guys is not being mapped to. Is this an injective function? Well no, because I have f of 5 and f of 4 both map to d. So this is what breaks its one to oneness, or its injectiveness. This is what breaks its surjectiveness."}, {"video_title": "Surjective (onto) and injective (one-to-one) functions Linear Algebra Khan Academy.mp3", "Sentence": "When I added this e here, we said this is not surjective anymore, because every one of these guys is not being mapped to. Is this an injective function? Well no, because I have f of 5 and f of 4 both map to d. So this is what breaks its one to oneness, or its injectiveness. This is what breaks its surjectiveness. If I wanted to make this a surjective and an injective function, I would delete that mapping and I would change f of 5 to be e. Now everything is one to one. I don't have the mapping from two elements of x going to the same element of y anymore. And everything in y now gets mapped too."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In this video, I want to kind of go back to basics and just give you a lot of examples and give you a more tangible sense for what vectors are and how we operate with them. So let me define a couple of vectors here. And most of my vectors I'm going to do in this video are going to be an r2. And that's because they're easy to draw. r2, remember, r2 was a set of all two tuples, ordered two tuples, where each of the numbers, so you could have x1, and let me see, my 1 looks like a comma, x1 and x2, where each of these are real numbers. So each of them, x1 is a member of the reals, x1, and x2 is a member of the reals. And just to give you a sense of what that means, if this right here is my coordinate axes and I wanted to plot all my x1's, x2's, you could view this as the first coordinate."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's because they're easy to draw. r2, remember, r2 was a set of all two tuples, ordered two tuples, where each of the numbers, so you could have x1, and let me see, my 1 looks like a comma, x1 and x2, where each of these are real numbers. So each of them, x1 is a member of the reals, x1, and x2 is a member of the reals. And just to give you a sense of what that means, if this right here is my coordinate axes and I wanted to plot all my x1's, x2's, you could view this as the first coordinate. We always imagine that as our x-axis. And then our second coordinate, we plot it on the vertical axis that traditionally is our y-axis, but we'll just call that the second number axis, whatever. You could visually represent all of r2 by literally every single point on this plane if we were to continue off in infinity in every direction."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And just to give you a sense of what that means, if this right here is my coordinate axes and I wanted to plot all my x1's, x2's, you could view this as the first coordinate. We always imagine that as our x-axis. And then our second coordinate, we plot it on the vertical axis that traditionally is our y-axis, but we'll just call that the second number axis, whatever. You could visually represent all of r2 by literally every single point on this plane if we were to continue off in infinity in every direction. That's what r2 is. r1 would just be points just along one of these number lines, that would be r1. So you could immediately see that r2 is kind of a bigger space."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could visually represent all of r2 by literally every single point on this plane if we were to continue off in infinity in every direction. That's what r2 is. r1 would just be points just along one of these number lines, that would be r1. So you could immediately see that r2 is kind of a bigger space. But anyway, I said that I wouldn't be too abstract, I would show you examples. So let's get some vectors going in r2. So let me define my vector a. I'll make it nice and bold."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you could immediately see that r2 is kind of a bigger space. But anyway, I said that I wouldn't be too abstract, I would show you examples. So let's get some vectors going in r2. So let me define my vector a. I'll make it nice and bold. My vector a is equal to, I'll make some numbers up, negative 1, 2, and my vector b, make it nice and bold, let me make that, I don't know, 3, 3, 1. Those are my two vectors. Now let's just add them up and see what we get."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me define my vector a. I'll make it nice and bold. My vector a is equal to, I'll make some numbers up, negative 1, 2, and my vector b, make it nice and bold, let me make that, I don't know, 3, 3, 1. Those are my two vectors. Now let's just add them up and see what we get. Just based on my definition of vector addition, and I'll just stay in one color for now, just so I don't have to keep switching back and forth. So a, nice deep a, plus bolded b is equal to, I just add up each of those terms, negative 1 plus 3, and then 2 plus 1. That was my definition of vector addition."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's just add them up and see what we get. Just based on my definition of vector addition, and I'll just stay in one color for now, just so I don't have to keep switching back and forth. So a, nice deep a, plus bolded b is equal to, I just add up each of those terms, negative 1 plus 3, and then 2 plus 1. That was my definition of vector addition. So that is going to be equal to 2 and 3. Fair enough. This just came out of my definition of vector addition."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That was my definition of vector addition. So that is going to be equal to 2 and 3. Fair enough. This just came out of my definition of vector addition. But how can we represent this vector? So we already know that if we have coordinates, if I have the coordinate, and this is just a convention, it's just the way that we do it, the way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes, the first point I go along the horizontal, what we traditionally call our x-axis, and I go 1 in that direction, and then the convention is the second point, I go 1 in the vertical direction."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This just came out of my definition of vector addition. But how can we represent this vector? So we already know that if we have coordinates, if I have the coordinate, and this is just a convention, it's just the way that we do it, the way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes, the first point I go along the horizontal, what we traditionally call our x-axis, and I go 1 in that direction, and then the convention is the second point, I go 1 in the vertical direction. So the point 1, 1, I would, oh sorry, let me be very clear, this is 2 and 2, so 1 is right here and 1 is right there. So the point 1, 1 would be right there. That's just the standard convention."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I wanted to plot the point 1, 1, I go to my coordinate axes, the first point I go along the horizontal, what we traditionally call our x-axis, and I go 1 in that direction, and then the convention is the second point, I go 1 in the vertical direction. So the point 1, 1, I would, oh sorry, let me be very clear, this is 2 and 2, so 1 is right here and 1 is right there. So the point 1, 1 would be right there. That's just the standard convention. Now, our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that, and I'll show you in a second. But the convention for vectors is that you can start at any point."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's just the standard convention. Now, our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that, and I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two-dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1 and x2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But the convention for vectors is that you can start at any point. Let's say we're dealing with two-dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1 and x2. This could be any point in R2. To represent this vector, what we do is we draw a line from that point to the point x1, and let me call this, let's say that we wanted to draw a. So x1 minus 1, so I'm representing a."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that you're starting at the point x1 and x2. This could be any point in R2. To represent this vector, what we do is we draw a line from that point to the point x1, and let me call this, let's say that we wanted to draw a. So x1 minus 1, so I'm representing a. So I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now, if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So x1 minus 1, so I'm representing a. So I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now, if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point, that one right there. That's my starting point. So minus 4, 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point, that one right there. That's my starting point. So minus 4, 4. That's minus 4, 4. Now, if I want to represent my vector a, what I just said is that I add the first term in vector a to my first coordinate, so x1 plus minus 1, or x1 minus 1. So my new one is going to be, so this is my x1 minus 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So minus 4, 4. That's minus 4, 4. Now, if I want to represent my vector a, what I just said is that I add the first term in vector a to my first coordinate, so x1 plus minus 1, or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, so let's see, I'm starting at the point minus 4, 4. If I want to represent a, what I do is I draw an arrow to minus 4 plus this first term, minus 1, and then 4 plus the second term, 4 plus 2. And so this is what?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So my new one is going to be, so this is my x1 minus 4. So now it's going to be, so let's see, I'm starting at the point minus 4, 4. If I want to represent a, what I do is I draw an arrow to minus 4 plus this first term, minus 1, and then 4 plus the second term, 4 plus 2. And so this is what? This is minus 5, 6. So I go to minus 5, 6. So I go to that point right there, and I just draw a line."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so this is what? This is minus 5, 6. So I go to minus 5, 6. So I go to that point right there, and I just draw a line. So my vector will look like this. I draw a line from there to there, and I draw an arrow at the end point. So that's one representation of the vector minus 1, 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I go to that point right there, and I just draw a line. So my vector will look like this. I draw a line from there to there, and I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually, let me do it a little bit better. Because minus 5 is actually more, it's a little closer to right here. Minus 5, 6 is right there."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's one representation of the vector minus 1, 2. Actually, let me do it a little bit better. Because minus 5 is actually more, it's a little closer to right here. Minus 5, 6 is right there. So I draw my vector like that. But remember, this point minus 4, 4 was an arbitrary place to draw my vector. I could have started at this point here."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 5, 6 is right there. So I draw my vector like that. But remember, this point minus 4, 4 was an arbitrary place to draw my vector. I could have started at this point here. I could have started at the point 4, 6 and done the same thing. I could have gone minus 1 in the horizontal direction. That's my movement in the horizontal direction."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could have started at this point here. I could have started at the point 4, 6 and done the same thing. I could have gone minus 1 in the horizontal direction. That's my movement in the horizontal direction. And then plus 2 in the vertical direction. So I could have drawn, so minus 1 in the horizontal, and then plus 2 in the vertical, gets me right there. So I could have just as easily drawn my vector like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's my movement in the horizontal direction. And then plus 2 in the vertical direction. So I could have drawn, so minus 1 in the horizontal, and then plus 2 in the vertical, gets me right there. So I could have just as easily drawn my vector like that. These are both interpretations of the same vector a. And I should draw them in the color of vector a. So vector a was this light blue color right there."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could have just as easily drawn my vector like that. These are both interpretations of the same vector a. And I should draw them in the color of vector a. So vector a was this light blue color right there. So this is vector a. Sometimes there'll be a little arrow notation over the vector, but either of those vectors. I could draw an infinite number of vector a's."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector a was this light blue color right there. So this is vector a. Sometimes there'll be a little arrow notation over the vector, but either of those vectors. I could draw an infinite number of vector a's. I could draw vector a here. I could draw it like that. And vector a, it goes back 1 and up 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could draw an infinite number of vector a's. I could draw vector a here. I could draw it like that. And vector a, it goes back 1 and up 2. So vector a could be right there. Similarly, vector b. What does vector b do?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And vector a, it goes back 1 and up 2. So vector a could be right there. Similarly, vector b. What does vector b do? I could pick some arbitrary point for vector b. It goes to the right 3. So it goes to the right 1, 2, 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What does vector b do? I could pick some arbitrary point for vector b. It goes to the right 3. So it goes to the right 1, 2, 3. And then it goes up 1. So vector b, one representation of vector b looks like this. Another representation, I could start."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it goes to the right 1, 2, 3. And then it goes up 1. So vector b, one representation of vector b looks like this. Another representation, I could start. I could do it right here. I could start it right here. I could go to the right 3, 1, 2, 3, and then up 1."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Another representation, I could start. I could do it right here. I could start it right here. I could go to the right 3, 1, 2, 3, and then up 1. This would be another representation of my vector b. There's an infinite number of representations of them, but the convention is to often put them in what's called the standard position. And that's to start them off at 0, 0."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could go to the right 3, 1, 2, 3, and then up 1. This would be another representation of my vector b. There's an infinite number of representations of them, but the convention is to often put them in what's called the standard position. And that's to start them off at 0, 0. So your initial point, so let me write this down. Standard position is just to start the vectors at 0, 0 and then draw them. So vector a in standard position, I would start at 0, 0 like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's to start them off at 0, 0. So your initial point, so let me write this down. Standard position is just to start the vectors at 0, 0 and then draw them. So vector a in standard position, I would start at 0, 0 like that. And I would go back 1 and then up 2. So this is vector a in standard position right there. And then vector b in standard position."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector a in standard position, I would start at 0, 0 like that. And I would go back 1 and then up 2. So this is vector a in standard position right there. And then vector b in standard position. Let me write that. That's a. And then vector b in standard position is 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then vector b in standard position. Let me write that. That's a. And then vector b in standard position is 3. Go to 3 right and then up 1. But these are the vectors in standard position, but any of these other things we drew are just as valid. Now let's see if we can get an interpretation of what happened when we added a plus b."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then vector b in standard position is 3. Go to 3 right and then up 1. But these are the vectors in standard position, but any of these other things we drew are just as valid. Now let's see if we can get an interpretation of what happened when we added a plus b. Well, if I draw that vector in standard position, I just calculated, and I go 2, 3. So I go to the right 2 and I go up 3. So if I just draw it in standard position, it looks like this."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's see if we can get an interpretation of what happened when we added a plus b. Well, if I draw that vector in standard position, I just calculated, and I go 2, 3. So I go to the right 2 and I go up 3. So if I just draw it in standard position, it looks like this. That vector right there. And at first when you look at it, so this vector right here is the vector a plus b in standard position. When you draw it like that, it's not clear what the relationship is when we added a and b."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I just draw it in standard position, it looks like this. That vector right there. And at first when you look at it, so this vector right here is the vector a plus b in standard position. When you draw it like that, it's not clear what the relationship is when we added a and b. But to see the relationship, what you do is you put a and b head to tail. So what does that mean? You put the tail end of b to the front end of a."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "When you draw it like that, it's not clear what the relationship is when we added a and b. But to see the relationship, what you do is you put a and b head to tail. So what does that mean? You put the tail end of b to the front end of a. So you can remember, all of these are valid representations of b. All of the representations of the vector b, they're all parallel to each other, but they can start from anywhere. So another equally valid representation of vector b is to start at this point right here, kind of the end point of vector a in standard position, and then draw vector b starting from there."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You put the tail end of b to the front end of a. So you can remember, all of these are valid representations of b. All of the representations of the vector b, they're all parallel to each other, but they can start from anywhere. So another equally valid representation of vector b is to start at this point right here, kind of the end point of vector a in standard position, and then draw vector b starting from there. So you go 3 to the right. So you go 1, 2, 3, and then you go up 1. So vector b could also be drawn just like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So another equally valid representation of vector b is to start at this point right here, kind of the end point of vector a in standard position, and then draw vector b starting from there. So you go 3 to the right. So you go 1, 2, 3, and then you go up 1. So vector b could also be drawn just like that. And then you should see something interesting had happened. And remember, this vector b representation is not in standard position, but it's just an equally valid way to represent my vector. Now what do you see?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector b could also be drawn just like that. And then you should see something interesting had happened. And remember, this vector b representation is not in standard position, but it's just an equally valid way to represent my vector. Now what do you see? When I add a, which is right here, to b, what do I get if I connect the starting point of a with the end point of b? I get the addition. I have added the two vectors."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what do you see? When I add a, which is right here, to b, what do I get if I connect the starting point of a with the end point of b? I get the addition. I have added the two vectors. And I could have done that anywhere. I could have started with a here, and then I could have done the end point of, I could have started b here and gone 3 to the right, 1, 2, 3, and then up 1. And I could have drawn b right there like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have added the two vectors. And I could have done that anywhere. I could have started with a here, and then I could have done the end point of, I could have started b here and gone 3 to the right, 1, 2, 3, and then up 1. And I could have drawn b right there like that. And then if I were to add a plus b, I go to the starting point of a and then the end point of b. And that should also be the visual representation of a plus b. And just to make sure it confirms with this number, what I did here is I went 2 to the right, 1, 2, and then I went 3 up, 1, 2, 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I could have drawn b right there like that. And then if I were to add a plus b, I go to the starting point of a and then the end point of b. And that should also be the visual representation of a plus b. And just to make sure it confirms with this number, what I did here is I went 2 to the right, 1, 2, and then I went 3 up, 1, 2, 3. And I got a plus b. Now let's think about what happens when we scale our vectors, when we multiply it by times some scalar factor. So let me pick new vectors."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And just to make sure it confirms with this number, what I did here is I went 2 to the right, 1, 2, and then I went 3 up, 1, 2, 3. And I got a plus b. Now let's think about what happens when we scale our vectors, when we multiply it by times some scalar factor. So let me pick new vectors. Those have gotten monotonous. Let me define vector v. v for vector. Let's say that it is equal to 1, 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me pick new vectors. Those have gotten monotonous. Let me define vector v. v for vector. Let's say that it is equal to 1, 2. So if I just wanted to draw vector v in standard position, I would just go 1 to the horizontal and then 2 to the vertical. That's the vector in standard position. If I wanted to do it in a non-standard position, I could do it right here, 1 to the right, up 2, just like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that it is equal to 1, 2. So if I just wanted to draw vector v in standard position, I would just go 1 to the horizontal and then 2 to the vertical. That's the vector in standard position. If I wanted to do it in a non-standard position, I could do it right here, 1 to the right, up 2, just like that. Equally valid way of drawing vector v. Equally valid way of doing it. Now what happens if I multiply vector v? What if I have 2 times v?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I wanted to do it in a non-standard position, I could do it right here, 1 to the right, up 2, just like that. Equally valid way of drawing vector v. Equally valid way of doing it. Now what happens if I multiply vector v? What if I have 2 times v? 2 times my vector v is now going to be equal to 2 times each of these terms. So it's going to be 2 times 1, which is 2, and then 2 times 2, which is 4. Now what does 2 times vector v look like?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What if I have 2 times v? 2 times my vector v is now going to be equal to 2 times each of these terms. So it's going to be 2 times 1, which is 2, and then 2 times 2, which is 4. Now what does 2 times vector v look like? Well, let me just start from an arbitrary position. Let me just start right over here. So I'm going to go 2 to the right, 1, 2, and I go up 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what does 2 times vector v look like? Well, let me just start from an arbitrary position. Let me just start right over here. So I'm going to go 2 to the right, 1, 2, and I go up 4. 1, 2, 3, 4. So this is what 2 times vector v looks like. This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction, but now it's twice as long."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to go 2 to the right, 1, 2, and I go up 4. 1, 2, 3, 4. So this is what 2 times vector v looks like. This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction, but now it's twice as long. And that makes sense, because we scaled it by a factor of 2. When you multiply it by a scalar, you're not changing its direction. Its direction is the exact same thing as it was before."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction, but now it's twice as long. And that makes sense, because we scaled it by a factor of 2. When you multiply it by a scalar, you're not changing its direction. Its direction is the exact same thing as it was before. You're just scaling it by that amount. And I could draw this anywhere. I could have drawn it right here."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Its direction is the exact same thing as it was before. You're just scaling it by that amount. And I could draw this anywhere. I could have drawn it right here. I could have drawn 2v right on top of v, and then you would have seen it, and I don't want to cover it, you would have seen that it's exactly, in this case, when I drew it in standard position, it's collinear. It's along the same line. It's just twice as far."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could have drawn it right here. I could have drawn 2v right on top of v, and then you would have seen it, and I don't want to cover it, you would have seen that it's exactly, in this case, when I drew it in standard position, it's collinear. It's along the same line. It's just twice as far. It's just twice as long, but they have the exact same direction. Now what happens if I were to multiply minus 4 times our vector v? Well, then that will be equal to minus 4 times 1, which is minus 4, and then minus 4 times 2, which is minus 8."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just twice as far. It's just twice as long, but they have the exact same direction. Now what happens if I were to multiply minus 4 times our vector v? Well, then that will be equal to minus 4 times 1, which is minus 4, and then minus 4 times 2, which is minus 8. So this is my new vector. Minus 4 minus 8. This is minus 4 times our vector v. So let's just start at some arbitrary point."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, then that will be equal to minus 4 times 1, which is minus 4, and then minus 4 times 2, which is minus 8. So this is my new vector. Minus 4 minus 8. This is minus 4 times our vector v. So let's just start at some arbitrary point. Let's just do it in standard position. So you go to the left 4, and then down 8. It'll look like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is minus 4 times our vector v. So let's just start at some arbitrary point. Let's just do it in standard position. So you go to the left 4, and then down 8. It'll look like that. So this new vector is going to go, and this is going to look like this. I want to make sure I can draw a relatively straight line. There you go."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It'll look like that. So this new vector is going to go, and this is going to look like this. I want to make sure I can draw a relatively straight line. There you go. So this is minus 4 times our vector v. I'll do an arrow on it to make sure you know it's a vector. Now what happened? Well, we're kind of in the same direction."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There you go. So this is minus 4 times our vector v. I'll do an arrow on it to make sure you know it's a vector. Now what happened? Well, we're kind of in the same direction. Actually, we're in the exact opposite direction, but we're still along the same line. We're just in the exact opposite direction. And it's this negative right there that flipped us around."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we're kind of in the same direction. Actually, we're in the exact opposite direction, but we're still along the same line. We're just in the exact opposite direction. And it's this negative right there that flipped us around. If we just multiplied negative 1 times this, we would have just flipped around to right there. But we multiplied it by negative 4. So we scale it by 4, so you make it 4 times as long, and then it's negative."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And it's this negative right there that flipped us around. If we just multiplied negative 1 times this, we would have just flipped around to right there. But we multiplied it by negative 4. So we scale it by 4, so you make it 4 times as long, and then it's negative. So then it flips around. It flips backwards. So now that we have that notion, we can start understanding the idea of subtracting vectors."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we scale it by 4, so you make it 4 times as long, and then it's negative. So then it flips around. It flips backwards. So now that we have that notion, we can start understanding the idea of subtracting vectors. Let me make up two new vectors right now. Let's say my vector x is equal to 2, 4. And let's say I have a vector y."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now that we have that notion, we can start understanding the idea of subtracting vectors. Let me make up two new vectors right now. Let's say my vector x is equal to 2, 4. And let's say I have a vector y. Make it nice and bold. And then that is equal to negative 1 minus 2. And I want to think about the notion of what x minus y is equal to."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And let's say I have a vector y. Make it nice and bold. And then that is equal to negative 1 minus 2. And I want to think about the notion of what x minus y is equal to. Well, we can say that this is the same thing as x plus minus 1 times our vector y. So x plus minus 1 times our vector y. Now we can use our definitions."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want to think about the notion of what x minus y is equal to. Well, we can say that this is the same thing as x plus minus 1 times our vector y. So x plus minus 1 times our vector y. Now we can use our definitions. We know how to multiply by a scalar. So we'll say that this is equal to our x vector is 2, 4. And then what's minus 1 times y?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now we can use our definitions. We know how to multiply by a scalar. So we'll say that this is equal to our x vector is 2, 4. And then what's minus 1 times y? So minus 1 times y is 1. And then minus 1 times minus 2 is 2. So x minus y is going to be these two vectors added to each other."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then what's minus 1 times y? So minus 1 times y is 1. And then minus 1 times minus 2 is 2. So x minus y is going to be these two vectors added to each other. I'm just adding the minus of y. This right here is minus vector y. So this x minus y is going to be equal to 3 and 6."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So x minus y is going to be these two vectors added to each other. I'm just adding the minus of y. This right here is minus vector y. So this x minus y is going to be equal to 3 and 6. So let's see what that looks like when we visually represent them. Our vector x was 2, 4. So 2, 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this x minus y is going to be equal to 3 and 6. So let's see what that looks like when we visually represent them. Our vector x was 2, 4. So 2, 4. In standard position, it looks like this. That's my vector x. And then vector y in standard position."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 2, 4. In standard position, it looks like this. That's my vector x. And then vector y in standard position. Let me do it in a different color. I'll do y in green. Vector y is minus 1 minus 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then vector y in standard position. Let me do it in a different color. I'll do y in green. Vector y is minus 1 minus 2. So minus 1 minus 2. So minus 1 minus 2. It looks just like this."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Vector y is minus 1 minus 2. So minus 1 minus 2. So minus 1 minus 2. It looks just like this. And actually I ended up inadvertently doing collinear vectors. But hey, this is interesting too. So this is vector y."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It looks just like this. And actually I ended up inadvertently doing collinear vectors. But hey, this is interesting too. So this is vector y. So then what's their difference? It says 3, 6. So it's the vector 3, 6."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is vector y. So then what's their difference? It says 3, 6. So it's the vector 3, 6. So it's this vector. Let me draw it someplace else. If I start here, I go 1, 2, 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's the vector 3, 6. So it's this vector. Let me draw it someplace else. If I start here, I go 1, 2, 3. And then I go up 6. So then up 6. It's a vector that looks like this."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I start here, I go 1, 2, 3. And then I go up 6. So then up 6. It's a vector that looks like this. That's the difference between the two vectors. So at first you say, well, this is x minus y. You're like, hey, how is this the difference of these two?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's a vector that looks like this. That's the difference between the two vectors. So at first you say, well, this is x minus y. You're like, hey, how is this the difference of these two? Well, if you overlay this, if you just shift this and over this, you could actually just start here and go straight up. And you'll see that it's really the difference between the endpoints. You're kind of connecting the endpoints."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're like, hey, how is this the difference of these two? Well, if you overlay this, if you just shift this and over this, you could actually just start here and go straight up. And you'll see that it's really the difference between the endpoints. You're kind of connecting the endpoints. But I actually didn't want to draw collinear vectors. So let me do another example. Although that one's kind of interesting."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're kind of connecting the endpoints. But I actually didn't want to draw collinear vectors. So let me do another example. Although that one's kind of interesting. You often don't see that one in a book. Let me define vector x, in this case, to be 2, 3. And let me define vector y to be minus 4, minus 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Although that one's kind of interesting. You often don't see that one in a book. Let me define vector x, in this case, to be 2, 3. And let me define vector y to be minus 4, minus 2. So what would be x in standard position? It would be 2, 3. It'd look like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And let me define vector y to be minus 4, minus 2. So what would be x in standard position? It would be 2, 3. It'd look like that. That is our vector x if we start at the origin. So this is x. And then what does vector y look like?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It'd look like that. That is our vector x if we start at the origin. So this is x. And then what does vector y look like? I'll do y in orange. Minus 4, minus 2. So vector y looks like this."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then what does vector y look like? I'll do y in orange. Minus 4, minus 2. So vector y looks like this. Now, what is x minus y? Well, we could view this 2 plus minus 1 times this. Or we could just say 2 minus minus 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector y looks like this. Now, what is x minus y? Well, we could view this 2 plus minus 1 times this. Or we could just say 2 minus minus 4. I think you get the idea now. But we just did it the first way the last time. Because I wanted to go for my basic definitions of scalar multiplication."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or we could just say 2 minus minus 4. I think you get the idea now. But we just did it the first way the last time. Because I wanted to go for my basic definitions of scalar multiplication. So x minus y is just going to be equal to 2 plus minus 1 times minus 4, or 2 minus minus 4. That's the same thing as 2 plus 4. So it's 6."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because I wanted to go for my basic definitions of scalar multiplication. So x minus y is just going to be equal to 2 plus minus 1 times minus 4, or 2 minus minus 4. That's the same thing as 2 plus 4. So it's 6. And then it's 3 minus minus 2. So it's 5. So the difference between the two is the vector 6, 5."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's 6. And then it's 3 minus minus 2. So it's 5. So the difference between the two is the vector 6, 5. So you could draw it out here again. So you could go, to add 6 to 4, you go up there. And then to 5, you'd go like that."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the difference between the two is the vector 6, 5. So you could draw it out here again. So you could go, to add 6 to 4, you go up there. And then to 5, you'd go like that. So the vector would look something like this. It shouldn't curve like that. So that's x minus y."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then to 5, you'd go like that. So the vector would look something like this. It shouldn't curve like that. So that's x minus y. But if we drew them between, like in the last example, I showed that you could draw it between their two heads. So if you do it here, what does it look like? Well, if you start at this point right there, and you go 6 to the right, and then up 5, you end up right there."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's x minus y. But if we drew them between, like in the last example, I showed that you could draw it between their two heads. So if you do it here, what does it look like? Well, if you start at this point right there, and you go 6 to the right, and then up 5, you end up right there. So the difference between the two vectors looks like that. Which kind of should make sense intuitively. x minus y."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, if you start at this point right there, and you go 6 to the right, and then up 5, you end up right there. So the difference between the two vectors looks like that. Which kind of should make sense intuitively. x minus y. That's the difference between the two vectors. You can view the difference as how do you get from one vector to another vector, right? Like, let's go back to our second grade world of just scalars."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x minus y. That's the difference between the two vectors. You can view the difference as how do you get from one vector to another vector, right? Like, let's go back to our second grade world of just scalars. If I say what 7 minus 5 is, and you say it's equal to 2, well, that just tells you that 5 plus 2 is equal to 7. Or the difference between 5 and 7 is 2. And here you're saying, look, the difference between x and y is this vector right there."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Like, let's go back to our second grade world of just scalars. If I say what 7 minus 5 is, and you say it's equal to 2, well, that just tells you that 5 plus 2 is equal to 7. Or the difference between 5 and 7 is 2. And here you're saying, look, the difference between x and y is this vector right there. It's equal to that vector right there. Or you could say, look, if I take 5 and add 2, I get 7. Or you could say, look, if I take vector y and I add vector x minus y, then I get vector x."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And here you're saying, look, the difference between x and y is this vector right there. It's equal to that vector right there. Or you could say, look, if I take 5 and add 2, I get 7. Or you could say, look, if I take vector y and I add vector x minus y, then I get vector x. Now, let's do something else that's interesting. Let's do what y minus x is equal to. y minus x."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or you could say, look, if I take vector y and I add vector x minus y, then I get vector x. Now, let's do something else that's interesting. Let's do what y minus x is equal to. y minus x. What is that equal to? Do another color right here. Well, we'll take minus 4 minus 2, which is minus 6."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "y minus x. What is that equal to? Do another color right here. Well, we'll take minus 4 minus 2, which is minus 6. And then you have minus 2 minus 3. It's minus 5. So y minus x is going to be, let's see, if we start here, we're going to go down 6, 1, 2, 3, 4, 5, 6, and then back 5."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we'll take minus 4 minus 2, which is minus 6. And then you have minus 2 minus 3. It's minus 5. So y minus x is going to be, let's see, if we start here, we're going to go down 6, 1, 2, 3, 4, 5, 6, and then back 5. So back 2, 4, 5. So y minus x looks like this. It's really the exact same vector."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So y minus x is going to be, let's see, if we start here, we're going to go down 6, 1, 2, 3, 4, 5, 6, and then back 5. So back 2, 4, 5. So y minus x looks like this. It's really the exact same vector. Remember, it doesn't matter where we start. It's just pointing in the opposite direction. So if we shifted it here, I could draw it right on top of this."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's really the exact same vector. Remember, it doesn't matter where we start. It's just pointing in the opposite direction. So if we shifted it here, I could draw it right on top of this. It would be the exact as x minus y, but just in the opposite direction. Which is just a general good thing to know. So you can kind of view them as the negatives of each other."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if we shifted it here, I could draw it right on top of this. It would be the exact as x minus y, but just in the opposite direction. Which is just a general good thing to know. So you can kind of view them as the negatives of each other. And actually, let me make that point very clear. We drew y. y I drew before. Actually, let me draw x. x we could draw as 2, 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you can kind of view them as the negatives of each other. And actually, let me make that point very clear. We drew y. y I drew before. Actually, let me draw x. x we could draw as 2, 3. So you go to the right 2 and then up 3. I've done this before. This is x in non-standard position."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me draw x. x we could draw as 2, 3. So you go to the right 2 and then up 3. I've done this before. This is x in non-standard position. That's x as well. What is negative x? Negative x is minus 2 minus 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is x in non-standard position. That's x as well. What is negative x? Negative x is minus 2 minus 3. So if I were to start here, I'd go minus 2 and then I'd go minus 3. So minus x would look just like this. It looks just like x."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Negative x is minus 2 minus 3. So if I were to start here, I'd go minus 2 and then I'd go minus 3. So minus x would look just like this. It looks just like x. It's parallel. It has the same magnitude. It's just pointing in the exact opposite direction."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It looks just like x. It's parallel. It has the same magnitude. It's just pointing in the exact opposite direction. This is just a good thing to kind of really get seared into your brain is to have an intuition for these things. Now, just to kind of finish up this kind of idea of adding and subtracting vectors, let me just do everything I did so far was in an R2. But I want to show you that we can generalize them."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just pointing in the exact opposite direction. This is just a good thing to kind of really get seared into your brain is to have an intuition for these things. Now, just to kind of finish up this kind of idea of adding and subtracting vectors, let me just do everything I did so far was in an R2. But I want to show you that we can generalize them. And we can even generalize them to vector spaces that aren't normally intuitive for us to actually visualize. So let me define a couple of vectors. Let me define vector a to be equal to 0, minus 1, 2, and 3."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But I want to show you that we can generalize them. And we can even generalize them to vector spaces that aren't normally intuitive for us to actually visualize. So let me define a couple of vectors. Let me define vector a to be equal to 0, minus 1, 2, and 3. Let me define vector b to be equal to 4, minus 2, 0, 5. We can do the same addition and subtraction operations with them. It's just it'll be hard to visualize."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me define vector a to be equal to 0, minus 1, 2, and 3. Let me define vector b to be equal to 4, minus 2, 0, 5. We can do the same addition and subtraction operations with them. It's just it'll be hard to visualize. But we can keep them in just vector form so that it's still useful to think in four dimensions. So if I were to say 4 times a, this is the vector a, minus 2 times b, what is this going to be equal to? And this is a vector."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just it'll be hard to visualize. But we can keep them in just vector form so that it's still useful to think in four dimensions. So if I were to say 4 times a, this is the vector a, minus 2 times b, what is this going to be equal to? And this is a vector. What is this going to be equal to? Well, we could rewrite this as 4 times this whole column vector, 0, minus 1, 2, and 3, minus 2 times b, minus 2 times 4, minus 2, 0, 5. And what is this going to be equal to?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is a vector. What is this going to be equal to? Well, we could rewrite this as 4 times this whole column vector, 0, minus 1, 2, and 3, minus 2 times b, minus 2 times 4, minus 2, 0, 5. And what is this going to be equal to? This term right here, 4 times this, you're going to get 4 times 0 is 0, minus 4, 8. 4 times 2 is 8, 4 times 3 is 12. And then minus, I'll do it in yellow, minus 2 times 4 is 8, 2 times minus 2 is minus 4, 2 times 0 is 0, 2 times 5 is 10."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what is this going to be equal to? This term right here, 4 times this, you're going to get 4 times 0 is 0, minus 4, 8. 4 times 2 is 8, 4 times 3 is 12. And then minus, I'll do it in yellow, minus 2 times 4 is 8, 2 times minus 2 is minus 4, 2 times 0 is 0, 2 times 5 is 10. Now this isn't a good part of my board, so let me just, it doesn't write well right over there. I haven't figured out the problem, but if I were to just write it over here, what do we get? With 0 minus 8, minus 4, minus 4, minus negative 4."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then minus, I'll do it in yellow, minus 2 times 4 is 8, 2 times minus 2 is minus 4, 2 times 0 is 0, 2 times 5 is 10. Now this isn't a good part of my board, so let me just, it doesn't write well right over there. I haven't figured out the problem, but if I were to just write it over here, what do we get? With 0 minus 8, minus 4, minus 4, minus negative 4. So that's minus 4 plus 4, so that's 0. 8 minus 0 is 8, minus 4, minus 4. 8, 12 minus, what was this?"}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "With 0 minus 8, minus 4, minus 4, minus negative 4. So that's minus 4 plus 4, so that's 0. 8 minus 0 is 8, minus 4, minus 4. 8, 12 minus, what was this? I can't even read it, what it says. This was a 10. Oh, now you can see it again."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "8, 12 minus, what was this? I can't even read it, what it says. This was a 10. Oh, now you can see it again. Something, it's very bizarre. 2 times 5 is 10. So it's 12 minus 10, so it's 2."}, {"video_title": "Vector examples Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Oh, now you can see it again. Something, it's very bizarre. 2 times 5 is 10. So it's 12 minus 10, so it's 2. So when we take this vector and multiply it by 4 and subtract 2 times this vector, we just get this vector. And even though you can't represent this in kind of an easy, kind of graphable format, this is a useful concept. And we're going to see this later when we apply some of these vectors to multidimensional space."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's called a column space. And you could probably guess what it means just based on what it's called. But let's say I have some matrix A. Let's say it's an m by n matrix. So I can write my matrix A, and we've seen this multiple times. I can write it as a collection of column vectors. So this first one, second one, and I'll have n of them."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's an m by n matrix. So I can write my matrix A, and we've seen this multiple times. I can write it as a collection of column vectors. So this first one, second one, and I'll have n of them. How do I know that I have n of them? Because I have n columns. And each of these column vectors are going to have how many components?"}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this first one, second one, and I'll have n of them. How do I know that I have n of them? Because I have n columns. And each of these column vectors are going to have how many components? So v1, v2, all the way to vn. This matrix has m rows, right? So each of these guys are going to have m components."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And each of these column vectors are going to have how many components? So v1, v2, all the way to vn. This matrix has m rows, right? So each of these guys are going to have m components. So they're all members of Rm. So the column space is defined as all of the possible linear combinations of these column vectors. So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So each of these guys are going to have m components. So they're all members of Rm. So the column space is defined as all of the possible linear combinations of these column vectors. So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors. Well, what's all of the linear combinations of a set of vectors? It's the span of those vectors. So it's the span of vector 1, vector 2, all the way to vector n. And we've done it before when we first talked about span and subspaces, but it's pretty easy to show that the span of any set of vectors is a legitimate subspace."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors. Well, what's all of the linear combinations of a set of vectors? It's the span of those vectors. So it's the span of vector 1, vector 2, all the way to vector n. And we've done it before when we first talked about span and subspaces, but it's pretty easy to show that the span of any set of vectors is a legitimate subspace. It definitely contains the 0 vector. If you multiply all of these guys by 0, which is a valid linear combination added up, you'll see that it contains a 0 vector. Let's say I have some vector A that is a member of the column space of A."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's the span of vector 1, vector 2, all the way to vector n. And we've done it before when we first talked about span and subspaces, but it's pretty easy to show that the span of any set of vectors is a legitimate subspace. It definitely contains the 0 vector. If you multiply all of these guys by 0, which is a valid linear combination added up, you'll see that it contains a 0 vector. Let's say I have some vector A that is a member of the column space of A. That means it can be represented as some linear combination. So A is equal to c1 times vector 1 plus c2 times vector 2, all the way to cn times vector n. Now the question is, is this closed under multiplication? Is if I multiply A times some new, let me call it, let me say I multiply it times some scalar s. I'm just picking a random letter."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have some vector A that is a member of the column space of A. That means it can be represented as some linear combination. So A is equal to c1 times vector 1 plus c2 times vector 2, all the way to cn times vector n. Now the question is, is this closed under multiplication? Is if I multiply A times some new, let me call it, let me say I multiply it times some scalar s. I'm just picking a random letter. So s times A. Is this in my span? Well, s times A would be equal to sc1v1 plus sc2v2, all the way to scnvn, which is once again just a linear combination of these column vectors."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Is if I multiply A times some new, let me call it, let me say I multiply it times some scalar s. I'm just picking a random letter. So s times A. Is this in my span? Well, s times A would be equal to sc1v1 plus sc2v2, all the way to scnvn, which is once again just a linear combination of these column vectors. So this sa would clearly be a member of the column space of A. And then finally, to make sure it's a valid subspace, and this actually doesn't apply just to column spaces. This applies to any span."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, s times A would be equal to sc1v1 plus sc2v2, all the way to scnvn, which is once again just a linear combination of these column vectors. So this sa would clearly be a member of the column space of A. And then finally, to make sure it's a valid subspace, and this actually doesn't apply just to column spaces. This applies to any span. This is actually a review of what we've done in the past. We just have to make sure it's closed under addition. So let's say A is a member of our column space."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This applies to any span. This is actually a review of what we've done in the past. We just have to make sure it's closed under addition. So let's say A is a member of our column space. Let's say B is also a member of our column space, or our span of all of these column vectors. Then B could be rewritten as, I don't know, let me say b1 times v1 plus b2 times v2, all the way to bn times vn. And my question is, is A plus B a member of our span, of our column space, the span of these vectors?"}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say A is a member of our column space. Let's say B is also a member of our column space, or our span of all of these column vectors. Then B could be rewritten as, I don't know, let me say b1 times v1 plus b2 times v2, all the way to bn times vn. And my question is, is A plus B a member of our span, of our column space, the span of these vectors? Well, sure. What's A plus B? A plus B is equal to c1 plus b1 times v1 plus c2 plus b2 times v2."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And my question is, is A plus B a member of our span, of our column space, the span of these vectors? Well, sure. What's A plus B? A plus B is equal to c1 plus b1 times v1 plus c2 plus b2 times v2. I'm just literally adding this term to that term to get that term, this term to this term to get this term, and then it goes all the way to bn plus cn times vn, which is clearly just another linear combination of these guys. So this guy is definitely within the span. So the span of, and this, what I just did, it doesn't have to be unique to a matrix."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A plus B is equal to c1 plus b1 times v1 plus c2 plus b2 times v2. I'm just literally adding this term to that term to get that term, this term to this term to get this term, and then it goes all the way to bn plus cn times vn, which is clearly just another linear combination of these guys. So this guy is definitely within the span. So the span of, and this, what I just did, it doesn't have to be unique to a matrix. I mean, a matrix is just really just a way of writing a set of column vectors. So this applies to any span. So this is clearly a valid subspace."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the span of, and this, what I just did, it doesn't have to be unique to a matrix. I mean, a matrix is just really just a way of writing a set of column vectors. So this applies to any span. So this is clearly a valid subspace. So the column space of A is clearly a valid subspace. Now let's think about other ways we can interpret this notion of a column space. Let's think about it in terms of the expression, let me get a good color."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is clearly a valid subspace. So the column space of A is clearly a valid subspace. Now let's think about other ways we can interpret this notion of a column space. Let's think about it in terms of the expression, let me get a good color. If I were to multiply my, let's think about this. Let's think about the set of all the values of, if I take my m by n matrix A and I multiply it by any vector x where x is a member of, remember, x has to be a member of Rn, it has to have n components in order for this multiplication to be well-defined. So x has to be a member of Rn."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's think about it in terms of the expression, let me get a good color. If I were to multiply my, let's think about this. Let's think about the set of all the values of, if I take my m by n matrix A and I multiply it by any vector x where x is a member of, remember, x has to be a member of Rn, it has to have n components in order for this multiplication to be well-defined. So x has to be a member of Rn. Let's think about what this means. This is the set, if this says, look, I can take any member, any n component vector and multiply it by a, and I care about all of the possible products that this could equal, all the possible values of Ax when I can pick and choose any possible x from Rn. Let's think about what that means."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So x has to be a member of Rn. Let's think about what this means. This is the set, if this says, look, I can take any member, any n component vector and multiply it by a, and I care about all of the possible products that this could equal, all the possible values of Ax when I can pick and choose any possible x from Rn. Let's think about what that means. Ax, if I write a like that and if I write x like this, let me write it a little bit better, let me write x like this, x1, x2, all the way to xn. What is Ax? Well, Ax could be rewritten as x1, and we've seen this before, Ax is equal to x1 times v1 plus x2 times v2 all the way to plus xn times vn."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's think about what that means. Ax, if I write a like that and if I write x like this, let me write it a little bit better, let me write x like this, x1, x2, all the way to xn. What is Ax? Well, Ax could be rewritten as x1, and we've seen this before, Ax is equal to x1 times v1 plus x2 times v2 all the way to plus xn times vn. We've seen this multiple times. This comes out of our definition of matrix vector products. Now, if Ax is equal to this, and I'm essentially saying that I can pick any vector x in Rn, I'm saying that I can pick all possible values of the entries here, all possible real values and all possible combinations of them."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, Ax could be rewritten as x1, and we've seen this before, Ax is equal to x1 times v1 plus x2 times v2 all the way to plus xn times vn. We've seen this multiple times. This comes out of our definition of matrix vector products. Now, if Ax is equal to this, and I'm essentially saying that I can pick any vector x in Rn, I'm saying that I can pick all possible values of the entries here, all possible real values and all possible combinations of them. So what is this equal to? What is the set of all possible? So I could rewrite this statement here as the set of all possible x1, v1, plus x2, v2, all the way to xn, vn, where x1, x2, all the way to xn are a member of the real numbers."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, if Ax is equal to this, and I'm essentially saying that I can pick any vector x in Rn, I'm saying that I can pick all possible values of the entries here, all possible real values and all possible combinations of them. So what is this equal to? What is the set of all possible? So I could rewrite this statement here as the set of all possible x1, v1, plus x2, v2, all the way to xn, vn, where x1, x2, all the way to xn are a member of the real numbers. That's all I'm saying here. This statement is the equivalent of this. When I say that the vector x can be any member of Rn, I'm saying that its components can be any members of the real numbers."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could rewrite this statement here as the set of all possible x1, v1, plus x2, v2, all the way to xn, vn, where x1, x2, all the way to xn are a member of the real numbers. That's all I'm saying here. This statement is the equivalent of this. When I say that the vector x can be any member of Rn, I'm saying that its components can be any members of the real numbers. So if I just take the set of all of the, essentially, the combinations of these column vectors where their real numbers, where their coefficients are members of the real numbers, what am I doing? This is all the possible linear combinations of the column vectors of A. So this is equal to the span of the v1, v2, all the way to vn, which is the same exact same thing as the column space of A."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "When I say that the vector x can be any member of Rn, I'm saying that its components can be any members of the real numbers. So if I just take the set of all of the, essentially, the combinations of these column vectors where their real numbers, where their coefficients are members of the real numbers, what am I doing? This is all the possible linear combinations of the column vectors of A. So this is equal to the span of the v1, v2, all the way to vn, which is the same exact same thing as the column space of A. So the column space of A, you could say, hey, what are all of the possible vectors, or the set of all vectors I can create by taking linear combinations of these guys, or the span of these guys? Or you can view it as, what are all of the possible values that Ax can take on if x is a member of Rn? So let's think about it this way."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is equal to the span of the v1, v2, all the way to vn, which is the same exact same thing as the column space of A. So the column space of A, you could say, hey, what are all of the possible vectors, or the set of all vectors I can create by taking linear combinations of these guys, or the span of these guys? Or you can view it as, what are all of the possible values that Ax can take on if x is a member of Rn? So let's think about it this way. Let's say that I were to tell you that I need to solve the equation Ax is equal to, let me say, well, the convention is to write a b there, but let me put a special b there. Let me put b1. Let's say that I need to solve this equation Ax is equal to b1."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's think about it this way. Let's say that I were to tell you that I need to solve the equation Ax is equal to, let me say, well, the convention is to write a b there, but let me put a special b there. Let me put b1. Let's say that I need to solve this equation Ax is equal to b1. And then I were to tell you, let's say that I were to figure out the column space of A, and I say b1 is not a member of the column space of A. Now what does that tell me? That tells me that this right here can never take on the value b1, right?"}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that I need to solve this equation Ax is equal to b1. And then I were to tell you, let's say that I were to figure out the column space of A, and I say b1 is not a member of the column space of A. Now what does that tell me? That tells me that this right here can never take on the value b1, right? Because all of the values that this can take on is the column space of A. So if b1 is not in this, it means that this cannot take on the value of b1. So this would imply that this equation we're trying to set up, Ax is equal to b1, has no solution."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That tells me that this right here can never take on the value b1, right? Because all of the values that this can take on is the column space of A. So if b1 is not in this, it means that this cannot take on the value of b1. So this would imply that this equation we're trying to set up, Ax is equal to b1, has no solution. If it had a solution, if it had some solution, so let's say that Ax equals b2. So let's say Ax equals b2 has at least one solution. Has at least one solution."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this would imply that this equation we're trying to set up, Ax is equal to b1, has no solution. If it had a solution, if it had some solution, so let's say that Ax equals b2. So let's say Ax equals b2 has at least one solution. Has at least one solution. What does this mean? Well, that means that this, for a particular x, or maybe for many different x's, you can definitely achieve this value. For there are some x's that when you multiply it by a, you definitely are able to get this value."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Has at least one solution. What does this mean? Well, that means that this, for a particular x, or maybe for many different x's, you can definitely achieve this value. For there are some x's that when you multiply it by a, you definitely are able to get this value. So this implies that b2 is definitely a member of the column space of A. Some of this stuff, on some level, it's almost obvious. This comes out of the definition of the column space, the column space is all of the linear combinations of the column vectors, which another interpretation is all of the values that Ax can take on."}, {"video_title": "Column space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For there are some x's that when you multiply it by a, you definitely are able to get this value. So this implies that b2 is definitely a member of the column space of A. Some of this stuff, on some level, it's almost obvious. This comes out of the definition of the column space, the column space is all of the linear combinations of the column vectors, which another interpretation is all of the values that Ax can take on. So if I try to set Ax to some value that it can't take on, clearly I'm not going to have some solution. If I set Ax, if I am able to find a solution, if I am able to find some x value where Ax is equal to b2, then b2 definitely is one of the values that Ax can take on. Anyway, I think I'll leave you there, now that you have at least a kind of abstract understanding of what a column space is."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's actually use that method in this video right here. So I'm going to use the same matrix that we started off with in the last video. And it seems like a fairly good matrix. We know that its reduced row echelon form is the identity matrix, so we know it's invertible. So let's find its inverse. So the technique is pretty straightforward. You literally just apply the same transformations you would apply to this guy to get you to the identity matrix."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know that its reduced row echelon form is the identity matrix, so we know it's invertible. So let's find its inverse. So the technique is pretty straightforward. You literally just apply the same transformations you would apply to this guy to get you to the identity matrix. You would apply those same transformations to the identity matrix. And that's because the collection of those transformations, if you represent them as matrices, are really just the inverse of this guy. So let's just do it."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You literally just apply the same transformations you would apply to this guy to get you to the identity matrix. You would apply those same transformations to the identity matrix. And that's because the collection of those transformations, if you represent them as matrices, are really just the inverse of this guy. So let's just do it. So I'll create an augmented matrix here. Maybe I'll do it right here. Let me make it a little bit neater."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's just do it. So I'll create an augmented matrix here. Maybe I'll do it right here. Let me make it a little bit neater. So first I'll write a. So it's 1, minus 1, 1. And then minus 1, 2, 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me make it a little bit neater. So first I'll write a. So it's 1, minus 1, 1. And then minus 1, 2, 1. Minus 1, 3, 4. And then I will augment it with the identity matrix. With 1, 0, 0, 0, 1, 0, 0, 0, 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then minus 1, 2, 1. Minus 1, 3, 4. And then I will augment it with the identity matrix. With 1, 0, 0, 0, 1, 0, 0, 0, 1. Now, if I want to get a into reduced row echelon form, maybe I'll replace the second row. So I'll keep the first row the same for now. So let me just draw it like this."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "With 1, 0, 0, 0, 1, 0, 0, 0, 1. Now, if I want to get a into reduced row echelon form, maybe I'll replace the second row. So I'll keep the first row the same for now. So let me just draw it like this. Let me keep my first row the same. The entire first row. 1, minus 1, minus 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me just draw it like this. Let me keep my first row the same. The entire first row. 1, minus 1, minus 1. It's going to be augmented with 1, 0, 0. Keep the whole first row the same. Let's replace the second row with the second row plus the first row."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1, minus 1, minus 1. It's going to be augmented with 1, 0, 0. Keep the whole first row the same. Let's replace the second row with the second row plus the first row. So minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's replace the second row with the second row plus the first row. So minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2. 0 plus 1 is 0. 1 plus, oh sorry, that was a tricky one. 0 plus 1 is 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "3 plus minus 1 is 2. 0 plus 1 is 0. 1 plus, oh sorry, that was a tricky one. 0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. All I did is I added these two rows up."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. All I did is I added these two rows up. Now, this third row, let me replace it. I want to get a 0 here. So let me replace the third row with the third row minus the first row."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I added these two rows up. Now, this third row, let me replace it. I want to get a 0 here. So let me replace the third row with the third row minus the first row. So 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me replace the third row with the third row minus the first row. So 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5. 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "4 minus minus 1 is 5. 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1. Just like that. Now, what do we want to do? Well, we've gotten this far."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then 1 minus 0 is 1. Just like that. Now, what do we want to do? Well, we've gotten this far. We want to zero out that guy and that guy. So let's keep our second row the same. Let me write it down here."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we've gotten this far. We want to zero out that guy and that guy. So let's keep our second row the same. Let me write it down here. Let's keep our second row the same. So it's 0, 1, 2. And then you augmented it with 1, 1, 0."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it down here. Let's keep our second row the same. So it's 0, 1, 2. And then you augmented it with 1, 1, 0. Just like that. And let's replace my first row with the first row plus the second row. So 1 plus 0 is 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you augmented it with 1, 1, 0. Just like that. And let's replace my first row with the first row plus the second row. So 1 plus 0 is 1. Minus 1 plus 1 is 0. That's why I did that, to get a 0 there. Minus 1 plus 2 is 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 1 plus 0 is 1. Minus 1 plus 1 is 0. That's why I did that, to get a 0 there. Minus 1 plus 2 is 1. 1 plus 1 is 2. 0 plus 1 is 1. 0 plus 0 is 0."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 plus 2 is 1. 1 plus 1 is 2. 0 plus 1 is 1. 0 plus 0 is 0. And now I also want to zero out this guy right here. So let's replace the third row with the third row minus 2 times the second row. So 0 minus 2 times 0 is 0."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0 plus 0 is 0. And now I also want to zero out this guy right here. So let's replace the third row with the third row minus 2 times the second row. So 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 5 minus 2 times 2 is 5 minus 4. That's 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 5 minus 2 times 2 is 5 minus 4. That's 1. Minus 1 minus 2 times 1 is minus 3. 0 minus 2 times 1, that's minus 2. And then 1 minus 2 times 0 is just 1 again."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's 1. Minus 1 minus 2 times 1 is minus 3. 0 minus 2 times 1, that's minus 2. And then 1 minus 2 times 0 is just 1 again. All right, home stretch. Now I just want to zero out these guys right here. I just want to zero out those guys right there."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then 1 minus 2 times 0 is just 1 again. All right, home stretch. Now I just want to zero out these guys right here. I just want to zero out those guys right there. So let me just keep my third row the same. Let me switch colors, keep things colorful. I'm going to keep my third row the same, so it's 0, 0, 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I just want to zero out those guys right there. So let me just keep my third row the same. Let me switch colors, keep things colorful. I'm going to keep my third row the same, so it's 0, 0, 1. I'm going to augment it with minus 3, minus 2, and 1. Now let's replace our first row with the first row minus the third row. So 1 minus 0 is 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to keep my third row the same, so it's 0, 0, 1. I'm going to augment it with minus 3, minus 2, and 1. Now let's replace our first row with the first row minus the third row. So 1 minus 0 is 1. 0 minus 0 is 0. 1 minus 1 is 0. 2 minus minus 3, that's 5."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 1 minus 0 is 1. 0 minus 0 is 0. 1 minus 1 is 0. 2 minus minus 3, that's 5. 1 minus minus 2 is 3. 0 minus 1 is minus 1. Now let's replace the second row with the second row minus 2 times the third row."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 minus minus 3, that's 5. 1 minus minus 2 is 3. 0 minus 1 is minus 1. Now let's replace the second row with the second row minus 2 times the third row. So 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 0. 2 minus 2 times 1 is 0."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's replace the second row with the second row minus 2 times the third row. So 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 1 minus 2 times 0 is 1. It's not 0. 2 minus 2 times 1 is 0."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 2 times 1 is 0. 1 minus 2 times 0 is 1. It's not 0. 2 minus 2 times 1 is 0. 1 minus 2 times minus 3. That is 1 plus 2 times 3. That is 7."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 2 times 1 is 0. 1 minus 2 times minus 3. That is 1 plus 2 times 3. That is 7. 1 minus 2 times minus 2. That's 1 plus 4, which is 5. And then 0 minus 2 times 1."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is 7. 1 minus 2 times minus 2. That's 1 plus 4, which is 5. And then 0 minus 2 times 1. So that's minus 2. And just like that, we've gotten the A part of our augmented matrix into reduced row echelon form. This is the reduced row echelon form of A."}, {"video_title": "Example of finding matrix inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then 0 minus 2 times 1. So that's minus 2. And just like that, we've gotten the A part of our augmented matrix into reduced row echelon form. This is the reduced row echelon form of A. And when you apply those exact same transformations, because if you think about it, that series of matrix products that got you from this to the identity matrix, that by definition is the identity matrix. So you apply those same transformations to the identity matrix, you're going to get the inverse of A. So this right here is A inverse."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I have some matrix, let's just call it B. If my matrix B looks like this, if its entries are A, B, C, D, we defined the determinant of B, which could also be written as B with these lines around it, which could also be written as the entries of the matrix with those lines around it. A, B, C, D. And I don't want you to get these confused. This is the matrix when you have the brackets. This is the determinant of the matrix when you just have these straight lines. And this, by definition, was equal to A, D minus B, C. And you saw in the last video, or maybe you saw in the last video, what the motivation for this came from. When we figured out the inverse of B, we determined that it was equal to the inverse of B was equal to 1 over A, D minus B, C times another matrix, which was essentially these two entries swapped."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the matrix when you have the brackets. This is the determinant of the matrix when you just have these straight lines. And this, by definition, was equal to A, D minus B, C. And you saw in the last video, or maybe you saw in the last video, what the motivation for this came from. When we figured out the inverse of B, we determined that it was equal to the inverse of B was equal to 1 over A, D minus B, C times another matrix, which was essentially these two entries swapped. So you got a D and an A. And then these two entries made negative. So minus C and minus B."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When we figured out the inverse of B, we determined that it was equal to the inverse of B was equal to 1 over A, D minus B, C times another matrix, which was essentially these two entries swapped. So you got a D and an A. And then these two entries made negative. So minus C and minus B. This was the inverse of B. And we said, well, when is this defined? This is defined as long as this character right here does not equal 0."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So minus C and minus B. This was the inverse of B. And we said, well, when is this defined? This is defined as long as this character right here does not equal 0. So we said, hey, this looks pretty important. Let's call this thing right there the determinant. Let's call this right here the determinant."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is defined as long as this character right here does not equal 0. So we said, hey, this looks pretty important. Let's call this thing right there the determinant. Let's call this right here the determinant. And then we could say that B is invertible if and only if the determinant of B does not equal 0. Because if it equals 0, then this formula for your inverse won't be well-defined. We just got this from our technique of creating an augmented matrix, whatnot."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's call this right here the determinant. And then we could say that B is invertible if and only if the determinant of B does not equal 0. Because if it equals 0, then this formula for your inverse won't be well-defined. We just got this from our technique of creating an augmented matrix, whatnot. But the big takeaway is we defined this notion of a determinant for a 2 by 2 matrix. Now, the next question is, well, that's just a 2 by 2. Everything we do in linear algebra, we like to generalize it to higher numbers of rows and columns."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just got this from our technique of creating an augmented matrix, whatnot. But the big takeaway is we defined this notion of a determinant for a 2 by 2 matrix. Now, the next question is, well, that's just a 2 by 2. Everything we do in linear algebra, we like to generalize it to higher numbers of rows and columns. So the next step, at least, let's just do baby steps. Let's start with a 3 by 3. Let's define what its determinant is."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Everything we do in linear algebra, we like to generalize it to higher numbers of rows and columns. So the next step, at least, let's just do baby steps. Let's start with a 3 by 3. Let's define what its determinant is. So let me construct a 3 by 3 matrix here. Let's say my matrix A is equal to, let me just write its entries, first row, first column, first row, second column, first row, third column. Then you have A21, A22, A23, then you have A31, third row, first column, A32, and then A33."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's define what its determinant is. So let me construct a 3 by 3 matrix here. Let's say my matrix A is equal to, let me just write its entries, first row, first column, first row, second column, first row, third column. Then you have A21, A22, A23, then you have A31, third row, first column, A32, and then A33. That is a 3 by 3 matrix. Clearly, three rows and three columns. This is 3 by 3."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then you have A21, A22, A23, then you have A31, third row, first column, A32, and then A33. That is a 3 by 3 matrix. Clearly, three rows and three columns. This is 3 by 3. I am going to define the determinant of A. So this is a definition. I'm going to define the determinant of this 3 by 3 matrix A as being equal to, and this is a little bit convoluted."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is 3 by 3. I am going to define the determinant of A. So this is a definition. I'm going to define the determinant of this 3 by 3 matrix A as being equal to, and this is a little bit convoluted. But you'll get the hang of it eventually. In the next several videos, we're just going to do a ton of determinants, so it just becomes a bit of second nature to you. It's a little computationally intensive sometimes."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to define the determinant of this 3 by 3 matrix A as being equal to, and this is a little bit convoluted. But you'll get the hang of it eventually. In the next several videos, we're just going to do a ton of determinants, so it just becomes a bit of second nature to you. It's a little computationally intensive sometimes. But it equals this first row, it equals A11 times the determinant of the matrix you get if you get rid of this guy's column and row. So if you get rid of this guy's column and row, you're left with this matrix here. So times the determinant of the matrix A22, A23, A32, and then A33, just like that."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's a little computationally intensive sometimes. But it equals this first row, it equals A11 times the determinant of the matrix you get if you get rid of this guy's column and row. So if you get rid of this guy's column and row, you're left with this matrix here. So times the determinant of the matrix A22, A23, A32, and then A33, just like that. So that's our first entry, and that's a plus this. And then I said it's a plus this because the next entry is going to be a minus. You have a minus this guy right here."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So times the determinant of the matrix A22, A23, A32, and then A33, just like that. So that's our first entry, and that's a plus this. And then I said it's a plus this because the next entry is going to be a minus. You have a minus this guy right here. So then you're going to have minus A12 times the matrix you get if you eliminate his column and his row. So times, you're going to get these entries right there. So A21, A21, A23, A23, A31, A31, and then you have A33."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have a minus this guy right here. So then you're going to have minus A12 times the matrix you get if you eliminate his column and his row. So times, you're going to get these entries right there. So A21, A21, A23, A23, A31, A31, and then you have A33. We're not quite done. You could probably guess what the next one's going to be. And you're going to have a plus, let me switch to a better color, plus this guy, plus A13 times the determinant of its, I guess you could call it its sub-matrix."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A21, A21, A23, A23, A31, A31, and then you have A33. We're not quite done. You could probably guess what the next one's going to be. And you're going to have a plus, let me switch to a better color, plus this guy, plus A13 times the determinant of its, I guess you could call it its sub-matrix. We'll call it that for now. So this matrix right here. So A21, A21, A22, A31, A32."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you're going to have a plus, let me switch to a better color, plus this guy, plus A13 times the determinant of its, I guess you could call it its sub-matrix. We'll call it that for now. So this matrix right here. So A21, A21, A22, A31, A32. This is our definition of the determinant of a 3 by 3 matrix. And the motivation is because when you take the determinant of a 3 by 3, it turns out, I haven't shown it to you yet, that the property is the same. That if the determinant of this is 0, you will not be able to find an inverse."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A21, A21, A22, A31, A32. This is our definition of the determinant of a 3 by 3 matrix. And the motivation is because when you take the determinant of a 3 by 3, it turns out, I haven't shown it to you yet, that the property is the same. That if the determinant of this is 0, you will not be able to find an inverse. And when I define the determinant in this way, if the determinant does not equal 0, you will be able to find an inverse. So that's where this came from. And I haven't shown you that yet."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That if the determinant of this is 0, you will not be able to find an inverse. And when I define the determinant in this way, if the determinant does not equal 0, you will be able to find an inverse. So that's where this came from. And I haven't shown you that yet. And I might not show you, because it's super computational. It'll take a long time. It'll be very hairy, and I'll make careless mistakes."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I haven't shown you that yet. And I might not show you, because it's super computational. It'll take a long time. It'll be very hairy, and I'll make careless mistakes. But the motivation comes from the exact same place as the 2 by 2 version. But I think what you probably want to see right now is at least just this thing applied to an actual matrix. Because this looks all abstract right now."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It'll be very hairy, and I'll make careless mistakes. But the motivation comes from the exact same place as the 2 by 2 version. But I think what you probably want to see right now is at least just this thing applied to an actual matrix. Because this looks all abstract right now. But if you do it with an actual matrix, you'll actually see it's not too bad. So let's leave the definition up there. And let's say that I have the matrix 1, 2, 4, 2, 2, minus 1, 3, and 4, 0, 1."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because this looks all abstract right now. But if you do it with an actual matrix, you'll actually see it's not too bad. So let's leave the definition up there. And let's say that I have the matrix 1, 2, 4, 2, 2, minus 1, 3, and 4, 0, 1. So by our definition of a determinant, the determinant of this guy right here, the determinant of this guy, so let's say I call that matrix, let's call that c. c is equal to that. So if we want to figure out the determinant of c, the determinant of c is equal to, I take this guy right here, so let me take that, 1 times the determinant of, let's just call it this sub-matrix right here. So we have a minus 1, minus 1, we have a, sorry, we have a minus 1, got to be careful, we have a 3, we have a 0, and we have a 1, just like that."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that I have the matrix 1, 2, 4, 2, 2, minus 1, 3, and 4, 0, 1. So by our definition of a determinant, the determinant of this guy right here, the determinant of this guy, so let's say I call that matrix, let's call that c. c is equal to that. So if we want to figure out the determinant of c, the determinant of c is equal to, I take this guy right here, so let me take that, 1 times the determinant of, let's just call it this sub-matrix right here. So we have a minus 1, minus 1, we have a, sorry, we have a minus 1, got to be careful, we have a 3, we have a 0, and we have a 1, just like that. Notice I got rid of this guy's column and this guy's row, and I was just left with minus 1, 3, 0, 1. Minus 1, 3, 0, and 1. Next, I take this guy."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we have a minus 1, minus 1, we have a, sorry, we have a minus 1, got to be careful, we have a 3, we have a 0, and we have a 1, just like that. Notice I got rid of this guy's column and this guy's row, and I was just left with minus 1, 3, 0, 1. Minus 1, 3, 0, and 1. Next, I take this guy. And this is the trick. You have to alternate signs. If you start with a positive here, this next one's going to be a minus."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Next, I take this guy. And this is the trick. You have to alternate signs. If you start with a positive here, this next one's going to be a minus. So you're going to have minus 2 times the sub-matrix, we can call it, if we get rid of this guy's column and this guy's row. So 2, 3, 4, 1. I just blanked this out."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you start with a positive here, this next one's going to be a minus. So you're going to have minus 2 times the sub-matrix, we can call it, if we get rid of this guy's column and this guy's row. So 2, 3, 4, 1. I just blanked this out. If I could videotape my finger, I would cover my finger over this column right here and over that row, and you'd just see a 2, a 3, a 4, and a 1. And that's what I put right there. And then finally, we'll have plus, we went plus, minus, plus, so finally we'll have plus 4 times the determinant of the sub-matrix, if you get rid of that row and that column."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I just blanked this out. If I could videotape my finger, I would cover my finger over this column right here and over that row, and you'd just see a 2, a 3, a 4, and a 1. And that's what I put right there. And then finally, we'll have plus, we went plus, minus, plus, so finally we'll have plus 4 times the determinant of the sub-matrix, if you get rid of that row and that column. So 2, minus 1, 4, 0. 2, minus 1, 4, and 0. Now these are pretty straightforward."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, we'll have plus, we went plus, minus, plus, so finally we'll have plus 4 times the determinant of the sub-matrix, if you get rid of that row and that column. So 2, minus 1, 4, 0. 2, minus 1, 4, and 0. Now these are pretty straightforward. These are not too bad to compute. Let's actually do it. So this is going to be equal to 1 times what?"}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now these are pretty straightforward. These are not too bad to compute. Let's actually do it. So this is going to be equal to 1 times what? Minus 1 times 1. Let me just write it out. Minus 1 times 1 minus 0 times 3."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 times what? Minus 1 times 1. Let me just write it out. Minus 1 times 1 minus 0 times 3. This just comes from the definition of a 2 by 2 determinant. We've already defined that. And then we're going to have a minus 2 times 2 times 1 minus 4 times 3."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 times 1 minus 0 times 3. This just comes from the definition of a 2 by 2 determinant. We've already defined that. And then we're going to have a minus 2 times 2 times 1 minus 4 times 3. And then finally, we're going to have a plus 4 times 2 times 0 minus minus 1 minus 1 times 4. Minus minus 1 times 4. I wrote it all out."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to have a minus 2 times 2 times 1 minus 4 times 3. And then finally, we're going to have a plus 4 times 2 times 0 minus minus 1 minus 1 times 4. Minus minus 1 times 4. I wrote it all out. So you can see that these are just, this thing right here is just this thing right here. And then you have the 4 out front. This thing right here was just this thing right here."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I wrote it all out. So you can see that these are just, this thing right here is just this thing right here. And then you have the 4 out front. This thing right here was just this thing right here. It's just the determinant of the 2 by 2 sub-matrix for each of these guys. And if we compute this, this is equal to minus 1 times 1 is minus 1 minus 0. That's 0."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This thing right here was just this thing right here. It's just the determinant of the 2 by 2 sub-matrix for each of these guys. And if we compute this, this is equal to minus 1 times 1 is minus 1 minus 0. That's 0. So this is a minus 1 times 1. So that's a minus 1. And then we get, what is this equal to?"}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's 0. So this is a minus 1 times 1. So that's a minus 1. And then we get, what is this equal to? This right here is 12. So you get 2 minus 12. You get 2 times 1 minus 4 times 3."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we get, what is this equal to? This right here is 12. So you get 2 minus 12. You get 2 times 1 minus 4 times 3. So it's minus 10. So that is equal to minus 10. And you have a minus 10 times a minus 2."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get 2 times 1 minus 4 times 3. So it's minus 10. So that is equal to minus 10. And you have a minus 10 times a minus 2. So that becomes a plus 20. Minus 2 times minus 10. And then finally in the green, we have 2 times 0."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you have a minus 10 times a minus 2. So that becomes a plus 20. Minus 2 times minus 10. And then finally in the green, we have 2 times 0. That's just a 0. And then you have minus 1 times 4, which is minus 4. Then you have a minus sign here."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then finally in the green, we have 2 times 0. That's just a 0. And then you have minus 1 times 4, which is minus 4. Then you have a minus sign here. So it's plus 4. And then so this all becomes a plus 4. Plus 4 times 4 is 16."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then you have a minus sign here. So it's plus 4. And then so this all becomes a plus 4. Plus 4 times 4 is 16. So plus 16. And what do we get when we add this up? We get 20 plus 16 minus 1."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Plus 4 times 4 is 16. So plus 16. And what do we get when we add this up? We get 20 plus 16 minus 1. It is equal to 35. We're done. We found the determinant of our 3 by 3 matrix."}, {"video_title": "3 x 3 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We get 20 plus 16 minus 1. It is equal to 35. We're done. We found the determinant of our 3 by 3 matrix. Not too bad. Right there. So that is equal to the determinant of c. So the fact that this isn't 0 tells you that c is invertible."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I had the vectors, let's say I had the set of vectors, I don't want to do it that thick. Let's say one of the vectors is the vector 2, 3, and then the other vector is the vector 4, 6. And I just want to answer the question, what is the span of these vectors? Let's assume that these are position vectors. What are all of the vectors that these two vectors can represent? Well, if you just look at it, and remember the span is just all of the vectors that can be represented by linear combinations of these. So it's the set of all the vectors, and if I have some constant times 2 times that vector, plus some other constant times this vector, it's all the possibilities that I can represent when I just put a bunch of different real numbers for c1 and c2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's assume that these are position vectors. What are all of the vectors that these two vectors can represent? Well, if you just look at it, and remember the span is just all of the vectors that can be represented by linear combinations of these. So it's the set of all the vectors, and if I have some constant times 2 times that vector, plus some other constant times this vector, it's all the possibilities that I can represent when I just put a bunch of different real numbers for c1 and c2. Now the first thing you might realize is that, look, this vector 2, this is just the same thing as 2 times this vector. So I could just rewrite it like this. I could just rewrite it as c1 times the vector 2, 3, plus c2 times the vector, and here I'm going to write, instead of writing the vector 4, 6, I'm going to write 2 times the vector 2, 3."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's the set of all the vectors, and if I have some constant times 2 times that vector, plus some other constant times this vector, it's all the possibilities that I can represent when I just put a bunch of different real numbers for c1 and c2. Now the first thing you might realize is that, look, this vector 2, this is just the same thing as 2 times this vector. So I could just rewrite it like this. I could just rewrite it as c1 times the vector 2, 3, plus c2 times the vector, and here I'm going to write, instead of writing the vector 4, 6, I'm going to write 2 times the vector 2, 3. Because this vector is just a multiple of that vector, so I could write c2 times 2 times 2, 3. I think you see that this is equivalent to the 4, 6. 2 times 2 is 4, 2 times 3 is 6."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could just rewrite it as c1 times the vector 2, 3, plus c2 times the vector, and here I'm going to write, instead of writing the vector 4, 6, I'm going to write 2 times the vector 2, 3. Because this vector is just a multiple of that vector, so I could write c2 times 2 times 2, 3. I think you see that this is equivalent to the 4, 6. 2 times 2 is 4, 2 times 3 is 6. Well then we can simplify this a little bit. We can rewrite this as just c1 plus 2c2, all of that, times 2, 3, times our vector 2, 3. And this is just some arbitrary constant."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times 2 is 4, 2 times 3 is 6. Well then we can simplify this a little bit. We can rewrite this as just c1 plus 2c2, all of that, times 2, 3, times our vector 2, 3. And this is just some arbitrary constant. It's some arbitrary constant plus 2 times some other arbitrary constant. So we could just call this c3 times my vector 2, 3. In this situation, even though we started with two vectors, and I said, well, you know, the span of these two vectors is equal to all of the vectors that can be constructed with some linear combination of these, any linear combination of these, if I just use this substitution right here, can be reduced to just a scalar multiple of my first vector."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is just some arbitrary constant. It's some arbitrary constant plus 2 times some other arbitrary constant. So we could just call this c3 times my vector 2, 3. In this situation, even though we started with two vectors, and I said, well, you know, the span of these two vectors is equal to all of the vectors that can be constructed with some linear combination of these, any linear combination of these, if I just use this substitution right here, can be reduced to just a scalar multiple of my first vector. And I could have gone the other way around. I could have substituted this vector as being 1 half times this, and just made any combination a scalar multiple of the second vector. But the fact is, is that instead of talking about linear combinations of two vectors, I can reduce this to just a scalar combination of one vector."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In this situation, even though we started with two vectors, and I said, well, you know, the span of these two vectors is equal to all of the vectors that can be constructed with some linear combination of these, any linear combination of these, if I just use this substitution right here, can be reduced to just a scalar multiple of my first vector. And I could have gone the other way around. I could have substituted this vector as being 1 half times this, and just made any combination a scalar multiple of the second vector. But the fact is, is that instead of talking about linear combinations of two vectors, I can reduce this to just a scalar combination of one vector. And we've seen in R2 a scalar combination of one vector, especially if they're position vectors. For example, this vector 2, 3, it's 2, 3, it looks like this. All the scalar combinations of that vector are just going to lie along this line."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But the fact is, is that instead of talking about linear combinations of two vectors, I can reduce this to just a scalar combination of one vector. And we've seen in R2 a scalar combination of one vector, especially if they're position vectors. For example, this vector 2, 3, it's 2, 3, it looks like this. All the scalar combinations of that vector are just going to lie along this line. So 2, 3, it's going to be right there. They're all just going to lie along that line right there. So 5 and then 4, along this line, going in both directions forever."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All the scalar combinations of that vector are just going to lie along this line. So 2, 3, it's going to be right there. They're all just going to lie along that line right there. So 5 and then 4, along this line, going in both directions forever. And if I take a negative values of 2, 3, I'm going to go down here. If I take positive values, I'm going to go here. If I get really large positive values, I'm going to go up here."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 5 and then 4, along this line, going in both directions forever. And if I take a negative values of 2, 3, I'm going to go down here. If I take positive values, I'm going to go here. If I get really large positive values, I'm going to go up here. But I can just represent the vectors, and when you put them in standard form, their arrows essentially would trace out this line. So you could say that the span of my set of vectors, let me put it over here, the span of the vectors, let me do it this way, the span of the set of vectors 2, 3, and 4, 6, is just this line here. Even though we have two vectors, they're essentially collinear, they're multiples of each other."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I get really large positive values, I'm going to go up here. But I can just represent the vectors, and when you put them in standard form, their arrows essentially would trace out this line. So you could say that the span of my set of vectors, let me put it over here, the span of the vectors, let me do it this way, the span of the set of vectors 2, 3, and 4, 6, is just this line here. Even though we have two vectors, they're essentially collinear, they're multiples of each other. I mean, 4, 6 is, if this is 2, 3, 4, 6 is just this right here. 4, 6, it's just that longer one right there. They're collinear."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Even though we have two vectors, they're essentially collinear, they're multiples of each other. I mean, 4, 6 is, if this is 2, 3, 4, 6 is just this right here. 4, 6, it's just that longer one right there. They're collinear. These two things are collinear. Now, so in this case, when we have two collinear vectors in R2, they essentially, their span just reduces to that line. You can't represent some vector like, you can't represent, let me do a new color, you can't represent this vector right there with some combination of those two vectors."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're collinear. These two things are collinear. Now, so in this case, when we have two collinear vectors in R2, they essentially, their span just reduces to that line. You can't represent some vector like, you can't represent, let me do a new color, you can't represent this vector right there with some combination of those two vectors. There's no way to kind of break out of this line. So there's no way that you can represent everything in R2. So the span is just that line there."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can't represent some vector like, you can't represent, let me do a new color, you can't represent this vector right there with some combination of those two vectors. There's no way to kind of break out of this line. So there's no way that you can represent everything in R2. So the span is just that line there. Now, a related idea to this, notice, you had two vectors, but it kind of reduced to one vector when you took its linear combinations. The related idea here is that we call this set, we call it linearly dependent. Let me write that down."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the span is just that line there. Now, a related idea to this, notice, you had two vectors, but it kind of reduced to one vector when you took its linear combinations. The related idea here is that we call this set, we call it linearly dependent. Let me write that down. Linearly dependent. This is a linearly dependent set. And linearly dependent just means that one of the vectors in the set can be represented by some combination of the other vectors in the set."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. Linearly dependent. This is a linearly dependent set. And linearly dependent just means that one of the vectors in the set can be represented by some combination of the other vectors in the set. And a way to think about it is, whichever vector you pick that can be represented by the others, it's not adding any new directionality or any new information, right? In this case, we already had a vector that went in this direction. And when you throw this 4, 6 on there, you're going in the same direction, just scaled up."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And linearly dependent just means that one of the vectors in the set can be represented by some combination of the other vectors in the set. And a way to think about it is, whichever vector you pick that can be represented by the others, it's not adding any new directionality or any new information, right? In this case, we already had a vector that went in this direction. And when you throw this 4, 6 on there, you're going in the same direction, just scaled up. So it's not adding us, it's not giving us any new dimension, letting us break out of this line, right? And you can imagine in 3-space, if you have one vector that looks like this, and another vector that looks like this, two vectors that aren't collinear, and they're going to define a kind of a two-dimensional space, they can define a two-dimensional space, let's say that this is the plane defined by those two vectors. In order to define R3, a third vector in that set can't be coplanar with those two."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And when you throw this 4, 6 on there, you're going in the same direction, just scaled up. So it's not adding us, it's not giving us any new dimension, letting us break out of this line, right? And you can imagine in 3-space, if you have one vector that looks like this, and another vector that looks like this, two vectors that aren't collinear, and they're going to define a kind of a two-dimensional space, they can define a two-dimensional space, let's say that this is the plane defined by those two vectors. In order to define R3, a third vector in that set can't be coplanar with those two. If this third vector is coplanar with these, it's not adding any more directionality. So this set of three vectors will also be linearly dependent. And another way to think about it is that these two purple vectors span this plane, span the plane that they define, essentially, right?"}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In order to define R3, a third vector in that set can't be coplanar with those two. If this third vector is coplanar with these, it's not adding any more directionality. So this set of three vectors will also be linearly dependent. And another way to think about it is that these two purple vectors span this plane, span the plane that they define, essentially, right? Anything in this plane going in any direction can be, any vector in this plane, when we say span it, that means that any vector can be represented by a linear combination of this vector and this vector, which means that if this vector is on that plane, it can be represented as a linear combination of that vector and that vector. This green vector I added isn't going to add anything to the span of our set of vectors, and that's because this is a linearly dependent set. This one can be represented by a sum of that one and that one, because this one and this one span this plane."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And another way to think about it is that these two purple vectors span this plane, span the plane that they define, essentially, right? Anything in this plane going in any direction can be, any vector in this plane, when we say span it, that means that any vector can be represented by a linear combination of this vector and this vector, which means that if this vector is on that plane, it can be represented as a linear combination of that vector and that vector. This green vector I added isn't going to add anything to the span of our set of vectors, and that's because this is a linearly dependent set. This one can be represented by a sum of that one and that one, because this one and this one span this plane. In order for the span of these three vectors to kind of get more dimensionality or start representing R3, the third vector will have to break out of that plane. And if a vector is breaking out of that plane, that means it's a vector that can't be represented anywhere on that plane, so it's outside of the span of those two vectors. Or it's outside, it can't be represented by a linear combination of this one and this one."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This one can be represented by a sum of that one and that one, because this one and this one span this plane. In order for the span of these three vectors to kind of get more dimensionality or start representing R3, the third vector will have to break out of that plane. And if a vector is breaking out of that plane, that means it's a vector that can't be represented anywhere on that plane, so it's outside of the span of those two vectors. Or it's outside, it can't be represented by a linear combination of this one and this one. So if you had a vector of this one, this one, and this one, and just those three, none of these other things that I drew, that would be linearly independent. Let me draw a couple more examples for you. That one might have been a little too abstract."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or it's outside, it can't be represented by a linear combination of this one and this one. So if you had a vector of this one, this one, and this one, and just those three, none of these other things that I drew, that would be linearly independent. Let me draw a couple more examples for you. That one might have been a little too abstract. For example, if I had the vectors 2, 3, and I have the vector 7, 2, and I have the vector 9, 5, and I were to ask you, are these linearly dependent or independent? So at first you say, well, you know, it's not trivial. Let's see, this isn't a scalar multiple of that."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That one might have been a little too abstract. For example, if I had the vectors 2, 3, and I have the vector 7, 2, and I have the vector 9, 5, and I were to ask you, are these linearly dependent or independent? So at first you say, well, you know, it's not trivial. Let's see, this isn't a scalar multiple of that. That doesn't look like a scalar multiple of either of the other two. Maybe they're linearly independent. But then if you kind of inspect them, you kind of see that if we call this v1, vector 1 plus vector 2 is equal to vector 3."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, this isn't a scalar multiple of that. That doesn't look like a scalar multiple of either of the other two. Maybe they're linearly independent. But then if you kind of inspect them, you kind of see that if we call this v1, vector 1 plus vector 2 is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of 2 space, it's just a general idea that or let me see, let me draw it in R2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But then if you kind of inspect them, you kind of see that if we call this v1, vector 1 plus vector 2 is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of 2 space, it's just a general idea that or let me see, let me draw it in R2. It's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there, draw it in our standard position, and I draw the vector 7, 2, I could show you that any point in R2 can be represented by some linear combination of these two vectors."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if we were to show it, draw it in kind of 2 space, it's just a general idea that or let me see, let me draw it in R2. It's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there, draw it in our standard position, and I draw the vector 7, 2, I could show you that any point in R2 can be represented by some linear combination of these two vectors. We could even do kind of a graphical representation. But I've done that in the previous video. So I could write that the span of v1 and v2 is equal to R2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there, draw it in our standard position, and I draw the vector 7, 2, I could show you that any point in R2 can be represented by some linear combination of these two vectors. We could even do kind of a graphical representation. But I've done that in the previous video. So I could write that the span of v1 and v2 is equal to R2. That means that every vector, every position here can be represented by some linear combination of these two guys. Now, the vector 9, 5, it's right there, it is in R2. Clearly, I just graphed it on this plane."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could write that the span of v1 and v2 is equal to R2. That means that every vector, every position here can be represented by some linear combination of these two guys. Now, the vector 9, 5, it's right there, it is in R2. Clearly, I just graphed it on this plane. It's in our two-dimensional real number space, or I guess we could call it a space, or in our set R2, it's there. It's right there. So we just said that anything in R2 can be represented by a linear combination of those two guys."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Clearly, I just graphed it on this plane. It's in our two-dimensional real number space, or I guess we could call it a space, or in our set R2, it's there. It's right there. So we just said that anything in R2 can be represented by a linear combination of those two guys. So clearly, this is in R2, so it can be represented as a linear combination. So hopefully you're starting to see the relationship between span and linear independence or linear dependence. Let me do another example."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we just said that anything in R2 can be represented by a linear combination of those two guys. So clearly, this is in R2, so it can be represented as a linear combination. So hopefully you're starting to see the relationship between span and linear independence or linear dependence. Let me do another example. Let's say I have the vectors. Let me do a new color. Let's say I have the vector, and this will be a little bit obvious, 7, 0."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do another example. Let's say I have the vectors. Let me do a new color. Let's say I have the vector, and this will be a little bit obvious, 7, 0. So that's my v1. And then I have my second vector, which is 0, minus 1. That's v2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have the vector, and this will be a little bit obvious, 7, 0. So that's my v1. And then I have my second vector, which is 0, minus 1. That's v2. Now, is this set linearly independent? Well, can I represent either of these as a combination of the other? And really, when I say as a combination, you would have to scale up one to get the other because there's only two vectors here."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's v2. Now, is this set linearly independent? Well, can I represent either of these as a combination of the other? And really, when I say as a combination, you would have to scale up one to get the other because there's only two vectors here. Now, trying to add up to this vector, the only thing I have to deal with is this one, so all I can do is scale it up. Well, there's nothing I can do. No matter what I multiply this vector by, some constant, and add it to itself or scale it up, this term right here is always going to be 0."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And really, when I say as a combination, you would have to scale up one to get the other because there's only two vectors here. Now, trying to add up to this vector, the only thing I have to deal with is this one, so all I can do is scale it up. Well, there's nothing I can do. No matter what I multiply this vector by, some constant, and add it to itself or scale it up, this term right here is always going to be 0. So nothing I can multiply this by is going to get me to this vector. Likewise, no matter what I multiply this vector by, the top term is always going to be 0, so there's no way I could get to this vector. So both of these vectors, there's no way that you can represent one as a combination of the other."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "No matter what I multiply this vector by, some constant, and add it to itself or scale it up, this term right here is always going to be 0. So nothing I can multiply this by is going to get me to this vector. Likewise, no matter what I multiply this vector by, the top term is always going to be 0, so there's no way I could get to this vector. So both of these vectors, there's no way that you can represent one as a combination of the other. So these two are linearly independent. And you can even see it in, if we graph it, 1 is 7, 0, which is like that. Let me do it in a non-yellow color."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So both of these vectors, there's no way that you can represent one as a combination of the other. So these two are linearly independent. And you can even see it in, if we graph it, 1 is 7, 0, which is like that. Let me do it in a non-yellow color. 7, 0, and 1 is 0, minus 1. And I think you can clearly see that if you take a linear combination of any of these two, you can represent anything in R2. So the span of these, just to kind of get used to our notion of span, of V1 and V2 is equal to R2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in a non-yellow color. 7, 0, and 1 is 0, minus 1. And I think you can clearly see that if you take a linear combination of any of these two, you can represent anything in R2. So the span of these, just to kind of get used to our notion of span, of V1 and V2 is equal to R2. Now this is another interesting point to make. I said the span of V1 and V2 is R2. What is the span of V1, V2, and V3 in this example up here?"}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the span of these, just to kind of get used to our notion of span, of V1 and V2 is equal to R2. Now this is another interesting point to make. I said the span of V1 and V2 is R2. What is the span of V1, V2, and V3 in this example up here? I already told you, I already showed you that this third vector can be represented as a linear combination of these two. It's actually just these two summed up. I could even draw it right here."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is the span of V1, V2, and V3 in this example up here? I already told you, I already showed you that this third vector can be represented as a linear combination of these two. It's actually just these two summed up. I could even draw it right here. It's just those two vectors summed up. So it clearly can be represented as a linear combination of those two. So what's its span?"}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could even draw it right here. It's just those two vectors summed up. So it clearly can be represented as a linear combination of those two. So what's its span? Well, the fact that this is redundant means that it doesn't change its span. It doesn't change all of the possible linear combinations. So its span is also going to be R2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what's its span? Well, the fact that this is redundant means that it doesn't change its span. It doesn't change all of the possible linear combinations. So its span is also going to be R2. It's just that this was more vectors than you needed to span R2. R2 is a two-dimensional space, and you needed two vectors. So this was kind of a more efficient way of providing a basis."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So its span is also going to be R2. It's just that this was more vectors than you needed to span R2. R2 is a two-dimensional space, and you needed two vectors. So this was kind of a more efficient way of providing a basis. And I haven't defined basis formally yet, but I just want to use it a little conversationally, and then it'll make sense to you when I define it formally. This provides a better basis, or this provides a basis, kind of a non-redundant set of vectors that can represent R2, while this one right here is redundant. So it's not a good basis for R2."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this was kind of a more efficient way of providing a basis. And I haven't defined basis formally yet, but I just want to use it a little conversationally, and then it'll make sense to you when I define it formally. This provides a better basis, or this provides a basis, kind of a non-redundant set of vectors that can represent R2, while this one right here is redundant. So it's not a good basis for R2. Let me give you one more example in three dimensions, and then in the next video I'm going to make a more formal definition of linear dependence or independence. So let's say that I had the vector, let me say, 2, 0, 0. I'm going to make a similar argument that I made up there."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's not a good basis for R2. Let me give you one more example in three dimensions, and then in the next video I'm going to make a more formal definition of linear dependence or independence. So let's say that I had the vector, let me say, 2, 0, 0. I'm going to make a similar argument that I made up there. The vector 2, 0, 0, the vector 0, 1, 0, and the vector 0, 0, 7. Now, we're now in R3, right? Each of these are three-dimensional vectors."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to make a similar argument that I made up there. The vector 2, 0, 0, the vector 0, 1, 0, and the vector 0, 0, 7. Now, we're now in R3, right? Each of these are three-dimensional vectors. Now, are these linearly dependent or independent? Well, there's no way with some combination of these two vectors that I can end up with a non-zero term right here to make this third vector, right? Because no matter what I multiply this one by and this one by, this last term is going to be 0."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Each of these are three-dimensional vectors. Now, are these linearly dependent or independent? Well, there's no way with some combination of these two vectors that I can end up with a non-zero term right here to make this third vector, right? Because no matter what I multiply this one by and this one by, this last term is going to be 0. So this is kind of adding a new direction to our set of vectors. Likewise, there's no combination of this guy and this guy that I can get a non-zero term here. And finally, no combination of this guy and this guy that I can get a non-zero term here."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because no matter what I multiply this one by and this one by, this last term is going to be 0. So this is kind of adding a new direction to our set of vectors. Likewise, there's no combination of this guy and this guy that I can get a non-zero term here. And finally, no combination of this guy and this guy that I can get a non-zero term here. So this set is linearly independent. And if you were to graph these in three dimensions, you would see that none of these, these three do not lie on the same plane. Obviously, any two of them lie on the same plane."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And finally, no combination of this guy and this guy that I can get a non-zero term here. So this set is linearly independent. And if you were to graph these in three dimensions, you would see that none of these, these three do not lie on the same plane. Obviously, any two of them lie on the same plane. But if you were to actually graph it, you get 2, 0, let me say that that's the x-axis, that's 2, 0, 0. Then you have this 0, 1, 0, maybe that's the y-axis. And then you have 0, 0, 7."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Obviously, any two of them lie on the same plane. But if you were to actually graph it, you get 2, 0, let me say that that's the x-axis, that's 2, 0, 0. Then you have this 0, 1, 0, maybe that's the y-axis. And then you have 0, 0, 7. It'll look something like this. So it almost looks like your three-dimensional axes. It almost looks like the vectors i, j, k, they're just scaled up a little bit."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you have 0, 0, 7. It'll look something like this. So it almost looks like your three-dimensional axes. It almost looks like the vectors i, j, k, they're just scaled up a little bit. But you can always correct that by just scaling them down, right? Because we care about any linear combination of these. So the span of these three vectors right here, because they're all adding new directionality, is R3."}, {"video_title": "Introduction to linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It almost looks like the vectors i, j, k, they're just scaled up a little bit. But you can always correct that by just scaling them down, right? Because we care about any linear combination of these. So the span of these three vectors right here, because they're all adding new directionality, is R3. Anyway, I thought I would leave you there in this video. I realize I've been making longer and longer videos, and I want to get back in the habit of making shorter ones. In the next video, I'm going to make a more formal definition of linear dependence, and we'll do a bunch more examples."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've done this before. Let's call this first column V1. And let's call the second column V2. So we can rewrite it here. We could say V1 is equal to AC. And we could write that V2 is equal to BD. And these are both members of R2."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can rewrite it here. We could say V1 is equal to AC. And we could write that V2 is equal to BD. And these are both members of R2. And just to have a nice visualization in our head, let's graph these two. Let me draw my axes. That's my vertical axis."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And these are both members of R2. And just to have a nice visualization in our head, let's graph these two. Let me draw my axes. That's my vertical axis. That's my horizontal axis. And maybe V1 looks something like this. V1 might look something like that."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's my vertical axis. That's my horizontal axis. And maybe V1 looks something like this. V1 might look something like that. So that is V1. Its horizontal component will be A. Its vertical coordinate, if you view this as maybe a position vector or just how we're drawing it, is C. And then V2, let's just say it looks something like this."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "V1 might look something like that. So that is V1. Its horizontal component will be A. Its vertical coordinate, if you view this as maybe a position vector or just how we're drawing it, is C. And then V2, let's just say it looks something like this. Let's say that they're not the same vector. So V2 looks like that. Its horizontal coordinate is going to be B."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Its vertical coordinate, if you view this as maybe a position vector or just how we're drawing it, is C. And then V2, let's just say it looks something like this. Let's say that they're not the same vector. So V2 looks like that. Its horizontal coordinate is going to be B. And its vertical coordinate is going to be D. Now, what we're going to concern ourselves with in this video is the parallelogram generated by these two guys. And what I mean by that is imagine that these two guys are position vectors that are specifying points on a parallelogram. And then, of course, the origin is also another point on the parallelogram."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Its horizontal coordinate is going to be B. And its vertical coordinate is going to be D. Now, what we're going to concern ourselves with in this video is the parallelogram generated by these two guys. And what I mean by that is imagine that these two guys are position vectors that are specifying points on a parallelogram. And then, of course, the origin is also another point on the parallelogram. So what will be the last point on the parallelogram? Well, you can imagine. A parallelogram, we already have two sides of it."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then, of course, the origin is also another point on the parallelogram. So what will be the last point on the parallelogram? Well, you can imagine. A parallelogram, we already have two sides of it. So the other two sides have to be parallel. So one side will look like that. It's parallel to V1, the way I've drawn it."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A parallelogram, we already have two sides of it. So the other two sides have to be parallel. So one side will look like that. It's parallel to V1, the way I've drawn it. And then the other side will look like this. That's our parallelogram. The parallelogram generated by V2 and V1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's parallel to V1, the way I've drawn it. And then the other side will look like this. That's our parallelogram. The parallelogram generated by V2 and V1. What we're going to concern ourselves with specifically is the area of the parallelogram generated by V1 and V2. So how do we figure that out? So in general, if I have just any parallelogram, this is kind of a tilted one."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The parallelogram generated by V2 and V1. What we're going to concern ourselves with specifically is the area of the parallelogram generated by V1 and V2. So how do we figure that out? So in general, if I have just any parallelogram, this is kind of a tilted one. But if I just have any parallelogram, let me just draw any parallelogram right there. The area is just equal to the base. So that could be the base times the height."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So in general, if I have just any parallelogram, this is kind of a tilted one. But if I just have any parallelogram, let me just draw any parallelogram right there. The area is just equal to the base. So that could be the base times the height. So it's equal to base. I'll write capital B, since we have a lowercase b there. Base times height."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that could be the base times the height. So it's equal to base. I'll write capital B, since we have a lowercase b there. Base times height. That's what the area of a parallelogram would be. Now, what are the base and the height in this situation? So let me write this down."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Base times height. That's what the area of a parallelogram would be. Now, what are the base and the height in this situation? So let me write this down. The area of our parallelogram is equal to the base times the height, and actually, well, let me just write it here. So what is the base here? The base here is going to be the length of our vector V. So this is our base."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me write this down. The area of our parallelogram is equal to the base times the height, and actually, well, let me just write it here. So what is the base here? The base here is going to be the length of our vector V. So this is our base. So this right here is going to be the length of vector V1, the length of this orange vector right here. And what's the height of this parallelogram going to be? So we could drop a perpendicular here, and the length of this line right here is going to be our height."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The base here is going to be the length of our vector V. So this is our base. So this right here is going to be the length of vector V1, the length of this orange vector right here. And what's the height of this parallelogram going to be? So we could drop a perpendicular here, and the length of this line right here is going to be our height. So how can we figure out the, you know, we know what V1 is, so we can figure out the base pretty easily, but how can we figure out the height? Well, one thing we can do is if we can figure out this guy right here, if we can figure out this guy, we could use a Pythagorean theorem, because the length of this vector squared plus h squared is going to be equal to the length of V2 squared. So let's see if we can do that."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could drop a perpendicular here, and the length of this line right here is going to be our height. So how can we figure out the, you know, we know what V1 is, so we can figure out the base pretty easily, but how can we figure out the height? Well, one thing we can do is if we can figure out this guy right here, if we can figure out this guy, we could use a Pythagorean theorem, because the length of this vector squared plus h squared is going to be equal to the length of V2 squared. So let's see if we can do that. What is this guy? What is this green guy right here? Well, if you imagine a line, let's imagine some line L. So let's say L is a line spanned by V1, which means you take all of the multiples of V1 and all of the positions that they specify will create a set of points, and that is my line L. So you take all the multiples of V1, you're going to get every point along this line, along a line that's like that."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can do that. What is this guy? What is this green guy right here? Well, if you imagine a line, let's imagine some line L. So let's say L is a line spanned by V1, which means you take all of the multiples of V1 and all of the positions that they specify will create a set of points, and that is my line L. So you take all the multiples of V1, you're going to get every point along this line, along a line that's like that. Now, if we have L defined that way, that line right there is L, I don't know if you can see it, let me do it a little bit better, and it goes through V1 and it just keeps going over there. What is this green line right there? This green line that we're concerned with, that's the projection onto L of what?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, if you imagine a line, let's imagine some line L. So let's say L is a line spanned by V1, which means you take all of the multiples of V1 and all of the positions that they specify will create a set of points, and that is my line L. So you take all the multiples of V1, you're going to get every point along this line, along a line that's like that. Now, if we have L defined that way, that line right there is L, I don't know if you can see it, let me do it a little bit better, and it goes through V1 and it just keeps going over there. What is this green line right there? This green line that we're concerned with, that's the projection onto L of what? Well, we have a perpendicular here, you can imagine the light source coming down, I don't know if that analogy helps you, but it's kind of the shadow of V2 onto that line. So it's a projection of V2, of your vector, V2 onto L is this green line right there. So if we want to figure out H, we can just use the Pythagorean theorem."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This green line that we're concerned with, that's the projection onto L of what? Well, we have a perpendicular here, you can imagine the light source coming down, I don't know if that analogy helps you, but it's kind of the shadow of V2 onto that line. So it's a projection of V2, of your vector, V2 onto L is this green line right there. So if we want to figure out H, we can just use the Pythagorean theorem. So we can say that the length of H squared, or we could just write, well, I'm just writing H as the length, I'm not even specifying as a vector. So we could say that H squared, where that is the length of this line, plus the length of this vector squared, and the length of that vector squared is the length of the projection onto L of V2. I'll do that in a different color."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to figure out H, we can just use the Pythagorean theorem. So we can say that the length of H squared, or we could just write, well, I'm just writing H as the length, I'm not even specifying as a vector. So we could say that H squared, where that is the length of this line, plus the length of this vector squared, and the length of that vector squared is the length of the projection onto L of V2. I'll do that in a different color. So the length of the projection onto L of V2 squared, we're just doing the Pythagorean theorem. This squared plus this squared is going to equal that squared. It's going to be equal to the length of V2 squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do that in a different color. So the length of the projection onto L of V2 squared, we're just doing the Pythagorean theorem. This squared plus this squared is going to equal that squared. It's going to be equal to the length of V2 squared. That's just Pythagorean theorem. Nothing fancy there. So how can we simplify?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be equal to the length of V2 squared. That's just Pythagorean theorem. Nothing fancy there. So how can we simplify? We want to figure out, we want to solve for H. And actually, let's just solve for H squared for now, because it'll keep things a little bit simpler. So we can say that H squared is equal to this guy. It's equal to the length of my vector V2 squared minus the length of the projection squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So how can we simplify? We want to figure out, we want to solve for H. And actually, let's just solve for H squared for now, because it'll keep things a little bit simpler. So we can say that H squared is equal to this guy. It's equal to the length of my vector V2 squared minus the length of the projection squared. So minus, I'll do that in purple, minus the length of the projection onto L of V2 squared. Remember, this thing is just this thing right here. We're just doing the Pythagorean theorem."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to the length of my vector V2 squared minus the length of the projection squared. So minus, I'll do that in purple, minus the length of the projection onto L of V2 squared. Remember, this thing is just this thing right here. We're just doing the Pythagorean theorem. So let's see if we can simplify this or write it in terms that we understand. So the length of a vector squared, this is just equal to V2 dot V2. You take a vector, you dot it with itself, and you get the length of that vector squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're just doing the Pythagorean theorem. So let's see if we can simplify this or write it in terms that we understand. So the length of a vector squared, this is just equal to V2 dot V2. You take a vector, you dot it with itself, and you get the length of that vector squared. We saw that many, many videos ago. And then what is this guy going to be equal to? Well, the projection, I'll do it over here."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You take a vector, you dot it with itself, and you get the length of that vector squared. We saw that many, many videos ago. And then what is this guy going to be equal to? Well, the projection, I'll do it over here. The projection onto L of V2 is going to be equal to V2 dot the spanning vector, which is V1. So it's V2 dot V1 over the spanning vector dotted with itself. V1 dot V1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, the projection, I'll do it over here. The projection onto L of V2 is going to be equal to V2 dot the spanning vector, which is V1. So it's V2 dot V1 over the spanning vector dotted with itself. V1 dot V1. We saw this several videos ago when we learned about projections. And this is just a number. This is just a number right there."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "V1 dot V1. We saw this several videos ago when we learned about projections. And this is just a number. This is just a number right there. And then it's going to be times the spanning vector itself, so times V1. That is what the projection is. So it's going to be this minus the projection, the length of the projection squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is just a number right there. And then it's going to be times the spanning vector itself, so times V1. That is what the projection is. So it's going to be this minus the projection, the length of the projection squared. So what is the length of the projection squared? Well, that's this guy dotted with himself. Let me write it this way."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be this minus the projection, the length of the projection squared. So what is the length of the projection squared? Well, that's this guy dotted with himself. Let me write it this way. Let me take it step by step. So this is going to be minus, I want to make sure I can still see that up there so I don't have to rewrite it, minus the projection is going to be, it looks a little complicated, but hopefully things will simplify, V2 dot V1 over V1 dot V1 times, switch colors, times the vector, this is all just going to end up being a number. Remember, you take dot products, you get numbers, times the vector V1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it this way. Let me take it step by step. So this is going to be minus, I want to make sure I can still see that up there so I don't have to rewrite it, minus the projection is going to be, it looks a little complicated, but hopefully things will simplify, V2 dot V1 over V1 dot V1 times, switch colors, times the vector, this is all just going to end up being a number. Remember, you take dot products, you get numbers, times the vector V1. And we're going to take the length of that whole thing squared, and all of this is going to be equal to h squared. Once again, just the Pythagorean theorem. This squared, which is the height squared, is equal to your hypotenuse squared, this is your hypotenuse squared, minus the other side squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, you take dot products, you get numbers, times the vector V1. And we're going to take the length of that whole thing squared, and all of this is going to be equal to h squared. Once again, just the Pythagorean theorem. This squared, which is the height squared, is equal to your hypotenuse squared, this is your hypotenuse squared, minus the other side squared. This is the other side squared. It looks a little complicated, but it's just the projection of this guy onto that right there. So let's see if we can simplify this a little bit."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This squared, which is the height squared, is equal to your hypotenuse squared, this is your hypotenuse squared, minus the other side squared. This is the other side squared. It looks a little complicated, but it's just the projection of this guy onto that right there. So let's see if we can simplify this a little bit. We're just going to have to break out some algebra, or let's see what we can do here. Well, actually not algebra, some linear algebra. So what is this guy?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can simplify this a little bit. We're just going to have to break out some algebra, or let's see what we can do here. Well, actually not algebra, some linear algebra. So what is this guy? Well, this guy is just the dot product of this with itself. So this is just equal to, we know, I mean, any vector, if you take the square of its length, it's just that vector dotted with itself. So this thing, if we're taking the square of this guy's length, is just equal to this guy dotted with himself."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is this guy? Well, this guy is just the dot product of this with itself. So this is just equal to, we know, I mean, any vector, if you take the square of its length, it's just that vector dotted with itself. So this thing, if we're taking the square of this guy's length, is just equal to this guy dotted with himself. So let me write everything over again. So we get h squared is equal to v2 dot v2, and then minus this guy dotted with himself. So minus v2 dot v1 over v1 dot v1 times the vector v1 dotted with itself, dotted with v2 dot v1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this thing, if we're taking the square of this guy's length, is just equal to this guy dotted with himself. So let me write everything over again. So we get h squared is equal to v2 dot v2, and then minus this guy dotted with himself. So minus v2 dot v1 over v1 dot v1 times the vector v1 dotted with itself, dotted with v2 dot v1. Remember, this green part is just a number. Over v1 dot v1 times v1. And what is this equal to?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So minus v2 dot v1 over v1 dot v1 times the vector v1 dotted with itself, dotted with v2 dot v1. Remember, this green part is just a number. Over v1 dot v1 times v1. And what is this equal to? Let's just simplify this. These are just scalar quantities, and we saw that the dot product is associative with respect to scalar quantities, so we can just change the order here. So this is going to be equal to the scalar quantity times itself."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? Let's just simplify this. These are just scalar quantities, and we saw that the dot product is associative with respect to scalar quantities, so we can just change the order here. So this is going to be equal to the scalar quantity times itself. So we could say this is equal to v2 dot v1. Let me write it this way. v2 dot v1, that's going to be, and we're going to multiply the numerator times itself, v2 dot v1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to the scalar quantity times itself. So we could say this is equal to v2 dot v1. Let me write it this way. v2 dot v1, that's going to be, and we're going to multiply the numerator times itself, v2 dot v1. And then all of that over v1 dot v1 times v1 dot v1. Remember, I'm just taking these two terms and multiplying them by each other. I'm just switching the order, and then we know that the scalars can kind of be taken out, times these two guys dot each other, times v1 dot v1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "v2 dot v1, that's going to be, and we're going to multiply the numerator times itself, v2 dot v1. And then all of that over v1 dot v1 times v1 dot v1. Remember, I'm just taking these two terms and multiplying them by each other. I'm just switching the order, and then we know that the scalars can kind of be taken out, times these two guys dot each other, times v1 dot v1. That's what this simplifies to. Now, this is now a number. We had vectors here, but when you take the dot product, you just get a number."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just switching the order, and then we know that the scalars can kind of be taken out, times these two guys dot each other, times v1 dot v1. That's what this simplifies to. Now, this is now a number. We had vectors here, but when you take the dot product, you just get a number. And this number is the same as this number. So we can simplify it a little bit, so we can cross those two guys out, and then we are left with that our height squared is equal to this expression times itself. We have it times itself twice, so it's equal to v2 dot v2 minus this guy times itself, so v2 dot v1 squared, all of that over just one of these guys, v1 dot v1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We had vectors here, but when you take the dot product, you just get a number. And this number is the same as this number. So we can simplify it a little bit, so we can cross those two guys out, and then we are left with that our height squared is equal to this expression times itself. We have it times itself twice, so it's equal to v2 dot v2 minus this guy times itself, so v2 dot v1 squared, all of that over just one of these guys, v1 dot v1. That is what the height squared is. Now, we have the height squared. We could take the square root if we just want to solve for the height, but to keep our math simple, we know that area is equal to base times height."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have it times itself twice, so it's equal to v2 dot v2 minus this guy times itself, so v2 dot v1 squared, all of that over just one of these guys, v1 dot v1. That is what the height squared is. Now, we have the height squared. We could take the square root if we just want to solve for the height, but to keep our math simple, we know that area is equal to base times height. Let's just say what the area squared is equal to, because then both of these terms will get squared. So we know that if the area is equal to base times height, we saw that at the beginning of the video, then the area squared is going to be equal to these two guys squared. It's going to be equal to base squared times height squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could take the square root if we just want to solve for the height, but to keep our math simple, we know that area is equal to base times height. Let's just say what the area squared is equal to, because then both of these terms will get squared. So we know that if the area is equal to base times height, we saw that at the beginning of the video, then the area squared is going to be equal to these two guys squared. It's going to be equal to base squared times height squared. Now, what is the base squared? Let me do it like this. So the base squared, we already saw, the base of our parallelogram is the length, the base of our entire parallelogram, remember this is the parallelogram in question."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be equal to base squared times height squared. Now, what is the base squared? Let me do it like this. So the base squared, we already saw, the base of our parallelogram is the length, the base of our entire parallelogram, remember this is the parallelogram in question. The base is the length of vector v1. Now, what is the base squared? The base squared is going to be the length of vector v1 squared, or another way of writing that is v1 dot v1."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the base squared, we already saw, the base of our parallelogram is the length, the base of our entire parallelogram, remember this is the parallelogram in question. The base is the length of vector v1. Now, what is the base squared? The base squared is going to be the length of vector v1 squared, or another way of writing that is v1 dot v1. And we already know what the height squared is. It's this expression right there. Let me write that down."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The base squared is going to be the length of vector v1 squared, or another way of writing that is v1 dot v1. And we already know what the height squared is. It's this expression right there. Let me write that down. The height squared is the height squared right there. So what is our area squared going to be? Our area squared, let me go down here while I have more space, our area squared is going to be equal to our base squared, which is v1 dot v1 times our height squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. The height squared is the height squared right there. So what is our area squared going to be? Our area squared, let me go down here while I have more space, our area squared is going to be equal to our base squared, which is v1 dot v1 times our height squared. Let me write it like this. Oh, I was using the wrong color. Times this guy over here."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our area squared, let me go down here while I have more space, our area squared is going to be equal to our base squared, which is v1 dot v1 times our height squared. Let me write it like this. Oh, I was using the wrong color. Times this guy over here. v2 dot v2 minus v2 dot v1 squared over v1 dot v1. Now, what does this simplify to? Well, this is just a number."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times this guy over here. v2 dot v2 minus v2 dot v1 squared over v1 dot v1. Now, what does this simplify to? Well, this is just a number. These are all just numbers. So if I multiply, if I distribute this out, this is equal to what? This times this is equal to v1, let me color code it, v1 dot v1 times this guy times v2 dot v2."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is just a number. These are all just numbers. So if I multiply, if I distribute this out, this is equal to what? This times this is equal to v1, let me color code it, v1 dot v1 times this guy times v2 dot v2. And then when I multiply this guy times that guy, what happens? Well, I have this guy in the numerator and that guy in the denominator, so they cancel out. So I'm just left with minus v2 dot v1 squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This times this is equal to v1, let me color code it, v1 dot v1 times this guy times v2 dot v2. And then when I multiply this guy times that guy, what happens? Well, I have this guy in the numerator and that guy in the denominator, so they cancel out. So I'm just left with minus v2 dot v1 squared. Now, let's remind ourselves what these two vectors were. v1 was the vector ac and v2 was the vector bd. Let me rewrite it down here so we have it to work with."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just left with minus v2 dot v1 squared. Now, let's remind ourselves what these two vectors were. v1 was the vector ac and v2 was the vector bd. Let me rewrite it down here so we have it to work with. So v1 was equal to the vector ac and v2 is equal to the vector bd. So what is v1 dot v1? What is v1 dot v1?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me rewrite it down here so we have it to work with. So v1 was equal to the vector ac and v2 is equal to the vector bd. So what is v1 dot v1? What is v1 dot v1? That is equal to a dot a, which is, or a times a, a squared plus c squared. That's this right there. And that's what's v2 dot v2."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is v1 dot v1? That is equal to a dot a, which is, or a times a, a squared plus c squared. That's this right there. And that's what's v2 dot v2. So we're going to have that times v2 dot v2. v2 dot v2 is v squared plus d squared. And then we're going to have minus v2 dot v1 squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that's what's v2 dot v2. So we're going to have that times v2 dot v2. v2 dot v2 is v squared plus d squared. And then we're going to have minus v2 dot v1 squared. So what's v2 dot v1? It is, well, it's b times a plus d times c, or a times b plus, we're just dotting these two guys. So it's ab plus cd."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to have minus v2 dot v1 squared. So what's v2 dot v1? It is, well, it's b times a plus d times c, or a times b plus, we're just dotting these two guys. So it's ab plus cd. And then we're squaring it. Let's see what this simplifies to. Hopefully it simplifies to something."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's ab plus cd. And then we're squaring it. Let's see what this simplifies to. Hopefully it simplifies to something. Let me switch colors. So if we just multiply this out, let me write it here. Our area squared is equal to a squared times b squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Hopefully it simplifies to something. Let me switch colors. So if we just multiply this out, let me write it here. Our area squared is equal to a squared times b squared. a squared b squared. a squared times d squared. Plus a squared times d squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our area squared is equal to a squared times b squared. a squared b squared. a squared times d squared. Plus a squared times d squared. Plus c squared times b squared. Plus c squared times b squared. Plus c squared times d squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Plus a squared times d squared. Plus c squared times b squared. Plus c squared times b squared. Plus c squared times d squared. Plus c squared times d squared. And then minus this guy squared. What is that going to be equal to?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Plus c squared times d squared. Plus c squared times d squared. And then minus this guy squared. What is that going to be equal to? ab squared is a squared b squared. And then I'm going to multiply these guys times each other twice. So it's going to be plus 2abcd."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is that going to be equal to? ab squared is a squared b squared. And then I'm going to multiply these guys times each other twice. So it's going to be plus 2abcd. And then you're going to have a plus c squared d squared. I just foiled this out. That's the best way you could kind of think about it."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be plus 2abcd. And then you're going to have a plus c squared d squared. I just foiled this out. That's the best way you could kind of think about it. And then if I distribute this negative sign, what do I have? Let me rewrite everything. It's equal to a squared b squared plus a squared d squared plus c squared b squared plus c squared d squared minus a squared b squared minus 2abcd minus c squared d squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the best way you could kind of think about it. And then if I distribute this negative sign, what do I have? Let me rewrite everything. It's equal to a squared b squared plus a squared d squared plus c squared b squared plus c squared d squared minus a squared b squared minus 2abcd minus c squared d squared. All I did is I distributed the minus sign. Now it looks like some things will simplify nicely. And remember, all of this was equal to our area squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to a squared b squared plus a squared d squared plus c squared b squared plus c squared d squared minus a squared b squared minus 2abcd minus c squared d squared. All I did is I distributed the minus sign. Now it looks like some things will simplify nicely. And remember, all of this was equal to our area squared. We have an ab squared. We have a minus ab squared. They cancel out."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And remember, all of this was equal to our area squared. We have an ab squared. We have a minus ab squared. They cancel out. We have a minus cd squared and a cd squared. So they cancel out. So all we're left with is that the area of our parallelogram squared is equal to a squared d squared minus 2abcd plus c squared plus c squared b squared."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They cancel out. We have a minus cd squared and a cd squared. So they cancel out. So all we're left with is that the area of our parallelogram squared is equal to a squared d squared minus 2abcd plus c squared plus c squared b squared. Now this might look a little bit bizarre to you. Well, if you made a substitution right here, if you said that x is equal to ad and if you said y is equal to cb, then what does this become? This is equal to x squared minus 2 times xy plus y squared, right?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So all we're left with is that the area of our parallelogram squared is equal to a squared d squared minus 2abcd plus c squared plus c squared b squared. Now this might look a little bit bizarre to you. Well, if you made a substitution right here, if you said that x is equal to ad and if you said y is equal to cb, then what does this become? This is equal to x squared minus 2 times xy plus y squared, right? Hopefully you recognize this. And this is just the same thing as x minus y squared. So this was our substitutions we made."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to x squared minus 2 times xy plus y squared, right? Hopefully you recognize this. And this is just the same thing as x minus y squared. So this was our substitutions we made. I did this just so you can visualize this a little bit better. So we have our area squared is equal to x minus y squared or ad minus cb or let me write it bc squared. That's what the area of our parallelogram squared is."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this was our substitutions we made. I did this just so you can visualize this a little bit better. So we have our area squared is equal to x minus y squared or ad minus cb or let me write it bc squared. That's what the area of our parallelogram squared is. And if you don't quite understand what I did here, I just made these substitutions so you can recognize it better, but just to understand that this is the same thing as this, if you want, you can just multiply this guy out and you'll get that right there. But what is this? What is this thing right here?"}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what the area of our parallelogram squared is. And if you don't quite understand what I did here, I just made these substitutions so you can recognize it better, but just to understand that this is the same thing as this, if you want, you can just multiply this guy out and you'll get that right there. But what is this? What is this thing right here? It's the determinant. This is the determinant of our original matrix up here. Well, I call that matrix A and then I used A again for area, so let me write it this way."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is this thing right here? It's the determinant. This is the determinant of our original matrix up here. Well, I call that matrix A and then I used A again for area, so let me write it this way. Area squared, let me write it like this. Area squared is equal to ad minus bc squared. So this is area."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, I call that matrix A and then I used A again for area, so let me write it this way. Area squared, let me write it like this. Area squared is equal to ad minus bc squared. So this is area. These a's are all area. But what is this? This is the determinant of my matrix."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is area. These a's are all area. But what is this? This is the determinant of my matrix. That is the determinant of my matrix A, my original matrix that I started the problem with, which is equal to the determinant of ABCD. The determinant of this is ad minus bc by definition. So your area, this is exciting, the area squared, so the area of your parallelogram squared, is equal to the determinant of the matrix whose column vectors construct that parallelogram."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the determinant of my matrix. That is the determinant of my matrix A, my original matrix that I started the problem with, which is equal to the determinant of ABCD. The determinant of this is ad minus bc by definition. So your area, this is exciting, the area squared, so the area of your parallelogram squared, is equal to the determinant of the matrix whose column vectors construct that parallelogram. Is equal to the determinant of your matrix squared. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. Which is a pretty neat outcome, especially considering how much hairy algebra we had to go through."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So your area, this is exciting, the area squared, so the area of your parallelogram squared, is equal to the determinant of the matrix whose column vectors construct that parallelogram. Is equal to the determinant of your matrix squared. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. Which is a pretty neat outcome, especially considering how much hairy algebra we had to go through. Let's go back all the way over here. Go back to the drawing. So if we want to figure out the area of this parallelogram right here that is defined or that is created by the two column vectors of a matrix, we literally just have to find the determinant of the matrix."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which is a pretty neat outcome, especially considering how much hairy algebra we had to go through. Let's go back all the way over here. Go back to the drawing. So if we want to figure out the area of this parallelogram right here that is defined or that is created by the two column vectors of a matrix, we literally just have to find the determinant of the matrix. The area of this is equal to the absolute value of the determinant of A. And you have to do that because this might be negative. You could imagine if you kind of swapped these guys around, if you swap some of the rows, this guy would be negative, but you can't have a negative area."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to figure out the area of this parallelogram right here that is defined or that is created by the two column vectors of a matrix, we literally just have to find the determinant of the matrix. The area of this is equal to the absolute value of the determinant of A. And you have to do that because this might be negative. You could imagine if you kind of swapped these guys around, if you swap some of the rows, this guy would be negative, but you can't have a negative area. And it wouldn't really change the definition, it really wouldn't change what's spanned. If you switch v1 and v2, you're still spanning the same parallelogram, you just might get the negative of the determinant. But that is a really neat outcome."}, {"video_title": "Determinant and area of a parallelogram Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could imagine if you kind of swapped these guys around, if you swap some of the rows, this guy would be negative, but you can't have a negative area. And it wouldn't really change the definition, it really wouldn't change what's spanned. If you switch v1 and v2, you're still spanning the same parallelogram, you just might get the negative of the determinant. But that is a really neat outcome. And when you first learn determinants in school, I mean, we learned the first motivation for a determinant was this idea of, well, when we take the inverse of a 2 by 2, this thing shows up in the denominator. We call that the determinant. But now there's this other interpretation here."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And I didn't speak of it much in terms of transformations, but it was a transformation. We defined a projection onto that line L as a transformation. In the video, we drew it as transformations within R2, but it could be in general a transformation from Rn to Rn. And we defined it as the projection of x onto L was equal to the dot product of x with this defining vector, x dot this defining vector, divided by that defining vector dotted with itself. That defining vector dotted with itself. All of that times the defining vector of the line. This was our definition."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And we defined it as the projection of x onto L was equal to the dot product of x with this defining vector, x dot this defining vector, divided by that defining vector dotted with itself. That defining vector dotted with itself. All of that times the defining vector of the line. This was our definition. Now a couple of things might have popped out at you right when we first saw this. This, when you dot a vector with itself, what's that equal to? We know that if I take some vector and I dot it with itself, that is equivalent to the length of the vector squared."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This was our definition. Now a couple of things might have popped out at you right when we first saw this. This, when you dot a vector with itself, what's that equal to? We know that if I take some vector and I dot it with itself, that is equivalent to the length of the vector squared. So we could rewrite this as being equal to x dot v over the length of v squared, all of that times v. Now, wouldn't it be nice if the length of v was 1? If the length of v was equal to 1. Because then if the length of v was 1, or this is another way of saying that v is a unit vector, then our formula for our projection would just simplify to x dot v, all of that times, this will just be some scalar number, that times v. You're saying, hey Sal, how do we know if this is a unit vector or not?"}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "We know that if I take some vector and I dot it with itself, that is equivalent to the length of the vector squared. So we could rewrite this as being equal to x dot v over the length of v squared, all of that times v. Now, wouldn't it be nice if the length of v was 1? If the length of v was equal to 1. Because then if the length of v was 1, or this is another way of saying that v is a unit vector, then our formula for our projection would just simplify to x dot v, all of that times, this will just be some scalar number, that times v. You're saying, hey Sal, how do we know if this is a unit vector or not? And what you can realize is that any, let me draw it this way, so when I drew it in the previous video, I just picked a line like that. And the line can be really defined, this vector v in the line, can be any of the vectors that's contained in the line. So the vector v could be like that."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Because then if the length of v was 1, or this is another way of saying that v is a unit vector, then our formula for our projection would just simplify to x dot v, all of that times, this will just be some scalar number, that times v. You're saying, hey Sal, how do we know if this is a unit vector or not? And what you can realize is that any, let me draw it this way, so when I drew it in the previous video, I just picked a line like that. And the line can be really defined, this vector v in the line, can be any of the vectors that's contained in the line. So the vector v could be like that. So let's say someone gives you a vector v that isn't a unit vector. So let's say that the length of v is not equal to 1. How can you define a line using some unit vector?"}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So the vector v could be like that. So let's say someone gives you a vector v that isn't a unit vector. So let's say that the length of v is not equal to 1. How can you define a line using some unit vector? Well, you can just normalize v. So you can define some unit vector right here. You could define some vector right there. Let's call it u, and I'll say it's a unit vector."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "How can you define a line using some unit vector? Well, you can just normalize v. So you can define some unit vector right here. You could define some vector right there. Let's call it u, and I'll say it's a unit vector. And let's just say that that is equal to 1 over the length of v times v. I showed you this in the unit vector video. You can construct a unit vector that goes in the same direction as any vector, essentially just by dividing, or I guess multiplying, that vector times 1 over its length. So in general, we can just always redefine the line."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Let's call it u, and I'll say it's a unit vector. And let's just say that that is equal to 1 over the length of v times v. I showed you this in the unit vector video. You can construct a unit vector that goes in the same direction as any vector, essentially just by dividing, or I guess multiplying, that vector times 1 over its length. So in general, we can just always redefine the line. All of the possible scalar multiples of v are going to be the same thing as all of the scalar multiples of our unit vector, u, which is just a scalar multiple of v. So we can redefine our line. If we redefine our line, L, as being equal to all of the possible scalar multiples of our unit vector, where the scalars are any members of the real numbers, then our projection definition simplifies a good bit. The projection of x onto L then just becomes x dot our unit vector times the unit vector times the unit vector itself."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So in general, we can just always redefine the line. All of the possible scalar multiples of v are going to be the same thing as all of the scalar multiples of our unit vector, u, which is just a scalar multiple of v. So we can redefine our line. If we redefine our line, L, as being equal to all of the possible scalar multiples of our unit vector, where the scalars are any members of the real numbers, then our projection definition simplifies a good bit. The projection of x onto L then just becomes x dot our unit vector times the unit vector times the unit vector itself. And so you can imagine a world. I mean, that case that I did in the previous video, where I had those two vectors, where I said the vector v that defined the line. I think it was the vector 2, 2, 1, and our vector x was equal to 3, I think it was equal to 2, 3."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "The projection of x onto L then just becomes x dot our unit vector times the unit vector times the unit vector itself. And so you can imagine a world. I mean, that case that I did in the previous video, where I had those two vectors, where I said the vector v that defined the line. I think it was the vector 2, 2, 1, and our vector x was equal to 3, I think it was equal to 2, 3. If you want to do this definition, we just have to turn this guy into a unit vector first. And the way you turn them into a unit vector is you just figure out the magnitude. So in this case, the magnitude of v is equal to what?"}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "I think it was the vector 2, 2, 1, and our vector x was equal to 3, I think it was equal to 2, 3. If you want to do this definition, we just have to turn this guy into a unit vector first. And the way you turn them into a unit vector is you just figure out the magnitude. So in this case, the magnitude of v is equal to what? 2 squared plus 1 squared is 1, and you take the square root of that. Let me just write this is equal to the square root of 2 squared plus 1 squared, which is equal to the square root of 5. And so you can define your u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, the magnitude of v is equal to what? 2 squared plus 1 squared is 1, and you take the square root of that. Let me just write this is equal to the square root of 2 squared plus 1 squared, which is equal to the square root of 5. And so you can define your u. Your unit vector could just be 1 over this times that guy. So it's 1 over the square root of 5 times 2, 1. And you can multiply it out or not."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And so you can define your u. Your unit vector could just be 1 over this times that guy. So it's 1 over the square root of 5 times 2, 1. And you can multiply it out or not. You could just leave it in this form. But you can always, for any vector v, you can always find a unit vector that goes in the same direction, assuming that we're dealing with non-zero vectors. So you can always reduce anything like this to some other definition like this, where this is the unit vector version of your vector v up there."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And you can multiply it out or not. You could just leave it in this form. But you can always, for any vector v, you can always find a unit vector that goes in the same direction, assuming that we're dealing with non-zero vectors. So you can always reduce anything like this to some other definition like this, where this is the unit vector version of your vector v up there. Now, I just said that, look, this is a transformation from Rn to Rn. The one thing that we're not sure of just yet is, is this a linear transformation? And I said we can always write it like this."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So you can always reduce anything like this to some other definition like this, where this is the unit vector version of your vector v up there. Now, I just said that, look, this is a transformation from Rn to Rn. The one thing that we're not sure of just yet is, is this a linear transformation? And I said we can always write it like this. So let's see if this is always going to be a linear transformation. So there's two conditions for it to be a linear transformation. The first is that the transformation, so let's see what happens if I take the projection onto L of two vectors, let's say the vector a plus the vector v. If I take the sum of their vectors, if this is a linear transformation, this should be equivalent to taking each of their projections individually and then summing."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And I said we can always write it like this. So let's see if this is always going to be a linear transformation. So there's two conditions for it to be a linear transformation. The first is that the transformation, so let's see what happens if I take the projection onto L of two vectors, let's say the vector a plus the vector v. If I take the sum of their vectors, if this is a linear transformation, this should be equivalent to taking each of their projections individually and then summing. Let's see if this is the case. So this is equal to, by our definition, we'll use a unit vector version because this is simpler. This is equal to a plus b, that's our x, dot u, dot u, and then all of that times our unit vector."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "The first is that the transformation, so let's see what happens if I take the projection onto L of two vectors, let's say the vector a plus the vector v. If I take the sum of their vectors, if this is a linear transformation, this should be equivalent to taking each of their projections individually and then summing. Let's see if this is the case. So this is equal to, by our definition, we'll use a unit vector version because this is simpler. This is equal to a plus b, that's our x, dot u, dot u, and then all of that times our unit vector. Now we know that the dot product has a distributive property. So then this is equal to a dot u plus b dot u. These are unit vectors."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to a plus b, that's our x, dot u, dot u, and then all of that times our unit vector. Now we know that the dot product has a distributive property. So then this is equal to a dot u plus b dot u. These are unit vectors. All of that times the vector u. These are just scalar numbers. So scalar multiplication has the distributive property."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "These are unit vectors. All of that times the vector u. These are just scalar numbers. So scalar multiplication has the distributive property. So this is equal to a dot u times our vector u. Remember, this is just going to be some scalar. Plus b dot u times our unit vector u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So scalar multiplication has the distributive property. So this is equal to a dot u times our vector u. Remember, this is just going to be some scalar. Plus b dot u times our unit vector u. What is this equal to? Well, this right here is equal to the projection of a. This is equal to the projection of a onto l by definition, right here, by this definition."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Plus b dot u times our unit vector u. What is this equal to? Well, this right here is equal to the projection of a. This is equal to the projection of a onto l by definition, right here, by this definition. If we assume that we're dealing with a unit vector definition for the line. And then this is equal to this whole thing right here. And then this whole thing right here is equal to plus the projection onto l of the vector b."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to the projection of a onto l by definition, right here, by this definition. If we assume that we're dealing with a unit vector definition for the line. And then this is equal to this whole thing right here. And then this whole thing right here is equal to plus the projection onto l of the vector b. So we see our first condition for this being a linear transformation holds. The projection of the sum of the vectors is equal to the sum of the projections of the vectors. Now our second condition is that the projection of a scalar multiple should be equal to a scalar multiple of the projection."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And then this whole thing right here is equal to plus the projection onto l of the vector b. So we see our first condition for this being a linear transformation holds. The projection of the sum of the vectors is equal to the sum of the projections of the vectors. Now our second condition is that the projection of a scalar multiple should be equal to a scalar multiple of the projection. Let me write that down. So what is the projection onto l of some scalar multiple of some vector a? Well, that is equal to c a dot our unit vector u times the unit vector u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Now our second condition is that the projection of a scalar multiple should be equal to a scalar multiple of the projection. Let me write that down. So what is the projection onto l of some scalar multiple of some vector a? Well, that is equal to c a dot our unit vector u times the unit vector u. And this one's a little bit more straightforward because this is the scalar multiple. We've seen it in our dot product properties. This is equal to c times a dot u times the vector u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Well, that is equal to c a dot our unit vector u times the unit vector u. And this one's a little bit more straightforward because this is the scalar multiple. We've seen it in our dot product properties. This is equal to c times a dot u times the vector u. And this is just equal to c times, this right here is the projection of a onto l. The projection of a onto l. So we've met both of our conditions for a linear transformation. So we know that our projection onto a line l in Rn is a linear transformation. So that tells us that we can represent it as a matrix transformation."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to c times a dot u times the vector u. And this is just equal to c times, this right here is the projection of a onto l. The projection of a onto l. So we've met both of our conditions for a linear transformation. So we know that our projection onto a line l in Rn is a linear transformation. So that tells us that we can represent it as a matrix transformation. So what I want to do, we know that the projection of x onto l, we already know this definition. It can be rewritten. It doesn't hurt to rewrite it."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So that tells us that we can represent it as a matrix transformation. So what I want to do, we know that the projection of x onto l, we already know this definition. It can be rewritten. It doesn't hurt to rewrite it. As x dot some unit vector that defines our line. Let me draw it with a little hat to show that it's a unit vector, times the unit vector itself so that we actually get a vector. And now what I want to do is, how can I write this as some matrix product, some matrix vector product."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "It doesn't hurt to rewrite it. As x dot some unit vector that defines our line. Let me draw it with a little hat to show that it's a unit vector, times the unit vector itself so that we actually get a vector. And now what I want to do is, how can I write this as some matrix product, some matrix vector product. I want to write it as a product of some matrix times x. And just to simplify things, since we're actually dealing with a matrix, let's limit ourselves to the case of R2. So I'm assuming that my projection onto l is going to be a mapping from R2 to R2."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And now what I want to do is, how can I write this as some matrix product, some matrix vector product. I want to write it as a product of some matrix times x. And just to simplify things, since we're actually dealing with a matrix, let's limit ourselves to the case of R2. So I'm assuming that my projection onto l is going to be a mapping from R2 to R2. But you could do what I'm doing here with an arbitrary dimension. So if we're doing an R2, then our matrix A right there is going to be a 2 by 2 matrix. And we've seen in multiple videos that to figure out the matrix A, we just take the identity matrix that just has the standard basis vectors as columns, 0, 1, or 1, 0, and then 0, 1."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So I'm assuming that my projection onto l is going to be a mapping from R2 to R2. But you could do what I'm doing here with an arbitrary dimension. So if we're doing an R2, then our matrix A right there is going to be a 2 by 2 matrix. And we've seen in multiple videos that to figure out the matrix A, we just take the identity matrix that just has the standard basis vectors as columns, 0, 1, or 1, 0, and then 0, 1. And we apply the transformation to each of these columns. So we could say that A is going to be equal to, its first column is going to be equal to the projection onto l of this thing right here. Let me do it in this orange color right here."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And we've seen in multiple videos that to figure out the matrix A, we just take the identity matrix that just has the standard basis vectors as columns, 0, 1, or 1, 0, and then 0, 1. And we apply the transformation to each of these columns. So we could say that A is going to be equal to, its first column is going to be equal to the projection onto l of this thing right here. Let me do it in this orange color right here. So what is that going to be? That is going to be this dot u. So let me write my u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in this orange color right here. So what is that going to be? That is going to be this dot u. So let me write my u. So my unit vector, let's just assume that u can be rewritten as my unit vector is equal to some u1 and u2, just like that, and so what I need to do is I need to take this dot my unit vector, let me write this down. So let me write this on the side. So the first thing I want to do is figure out what the projection, I'll just write it here."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So let me write my u. So my unit vector, let's just assume that u can be rewritten as my unit vector is equal to some u1 and u2, just like that, and so what I need to do is I need to take this dot my unit vector, let me write this down. So let me write this on the side. So the first thing I want to do is figure out what the projection, I'll just write it here. The projection onto l, let me write it this way. So writing the projection, we know the projection is just equal to this dot this times that vector. So let me write that."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing I want to do is figure out what the projection, I'll just write it here. The projection onto l, let me write it this way. So writing the projection, we know the projection is just equal to this dot this times that vector. So let me write that. So it's the vector 1, 0 dot the unit vector u, which is just u1, u2. And we're going to have that times my unit vector. Or maybe I'll write it like this, times the vector u1, u2."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that. So it's the vector 1, 0 dot the unit vector u, which is just u1, u2. And we're going to have that times my unit vector. Or maybe I'll write it like this, times the vector u1, u2. This is going to be my first column in my transformation matrix. My second column is going to be the same thing, but I'm now going to take the projection of this guy. The definition of our projection is you dot this guy with our unit vector."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Or maybe I'll write it like this, times the vector u1, u2. This is going to be my first column in my transformation matrix. My second column is going to be the same thing, but I'm now going to take the projection of this guy. The definition of our projection is you dot this guy with our unit vector. So we dot it, we're taking the dot product of 0, 1, 0, 1 dot my unit vector, dot u1, u2. And I'm going to multiply that times my unit vector, times u1, u2. This seems very complicated, but it should simplify when we actually try to work out our transformation matrix."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "The definition of our projection is you dot this guy with our unit vector. So we dot it, we're taking the dot product of 0, 1, 0, 1 dot my unit vector, dot u1, u2. And I'm going to multiply that times my unit vector, times u1, u2. This seems very complicated, but it should simplify when we actually try to work out our transformation matrix. Let's do it. When I dot these two guys, what do I get? Let me write it here."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This seems very complicated, but it should simplify when we actually try to work out our transformation matrix. Let's do it. When I dot these two guys, what do I get? Let me write it here. So my matrix A will become 1 times u1 plus 0 times u2. Well, that's just u1. This whole thing just simplifies to u1 when I take the dot product of these two things, times u1, u2."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it here. So my matrix A will become 1 times u1 plus 0 times u2. Well, that's just u1. This whole thing just simplifies to u1 when I take the dot product of these two things, times u1, u2. That's going to be my first column. And then my second column, if I dot these two guys, I get 0 times u1 plus 1 times u2. So I'm going to get u2 times my unit vector, u1, u2."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "This whole thing just simplifies to u1 when I take the dot product of these two things, times u1, u2. That's going to be my first column. And then my second column, if I dot these two guys, I get 0 times u1 plus 1 times u2. So I'm going to get u2 times my unit vector, u1, u2. And then if I multiply that out, this will be equal to what? I can just write them as columns. u1 times u1 is u1 squared."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to get u2 times my unit vector, u1, u2. And then if I multiply that out, this will be equal to what? I can just write them as columns. u1 times u1 is u1 squared. u1 times u2 is u1, u2. u2 times u1 is just u2 times u1. And then u2 times u2 is u2 squared."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "u1 times u1 is u1 squared. u1 times u2 is u1, u2. u2 times u1 is just u2 times u1. And then u2 times u2 is u2 squared. So you give me any unit vector, and I will give you the transformation that gives you any projection of some other vector onto the line defined by that. I know that was kind of a very long way of saying that. But let's go back to what I did before."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "And then u2 times u2 is u2 squared. So you give me any unit vector, and I will give you the transformation that gives you any projection of some other vector onto the line defined by that. I know that was kind of a very long way of saying that. But let's go back to what I did before. Let's say we want to find any projection onto the line, onto the vector. Let me draw it here. We'll do the same example that we did in the last video."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "But let's go back to what I did before. Let's say we want to find any projection onto the line, onto the vector. Let me draw it here. We'll do the same example that we did in the last video. That if I have some vector v that looks like that, and we said the vector v was equal to the vector 2, 1. That was my vector v. How can we find some transformation for the projection onto the line defined by v? So onto this line right here."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "We'll do the same example that we did in the last video. That if I have some vector v that looks like that, and we said the vector v was equal to the vector 2, 1. That was my vector v. How can we find some transformation for the projection onto the line defined by v? So onto this line right here. The line defined by v. Well, what we can first do is convert v into a unit vector. So we can convert v into a unit vector that goes in the same direction. So unit vector u."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So onto this line right here. The line defined by v. Well, what we can first do is convert v into a unit vector. So we can convert v into a unit vector that goes in the same direction. So unit vector u. And we did that already up here, where we essentially just divided v by its length. So let's take v and divide by its length. The unit vector is this, 1 over the square root of 5 times our vector v. So it was this."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So unit vector u. And we did that already up here, where we essentially just divided v by its length. So let's take v and divide by its length. The unit vector is this, 1 over the square root of 5 times our vector v. So it was this. It was 1 over the square root of 5 times our vector v right there, so you start with a unit vector there. And then you just create this matrix, and then we will have our transformation matrix. So if this is our u, what will our matrix be equal to?"}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "The unit vector is this, 1 over the square root of 5 times our vector v. So it was this. It was 1 over the square root of 5 times our vector v right there, so you start with a unit vector there. And then you just create this matrix, and then we will have our transformation matrix. So if this is our u, what will our matrix be equal to? If this is u, then our matrix would be equal to u1 squared. Well, what is u1 squared? Let me actually rewrite our u a little bit, not at an angle."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So if this is our u, what will our matrix be equal to? If this is u, then our matrix would be equal to u1 squared. Well, what is u1 squared? Let me actually rewrite our u a little bit, not at an angle. So our vector u, our unit vector that defines this line, is equal to the vector 2 over the square root of 5 and 1 over the square root of 5. I just multiplied out this scalar. So if we want to construct this matrix, we get a is equal to u1 squared."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Let me actually rewrite our u a little bit, not at an angle. So our vector u, our unit vector that defines this line, is equal to the vector 2 over the square root of 5 and 1 over the square root of 5. I just multiplied out this scalar. So if we want to construct this matrix, we get a is equal to u1 squared. What's this squared? It becomes 2 squared 4 over the square root of 5 squared, which is just 5. Equals 4 over 5."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to construct this matrix, we get a is equal to u1 squared. What's this squared? It becomes 2 squared 4 over the square root of 5 squared, which is just 5. Equals 4 over 5. And then what is u1 times u2? 2 times 1 over the square root of 5 times the square root of 5, so 2 fifths. I just multiplied these two."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Equals 4 over 5. And then what is u1 times u2? 2 times 1 over the square root of 5 times the square root of 5, so 2 fifths. I just multiplied these two. What is 2 times u1? Well, same thing. Order doesn't matter when you multiply."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "I just multiplied these two. What is 2 times u1? Well, same thing. Order doesn't matter when you multiply. So this will also be 2 fifths. And then what is u2 squared? 1 squared over the square root of 5 squared is just 1 fifth."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "Order doesn't matter when you multiply. So this will also be 2 fifths. And then what is u2 squared? 1 squared over the square root of 5 squared is just 1 fifth. So now we can say, and that's the neat thing about creating these matrices, that the projection, let's say we have some, let's say this is the origin right here, and we have some other vector x right here. We can now define our transformation. The projection of onto L, where L is equal to any scalar multiple of our unit vector u, that's right here, where it's a member of the reals."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "1 squared over the square root of 5 squared is just 1 fifth. So now we can say, and that's the neat thing about creating these matrices, that the projection, let's say we have some, let's say this is the origin right here, and we have some other vector x right here. We can now define our transformation. The projection of onto L, where L is equal to any scalar multiple of our unit vector u, that's right here, where it's a member of the reals. That is our line L. The projection onto L of any vector x is equal to this matrix, is equal to the matrix 4, 5, 2 5, 2 fifths, 2 fifths, 1 fifth, times x. Which is a pretty neat result, at least for me, because we once again reduced everything to just a matrix multiplication. So if you take this x and you multiply it by this matrix, you're going to get its projection onto the L, onto the line."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "The projection of onto L, where L is equal to any scalar multiple of our unit vector u, that's right here, where it's a member of the reals. That is our line L. The projection onto L of any vector x is equal to this matrix, is equal to the matrix 4, 5, 2 5, 2 fifths, 2 fifths, 1 fifth, times x. Which is a pretty neat result, at least for me, because we once again reduced everything to just a matrix multiplication. So if you take this x and you multiply it by this matrix, you're going to get its projection onto the L, onto the line. If you take this vector, let's say a, and you multiply it times this matrix right there, you're going to get its projection onto the line. If you take this vector, no, I should go through the origin. I want to draw it in standard position."}, {"video_title": "Expressing a projection on to a line as a matrix vector prod Linear Algebra Khan Academy.mp3", "Sentence": "So if you take this x and you multiply it by this matrix, you're going to get its projection onto the L, onto the line. If you take this vector, let's say a, and you multiply it times this matrix right there, you're going to get its projection onto the line. If you take this vector, no, I should go through the origin. I want to draw it in standard position. If you take this vector right there and multiply it times this matrix, you're going to get this vector right here that is contained in the line. And when you subtract it from this, it's orthogonal. We know the definition."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I've got some subspace of Rn called v. So v is a subspace of Rn. And let's say that I know its basis. Let's say the set. So I have a bunch of, let me make that bracket a little nicer, so let's say the set of the vectors v1, v2, all the way to vk. Let's say that this is equal to, this is the basis for v. And just as a reminder, that means that these vectors both span v and they're linearly independent. You can kind of see this is the minimum set of vectors in Rn that span v. So if I were to ask you what the dimension of v is, that's just the number of vectors you have in your basis for the subspace. So we have 1, 2, and we count to k vectors."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So I have a bunch of, let me make that bracket a little nicer, so let's say the set of the vectors v1, v2, all the way to vk. Let's say that this is equal to, this is the basis for v. And just as a reminder, that means that these vectors both span v and they're linearly independent. You can kind of see this is the minimum set of vectors in Rn that span v. So if I were to ask you what the dimension of v is, that's just the number of vectors you have in your basis for the subspace. So we have 1, 2, and we count to k vectors. So it is equal to k. Now let's think about if we can somehow figure out what the dimension of the orthogonal complement of v can be. To do that, let's construct a matrix. Let's construct a matrix whose column vectors are these basis vectors."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So we have 1, 2, and we count to k vectors. So it is equal to k. Now let's think about if we can somehow figure out what the dimension of the orthogonal complement of v can be. To do that, let's construct a matrix. Let's construct a matrix whose column vectors are these basis vectors. So let's construct a matrix A. And let's say it looks like this. First column is v1, this first basis vector right there."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Let's construct a matrix whose column vectors are these basis vectors. So let's construct a matrix A. And let's say it looks like this. First column is v1, this first basis vector right there. v2 is the second one. And then you go all the way to vk. You go all the way to vk like that."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "First column is v1, this first basis vector right there. v2 is the second one. And then you go all the way to vk. You go all the way to vk like that. Just to make sure we remember the dimensions, we have k of these vectors, so we're going to have k columns. And then how many rows are we going to have? Well, there's a member of Rn, so these are all going to have n entries in each of these vectors."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "You go all the way to vk like that. Just to make sure we remember the dimensions, we have k of these vectors, so we're going to have k columns. And then how many rows are we going to have? Well, there's a member of Rn, so these are all going to have n entries in each of these vectors. There's going to be an n, we're going to have n rows and k columns. It's an n by k matrix. Now, what's another way of expressing the subspace v?"}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Well, there's a member of Rn, so these are all going to have n entries in each of these vectors. There's going to be an n, we're going to have n rows and k columns. It's an n by k matrix. Now, what's another way of expressing the subspace v? Well, the basis for v is spanned by these basis vectors, which is the columns of these. So if I talk about the span, so let me write it this. v is equal to the span of these guys, v1, v2, all the way to vk."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Now, what's another way of expressing the subspace v? Well, the basis for v is spanned by these basis vectors, which is the columns of these. So if I talk about the span, so let me write it this. v is equal to the span of these guys, v1, v2, all the way to vk. And that's just the same thing as the column space of A. These are the column vectors and the span of them, that's equal to the column space of A. Now, I said a little while ago, we want to somehow relate to the orthogonal complement of v. Well, what's the orthogonal complement of the column space of A?"}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "v is equal to the span of these guys, v1, v2, all the way to vk. And that's just the same thing as the column space of A. These are the column vectors and the span of them, that's equal to the column space of A. Now, I said a little while ago, we want to somehow relate to the orthogonal complement of v. Well, what's the orthogonal complement of the column space of A? The orthogonal complement of the column space of A, I showed you, I think it was two or three videos ago, that the column space of A's orthogonal complement is equal to, you could either view it as the null space of A transpose, or another way you could call it, this is the left null space of A. This is equivalent to the orthogonal complement of the column space of A, which is also going to be equal to, which is also, since this piece right here is the same thing as v, if you take its orthogonal complement, that's the same thing as v's orthogonal complement. So if we want to figure out the dimension of the orthogonal complement of v, we just need to figure out the dimension of the left null space of A, or the null space of A's transpose."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Now, I said a little while ago, we want to somehow relate to the orthogonal complement of v. Well, what's the orthogonal complement of the column space of A? The orthogonal complement of the column space of A, I showed you, I think it was two or three videos ago, that the column space of A's orthogonal complement is equal to, you could either view it as the null space of A transpose, or another way you could call it, this is the left null space of A. This is equivalent to the orthogonal complement of the column space of A, which is also going to be equal to, which is also, since this piece right here is the same thing as v, if you take its orthogonal complement, that's the same thing as v's orthogonal complement. So if we want to figure out the dimension of the orthogonal complement of v, we just need to figure out the dimension of the left null space of A, or the null space of A's transpose. Let me write that down. So the dimension, I get you tongue tied sometimes, the dimension of the orthogonal complement of v is going to be equal to the dimension of A transpose. Or another way to think of it is, sorry, not just the dimension of A transpose, the dimension of the null space of A transpose."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to figure out the dimension of the orthogonal complement of v, we just need to figure out the dimension of the left null space of A, or the null space of A's transpose. Let me write that down. So the dimension, I get you tongue tied sometimes, the dimension of the orthogonal complement of v is going to be equal to the dimension of A transpose. Or another way to think of it is, sorry, not just the dimension of A transpose, the dimension of the null space of A transpose. And if you have a good memory, I don't use the word a lot, this thing is the nullity. This is the nullity of A transpose. The dimension of your null space is nullity, the dimension of your column space is your rank."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to think of it is, sorry, not just the dimension of A transpose, the dimension of the null space of A transpose. And if you have a good memory, I don't use the word a lot, this thing is the nullity. This is the nullity of A transpose. The dimension of your null space is nullity, the dimension of your column space is your rank. Now let's see what we can do here. So let's just take A transpose. So you could just imagine A transpose for a second."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "The dimension of your null space is nullity, the dimension of your column space is your rank. Now let's see what we can do here. So let's just take A transpose. So you could just imagine A transpose for a second. So A transpose, I could just even draw it out, it's going to be a k by n matrix. It looks like this. These columns are going to turn into rows."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So you could just imagine A transpose for a second. So A transpose, I could just even draw it out, it's going to be a k by n matrix. It looks like this. These columns are going to turn into rows. So this is going to be v1 transpose, v2 transpose, all the way down to vk transpose. It's vectors, so these are all now row vectors. So we know one thing."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "These columns are going to turn into rows. So this is going to be v1 transpose, v2 transpose, all the way down to vk transpose. It's vectors, so these are all now row vectors. So we know one thing. We know one relationship between the rank and nullity of any matrix. We know that they're equal to the number of columns we have. We know that the rank of A transpose plus the nullity of A transpose is equal to the number of columns of A transpose."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So we know one thing. We know one relationship between the rank and nullity of any matrix. We know that they're equal to the number of columns we have. We know that the rank of A transpose plus the nullity of A transpose is equal to the number of columns of A transpose. So we have n columns. Each of these have n entries. It is equal to n. We saw this a while ago."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "We know that the rank of A transpose plus the nullity of A transpose is equal to the number of columns of A transpose. So we have n columns. Each of these have n entries. It is equal to n. We saw this a while ago. If you want just a bit of a reminder of where that comes from, if I wrote A transpose as a bunch of column vectors, which I can, or maybe let me take some other vector b, because I want to just remind you why this made sense. If I take some vector b here, and it's got a bunch of column vectors, b1, b2, all the way to bn, and I put it into reduced row echelon form, you're going to have some pivot columns and some non-pivot columns. So let's say this is a pivot column."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "It is equal to n. We saw this a while ago. If you want just a bit of a reminder of where that comes from, if I wrote A transpose as a bunch of column vectors, which I can, or maybe let me take some other vector b, because I want to just remind you why this made sense. If I take some vector b here, and it's got a bunch of column vectors, b1, b2, all the way to bn, and I put it into reduced row echelon form, you're going to have some pivot columns and some non-pivot columns. So let's say this is a pivot column. You know, I've got a 1 and a bunch of 0's. Let's say that this is one of them. And then let's say I've got one other one that's out."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So let's say this is a pivot column. You know, I've got a 1 and a bunch of 0's. Let's say that this is one of them. And then let's say I've got one other one that's out. It would be a 0 there. Let's say it's a 1 down there. And everything else is a non-pivot column."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "And then let's say I've got one other one that's out. It would be a 0 there. Let's say it's a 1 down there. And everything else is a non-pivot column. I showed you in the last video that your basis for your column space is the number of pivot columns you have. So these guys are pivot columns. The corresponding column vectors form a basis for your column space."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "And everything else is a non-pivot column. I showed you in the last video that your basis for your column space is the number of pivot columns you have. So these guys are pivot columns. The corresponding column vectors form a basis for your column space. Showed you that in the last video. And so if you want to know the dimension of your column space, you just have to count these things. So you just count these things."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "The corresponding column vectors form a basis for your column space. Showed you that in the last video. And so if you want to know the dimension of your column space, you just have to count these things. So you just count these things. This was equal to the number of, well, for this b's case, the rank of b is just equal to the number of pivot columns I have. Now the nullity is the dimension of your null space. And we've done multiple problems where we found the null space of matrices."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "So you just count these things. This was equal to the number of, well, for this b's case, the rank of b is just equal to the number of pivot columns I have. Now the nullity is the dimension of your null space. And we've done multiple problems where we found the null space of matrices. And every time, the dimension, it's a bit obvious, and I actually showed you this proof, it's related to the number of free columns you have, or non-pivot columns. So if you have no pivot columns, and if all of your columns are pivot columns, and none of them have free variables, or are associated with free variables, then your null space is going to be trivial. It's just going to have the 0 vector."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "And we've done multiple problems where we found the null space of matrices. And every time, the dimension, it's a bit obvious, and I actually showed you this proof, it's related to the number of free columns you have, or non-pivot columns. So if you have no pivot columns, and if all of your columns are pivot columns, and none of them have free variables, or are associated with free variables, then your null space is going to be trivial. It's just going to have the 0 vector. But the more free variables you have, the more dimensionality your null space has. So the free columns correspond to the null space. And they form, actually, a basis for your null space."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "It's just going to have the 0 vector. But the more free variables you have, the more dimensionality your null space has. So the free columns correspond to the null space. And they form, actually, a basis for your null space. And because of that, the basis for your null space vectors plus the basis for your column space is equal to the total number of columns you add. I've shown you that too in the past, but it's always good to remind ourselves where things come from. But this was just a bit of a side."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "And they form, actually, a basis for your null space. And because of that, the basis for your null space vectors plus the basis for your column space is equal to the total number of columns you add. I've shown you that too in the past, but it's always good to remind ourselves where things come from. But this was just a bit of a side. I did this with a separate vector b, just to remind ourselves where this thing right here came from. Now, in the last video, I showed you that the rank of a transpose is the same thing as the rank of a. This is equal to, this part right here is the same thing as the rank of a. I showed you that in the last video."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "But this was just a bit of a side. I did this with a separate vector b, just to remind ourselves where this thing right here came from. Now, in the last video, I showed you that the rank of a transpose is the same thing as the rank of a. This is equal to, this part right here is the same thing as the rank of a. I showed you that in the last video. When you transpose a matrix, it doesn't change its rank, or it doesn't change the dimension of its column space. So we can rewrite this statement right here as the rank of a plus the nullity of a transpose is equal to n. And the rank of a is the same thing as the dimension of the column space of a. And then the nullity of a transpose is the same thing as the dimension of the null space of a transpose."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, this part right here is the same thing as the rank of a. I showed you that in the last video. When you transpose a matrix, it doesn't change its rank, or it doesn't change the dimension of its column space. So we can rewrite this statement right here as the rank of a plus the nullity of a transpose is equal to n. And the rank of a is the same thing as the dimension of the column space of a. And then the nullity of a transpose is the same thing as the dimension of the null space of a transpose. That's just the definition of nullity. They're going to be equal to n. And what's the dimension, what's the column space of a? The column space of a, that's what's spanned by these vectors right here, which were the basis for v. So this is the same thing as the dimension of v. The column space of a is the same thing as the dimension of my subspace v that I started this video with."}, {"video_title": "dim(v) + dim(orthogonal complement of v) = n Linear Algebra Khan Academy.mp3", "Sentence": "And then the nullity of a transpose is the same thing as the dimension of the null space of a transpose. That's just the definition of nullity. They're going to be equal to n. And what's the dimension, what's the column space of a? The column space of a, that's what's spanned by these vectors right here, which were the basis for v. So this is the same thing as the dimension of v. The column space of a is the same thing as the dimension of my subspace v that I started this video with. And what is the null space of a transpose? The null space of a transpose, we saw already, that's the orthogonal complement of v. So I could write this as plus the dimension of the orthogonal complement of v is equal to n. And that's the result we wanted. If v is a subspace of Rn, so v subspace of Rn, that n is the same thing as that n, then the dimension of v plus the dimension of the orthogonal complement of v is going to be equal to n."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But the cross product is actually much more limited than the dot product. It's useful, but it's much more limited. The dot product is defined in any dimension. So this is defined for any two vectors that are in Rn. You could take the dot product of vectors that have two components. You could take the dot product of vectors that have a million components. The cross product is only defined in R3."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is defined for any two vectors that are in Rn. You could take the dot product of vectors that have two components. You could take the dot product of vectors that have a million components. The cross product is only defined in R3. The other major difference is the dot product. We're going to see this in a second when I define the dot product for you. I haven't defined it yet."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The cross product is only defined in R3. The other major difference is the dot product. We're going to see this in a second when I define the dot product for you. I haven't defined it yet. The dot product results in a scalar. When you take the dot product of two vectors, you just get a number. But with the cross product, you're going to see that we're going to get another vector."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I haven't defined it yet. The dot product results in a scalar. When you take the dot product of two vectors, you just get a number. But with the cross product, you're going to see that we're going to get another vector. The vector we're going to get is actually going to be a vector that's orthogonal to the two vectors that we're taking the cross product of. Now that I have you excited with anticipation, let me define it for you. You probably already have seen this once or twice in your mathematical careers."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But with the cross product, you're going to see that we're going to get another vector. The vector we're going to get is actually going to be a vector that's orthogonal to the two vectors that we're taking the cross product of. Now that I have you excited with anticipation, let me define it for you. You probably already have seen this once or twice in your mathematical careers. Let's say I have the vector A. It has to be in R3, so it only has three components, A1, A2, and A3. I'm going to cross that with the vector B."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You probably already have seen this once or twice in your mathematical careers. Let's say I have the vector A. It has to be in R3, so it only has three components, A1, A2, and A3. I'm going to cross that with the vector B. It has three components, B1, B2, and B3. A cross B is defined as a third vector. This is going to seem a little bit bizarre and hard to memorize because this is a definition."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to cross that with the vector B. It has three components, B1, B2, and B3. A cross B is defined as a third vector. This is going to seem a little bit bizarre and hard to memorize because this is a definition. I'll show you how I think about it when I have my vectors written in this column form. If you watch the physics playlist, I have a bunch of videos on the cross product. I show you how I think about the cross product when I have it in the ijk form."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is going to seem a little bit bizarre and hard to memorize because this is a definition. I'll show you how I think about it when I have my vectors written in this column form. If you watch the physics playlist, I have a bunch of videos on the cross product. I show you how I think about the cross product when I have it in the ijk form. When I have it like this, the way you think about it, this first term up here, this is going to be another three vector or another vector in R3. It's going to have one, two, three terms. For the first term, what you do is you ignore these top two terms of this vector."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I show you how I think about the cross product when I have it in the ijk form. When I have it like this, the way you think about it, this first term up here, this is going to be another three vector or another vector in R3. It's going to have one, two, three terms. For the first term, what you do is you ignore these top two terms of this vector. Then you look at the bottom two and you say A2 times B3 minus A3 times B2. I've made a few videos on determinants, although I haven't formally done them in this linear algebra playlist yet. If you remember finding out the cofactor terms when you're determining the determinant, or if you're just taking the determinant for a two by two matrix, this might seem very familiar."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For the first term, what you do is you ignore these top two terms of this vector. Then you look at the bottom two and you say A2 times B3 minus A3 times B2. I've made a few videos on determinants, although I haven't formally done them in this linear algebra playlist yet. If you remember finding out the cofactor terms when you're determining the determinant, or if you're just taking the determinant for a two by two matrix, this might seem very familiar. This first term right here is essentially the determinant of, if you get rid of this first row out of both of these guys right here, you should take A2 times B3 minus A3 times B2. It's A2 times B3 minus A3 times B2. That was hopefully pretty straightforward."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you remember finding out the cofactor terms when you're determining the determinant, or if you're just taking the determinant for a two by two matrix, this might seem very familiar. This first term right here is essentially the determinant of, if you get rid of this first row out of both of these guys right here, you should take A2 times B3 minus A3 times B2. It's A2 times B3 minus A3 times B2. That was hopefully pretty straightforward. Now, not to make your life any more complicated, when you do the middle row, when you do this one right here, so you cross that out, and you might want to do A1 times B3 minus A3 times B1. That would be natural because that's what we did up there. But the middle row, you do the opposite."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That was hopefully pretty straightforward. Now, not to make your life any more complicated, when you do the middle row, when you do this one right here, so you cross that out, and you might want to do A1 times B3 minus A3 times B1. That would be natural because that's what we did up there. But the middle row, you do the opposite. You do A3 times B1 minus A1 times B3, or you can kind of view it as the negative of what you would have done naturally. You would have done A1B3 minus A3B1. Now we're going to do A3B1 minus A1B3."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But the middle row, you do the opposite. You do A3 times B1 minus A1 times B3, or you can kind of view it as the negative of what you would have done naturally. You would have done A1B3 minus A3B1. Now we're going to do A3B1 minus A1B3. That was only for that middle row. Then for the bottom row, we cross that out again or ignore it, and we do A1 times B2 just like we did with the first row. A1B2 times A2B1 or minus A2B1."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now we're going to do A3B1 minus A1B3. That was only for that middle row. Then for the bottom row, we cross that out again or ignore it, and we do A1 times B2 just like we did with the first row. A1B2 times A2B1 or minus A2B1. This seems all hard to, and it is hard to remember. That's why I kind of have to get that system in place like I just talked to you about. But this might seem pretty bizarre and hairy."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A1B2 times A2B1 or minus A2B1. This seems all hard to, and it is hard to remember. That's why I kind of have to get that system in place like I just talked to you about. But this might seem pretty bizarre and hairy. Let me do a couple of examples with you just so you get the hang of our definition of the dot product in R3. Let's say I'm crossing the vector 1 minus 7 and 1, and I'm going to cross that with the vector 5, 2, 4. This is going to be equal to a third vector."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But this might seem pretty bizarre and hairy. Let me do a couple of examples with you just so you get the hang of our definition of the dot product in R3. Let's say I'm crossing the vector 1 minus 7 and 1, and I'm going to cross that with the vector 5, 2, 4. This is going to be equal to a third vector. Let me get some space to do my mathematics. For the first element in this vector, the first component, we just ignore the first components of these vectors. We say minus 7 times 4 minus 1 times 2."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to a third vector. Let me get some space to do my mathematics. For the first element in this vector, the first component, we just ignore the first components of these vectors. We say minus 7 times 4 minus 1 times 2. These are just regular multiplication. I'm not taking a dot product. These are just regular numbers."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We say minus 7 times 4 minus 1 times 2. These are just regular multiplication. I'm not taking a dot product. These are just regular numbers. Then for the middle term, we ignore the middle terms here, and then we do the opposite. We do 1 times 5 minus 1 times 4. Remember, the first, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the first term."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These are just regular numbers. Then for the middle term, we ignore the middle terms here, and then we do the opposite. We do 1 times 5 minus 1 times 4. Remember, the first, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the first term. But the middle term is the opposite. Then finally, the third term, you ignore the third terms here, and then you do it just like the first term. You start on the top left."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, the first, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the first term. But the middle term is the opposite. Then finally, the third term, you ignore the third terms here, and then you do it just like the first term. You start on the top left. 1 times 2 minus 7 times 5. That is equal to minus 7 times 4. I don't want to make a careless mistake here."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You start on the top left. 1 times 2 minus 7 times 5. That is equal to minus 7 times 4. I don't want to make a careless mistake here. That's minus 28 minus 2. This is minus 30 for that first term. This one is 5 minus 4, so it's minus 5 minus 4."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to make a careless mistake here. That's minus 28 minus 2. This is minus 30 for that first term. This one is 5 minus 4, so it's minus 5 minus 4. Sorry, 5 minus 4 is just 1. Then 2 minus minus 35. 2 minus minus 35."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This one is 5 minus 4, so it's minus 5 minus 4. Sorry, 5 minus 4 is just 1. Then 2 minus minus 35. 2 minus minus 35. That's 2 plus 35. That's 37. There you go."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 minus minus 35. That's 2 plus 35. That's 37. There you go. Hopefully, you understand at least the mechanics of the cross product. The next thing you're saying, well, okay, I can find the cross product of two things, but what is this good for? What does this do for me?"}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There you go. Hopefully, you understand at least the mechanics of the cross product. The next thing you're saying, well, okay, I can find the cross product of two things, but what is this good for? What does this do for me? The answer is that this third vector right here, depending on whether I stay in the abstract case or whether this is the case with numbers, this is orthogonal to the two vectors that we took the cross product of. This vector right here is orthogonal to A and B, which is pretty neat. If you just go think about the last video when we were talking about normal vectors to a plane, we can define a plane by two vectors."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What does this do for me? The answer is that this third vector right here, depending on whether I stay in the abstract case or whether this is the case with numbers, this is orthogonal to the two vectors that we took the cross product of. This vector right here is orthogonal to A and B, which is pretty neat. If you just go think about the last video when we were talking about normal vectors to a plane, we can define a plane by two vectors. If we define a plane, let's say that I have vector A right there, and then I have vector B like this, those define a plane in R3. Let me define your plane. All the linear combinations of those two guys, that's a plane in R3."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you just go think about the last video when we were talking about normal vectors to a plane, we can define a plane by two vectors. If we define a plane, let's say that I have vector A right there, and then I have vector B like this, those define a plane in R3. Let me define your plane. All the linear combinations of those two guys, that's a plane in R3. You can kind of view it as they might form a subspace in R3. That forms a plane. If you take A cross B, you get a third vector that's orthogonal to those two."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All the linear combinations of those two guys, that's a plane in R3. You can kind of view it as they might form a subspace in R3. That forms a plane. If you take A cross B, you get a third vector that's orthogonal to those two. A cross B will pop out like this. It will be orthogonal to both of them. It will look like that."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you take A cross B, you get a third vector that's orthogonal to those two. A cross B will pop out like this. It will be orthogonal to both of them. It will look like that. This vector right there is A cross B. You might say, Sal, how did you know? There are multiple vectors that are orthogonal."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It will look like that. This vector right there is A cross B. You might say, Sal, how did you know? There are multiple vectors that are orthogonal. Obviously, the length of the vector, and I didn't specify that there, but it could pop straight up like that, or you just as easily could have popped straight down like that. That also would be orthogonal to A and B. The way that A cross B is defined, you can essentially figure out the direction visually by using what's called the right-hand rule."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There are multiple vectors that are orthogonal. Obviously, the length of the vector, and I didn't specify that there, but it could pop straight up like that, or you just as easily could have popped straight down like that. That also would be orthogonal to A and B. The way that A cross B is defined, you can essentially figure out the direction visually by using what's called the right-hand rule. The way I think about it is you take your right hand, and let me see if I can draw a suitable right hand, point your index finger in the direction of A. If your index finger is in the direction of A, and then I point my middle finger in the direction of B, so my middle finger in this case is going to go something like that. My middle finger is going to do something like that, and then my other fingers do nothing."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The way that A cross B is defined, you can essentially figure out the direction visually by using what's called the right-hand rule. The way I think about it is you take your right hand, and let me see if I can draw a suitable right hand, point your index finger in the direction of A. If your index finger is in the direction of A, and then I point my middle finger in the direction of B, so my middle finger in this case is going to go something like that. My middle finger is going to do something like that, and then my other fingers do nothing. Then my thumb will go in the direction of A cross B. You can see that there. My thumb is in the direction of A cross B."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My middle finger is going to do something like that, and then my other fingers do nothing. Then my thumb will go in the direction of A cross B. You can see that there. My thumb is in the direction of A cross B. Assuming that you are anatomically similar to me, then you'll get the same result. Let me draw it all. This is vector A, vector B goes in that direction."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My thumb is in the direction of A cross B. Assuming that you are anatomically similar to me, then you'll get the same result. Let me draw it all. This is vector A, vector B goes in that direction. Then you know, hopefully you don't have a thumb hanging down here, you know that A cross B in this example will point up, and it's orthogonal to both. Just to satisfy you a little bit, that the vector is definitely orthogonal, or that this thing is definitely orthogonal to both of these, let's just play with it and see that that definitely is the case. What is orthogonal?"}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is vector A, vector B goes in that direction. Then you know, hopefully you don't have a thumb hanging down here, you know that A cross B in this example will point up, and it's orthogonal to both. Just to satisfy you a little bit, that the vector is definitely orthogonal, or that this thing is definitely orthogonal to both of these, let's just play with it and see that that definitely is the case. What is orthogonal? What is really in our context the definition of orthogonal? Orthogonal vectors, orthogonal if A and B are orthogonal, that means that A dot B is equal to 0. Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is orthogonal? What is really in our context the definition of orthogonal? Orthogonal vectors, orthogonal if A and B are orthogonal, that means that A dot B is equal to 0. Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors. These could also be 0 vectors. Notice that I didn't say that any of these guys up here had to be non-zero. In a little bit, we'll talk about the angle between vectors, and then you have to assume non-zero."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors. These could also be 0 vectors. Notice that I didn't say that any of these guys up here had to be non-zero. In a little bit, we'll talk about the angle between vectors, and then you have to assume non-zero. But if you're just taking the cross product, nothing to stop you from taking, no reason why any of these numbers can't be 0. Let me show you that A cross B is definitely orthogonal to both A and B. I think that might be somewhat satisfying to you. Let me copy A cross B here."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In a little bit, we'll talk about the angle between vectors, and then you have to assume non-zero. But if you're just taking the cross product, nothing to stop you from taking, no reason why any of these numbers can't be 0. Let me show you that A cross B is definitely orthogonal to both A and B. I think that might be somewhat satisfying to you. Let me copy A cross B here. I don't feel like rewriting it. Let me paste it. I brought a little other stuff with it."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me copy A cross B here. I don't feel like rewriting it. Let me paste it. I brought a little other stuff with it. But let me take the dot product of that with just my vector A, which was just A1, A2, A3. What does the dot product look like? It's that term times that, so it's A1 times A2B3 minus A1 times that, minus A1 times A3B2."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I brought a little other stuff with it. But let me take the dot product of that with just my vector A, which was just A1, A2, A3. What does the dot product look like? It's that term times that, so it's A1 times A2B3 minus A1 times that, minus A1 times A3B2. Then you have plus this times this, so plus A2 times A3 times B1, and then minus A2A1B3. Then finally plus, I'll just continue it down here, plus A3 times A1B2 minus A3 times A2B1. All I did is I just took the dot product of these two things."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's that term times that, so it's A1 times A2B3 minus A1 times that, minus A1 times A3B2. Then you have plus this times this, so plus A2 times A3 times B1, and then minus A2A1B3. Then finally plus, I'll just continue it down here, plus A3 times A1B2 minus A3 times A2B1. All I did is I just took the dot product of these two things. I just took each of this. This guy times that was equal to those two terms. That guy times that was equal to the next two terms, equal to those two terms."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I just took the dot product of these two things. I just took each of this. This guy times that was equal to those two terms. That guy times that was equal to the next two terms, equal to those two terms. Then this guy times that was equal to those two terms. If these guys are really orthogonal, then this should be equal to zero. Let's see if that's the case."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That guy times that was equal to the next two terms, equal to those two terms. Then this guy times that was equal to those two terms. If these guys are really orthogonal, then this should be equal to zero. Let's see if that's the case. I have an A1A2B3, a positive here. Then I'm subtracting the same thing here. This is the same thing as A1A2B3, but it's just a minus."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if that's the case. I have an A1A2B3, a positive here. Then I'm subtracting the same thing here. This is the same thing as A1A2B3, but it's just a minus. That will cancel out with that. Let's see what else we have. We have a minus A1A3B2."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is the same thing as A1A2B3, but it's just a minus. That will cancel out with that. Let's see what else we have. We have a minus A1A3B2. We have a plus A1A3B2 there. These two are going to cancel out. I think you see where this is going."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have a minus A1A3B2. We have a plus A1A3B2 there. These two are going to cancel out. I think you see where this is going. You have a positive A2A3B1. Then you have a negative A2A3B1 there. These will also cancel out."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I think you see where this is going. You have a positive A2A3B1. Then you have a negative A2A3B1 there. These will also cancel out. I just showed you that it's orthogonal to A. Let me show you that it's orthogonal to B. Let me get another version of the cross product of the two vectors."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These will also cancel out. I just showed you that it's orthogonal to A. Let me show you that it's orthogonal to B. Let me get another version of the cross product of the two vectors. I'll probably scroll down a little bit. Let me go back. Let me multiply that times the vector B. B1, B2, and B3."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me get another version of the cross product of the two vectors. I'll probably scroll down a little bit. Let me go back. Let me multiply that times the vector B. B1, B2, and B3. I'll do it here, just to have some space. B1 times this whole thing right here is B1A2B3 minus B1 times this. Minus B1A3B2."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me multiply that times the vector B. B1, B2, and B3. I'll do it here, just to have some space. B1 times this whole thing right here is B1A2B3 minus B1 times this. Minus B1A3B2. Let me switch colors. Then B2 times this thing here is going to be plus. This is all really one expression."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus B1A3B2. Let me switch colors. Then B2 times this thing here is going to be plus. This is all really one expression. I'm just writing it on multiple lines. This isn't a vector. Remember, when you take the dot product of two things, you get a scalar quantity."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is all really one expression. I'm just writing it on multiple lines. This isn't a vector. Remember, when you take the dot product of two things, you get a scalar quantity. Plus B2 times this thing. B2A3B1 minus B2A1B3. Then finally, B3 times this."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, when you take the dot product of two things, you get a scalar quantity. Plus B2 times this thing. B2A3B1 minus B2A1B3. Then finally, B3 times this. Plus B3A1B2 minus B3A2B1. If these guys are definitely orthogonal, then this thing needs to equal zero. Let's see if that's the case."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then finally, B3 times this. Plus B3A1B2 minus B3A2B1. If these guys are definitely orthogonal, then this thing needs to equal zero. Let's see if that's the case. We have a B1A2B3. B1 and a B3. B1A2B3, that's a positive one."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if that's the case. We have a B1A2B3. B1 and a B3. B1A2B3, that's a positive one. Then this is a negative one. You have a B3, an A2, and a B1. That and that cancel out."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "B1A2B3, that's a positive one. Then this is a negative one. You have a B3, an A2, and a B1. That and that cancel out. Here you have a minus B1A3B2. You have a B1 and a B2. You have a minus B1A3B2."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That and that cancel out. Here you have a minus B1A3B2. You have a B1 and a B2. You have a minus B1A3B2. This is a plus, the same thing. It's just switching the order of the multiplication. These two are the same term."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You have a minus B1A3B2. This is a plus, the same thing. It's just switching the order of the multiplication. These two are the same term. They're just opposites of each other. They cancel out. Then finally, you have a B2, an A1, and a B3."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These two are the same term. They're just opposites of each other. They cancel out. Then finally, you have a B2, an A1, and a B3. It's a negative. Then you have a positive version of the same thing. These two guys cancel out."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then finally, you have a B2, an A1, and a B3. It's a negative. Then you have a positive version of the same thing. These two guys cancel out. You see that this is also equal to zero. Hopefully you're satisfied that this vector right here is definitely orthogonal to both A and B. That's because that's how it was designed."}, {"video_title": "Cross product introduction Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These two guys cancel out. You see that this is also equal to zero. Hopefully you're satisfied that this vector right here is definitely orthogonal to both A and B. That's because that's how it was designed. This is a definition. You could do a little bit of algebra and you could have, without me explaining this definition to you, you could have actually come up with this definition on your own. Obviously, this was designed to have other interesting properties to it."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, we started with a linear transformation S that was a mapping between the set X, that was a subset of Rn, to the set Y. And then we had another transformation that was a mapping from the set Y to the set Z. And we asked ourselves, given these two linear transformations, could we construct a linear transformation that goes all the way from X to Z? And what we did is we made a definition. We said, well, let's create something called the composition of T with S. And what that is, is first you apply S to some vector in X to get some vector in Y. And that's your vector right there. And then you apply T to that to get to Z."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what we did is we made a definition. We said, well, let's create something called the composition of T with S. And what that is, is first you apply S to some vector in X to get some vector in Y. And that's your vector right there. And then you apply T to that to get to Z. And so we defined it that way. And our next question was, was that a linear transformation? We showed that it was."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you apply T to that to get to Z. And so we defined it that way. And our next question was, was that a linear transformation? We showed that it was. It met the two requirements for them. And because it is a linear transformation, I left off in the last video saying that it should be able to be represented by some matrix vector product, where this will have to be an L by N matrix. Because it's a mapping from an N dimensional space, which was X, it was a subset of Rn, to an L dimensional space."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We showed that it was. It met the two requirements for them. And because it is a linear transformation, I left off in the last video saying that it should be able to be represented by some matrix vector product, where this will have to be an L by N matrix. Because it's a mapping from an N dimensional space, which was X, it was a subset of Rn, to an L dimensional space. Because Z is a subset of RL. Now, in this video, let's try to actually construct this matrix. So at the beginning of the last video, I told you that T of X could be written as some matrix product B times X."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because it's a mapping from an N dimensional space, which was X, it was a subset of Rn, to an L dimensional space. Because Z is a subset of RL. Now, in this video, let's try to actually construct this matrix. So at the beginning of the last video, I told you that T of X could be written as some matrix product B times X. Let me write that. Let me rewrite it down here. So I told you that the linear transformation T, applied to some vector X, could be written as the matrix vector product B times the vector X."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So at the beginning of the last video, I told you that T of X could be written as some matrix product B times X. Let me write that. Let me rewrite it down here. So I told you that the linear transformation T, applied to some vector X, could be written as the matrix vector product B times the vector X. And since it was a mapping from a M dimensional space to an L dimensional space, we know that this is going to be an L by M matrix. Now, similarly, I told you that the transformation S can also be written as a matrix vector product, where we could say maybe A is its matrix representation times the vector X. And since this is a mapping from a N dimensional space to an M dimensional space, this will be an M by N matrix."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I told you that the linear transformation T, applied to some vector X, could be written as the matrix vector product B times the vector X. And since it was a mapping from a M dimensional space to an L dimensional space, we know that this is going to be an L by M matrix. Now, similarly, I told you that the transformation S can also be written as a matrix vector product, where we could say maybe A is its matrix representation times the vector X. And since this is a mapping from a N dimensional space to an M dimensional space, this will be an M by N matrix. Now, by definition, what was the composition of T of S? T with S. What is this? By definition, we said that this is equal to, you first apply the linear transformation S to X, and I'll arbitrarily switch colors."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And since this is a mapping from a N dimensional space to an M dimensional space, this will be an M by N matrix. Now, by definition, what was the composition of T of S? T with S. What is this? By definition, we said that this is equal to, you first apply the linear transformation S to X, and I'll arbitrarily switch colors. So you first apply the transformation S to X, and that essentially gets you a vector right there. This is just a vector in Rm, or it's really a vector in Y, which is a subset of Rm. And then you apply the transformation T to that vector to get you into Z."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "By definition, we said that this is equal to, you first apply the linear transformation S to X, and I'll arbitrarily switch colors. So you first apply the transformation S to X, and that essentially gets you a vector right there. This is just a vector in Rm, or it's really a vector in Y, which is a subset of Rm. And then you apply the transformation T to that vector to get you into Z. So then you apply T to that. Well, given this, we can use our matrix representations to replace this kind of transformation representation, although they're really the same thing. What is the transformation of S applied to X?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you apply the transformation T to that vector to get you into Z. So then you apply T to that. Well, given this, we can use our matrix representations to replace this kind of transformation representation, although they're really the same thing. What is the transformation of S applied to X? Well, it's just this right here is just A times X. That is just A times X, where this is an M by N matrix. So we could say that this is equal to the transformation applied to A times X."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is the transformation of S applied to X? Well, it's just this right here is just A times X. That is just A times X, where this is an M by N matrix. So we could say that this is equal to the transformation applied to A times X. Now, what is the T transformation applied to any vector X? Well, that's the matrix B times your vector X. So this thing right here is going to be equal to B times whatever I put in there."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that this is equal to the transformation applied to A times X. Now, what is the T transformation applied to any vector X? Well, that's the matrix B times your vector X. So this thing right here is going to be equal to B times whatever I put in there. So the matrix B times the matrix A times the vector X right there. This is what our composition transformation is. The composition of T with S applied to the vector X, which takes us from the set X all the way to the set Z, is this."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here is going to be equal to B times whatever I put in there. So the matrix B times the matrix A times the vector X right there. This is what our composition transformation is. The composition of T with S applied to the vector X, which takes us from the set X all the way to the set Z, is this. If we use the matrix forms of the two transformations. Now, we know that we can, at the end of last video, I said I wanted to find just some matrix that if I multiply it times this vector, that is equivalent to this transformation. And I know that I can find this matrix."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The composition of T with S applied to the vector X, which takes us from the set X all the way to the set Z, is this. If we use the matrix forms of the two transformations. Now, we know that we can, at the end of last video, I said I wanted to find just some matrix that if I multiply it times this vector, that is equivalent to this transformation. And I know that I can find this matrix. I know that this exists because this is a linear transformation. So how can we do that? Well, we just do what we've always done in the past."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I know that I can find this matrix. I know that this exists because this is a linear transformation. So how can we do that? Well, we just do what we've always done in the past. We find, we start with the identity matrix, and we apply the transformation to every column of the identity matrix. And then you end up with your matrix representation of the transformation itself. So first of all, how big is the identity matrix going to be?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we just do what we've always done in the past. We find, we start with the identity matrix, and we apply the transformation to every column of the identity matrix. And then you end up with your matrix representation of the transformation itself. So first of all, how big is the identity matrix going to be? Well, these guys that we're inputting into our transformation, they are subsets of X, or they're members of X, which is an n-dimensional space. It's a subset of Rn. So these, let me write this."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So first of all, how big is the identity matrix going to be? Well, these guys that we're inputting into our transformation, they are subsets of X, or they're members of X, which is an n-dimensional space. It's a subset of Rn. So these, let me write this. X is a member of Rn. So all we do to figure out C is we apply, so we start off with the identity matrix. So let's see."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these, let me write this. X is a member of Rn. So all we do to figure out C is we apply, so we start off with the identity matrix. So let's see. We start with the identity matrix, the n-dimensional identity matrix, because we're starting, our domain is Rn. And of course, we know what that looks like. We draw a little straighter."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see. We start with the identity matrix, the n-dimensional identity matrix, because we're starting, our domain is Rn. And of course, we know what that looks like. We draw a little straighter. We have 1, 0, all the way down. It's going to be an n by n matrix. And then 0, 1, all the way down."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We draw a little straighter. We have 1, 0, all the way down. It's going to be an n by n matrix. And then 0, 1, all the way down. 0's, these are 0's right here. And then you have 1's go all the way down the columns, and everything else is 0. We've seen this multiple times."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then 0, 1, all the way down. 0's, these are 0's right here. And then you have 1's go all the way down the columns, and everything else is 0. We've seen this multiple times. That's what your identity matrix looks like. Just 1's down the column from the top left to the bottom right. Now, to figure out C, the matrix representation of our transformation, all we do is we apply the transformation to each of these columns."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've seen this multiple times. That's what your identity matrix looks like. Just 1's down the column from the top left to the bottom right. Now, to figure out C, the matrix representation of our transformation, all we do is we apply the transformation to each of these columns. So we can write that our matrix C is equal to, it's going to take some space here, the transformation applied to this first column. But what is the transformation? It is the matrix B times the matrix A times whatever you're taking the transformation of."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, to figure out C, the matrix representation of our transformation, all we do is we apply the transformation to each of these columns. So we can write that our matrix C is equal to, it's going to take some space here, the transformation applied to this first column. But what is the transformation? It is the matrix B times the matrix A times whatever you're taking the transformation of. In this case, we're taking the transformation of that. So we're taking the transformation of 1, 0, 0, all the way down. There's a bunch of 0's."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It is the matrix B times the matrix A times whatever you're taking the transformation of. In this case, we're taking the transformation of that. So we're taking the transformation of 1, 0, 0, all the way down. There's a bunch of 0's. 1 followed by a bunch of 0's. That's going to be our first column of C. Our second column of C is going to be B times A times the second column of our identity matrix. And of course, you remember these are each the standard basis vectors for Rn."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "There's a bunch of 0's. 1 followed by a bunch of 0's. That's going to be our first column of C. Our second column of C is going to be B times A times the second column of our identity matrix. And of course, you remember these are each the standard basis vectors for Rn. So this is e1, it's a vector. e2, I could actually put a hat there because they're unit vectors, but you know that already. So this is going to be times e2, which is 0, 1, 0, all the way down, a bunch of 0's."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And of course, you remember these are each the standard basis vectors for Rn. So this is e1, it's a vector. e2, I could actually put a hat there because they're unit vectors, but you know that already. So this is going to be times e2, which is 0, 1, 0, all the way down, a bunch of 0's. And then we're going to keep doing that until we get to the last column, which is B times A times a bunch of 0's, all the way down and you get a 1. The nth term is just a 1 right there. Now what is this going to be equal to?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be times e2, which is 0, 1, 0, all the way down, a bunch of 0's. And then we're going to keep doing that until we get to the last column, which is B times A times a bunch of 0's, all the way down and you get a 1. The nth term is just a 1 right there. Now what is this going to be equal to? It looks fairly complicated right now. But all you have to do is make the realization, and we've seen this multiple times, that look, if we write our vector A or we write our matrix A as just a bunch of column vectors, so this is a column vector a1, a2, all the way to an, right? We already learned that this was an n by m matrix."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what is this going to be equal to? It looks fairly complicated right now. But all you have to do is make the realization, and we've seen this multiple times, that look, if we write our vector A or we write our matrix A as just a bunch of column vectors, so this is a column vector a1, a2, all the way to an, right? We already learned that this was an n by m matrix. Then what is the vector A times, for example, 1, well actually let me write it this way, times x1, x2, all the way down to xn. We've seen this multiple times. This is equivalent to x1 times a1 plus x2 times a2, all the way to plus xn times an."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We already learned that this was an n by m matrix. Then what is the vector A times, for example, 1, well actually let me write it this way, times x1, x2, all the way down to xn. We've seen this multiple times. This is equivalent to x1 times a1 plus x2 times a2, all the way to plus xn times an. We've seen this multiple times. It's a linear combination of these column vectors where the weighting factors are the terms in our vector that we're taking the product of. So given that, what is this guy going to reduce to?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equivalent to x1 times a1 plus x2 times a2, all the way to plus xn times an. We've seen this multiple times. It's a linear combination of these column vectors where the weighting factors are the terms in our vector that we're taking the product of. So given that, what is this guy going to reduce to? This is going to be a1 times this first entry right here, times x1, plus a2 times the second entry, plus a3 times the third entry, but all of these other entries are 0. The x2's all the way to the xn are 0. So you're only going to end up with 1 times the first column here in A."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So given that, what is this guy going to reduce to? This is going to be a1 times this first entry right here, times x1, plus a2 times the second entry, plus a3 times the third entry, but all of these other entries are 0. The x2's all the way to the xn are 0. So you're only going to end up with 1 times the first column here in A. So this will reduce to, let me write this, so c will be equal to, the first column is going to be b times, now a times this e1 vector, I guess we could call it the standard basis vector right there. We already said it's just going to be 1 times the first column in A, plus 0 times the second column in A, plus 0 times the third column, and so on and so forth. So it's just 1 times the first column in A."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you're only going to end up with 1 times the first column here in A. So this will reduce to, let me write this, so c will be equal to, the first column is going to be b times, now a times this e1 vector, I guess we could call it the standard basis vector right there. We already said it's just going to be 1 times the first column in A, plus 0 times the second column in A, plus 0 times the third column, and so on and so forth. So it's just 1 times the first column in A. So it's just a1. That simple. Now, what is this one going to be equal to?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's just 1 times the first column in A. So it's just a1. That simple. Now, what is this one going to be equal to? Well, it's going to be 0 times the first column in A, plus 1 times the second column in A, plus 0 times the third column in A, and the rest are going to be 0. So it's just going to be 1 times the second column in A. So this second column in our transformation matrix is just going to be b times a2."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this one going to be equal to? Well, it's going to be 0 times the first column in A, plus 1 times the second column in A, plus 0 times the third column in A, and the rest are going to be 0. So it's just going to be 1 times the second column in A. So this second column in our transformation matrix is just going to be b times a2. And I think you get the idea here. Each of these, the next one's going to be b times a3, and all the way until you get b times an. All the way until you get b times an."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this second column in our transformation matrix is just going to be b times a2. And I think you get the idea here. Each of these, the next one's going to be b times a3, and all the way until you get b times an. All the way until you get b times an. And so that's how you would solve for your transformation matrix. Remember what we were trying to do. We were trying to find some, let me write down, kind of summarize everything that we've done so far."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All the way until you get b times an. And so that's how you would solve for your transformation matrix. Remember what we were trying to do. We were trying to find some, let me write down, kind of summarize everything that we've done so far. We had a mapping, S, that was a mapping from x to y. But x was a subset of Rn. y was a subset of Rm."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We were trying to find some, let me write down, kind of summarize everything that we've done so far. We had a mapping, S, that was a mapping from x to y. But x was a subset of Rn. y was a subset of Rm. And so we said that this transformation, this linear transformation, could be represented as some matrix A, where A is a m by n matrix times a vector x. Then I showed you another transformation. Let's call that, well, we already called it T, which was a mapping from y to z. z is a subset of Rl."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "y was a subset of Rm. And so we said that this transformation, this linear transformation, could be represented as some matrix A, where A is a m by n matrix times a vector x. Then I showed you another transformation. Let's call that, well, we already called it T, which was a mapping from y to z. z is a subset of Rl. And of course, T, the transformation T, applied to some vector in y can be represented as some matrix B times that vector. I should have drawn parentheses there, but you get the idea. And this, since it's a mapping from a subset of Rm to Rl, this will be an l by m matrix."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that, well, we already called it T, which was a mapping from y to z. z is a subset of Rl. And of course, T, the transformation T, applied to some vector in y can be represented as some matrix B times that vector. I should have drawn parentheses there, but you get the idea. And this, since it's a mapping from a subset of Rm to Rl, this will be an l by m matrix. And then we said, look, if we actually just take the definition of, if we take the composition of T with S, this reduced to, of some vector in x, this reduced to B. So first we applied the S transformation. So we multiplied the matrix A times x."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this, since it's a mapping from a subset of Rm to Rl, this will be an l by m matrix. And then we said, look, if we actually just take the definition of, if we take the composition of T with S, this reduced to, of some vector in x, this reduced to B. So first we applied the S transformation. So we multiplied the matrix A times x. We multiplied the matrix A times x. And then we apply the T transformation to this. So we just multiply B times that."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we multiplied the matrix A times x. We multiplied the matrix A times x. And then we apply the T transformation to this. So we just multiply B times that. We've multiplied B times that. Now we know that this is a linear transformation, which means it can be represented as a matrix vector product. And we just figured out what the matrix vector product is."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we just multiply B times that. We've multiplied B times that. Now we know that this is a linear transformation, which means it can be represented as a matrix vector product. And we just figured out what the matrix vector product is. So this thing is going to be equal to, this is equivalent to, let me switch colors, c times x, which is equal to this thing right there. Let me write it this way. B, B, A1, I'll try to be true to the colors, B, A1, where A1 is the first column vector in our matrix A."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we just figured out what the matrix vector product is. So this thing is going to be equal to, this is equivalent to, let me switch colors, c times x, which is equal to this thing right there. Let me write it this way. B, B, A1, I'll try to be true to the colors, B, A1, where A1 is the first column vector in our matrix A. And then the second column here is going to be B. And then we have A2, where this is the second column vector in A. And then you could keep going all the way until you have B times An."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "B, B, A1, I'll try to be true to the colors, B, A1, where A1 is the first column vector in our matrix A. And then the second column here is going to be B. And then we have A2, where this is the second column vector in A. And then you could keep going all the way until you have B times An. Right there. At times x, of course. Maybe make it purple."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you could keep going all the way until you have B times An. Right there. At times x, of course. Maybe make it purple. Times x, like that. Now, this is fair enough. We can always do this."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe make it purple. Times x, like that. Now, this is fair enough. We can always do this. If you give me some matrix, remember this is an L by M matrix, and you give me another matrix right here that is a M by N matrix, I can always do this. And how do I know I can always do that? Because how many entries are each of these As are going to have M entries, right?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can always do this. If you give me some matrix, remember this is an L by M matrix, and you give me another matrix right here that is a M by N matrix, I can always do this. And how do I know I can always do that? Because how many entries are each of these As are going to have M entries, right? They're going to be, let's see, AI, all of them are going to be members of RM. So this is well defined. This has M columns, this has M entries, so each of these matrix vector products are well defined."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because how many entries are each of these As are going to have M entries, right? They're going to be, let's see, AI, all of them are going to be members of RM. So this is well defined. This has M columns, this has M entries, so each of these matrix vector products are well defined. Now, this is an interesting thing, because we were able to figure out the actual matrix representation of this composition transformation. But let's extend it a little bit further. Wouldn't it be nice if this were the same thing as the matrices B times A?"}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This has M columns, this has M entries, so each of these matrix vector products are well defined. Now, this is an interesting thing, because we were able to figure out the actual matrix representation of this composition transformation. But let's extend it a little bit further. Wouldn't it be nice if this were the same thing as the matrices B times A? Wouldn't this be nice if this were the same things as the matrix B times A, all of that times X? Wouldn't it be nice if these were the same thing? Because then we could say that the composition of T with S of X is equal to the matrix representation of B times the matrix representation of S, and you take the product of those two, and then that will create a new matrix representation, which you could call C, that you can then multiply times X."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Wouldn't it be nice if this were the same thing as the matrices B times A? Wouldn't this be nice if this were the same things as the matrix B times A, all of that times X? Wouldn't it be nice if these were the same thing? Because then we could say that the composition of T with S of X is equal to the matrix representation of B times the matrix representation of S, and you take the product of those two, and then that will create a new matrix representation, which you could call C, that you can then multiply times X. So you won't have to do it individually every time, or do it this way. And I guess the truth of the matter is, there is nothing to stop us from defining this to be equal to B times A. We have not defined what a matrix times a matrix is yet."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because then we could say that the composition of T with S of X is equal to the matrix representation of B times the matrix representation of S, and you take the product of those two, and then that will create a new matrix representation, which you could call C, that you can then multiply times X. So you won't have to do it individually every time, or do it this way. And I guess the truth of the matter is, there is nothing to stop us from defining this to be equal to B times A. We have not defined what a matrix times a matrix is yet. So we might as well, this is good enough motivation for us to define it in this way. So let's throw in this definition. Let us define."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have not defined what a matrix times a matrix is yet. So we might as well, this is good enough motivation for us to define it in this way. So let's throw in this definition. Let us define. So if we have some matrix B, and we could, well, I don't have to draw what it looks like, but B is a L by M matrix, and then we have some other matrix A, and I'll actually show what A looks like, where these are its column vectors, A1, A2, all the way to An. Then we're going to define the product. So this is a definition."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let us define. So if we have some matrix B, and we could, well, I don't have to draw what it looks like, but B is a L by M matrix, and then we have some other matrix A, and I'll actually show what A looks like, where these are its column vectors, A1, A2, all the way to An. Then we're going to define the product. So this is a definition. We're going to define the product BA as being equal to the matrix B times each of the column vectors of A. So it's B times A1, that's going to be its first column. This is going to be B times A2, all the way to B times An."}, {"video_title": "Compositions of linear transformations 2 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is a definition. We're going to define the product BA as being equal to the matrix B times each of the column vectors of A. So it's B times A1, that's going to be its first column. This is going to be B times A2, all the way to B times An. And you've seen this before in algebra 2, but the reason why I did, when I went through kind of almost two videos to get to here, is to show you the motivation for why matrix products are defined this way. Because it makes the notion of compositions of transformations kind of natural. If you take the composition of one linear transformation with another, the resulting transformation matrix is just the product, as we've just defined it, of their two transformation matrices."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll this over a little bit. Let's say my first position vector is x-naught, and it is equal to, let's say it's equal to minus 2, minus 2. So if I were to graph x-naught, I'd go minus 2, minus 2. x-naught looks like that. x-naught. My next position vector I have is x1, and I'll say that's equal to minus 2, 2. So if I were to graph here, minus 2, 2. That's my next position vector right there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "x-naught. My next position vector I have is x1, and I'll say that's equal to minus 2, 2. So if I were to graph here, minus 2, 2. That's my next position vector right there. And this is x1. And when I say it's a position vector, they specify a specific coordinate in R2. And let me draw a third one just for fun."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's my next position vector right there. And this is x1. And when I say it's a position vector, they specify a specific coordinate in R2. And let me draw a third one just for fun. x2. Let's say that that is equal to, that is equal to 2, minus 2. So if I were to draw this, it'd be 2, minus 2."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let me draw a third one just for fun. x2. Let's say that that is equal to, that is equal to 2, minus 2. So if I were to draw this, it'd be 2, minus 2. It goes right here. So that vector right there is x2. Now, what I'm curious about, or I guess not curious about, what I want to do here is define the line segments that connect these points."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I were to draw this, it'd be 2, minus 2. It goes right here. So that vector right there is x2. Now, what I'm curious about, or I guess not curious about, what I want to do here is define the line segments that connect these points. So let's say I have my first line segment. Let me call it L1. Let me call it L-naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what I'm curious about, or I guess not curious about, what I want to do here is define the line segments that connect these points. So let's say I have my first line segment. Let me call it L1. Let me call it L-naught. L-naught. And I want it to be the line segment that connects x-naught to x1. How can I construct that?"}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me call it L-naught. L-naught. And I want it to be the line segment that connects x-naught to x1. How can I construct that? Well, that's going to be so that I want to construct this line segment right here. This little, let me do it in a different color, actually. Let me do it in orange."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How can I construct that? Well, that's going to be so that I want to construct this line segment right here. This little, let me do it in a different color, actually. Let me do it in orange. L-naught. So what I want to do is I want to find the set of all of these values right here, all of the position vectors that define points on this line right there. Well, we could define it as, we could start off as at x-naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in orange. L-naught. So what I want to do is I want to find the set of all of these values right here, all of the position vectors that define points on this line right there. Well, we could define it as, we could start off as at x-naught. We could say that orange line is x-naught plus scaled versions of the difference of x1 and x-naught, right? If you take x1 minus x-naught, you get this vector right there. That's x1 minus x-naught, that orange vector."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we could define it as, we could start off as at x-naught. We could say that orange line is x-naught plus scaled versions of the difference of x1 and x-naught, right? If you take x1 minus x-naught, you get this vector right there. That's x1 minus x-naught, that orange vector. I know I wrote it right there, so it's hard to read. But if you just take x1 minus x-naught, you get that. And that makes sense."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's x1 minus x-naught, that orange vector. I know I wrote it right there, so it's hard to read. But if you just take x1 minus x-naught, you get that. And that makes sense. x-naught plus this orange vector is equal to that blue vector. So if you just take different scaled-up versions of this guy, you're going to end up at different points in this direction, right? We're starting at x-naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that makes sense. x-naught plus this orange vector is equal to that blue vector. So if you just take different scaled-up versions of this guy, you're going to end up at different points in this direction, right? We're starting at x-naught. Maybe I should do that in green. We're starting at x-naught, and then we're going to add up scaled-up versions of this orange vector, which is just the difference of x1 and x-naught. So let me write that."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're starting at x-naught. Maybe I should do that in green. We're starting at x-naught, and then we're going to add up scaled-up versions of this orange vector, which is just the difference of x1 and x-naught. So let me write that. So scaled-up versions of x1 minus x-naught. Now, we have to constrict. If we just want to be in this line segment, we have to constrict our t. If we said t was a member of real numbers, if it was any real number, then we would essentially be defining the set of this entire vertical line going up and down to infinity in both the upwards direction and the downwards direction."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that. So scaled-up versions of x1 minus x-naught. Now, we have to constrict. If we just want to be in this line segment, we have to constrict our t. If we said t was a member of real numbers, if it was any real number, then we would essentially be defining the set of this entire vertical line going up and down to infinity in both the upwards direction and the downwards direction. But we just want to restrict it to start here and then go up here. And it doesn't necessarily have to have any direction. We could say this is true."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we just want to be in this line segment, we have to constrict our t. If we said t was a member of real numbers, if it was any real number, then we would essentially be defining the set of this entire vertical line going up and down to infinity in both the upwards direction and the downwards direction. But we just want to restrict it to start here and then go up here. And it doesn't necessarily have to have any direction. We could say this is true. Our little line segment here is true for t. Let me write it this way. t is greater than or equal to 0. So when t equals 0, this term cancels out, and we just have this point or this position right here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could say this is true. Our little line segment here is true for t. Let me write it this way. t is greater than or equal to 0. So when t equals 0, this term cancels out, and we just have this point or this position right here. Let me draw it in green. We just have that position there. And then t is going to be less than or equal to 1."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So when t equals 0, this term cancels out, and we just have this point or this position right here. Let me draw it in green. We just have that position there. And then t is going to be less than or equal to 1. What happens when t is equal to 1? When t is equal to 1, this becomes x1 minus x-naught. You have an x-naught there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then t is going to be less than or equal to 1. What happens when t is equal to 1? When t is equal to 1, this becomes x1 minus x-naught. You have an x-naught there. This x-naught and that x-naught cancel out, and you're just at this point right there. When t is equal to 1 half, just to make sure this all makes sense to you, what happens? You have x1 minus x-naught, which is this orange vector right here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have an x-naught there. This x-naught and that x-naught cancel out, and you're just at this point right there. When t is equal to 1 half, just to make sure this all makes sense to you, what happens? You have x1 minus x-naught, which is this orange vector right here. When t is equal to 1 half, you're essentially scaling that orange vector by half, and you end up right at that point, which is exactly where you want to be. You want halfway along that line segment. At t is equal to 0.25, you're going to be here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have x1 minus x-naught, which is this orange vector right here. When t is equal to 1 half, you're essentially scaling that orange vector by half, and you end up right at that point, which is exactly where you want to be. You want halfway along that line segment. At t is equal to 0.25, you're going to be here. t is equal to 0.75, you're going to be there. So at any value for t, being any real number between 0 and 1, you're going to end up at all of the points along that line segment. So that's our L-naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "At t is equal to 0.25, you're going to be here. t is equal to 0.75, you're going to be there. So at any value for t, being any real number between 0 and 1, you're going to end up at all of the points along that line segment. So that's our L-naught. That's our L-naught. It's just a set of vectors. Now, we could do the same exercise if we wanted to find out the line, the equation of the line that goes between x1 and x2, if we wanted to find the equation of that line."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's our L-naught. That's our L-naught. It's just a set of vectors. Now, we could do the same exercise if we wanted to find out the line, the equation of the line that goes between x1 and x2, if we wanted to find the equation of that line. We could call this L1. We call that L1. And L1 would be equal to x1 plus t times x2 minus x1 for 0 is less than or equal to t is less than or equal to 1."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we could do the same exercise if we wanted to find out the line, the equation of the line that goes between x1 and x2, if we wanted to find the equation of that line. We could call this L1. We call that L1. And L1 would be equal to x1 plus t times x2 minus x1 for 0 is less than or equal to t is less than or equal to 1. That's L1. And then finally, if we want to make a triangle out of this, let's define this line right here. Let's define that as L2."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And L1 would be equal to x1 plus t times x2 minus x1 for 0 is less than or equal to t is less than or equal to 1. That's L1. And then finally, if we want to make a triangle out of this, let's define this line right here. Let's define that as L2. L2 would be equal to the set of all vectors where you would start off at x2, set of all vectors that are x2 plus some scaled-up sum of x-naught minus x2. x-naught minus x2 is this vector right here. So x-naught minus x2 such that 0 is less than or equal to t is less than or equal to 1."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's define that as L2. L2 would be equal to the set of all vectors where you would start off at x2, set of all vectors that are x2 plus some scaled-up sum of x-naught minus x2. x-naught minus x2 is this vector right here. So x-naught minus x2 such that 0 is less than or equal to t is less than or equal to 1. And so if you take the combination, if you were to define kind of a superset, let me say, you know, if I could define my shape as, let's say, it's the union of all of those guys, so it's, you know, I could, well, let me just write it, L-naught, L1, and L2, then you'd have a nice triangle here. If you take the union of all of these three sets, you get that nice triangle there. Now, what I want to do in this video, I think this is all a bit of a review for you, but it's maybe a different way of looking at things than we've done in the past, is I want to understand what happens to this set right here when I take a transformation, a linear transformation of it."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So x-naught minus x2 such that 0 is less than or equal to t is less than or equal to 1. And so if you take the combination, if you were to define kind of a superset, let me say, you know, if I could define my shape as, let's say, it's the union of all of those guys, so it's, you know, I could, well, let me just write it, L-naught, L1, and L2, then you'd have a nice triangle here. If you take the union of all of these three sets, you get that nice triangle there. Now, what I want to do in this video, I think this is all a bit of a review for you, but it's maybe a different way of looking at things than we've done in the past, is I want to understand what happens to this set right here when I take a transformation, a linear transformation of it. So let me define a transformation. I'll make it a fairly straightforward transformation. Let me define my transformation of x, of any x, to be equal to the matrix 1, minus 1, 2, 0, times whatever vector x, so times x1, x2, and we know that any linear transformation can actually be written as a matrix, and vice versa."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what I want to do in this video, I think this is all a bit of a review for you, but it's maybe a different way of looking at things than we've done in the past, is I want to understand what happens to this set right here when I take a transformation, a linear transformation of it. So let me define a transformation. I'll make it a fairly straightforward transformation. Let me define my transformation of x, of any x, to be equal to the matrix 1, minus 1, 2, 0, times whatever vector x, so times x1, x2, and we know that any linear transformation can actually be written as a matrix, and vice versa. So you might have said, hey, you know, you're giving an example of the matrix, what about all those other ways to write a linear transformation? You can write all of those as a matrix. So let's translate, let's try to figure out what this is going to look like, what our triangle is going to look like when we transform every point in it."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me define my transformation of x, of any x, to be equal to the matrix 1, minus 1, 2, 0, times whatever vector x, so times x1, x2, and we know that any linear transformation can actually be written as a matrix, and vice versa. So you might have said, hey, you know, you're giving an example of the matrix, what about all those other ways to write a linear transformation? You can write all of those as a matrix. So let's translate, let's try to figure out what this is going to look like, what our triangle is going to look like when we transform every point in it. So what is, let me take the transformation first, the transformation of L naught, the transformation of L naught is equal to the transformation of this thing, right? This is just one of the particular members, for a particular t, this is one of the particular members of L naught, so it's going to be equal to the transformation, the transformation of x naught, minus the transformation of x1 minus x naught, such that, sorry, minus t times x1 minus x naught, that's a lowercase t, not the transformation, such that 0 is less than or equal to t is less than or equal to 1. Let me switch colors."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's translate, let's try to figure out what this is going to look like, what our triangle is going to look like when we transform every point in it. So what is, let me take the transformation first, the transformation of L naught, the transformation of L naught is equal to the transformation of this thing, right? This is just one of the particular members, for a particular t, this is one of the particular members of L naught, so it's going to be equal to the transformation, the transformation of x naught, minus the transformation of x1 minus x naught, such that, sorry, minus t times x1 minus x naught, that's a lowercase t, not the transformation, such that 0 is less than or equal to t is less than or equal to 1. Let me switch colors. This, just by the properties of linear transformations, this is equal to the transformation, let me put the brackets out, of x naught minus the transformation of our scalar t times x1 minus x naught, for all t's between 0 and 1, that part is getting a little redundant to keep saying it, and then what is this equal to? I take the transformation of a scaled up vector, that's just a scaled up transformation of that vector, so this is going to be equal to this part, the transformation of x naught minus t, our scalar multiple t, times the transformation of the vector x1 minus x naught, and then let me make sure I get my parentheses right, such that 0 is less than or equal to t is less than or equal to 1, and then the transformation of the sum of two vectors is equal to the sum of their transformations. We've all seen this before."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me switch colors. This, just by the properties of linear transformations, this is equal to the transformation, let me put the brackets out, of x naught minus the transformation of our scalar t times x1 minus x naught, for all t's between 0 and 1, that part is getting a little redundant to keep saying it, and then what is this equal to? I take the transformation of a scaled up vector, that's just a scaled up transformation of that vector, so this is going to be equal to this part, the transformation of x naught minus t, our scalar multiple t, times the transformation of the vector x1 minus x naught, and then let me make sure I get my parentheses right, such that 0 is less than or equal to t is less than or equal to 1, and then the transformation of the sum of two vectors is equal to the sum of their transformations. We've all seen this before. So our transformation of our first line, this one right here, L naught, is equal to the set where it's a transformation of x naught minus t times the transformation of x1 minus the transformation of x naught. And we've just done our first line so far, and I have a parentheses there, for 0 is less than t is less than or equal to 1. Now this is a pretty neat result, and it's going to simplify our lives a lot."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've all seen this before. So our transformation of our first line, this one right here, L naught, is equal to the set where it's a transformation of x naught minus t times the transformation of x1 minus the transformation of x naught. And we've just done our first line so far, and I have a parentheses there, for 0 is less than t is less than or equal to 1. Now this is a pretty neat result, and it's going to simplify our lives a lot. The transformation of the line segment that goes from x naught to x1 ends up just being the line segment that goes from the transformation of x naught to the transformation of x1. Let me make this clear. What is the transformation of x naught?"}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now this is a pretty neat result, and it's going to simplify our lives a lot. The transformation of the line segment that goes from x naught to x1 ends up just being the line segment that goes from the transformation of x naught to the transformation of x1. Let me make this clear. What is the transformation of x naught? Let's calculate these things. So x naught was minus 2 minus 2. Let me write out the transformation of x naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is the transformation of x naught? Let's calculate these things. So x naught was minus 2 minus 2. Let me write out the transformation of x naught. So the transformation of x naught is equal to let me write it out so I don't make any careless mistakes, 1, 2, minus 1, 0, times minus 2 minus 2. And so what would this be equal to? 1 minus 2, minus, so it's 1 minus 2 plus minus 1 times minus 2, that's going to be a plus 2."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write out the transformation of x naught. So the transformation of x naught is equal to let me write it out so I don't make any careless mistakes, 1, 2, minus 1, 0, times minus 2 minus 2. And so what would this be equal to? 1 minus 2, minus, so it's 1 minus 2 plus minus 1 times minus 2, that's going to be a plus 2. So it's minus 2 plus a plus 2, so it's equal to 0. And then we have 2 times minus 2, which is minus 4, 2 times minus 4, and then plus 0 times, so it's minus 4. So that's the transformation of x naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2, minus, so it's 1 minus 2 plus minus 1 times minus 2, that's going to be a plus 2. So it's minus 2 plus a plus 2, so it's equal to 0. And then we have 2 times minus 2, which is minus 4, 2 times minus 4, and then plus 0 times, so it's minus 4. So that's the transformation of x naught. Let me graph it. So it's 0 minus 4. So our x naught vector, so this is the transformation of x naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's the transformation of x naught. Let me graph it. So it's 0 minus 4. So our x naught vector, so this is the transformation of x naught. So the x naught, the transformation associated this vector with this vector down here, the one that goes straight down. Now let me take the transformation of the other guys. The transformation of x1, I'll just do it right here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So our x naught vector, so this is the transformation of x naught. So the x naught, the transformation associated this vector with this vector down here, the one that goes straight down. Now let me take the transformation of the other guys. The transformation of x1, I'll just do it right here. I'm running out of space. The transformation of x1 is equal to 1, 2, minus 1, 0 times minus 2, 2. So what is this equal to?"}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation of x1, I'll just do it right here. I'm running out of space. The transformation of x1 is equal to 1, 2, minus 1, 0 times minus 2, 2. So what is this equal to? This is equal to 1 times minus 2 plus minus 1 times 2, so that's minus 4. And then 2 times minus 2 is minus 4 plus 0, so minus 4, minus 4. So x1 is minus 4, minus 4."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is this equal to? This is equal to 1 times minus 2 plus minus 1 times 2, so that's minus 4. And then 2 times minus 2 is minus 4 plus 0, so minus 4, minus 4. So x1 is minus 4, minus 4. So our x1 looks like this. Our transformation of x1 is this vector right here in R2. Our transformation is going from R2 to R2, so that's why I'm able to draw them both on this nice Cartesian coordinate plane."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So x1 is minus 4, minus 4. So our x1 looks like this. Our transformation of x1 is this vector right here in R2. Our transformation is going from R2 to R2, so that's why I'm able to draw them both on this nice Cartesian coordinate plane. And we have one left. Let's take our transformation of x2. So the transformation of x2 is equal to our transformation matrix, 1, 2, minus 1, 0 times 2, minus 2."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our transformation is going from R2 to R2, so that's why I'm able to draw them both on this nice Cartesian coordinate plane. And we have one left. Let's take our transformation of x2. So the transformation of x2 is equal to our transformation matrix, 1, 2, minus 1, 0 times 2, minus 2. And so this will be equal to 1 times 2 is 2 plus minus 1 times minus 2, so it's 2 plus 2 is 4. And then I have 2 times 2 is 4 plus 0 times minus 2, so it's 4, 4. x2 is 4, 4, so it's this point right here. 4, 4 right there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the transformation of x2 is equal to our transformation matrix, 1, 2, minus 1, 0 times 2, minus 2. And so this will be equal to 1 times 2 is 2 plus minus 1 times minus 2, so it's 2 plus 2 is 4. And then I have 2 times 2 is 4 plus 0 times minus 2, so it's 4, 4. x2 is 4, 4, so it's this point right here. 4, 4 right there. x2 is, so the transformation of x2 is that vector right there. And so, you know, we are able to take the transformation of each of these points of this triangle, but who knows what the transformation does to everything in between, to all of these other, the actual sides of the triangle. But we're able to do a little math, and we just did the first side, when we just did L naught right there, and we found, just using the properties of a linear transformation, the definition of a linear transformation, actually, we're able to find that the transformation of L naught of this vertical line here, it just ends up becoming the line where we can start off at the transformation of x naught, that's this point, that's the point specified by this vector right here, and to that I add up scaled multiples of the transformation of x1 minus the transformation of x naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "4, 4 right there. x2 is, so the transformation of x2 is that vector right there. And so, you know, we are able to take the transformation of each of these points of this triangle, but who knows what the transformation does to everything in between, to all of these other, the actual sides of the triangle. But we're able to do a little math, and we just did the first side, when we just did L naught right there, and we found, just using the properties of a linear transformation, the definition of a linear transformation, actually, we're able to find that the transformation of L naught of this vertical line here, it just ends up becoming the line where we can start off at the transformation of x naught, that's this point, that's the point specified by this vector right here, and to that I add up scaled multiples of the transformation of x1 minus the transformation of x naught. What is this? The transformation of x1 minus the transformation of x naught. The transformation of x1 is just this vector right here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But we're able to do a little math, and we just did the first side, when we just did L naught right there, and we found, just using the properties of a linear transformation, the definition of a linear transformation, actually, we're able to find that the transformation of L naught of this vertical line here, it just ends up becoming the line where we can start off at the transformation of x naught, that's this point, that's the point specified by this vector right here, and to that I add up scaled multiples of the transformation of x1 minus the transformation of x naught. What is this? The transformation of x1 minus the transformation of x naught. The transformation of x1 is just this vector right here. The transformation of x naught is just that vector. So this whole term right here is just this vector minus that vector, or it's this vector right there. It's just that vector right there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation of x1 is just this vector right here. The transformation of x naught is just that vector. So this whole term right here is just this vector minus that vector, or it's this vector right there. It's just that vector right there. And so, what we essentially have, we've defined the same way that we did it in the first part of this video. This is just the same thing as the line segment that connects the point defined here and the point defined there. We took the difference of the two and we have scaled up versions of that between t is equal to zero and one."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's just that vector right there. And so, what we essentially have, we've defined the same way that we did it in the first part of this video. This is just the same thing as the line segment that connects the point defined here and the point defined there. We took the difference of the two and we have scaled up versions of that between t is equal to zero and one. So the transformation of L naught really just became the transformation, is just the line between the transformations of both of the endpoints, which is a pretty neat result. It makes our life simple. We can do the exact same logic to say, you know what, what's going to be the transformation, what's going to be the transformation of L1?"}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We took the difference of the two and we have scaled up versions of that between t is equal to zero and one. So the transformation of L naught really just became the transformation, is just the line between the transformations of both of the endpoints, which is a pretty neat result. It makes our life simple. We can do the exact same logic to say, you know what, what's going to be the transformation, what's going to be the transformation of L1? Well, L1 was the point, was between the points x1 and x2. It was between that point and that point. That was L1."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can do the exact same logic to say, you know what, what's going to be the transformation, what's going to be the transformation of L1? Well, L1 was the point, was between the points x1 and x2. It was between that point and that point. That was L1. So using the same logic, we can do the math all over again, but it applies to any line. I did it all abstractly here. The transformation of L1 is going to be the line that connects the transformation of the two endpoints."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That was L1. So using the same logic, we can do the math all over again, but it applies to any line. I did it all abstractly here. The transformation of L1 is going to be the line that connects the transformation of the two endpoints. So it's going to be the line that connects the transformation of x1 and the transformation of x2. So it's going to be, so this is, so we make this right here, is the transformation of L1. This right here is the transformation of L naught."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation of L1 is going to be the line that connects the transformation of the two endpoints. So it's going to be the line that connects the transformation of x1 and the transformation of x2. So it's going to be, so this is, so we make this right here, is the transformation of L1. This right here is the transformation of L naught. And then finally, what's the transformation of L2? The transformation of L2, L2 connects the points x2 and x naught. So that is L2 right there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This right here is the transformation of L naught. And then finally, what's the transformation of L2? The transformation of L2, L2 connects the points x2 and x naught. So that is L2 right there. So the transformation of it, using the same math that we've done before, is really just the line connecting the transformations of those two points. So the transformation of L2 is going to be equal to the line that connects the transformation of x2 to the transformation of x naught. So it's going to be that line right there."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that is L2 right there. So the transformation of it, using the same math that we've done before, is really just the line connecting the transformations of those two points. So the transformation of L2 is going to be equal to the line that connects the transformation of x2 to the transformation of x naught. So it's going to be that line right there. So this is the transformation of L2. Or if we defined our whole shape or our whole triangle as the set of all of these, the transformation of that, so the transformation of our whole shape, is now this skewed triangle. I think you're now getting a sense of why this might be useful in things like computer graphics or game development."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be that line right there. So this is the transformation of L2. Or if we defined our whole shape or our whole triangle as the set of all of these, the transformation of that, so the transformation of our whole shape, is now this skewed triangle. I think you're now getting a sense of why this might be useful in things like computer graphics or game development. Because when you look at things from different angles, you start to skew them and whatever else. But taking this transformation, we were able to turn this set of vectors, or the positions, or I guess this shape, which is specified by this set of vectors right here, we were able to change it into this shape in R2 specified by a different set of vectors. But the whole, I guess, takeaway from this video is you don't have to individually figure out what does this point right here translate into over here."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think you're now getting a sense of why this might be useful in things like computer graphics or game development. Because when you look at things from different angles, you start to skew them and whatever else. But taking this transformation, we were able to turn this set of vectors, or the positions, or I guess this shape, which is specified by this set of vectors right here, we were able to change it into this shape in R2 specified by a different set of vectors. But the whole, I guess, takeaway from this video is you don't have to individually figure out what does this point right here translate into over here. All you have to say is what were my endpoints, figure out their transformations, and then connect the dots in the same order. That's what is essentially the big takeaway from here. And this idea of when you transform one set into another set, they have some terminology associated with it."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But the whole, I guess, takeaway from this video is you don't have to individually figure out what does this point right here translate into over here. All you have to say is what were my endpoints, figure out their transformations, and then connect the dots in the same order. That's what is essentially the big takeaway from here. And this idea of when you transform one set into another set, they have some terminology associated with it. Let's say the transformation of L0, L0 was the set of vectors that specified this line right here. The transformation of L0, which is this set, is a set of vectors, sorry, L0 was this set. It was the set of vectors in our codomain that specified these points."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this idea of when you transform one set into another set, they have some terminology associated with it. Let's say the transformation of L0, L0 was the set of vectors that specified this line right here. The transformation of L0, which is this set, is a set of vectors, sorry, L0 was this set. It was the set of vectors in our codomain that specified these points. This is called the image of L0 under T. And it kind of makes sense. Why do they call it the image? Because T is taking this thing right here, this L0, and kind of distorting it or creating a new image of it in the codomain."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It was the set of vectors in our codomain that specified these points. This is called the image of L0 under T. And it kind of makes sense. Why do they call it the image? Because T is taking this thing right here, this L0, and kind of distorting it or creating a new image of it in the codomain. It's taking a set in the domain and creating a new image of it in the codomain right there. And we could say that T of the transformation of our entire shape, I defined our entire shape up here as this whole triangle right there, that's the image, and that turned into this purple triangle here, that's the image of S under T. So hopefully you found that pretty interesting. This is actually a super useful takeaway if you ever want to become a 3D programmer of subtype."}, {"video_title": "Image of a subset under a transformation Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because T is taking this thing right here, this L0, and kind of distorting it or creating a new image of it in the codomain. It's taking a set in the domain and creating a new image of it in the codomain right there. And we could say that T of the transformation of our entire shape, I defined our entire shape up here as this whole triangle right there, that's the image, and that turned into this purple triangle here, that's the image of S under T. So hopefully you found that pretty interesting. This is actually a super useful takeaway if you ever want to become a 3D programmer of subtype. In the next video, we'll explore what happens when S is no longer just a subset of our domain. Everything we've been dealing so far, L0, L1, and L2, our entire triangle, these were all subsets of Rn. In the next video, we'll talk about what happens when you take the transformation of all of Rn."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I take the cross product of A and then B cross C. And what we're going to do is we're going to express this, we can express this really as sum and differences of dot products. Well not just dot products, dot products scaling different vectors. You're going to see what I mean. But it simplifies this expression a good bit because cross products are hard to take. They're computationally intensive and at least in my mind they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors. But it's useful to know."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But it simplifies this expression a good bit because cross products are hard to take. They're computationally intensive and at least in my mind they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors. But it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula or the triple product expansion. So let's see how we can simplify this. So to do that let's start taking the cross product of B and C. And in all of these situations I'm just going to assume, let's say I have vector A. I'm just going to call that, that's going to be A, the X component of vector A times the unit vector I plus the Y component, not B, plus the Y component of vector A times the unit vector J plus the Z component of vector A times unit vector K. And I could do the same things for B and C. So if I say B sub Y I'm talking about what's scaling the J component in the B vector."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula or the triple product expansion. So let's see how we can simplify this. So to do that let's start taking the cross product of B and C. And in all of these situations I'm just going to assume, let's say I have vector A. I'm just going to call that, that's going to be A, the X component of vector A times the unit vector I plus the Y component, not B, plus the Y component of vector A times the unit vector J plus the Z component of vector A times unit vector K. And I could do the same things for B and C. So if I say B sub Y I'm talking about what's scaling the J component in the B vector. So let's first take this cross product over here. And if you've seen me take cross products you know that I like to do these little determinants. And let me just take it over here."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So to do that let's start taking the cross product of B and C. And in all of these situations I'm just going to assume, let's say I have vector A. I'm just going to call that, that's going to be A, the X component of vector A times the unit vector I plus the Y component, not B, plus the Y component of vector A times the unit vector J plus the Z component of vector A times unit vector K. And I could do the same things for B and C. So if I say B sub Y I'm talking about what's scaling the J component in the B vector. So let's first take this cross product over here. And if you've seen me take cross products you know that I like to do these little determinants. And let me just take it over here. So B cross C is going to be equal to the determinant, and I put an IJK up here. This is actually the definition of the cross product so no proof necessary to show you why this is true. This is just one way to remember the dot product if you remember how to take determinants of 3 by 3s."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And let me just take it over here. So B cross C is going to be equal to the determinant, and I put an IJK up here. This is actually the definition of the cross product so no proof necessary to show you why this is true. This is just one way to remember the dot product if you remember how to take determinants of 3 by 3s. And we'll put B's X term, B's Y coefficient, and B's Z component. And then you do the same thing for the C's, CX, CY, CZ. And then this is going to be equal to, so first you have the I component."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is just one way to remember the dot product if you remember how to take determinants of 3 by 3s. And we'll put B's X term, B's Y coefficient, and B's Z component. And then you do the same thing for the C's, CX, CY, CZ. And then this is going to be equal to, so first you have the I component. So it's going to be the I component times B. So you ignore this column and this row, so BYCZ minus BZCY. So I'm just ignoring all of this and I'm looking at this 2 by 2 over here, minus BZCY."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then this is going to be equal to, so first you have the I component. So it's going to be the I component times B. So you ignore this column and this row, so BYCZ minus BZCY. So I'm just ignoring all of this and I'm looking at this 2 by 2 over here, minus BZCY. And then we want to subtract the J component. Remember we alternate signs when we take our determinant. Subtract that and then we take out that column and that row, so it's going to be BXCZ minus BZCX."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just ignoring all of this and I'm looking at this 2 by 2 over here, minus BZCY. And then we want to subtract the J component. Remember we alternate signs when we take our determinant. Subtract that and then we take out that column and that row, so it's going to be BXCZ minus BZCX. And then finally plus the K component, K we're going to have BX times CY, BXCY minus BYCX. So this is, we just did the dot product. And now we want to take the, sorry we just did the cross product, I don't want to get you confused."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Subtract that and then we take out that column and that row, so it's going to be BXCZ minus BZCX. And then finally plus the K component, K we're going to have BX times CY, BXCY minus BYCX. So this is, we just did the dot product. And now we want to take the, sorry we just did the cross product, I don't want to get you confused. We just took the cross product of B and C. And now we need to take the cross product of that with A, or the cross product of A with this thing right over here. So let's do that. Instead of rewriting the vector, let me just set up another matrix here."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now we want to take the, sorry we just did the cross product, I don't want to get you confused. We just took the cross product of B and C. And now we need to take the cross product of that with A, or the cross product of A with this thing right over here. So let's do that. Instead of rewriting the vector, let me just set up another matrix here. So let me write my IJK up here and then let me write A's component. So we have A sub X, A sub Y, A sub Z. And then let's clean this up a little bit."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Instead of rewriting the vector, let me just set up another matrix here. So let me write my IJK up here and then let me write A's component. So we have A sub X, A sub Y, A sub Z. And then let's clean this up a little bit. Let's ignore this, we're just looking at, I want to do that in black. Let's do this in black so that we can erase that. Now this is a minus J times that."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then let's clean this up a little bit. Let's ignore this, we're just looking at, I want to do that in black. Let's do this in black so that we can erase that. Now this is a minus J times that. So what I'm going to do is, I'm going to get rid of the minus and the J, but I'm going to rewrite this with the signs swapped. So this is going to be, if we swap the signs, it's actually BZCX minus BXCZ. So let me delete everything else."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now this is a minus J times that. So what I'm going to do is, I'm going to get rid of the minus and the J, but I'm going to rewrite this with the signs swapped. So this is going to be, if we swap the signs, it's actually BZCX minus BXCZ. So let me delete everything else. So I just took the negative and I multiplied it by this. I'm not making any careless mistakes here, so let me just make my brush size a little bit bigger so I can erase that a little more efficiently. There you go."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me delete everything else. So I just took the negative and I multiplied it by this. I'm not making any careless mistakes here, so let me just make my brush size a little bit bigger so I can erase that a little more efficiently. There you go. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size. Alright, so now let's just take this cross product."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There you go. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size. Alright, so now let's just take this cross product. So once again, set it up as a determinant. And what I'm only going to focus on, because it will take the video, or it will take me forever, if I were to do the I, J, and K components, I'm just going to focus on the I component, just on the X component of this cross product. And then we can see that we'll get the same result for the J and the K, and then we can see what hopefully this simplifies down to."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Alright, so now let's just take this cross product. So once again, set it up as a determinant. And what I'm only going to focus on, because it will take the video, or it will take me forever, if I were to do the I, J, and K components, I'm just going to focus on the I component, just on the X component of this cross product. And then we can see that we'll get the same result for the J and the K, and then we can see what hopefully this simplifies down to. So if we just focus on the I component here, this is going to be I times, and we just look at this 2 by 2 matrix right over here, we ignore I's column, I's row, and we have Ay times all of this. So let me just multiply it out. So it's Ay times Bx Cy minus Ay times By Cx."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we can see that we'll get the same result for the J and the K, and then we can see what hopefully this simplifies down to. So if we just focus on the I component here, this is going to be I times, and we just look at this 2 by 2 matrix right over here, we ignore I's column, I's row, and we have Ay times all of this. So let me just multiply it out. So it's Ay times Bx Cy minus Ay times By Cx. And then we're going to want to subtract, we're going to have minus Az times this. So let's just do that. So it's minus or negative Az Bz Cx, and then we have a negative Az times this, so it's plus Az Bx Cz."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's Ay times Bx Cy minus Ay times By Cx. And then we're going to want to subtract, we're going to have minus Az times this. So let's just do that. So it's minus or negative Az Bz Cx, and then we have a negative Az times this, so it's plus Az Bx Cz. And now what I'm going to do, this is a little bit of a trick for this proof right here, just so that we get the results that I want. I'm just going to add and subtract the exact same thing. So I'm going to add an Ax Bx Cx, and then I'm going to subtract an Ax Bx Cx, minus Ax Bx Cx."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's minus or negative Az Bz Cx, and then we have a negative Az times this, so it's plus Az Bx Cz. And now what I'm going to do, this is a little bit of a trick for this proof right here, just so that we get the results that I want. I'm just going to add and subtract the exact same thing. So I'm going to add an Ax Bx Cx, and then I'm going to subtract an Ax Bx Cx, minus Ax Bx Cx. So clearly I have not changed this expression, I have just added and subtracted the same thing. And let's see what we can simplify. Remember, this is just the x component of our triple product, just the x component."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to add an Ax Bx Cx, and then I'm going to subtract an Ax Bx Cx, minus Ax Bx Cx. So clearly I have not changed this expression, I have just added and subtracted the same thing. And let's see what we can simplify. Remember, this is just the x component of our triple product, just the x component. But to do this, let me factor out. I'm going to factor out a Bx. So let me do this."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, this is just the x component of our triple product, just the x component. But to do this, let me factor out. I'm going to factor out a Bx. So let me do this. Let me get the Bx. So if I were to factor it out, and I'm going to factor out of this term, that has a Bx. I'm going to factor out of this term, and then I'm going to factor it out of this term."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me do this. Let me get the Bx. So if I were to factor it out, and I'm going to factor out of this term, that has a Bx. I'm going to factor out of this term, and then I'm going to factor it out of this term. So if I take the Bx out, I'm going to have an Ay Cy. Actually, let me write it a little bit differently. Let me factor it out of this one first."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to factor out of this term, and then I'm going to factor it out of this term. So if I take the Bx out, I'm going to have an Ay Cy. Actually, let me write it a little bit differently. Let me factor it out of this one first. So then it's going to have an Ax Cx. A sub x, C sub x. So I use this one up."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me factor it out of this one first. So then it's going to have an Ax Cx. A sub x, C sub x. So I use this one up. And then I'm going to have a, I'll do this one now, plus, if I factor the Bx out, I get A sub y, C sub y. I've used that one now, and now I have this one. I'm going to factor the Bx out. So I'm left with a plus A sub z, C sub z."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I use this one up. And then I'm going to have a, I'll do this one now, plus, if I factor the Bx out, I get A sub y, C sub y. I've used that one now, and now I have this one. I'm going to factor the Bx out. So I'm left with a plus A sub z, C sub z. So that's all of those. So I factored that out. And now from these right over here, let me factor out a negative C sub x."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm left with a plus A sub z, C sub z. So that's all of those. So I factored that out. And now from these right over here, let me factor out a negative C sub x. And so if I do that, let me go to this term right over here. I'm going to have an Ax Bx when I factor it out. So an Ax Bx, cross that out."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now from these right over here, let me factor out a negative C sub x. And so if I do that, let me go to this term right over here. I'm going to have an Ax Bx when I factor it out. So an Ax Bx, cross that out. And then over here I'm going to have an Ay By. Remember, I'm factoring out a negative C sub x. So I'm going to have a plus A sub y, B sub y."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So an Ax Bx, cross that out. And then over here I'm going to have an Ay By. Remember, I'm factoring out a negative C sub x. So I'm going to have a plus A sub y, B sub y. And then finally I'm going to have a plus A sub z, A sub z, B sub z. And what is this? Well, this right here in green, this is the exact same thing as the dot products of A and C. This is the dot product of the vectors A and C. It's the dot product of this vector and that vector."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to have a plus A sub y, B sub y. And then finally I'm going to have a plus A sub z, A sub z, B sub z. And what is this? Well, this right here in green, this is the exact same thing as the dot products of A and C. This is the dot product of the vectors A and C. It's the dot product of this vector and that vector. So that's the dot of A and C times the x component of B minus, I'll do this in the same, minus, once again, this is the dot product of A and B now, minus A dot B times the x component of C. And we can't forget, all of this was multiplied by the unit vector I. We're looking at the x component or the I component of that whole triple product. So that's going to be all of this, it's all of this is being, is times the unit vector I."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this right here in green, this is the exact same thing as the dot products of A and C. This is the dot product of the vectors A and C. It's the dot product of this vector and that vector. So that's the dot of A and C times the x component of B minus, I'll do this in the same, minus, once again, this is the dot product of A and B now, minus A dot B times the x component of C. And we can't forget, all of this was multiplied by the unit vector I. We're looking at the x component or the I component of that whole triple product. So that's going to be all of this, it's all of this is being, is times the unit vector I. Now, if we do this exact same thing, and I'm not going to do it because it's computationally intensive, but I think it won't be a huge leap of faith for you. This is for the x component. If I were to do the exact same thing for the y component, for the j component, so it'll be plus, if I do the same thing for the j component, we can really just pattern match."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be all of this, it's all of this is being, is times the unit vector I. Now, if we do this exact same thing, and I'm not going to do it because it's computationally intensive, but I think it won't be a huge leap of faith for you. This is for the x component. If I were to do the exact same thing for the y component, for the j component, so it'll be plus, if I do the same thing for the j component, we can really just pattern match. We have B sub x, C sub x, that's for the x component. Well, we'll have B sub y and C sub y for the j component. And then this is not component specific, so it'll be A, it will be A dot C over here, and minus A dot B over here."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to do the exact same thing for the y component, for the j component, so it'll be plus, if I do the same thing for the j component, we can really just pattern match. We have B sub x, C sub x, that's for the x component. Well, we'll have B sub y and C sub y for the j component. And then this is not component specific, so it'll be A, it will be A dot C over here, and minus A dot B over here. You can verify any of these for yourself if you don't believe me, but it's the exact same process we just did. And then finally for the z component, or the k component, let me put parentheses over here, same idea. You're going to have B sub z, C sub z, and then you're going to have A dot B over there, and then you're going to have A dot C over here."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then this is not component specific, so it'll be A, it will be A dot C over here, and minus A dot B over here. You can verify any of these for yourself if you don't believe me, but it's the exact same process we just did. And then finally for the z component, or the k component, let me put parentheses over here, same idea. You're going to have B sub z, C sub z, and then you're going to have A dot B over there, and then you're going to have A dot C over here. Now what does this become? How can we simplify this? Well, this right over here, we can expand this out, we can factor out an A dot C from all of these terms over here."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're going to have B sub z, C sub z, and then you're going to have A dot B over there, and then you're going to have A dot C over here. Now what does this become? How can we simplify this? Well, this right over here, we can expand this out, we can factor out an A dot C from all of these terms over here. Remember, this is going to be multiplied times i. Actually, let me not skip too many steps, just because I want you to believe what I'm doing. So if we expand the i here, instead of rewriting it, let me just do it like this."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this right over here, we can expand this out, we can factor out an A dot C from all of these terms over here. Remember, this is going to be multiplied times i. Actually, let me not skip too many steps, just because I want you to believe what I'm doing. So if we expand the i here, instead of rewriting it, let me just do it like this. It's a little bit messier, but let me just... So I could write this i there, and that i there. I'm kind of just distributing that x unit vector, or the i unit vector."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if we expand the i here, instead of rewriting it, let me just do it like this. It's a little bit messier, but let me just... So I could write this i there, and that i there. I'm kind of just distributing that x unit vector, or the i unit vector. And let me do the same thing for j. So I could put the j there, and I could put the j right over there. And then I could do the same thing for the k. Put the k there, and then put the k there."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm kind of just distributing that x unit vector, or the i unit vector. And let me do the same thing for j. So I could put the j there, and I could put the j right over there. And then I could do the same thing for the k. Put the k there, and then put the k there. And now what are these? Well, this part right over here is exactly the same thing as A dot C times, and I'll write it out here, B sub x times i plus B sub y times j plus B sub z times k. And then from that, we're going to subtract all of this, A dot B, times the exact same thing. And you're going to notice this right here is the same thing as vector B."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I could do the same thing for the k. Put the k there, and then put the k there. And now what are these? Well, this part right over here is exactly the same thing as A dot C times, and I'll write it out here, B sub x times i plus B sub y times j plus B sub z times k. And then from that, we're going to subtract all of this, A dot B, times the exact same thing. And you're going to notice this right here is the same thing as vector B. That is vector B. When you do it over here, you're going to get vector C. So I'll just write it over here. You're just going to get vector C. So just like that, we have a simplification for our triple product."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you're going to notice this right here is the same thing as vector B. That is vector B. When you do it over here, you're going to get vector C. So I'll just write it over here. You're just going to get vector C. So just like that, we have a simplification for our triple product. I know it took us a long time to get here, but this is a simplification. It might not look like one, but computationally it is. It's easier to do."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're just going to get vector C. So just like that, we have a simplification for our triple product. I know it took us a long time to get here, but this is a simplification. It might not look like one, but computationally it is. It's easier to do. If I have... I'll try to color-code it. A cross B cross C, we just saw that this is going to be equivalent to, and one way to think about it is, it's going to be, you take the first vector times the dot product of the... the first vector in this second dot product, the one that we have our parentheses around, the one that we would have to do first."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's easier to do. If I have... I'll try to color-code it. A cross B cross C, we just saw that this is going to be equivalent to, and one way to think about it is, it's going to be, you take the first vector times the dot product of the... the first vector in this second dot product, the one that we have our parentheses around, the one that we would have to do first. You take your first vector there, so it's vector B, and you multiply that times the dot product of the other two vectors. So A dot C, and from that you subtract the second vector, multiplied by the dot product of the other two vectors. Of A dot B."}, {"video_title": "Vector triple product expansion (very optional) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A cross B cross C, we just saw that this is going to be equivalent to, and one way to think about it is, it's going to be, you take the first vector times the dot product of the... the first vector in this second dot product, the one that we have our parentheses around, the one that we would have to do first. You take your first vector there, so it's vector B, and you multiply that times the dot product of the other two vectors. So A dot C, and from that you subtract the second vector, multiplied by the dot product of the other two vectors. Of A dot B. And we're done. This is our triple product expansion. Now once again, this isn't something that you really have to know."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I've got this matrix A here. It's a 2 by 3 matrix. And just as a bit of review, let's figure out its null space and its column space. So the null space of A is the set of all vectors x that are a member of, let's see, we have three columns here. So a member of R3, such that A times the vector are going to be equal to the 0 vector. So we could just set this up. We just need to figure out all of the x's that satisfy this in R3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the null space of A is the set of all vectors x that are a member of, let's see, we have three columns here. So a member of R3, such that A times the vector are going to be equal to the 0 vector. So we could just set this up. We just need to figure out all of the x's that satisfy this in R3. So we take our matrix A, 2, minus 1, minus 3, minus 4, 2, 6, multiply them times some arbitrary vector in R3 here. So you get x1, x2, x3. And you set them equal to the 0 vector."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just need to figure out all of the x's that satisfy this in R3. So we take our matrix A, 2, minus 1, minus 3, minus 4, 2, 6, multiply them times some arbitrary vector in R3 here. So you get x1, x2, x3. And you set them equal to the 0 vector. It's going to be the 0 vector in R2, because we have two rows here. You multiply a 2 by 3 matrix times a vector in R3, you're going to get a 2 by 1 vector, or a 2 by 1 matrix. So you're going to get the 0 vector in R3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you set them equal to the 0 vector. It's going to be the 0 vector in R2, because we have two rows here. You multiply a 2 by 3 matrix times a vector in R3, you're going to get a 2 by 1 vector, or a 2 by 1 matrix. So you're going to get the 0 vector in R3. And to solve what is essentially a system of equations, you get 2x1 minus x2 minus 3x3 is equal to 0, and so on and so forth. We can just set up an augmented matrix. So we can just set up this augmented matrix right here."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to get the 0 vector in R3. And to solve what is essentially a system of equations, you get 2x1 minus x2 minus 3x3 is equal to 0, and so on and so forth. We can just set up an augmented matrix. So we can just set up this augmented matrix right here. 2, minus 1, minus 3, minus 4, 2, 6. And then augment it with what we're trying to set it equal to to solve the system. And you know, we're going to perform a bunch of row operations here to put this in reduced row echelon form."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can just set up this augmented matrix right here. 2, minus 1, minus 3, minus 4, 2, 6. And then augment it with what we're trying to set it equal to to solve the system. And you know, we're going to perform a bunch of row operations here to put this in reduced row echelon form. And they're not going to change the right-hand side of this augmented matrix. And that's essentially the argument as to why the null space of the reduced row echelon form of A is the same thing as the null space of A. But anyway, that's just a bit of review."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you know, we're going to perform a bunch of row operations here to put this in reduced row echelon form. And they're not going to change the right-hand side of this augmented matrix. And that's essentially the argument as to why the null space of the reduced row echelon form of A is the same thing as the null space of A. But anyway, that's just a bit of review. So let's perform some row operations to solve this a little bit better. So the first thing I might want to do is divide the first row by 2. So if I divide the first row by 2, I get a 1 minus 1 half minus 3 halves."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But anyway, that's just a bit of review. So let's perform some row operations to solve this a little bit better. So the first thing I might want to do is divide the first row by 2. So if I divide the first row by 2, I get a 1 minus 1 half minus 3 halves. And then of course, 0 divided by 2 is 0. And let's just divide this row right here. Let's just divide it by, I don't know, let's divide it, just to simplify things, let's divide it by 4."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I divide the first row by 2, I get a 1 minus 1 half minus 3 halves. And then of course, 0 divided by 2 is 0. And let's just divide this row right here. Let's just divide it by, I don't know, let's divide it, just to simplify things, let's divide it by 4. So I'm doing 2 row operations in one step. But you can do that. I could have done it in 2 separate steps."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's just divide it by, I don't know, let's divide it, just to simplify things, let's divide it by 4. So I'm doing 2 row operations in one step. But you can do that. I could have done it in 2 separate steps. So if we divide it by 4, this becomes minus 1, 1 half, and then you get 3 halves, and then you get 0. And now let's keep my first row the same. I'm going to keep my first row the same."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I could have done it in 2 separate steps. So if we divide it by 4, this becomes minus 1, 1 half, and then you get 3 halves, and then you get 0. And now let's keep my first row the same. I'm going to keep my first row the same. It's 1 minus 1 half minus 3 halves. And of course, the 0 is the right-hand side. And let's replace my second row with my second row plus my first row."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to keep my first row the same. It's 1 minus 1 half minus 3 halves. And of course, the 0 is the right-hand side. And let's replace my second row with my second row plus my first row. So these are just linear operations on these guys. So negative 1 plus 1 is 0. 1 half plus minus 1 half is 0."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's replace my second row with my second row plus my first row. So these are just linear operations on these guys. So negative 1 plus 1 is 0. 1 half plus minus 1 half is 0. 3 halves plus minus 3 halves is 0. And of course, 0 plus 0 is 0. So what are we left with?"}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1 half plus minus 1 half is 0. 3 halves plus minus 3 halves is 0. And of course, 0 plus 0 is 0. So what are we left with? We're left with this right here. Or this is another way of saying that x1, let me write it this way. x1, I guess the easiest way to think about it is, you're multiplying the reduced row echelon form of A now, 1 minus 1 half minus 3 halves."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what are we left with? We're left with this right here. Or this is another way of saying that x1, let me write it this way. x1, I guess the easiest way to think about it is, you're multiplying the reduced row echelon form of A now, 1 minus 1 half minus 3 halves. You have a bunch of 0's here. Times x1, x2, x3 is equal to the r2 0 vector. This is another interpretation of this augmented matrix."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "x1, I guess the easiest way to think about it is, you're multiplying the reduced row echelon form of A now, 1 minus 1 half minus 3 halves. You have a bunch of 0's here. Times x1, x2, x3 is equal to the r2 0 vector. This is another interpretation of this augmented matrix. So this is just saying, this is useless. This is saying 0 times that plus 0 times that plus 0 times that is equal to 0. So it's giving us no information."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is another interpretation of this augmented matrix. So this is just saying, this is useless. This is saying 0 times that plus 0 times that plus 0 times that is equal to 0. So it's giving us no information. But this first row tells us that, let me switch colors, 1 times x1 minus 1 half times x2 minus 3 halves times x3 is equal to 0. All of the vectors whose components satisfy this are in my null space. If I want to write it a little bit differently, I could write it as x1 is equal to 1 half x2 plus 3 halves x3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's giving us no information. But this first row tells us that, let me switch colors, 1 times x1 minus 1 half times x2 minus 3 halves times x3 is equal to 0. All of the vectors whose components satisfy this are in my null space. If I want to write it a little bit differently, I could write it as x1 is equal to 1 half x2 plus 3 halves x3. Or if I wanted to write my solution set in vector form, I could write that my null space is going to be the set of all the vectors x1, x2, x3 that satisfy these conditions that are equal to what? Well, x2 and x3 are free variables. They're associated with the non-pivot entries or the non-pivot columns in our reduced row echelon form."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I want to write it a little bit differently, I could write it as x1 is equal to 1 half x2 plus 3 halves x3. Or if I wanted to write my solution set in vector form, I could write that my null space is going to be the set of all the vectors x1, x2, x3 that satisfy these conditions that are equal to what? Well, x2 and x3 are free variables. They're associated with the non-pivot entries or the non-pivot columns in our reduced row echelon form. That is a pivot column. So what I could set, so let me write it this way. It's going to be x2 times something plus x3 times something."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They're associated with the non-pivot entries or the non-pivot columns in our reduced row echelon form. That is a pivot column. So what I could set, so let me write it this way. It's going to be x2 times something plus x3 times something. Those are my two free variables. And we have here x1 is 1 half x2 plus 3 halves times x3. x2 is just going to be x2 times 1 plus 0 times x3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be x2 times something plus x3 times something. Those are my two free variables. And we have here x1 is 1 half x2 plus 3 halves times x3. x2 is just going to be x2 times 1 plus 0 times x3. x3 is going to be 0 times x2 plus 1 times x3. So our null space, these can be any real numbers right here. They're free variables."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "x2 is just going to be x2 times 1 plus 0 times x3. x3 is going to be 0 times x2 plus 1 times x3. So our null space, these can be any real numbers right here. They're free variables. So our null space is essentially all of the linear combinations of this guy and that guy. Or another way to write it, the null space of A is equal to the span, which is the same thing as all the linear combinations of, the span of 1 half 1 0. Notice these are vectors in R3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They're free variables. So our null space is essentially all of the linear combinations of this guy and that guy. Or another way to write it, the null space of A is equal to the span, which is the same thing as all the linear combinations of, the span of 1 half 1 0. Notice these are vectors in R3. And that makes sense because the null space is going to be a set of vectors in R3. So it's a span of that and that right there. And that right there."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Notice these are vectors in R3. And that makes sense because the null space is going to be a set of vectors in R3. So it's a span of that and that right there. And that right there. So 3 halves, 0, and 1. Just like that. And what is the column space of our original matrix A?"}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that right there. So 3 halves, 0, and 1. Just like that. And what is the column space of our original matrix A? So the column space of A is equal to just all of the linear, the subspace created by all of the linear combinations of these guys. Or essentially the span of the column vectors is equal to the span of 2 minus 4 minus 1 2 minus 3 6. These are all each separate vectors."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is the column space of our original matrix A? So the column space of A is equal to just all of the linear, the subspace created by all of the linear combinations of these guys. Or essentially the span of the column vectors is equal to the span of 2 minus 4 minus 1 2 minus 3 6. These are all each separate vectors. So it's the span of these three vectors. Now these guys might not be linearly independent. And actually when you put this guy in reduced row echelon form, you know that the basis vectors for this are the vectors that are associated with our pivot column."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are all each separate vectors. So it's the span of these three vectors. Now these guys might not be linearly independent. And actually when you put this guy in reduced row echelon form, you know that the basis vectors for this are the vectors that are associated with our pivot column. So we have one pivot column here. It's our first column. So we could say that we could use this as a basis vector."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And actually when you put this guy in reduced row echelon form, you know that the basis vectors for this are the vectors that are associated with our pivot column. So we have one pivot column here. It's our first column. So we could say that we could use this as a basis vector. And it makes sense because this guy right here is minus 2 times this guy. This guy right here is what? Minus 3 halves times that guy."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that we could use this as a basis vector. And it makes sense because this guy right here is minus 2 times this guy. This guy right here is what? Minus 3 halves times that guy. So these two guys can definitely be represented as linear combinations of that guy. So it's equal to the span of just the vector 2 minus 4. So if you were to ask me, and this is the basis for our column space, so if you want to know the rank, and this is all a bit of review, the rank of A is equal to the number of vectors in our basis for our column space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 3 halves times that guy. So these two guys can definitely be represented as linear combinations of that guy. So it's equal to the span of just the vector 2 minus 4. So if you were to ask me, and this is the basis for our column space, so if you want to know the rank, and this is all a bit of review, the rank of A is equal to the number of vectors in our basis for our column space. So it's going to be equal to 1. Now everything I just did is a bit of review. But with the last couple of videos, we've been dealing with transposes."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you were to ask me, and this is the basis for our column space, so if you want to know the rank, and this is all a bit of review, the rank of A is equal to the number of vectors in our basis for our column space. So it's going to be equal to 1. Now everything I just did is a bit of review. But with the last couple of videos, we've been dealing with transposes. So let's actually figure out the same ideas for the transpose of A. So A transpose looks like this. A transpose is equal to the vector 2 minus 1, or the matrix, 2 minus 1 minus 3 is the first column right there."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But with the last couple of videos, we've been dealing with transposes. So let's actually figure out the same ideas for the transpose of A. So A transpose looks like this. A transpose is equal to the vector 2 minus 1, or the matrix, 2 minus 1 minus 3 is the first column right there. And then the second column is going to be minus 4, 2, and 6. That is our transpose. So let's figure out the null space and the column space of our transpose."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A transpose is equal to the vector 2 minus 1, or the matrix, 2 minus 1 minus 3 is the first column right there. And then the second column is going to be minus 4, 2, and 6. That is our transpose. So let's figure out the null space and the column space of our transpose. Well, let me put this in reduced row echelon form so we can get the null space. Let me get the null space of this guy. So we could do the exact same exercise."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's figure out the null space and the column space of our transpose. Well, let me put this in reduced row echelon form so we can get the null space. Let me get the null space of this guy. So we could do the exact same exercise. Let me write it this way. The null space of A transpose. A transpose is a 3 by 2 matrix."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could do the exact same exercise. Let me write it this way. The null space of A transpose. A transpose is a 3 by 2 matrix. So it's equal to all of the vectors x that are members of R2, not R3 anymore, because now we're taking the transposes null space. Such that A transpose times R vectors are equal to the 0 vector in R3. And we can do that the same exact way we did it before."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A transpose is a 3 by 2 matrix. So it's equal to all of the vectors x that are members of R2, not R3 anymore, because now we're taking the transposes null space. Such that A transpose times R vectors are equal to the 0 vector in R3. And we can do that the same exact way we did it before. We set up an augmented matrix. We could just put it in reduced row echelon form and set them all equal to 0. So let's just do that."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we can do that the same exact way we did it before. We set up an augmented matrix. We could just put it in reduced row echelon form and set them all equal to 0. So let's just do that. So let me just put it in reduced row echelon form. So let me divide my first row by 2. Let's divide the first row by 2."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's just do that. So let me just put it in reduced row echelon form. So let me divide my first row by 2. Let's divide the first row by 2. Let's put it in reduced row echelon form. The first row divided by 2 is 1 minus 2. And then the second row, let me divide it."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's divide the first row by 2. Let's put it in reduced row echelon form. The first row divided by 2 is 1 minus 2. And then the second row, let me divide it. Well, let me just keep it the same. So minus 1, 2. And then this last row, let me divide it by 3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the second row, let me divide it. Well, let me just keep it the same. So minus 1, 2. And then this last row, let me divide it by 3. So it becomes minus 1 and 2. And now let me keep my first row the same. 1 minus 2."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then this last row, let me divide it by 3. So it becomes minus 1 and 2. And now let me keep my first row the same. 1 minus 2. And now let me replace my second row with my second row plus my first row. So minus 1 plus 1 is 0. 2 plus minus 2 is 0."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2. And now let me replace my second row with my second row plus my first row. So minus 1 plus 1 is 0. 2 plus minus 2 is 0. So you get some 0's. I'm going to do the same thing with the third row. Replace it with it plus the first row."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 plus minus 2 is 0. So you get some 0's. I'm going to do the same thing with the third row. Replace it with it plus the first row. Once again, you're going to get some 0's. So this is the reduced row echelon form of A transpose. So this is the reduced row echelon form of A transpose."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Replace it with it plus the first row. Once again, you're going to get some 0's. So this is the reduced row echelon form of A transpose. So this is the reduced row echelon form of A transpose. And its null space is the same as A transpose's null space. So we could say to find this null space, we can find all of the solutions to this equation. Times the vectors x1 and x2 is equal to 0, 0, and 0."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the reduced row echelon form of A transpose. And its null space is the same as A transpose's null space. So we could say to find this null space, we can find all of the solutions to this equation. Times the vectors x1 and x2 is equal to 0, 0, and 0. These aren't vectors. These are just entries right here. 0, 0, 0."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times the vectors x1 and x2 is equal to 0, 0, and 0. These aren't vectors. These are just entries right here. 0, 0, 0. So these two lines give us no information. But this first one does. So we get 1 times x1."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0, 0, 0. So these two lines give us no information. But this first one does. So we get 1 times x1. And notice, this is the pivot column right here. So x1 is going to be a pivot variable. This is a x2 will be a free variable."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we get 1 times x1. And notice, this is the pivot column right here. So x1 is going to be a pivot variable. This is a x2 will be a free variable. And just to be clear, the first column is our pivot column. So if we go back to A transpose, it's this first column here that is associated with the pivot column. So when we talk about its column space, this by itself will span the column space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a x2 will be a free variable. And just to be clear, the first column is our pivot column. So if we go back to A transpose, it's this first column here that is associated with the pivot column. So when we talk about its column space, this by itself will span the column space. This is all a review of what we did before. We're just applying it to the transpose. Let's go back to our null space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So when we talk about its column space, this by itself will span the column space. This is all a review of what we did before. We're just applying it to the transpose. Let's go back to our null space. So this tells us that 1 times x1, so x1 minus 2 times x2 is equal to 0. Or we could say that x1 is equal to 2 times x2. So all of the vectors in R2 that satisfy these conditions with these entries will be in the null space of A transpose."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's go back to our null space. So this tells us that 1 times x1, so x1 minus 2 times x2 is equal to 0. Or we could say that x1 is equal to 2 times x2. So all of the vectors in R2 that satisfy these conditions with these entries will be in the null space of A transpose. Let me write it this way. So the null space of A is going to be the set of all the vectors, let me write it here, the set of all the vectors x1, x2 that are a member of R2 clearly, such that x1, x2 is going to be equal to, well, our free variable is x2. So it's x2 times the vector."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So all of the vectors in R2 that satisfy these conditions with these entries will be in the null space of A transpose. Let me write it this way. So the null space of A is going to be the set of all the vectors, let me write it here, the set of all the vectors x1, x2 that are a member of R2 clearly, such that x1, x2 is going to be equal to, well, our free variable is x2. So it's x2 times the vector. So x1 has to be 2 times x2. And obviously x2, that's a 2, x2 is going to be 1 times x2. So what is this going to be?"}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's x2 times the vector. So x1 has to be 2 times x2. And obviously x2, that's a 2, x2 is going to be 1 times x2. So what is this going to be? Well, this is all of the linear combinations of this vector right here. So we could say it's equal to the span of our vector to 1. Now, that's the null space, sorry, this was the null space of A transpose."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is this going to be? Well, this is all of the linear combinations of this vector right here. So we could say it's equal to the span of our vector to 1. Now, that's the null space, sorry, this was the null space of A transpose. Have to be very careful there. Now what is the column space of A transpose? Well, the column space of A transpose is the set of all vectors spanned by the columns of A."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, that's the null space, sorry, this was the null space of A transpose. Have to be very careful there. Now what is the column space of A transpose? Well, the column space of A transpose is the set of all vectors spanned by the columns of A. So you could just say the span of this column vector and this column vector. But we know when we put it into reduced row echelon form, only this column vector was associated with a pivot column. So this by itself, this guy is a linear combination of this guy."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, the column space of A transpose is the set of all vectors spanned by the columns of A. So you could just say the span of this column vector and this column vector. But we know when we put it into reduced row echelon form, only this column vector was associated with a pivot column. So this by itself, this guy is a linear combination of this guy. If you multiply this, if you multiply him by minus 2, you get that guy right there. So it's consistent with everything we've learned. So it equals the span of just this guy right here."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this by itself, this guy is a linear combination of this guy. If you multiply this, if you multiply him by minus 2, you get that guy right there. So it's consistent with everything we've learned. So it equals the span of just this guy right here. Of just the vector 2, minus 1, and minus 3. Now, that's just a nice, neat exercise that we did. Notice that your span here, it's in R3, but it's just going to be a line in R3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it equals the span of just this guy right here. Of just the vector 2, minus 1, and minus 3. Now, that's just a nice, neat exercise that we did. Notice that your span here, it's in R3, but it's just going to be a line in R3. Maybe in the next video I'll do a more graphical representation of it. But I did this whole exercise to introduce you to the ideas of the null space of your transpose and the column space of your transpose. Think about what the column space of your transpose is."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Notice that your span here, it's in R3, but it's just going to be a line in R3. Maybe in the next video I'll do a more graphical representation of it. But I did this whole exercise to introduce you to the ideas of the null space of your transpose and the column space of your transpose. Think about what the column space of your transpose is. It's the subspace spanned by this vector and that vector. And it turns out that this guy is a multiple of that guy. So we could say just by that guy."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Think about what the column space of your transpose is. It's the subspace spanned by this vector and that vector. And it turns out that this guy is a multiple of that guy. So we could say just by that guy. But these were the rows of our original matrix A. So we could also view this as the span of the row vectors of our original guy, right? This is that column that is the basis for the column span of the R transpose matrix."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say just by that guy. But these were the rows of our original matrix A. So we could also view this as the span of the row vectors of our original guy, right? This is that column that is the basis for the column span of the R transpose matrix. And of course, this guy was a linear combination of that. So we could also view the column span of our transpose matrix. It's equivalent to the subspace spanned by these rows, or we could call that the row space of A."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is that column that is the basis for the column span of the R transpose matrix. And of course, this guy was a linear combination of that. So we could also view the column span of our transpose matrix. It's equivalent to the subspace spanned by these rows, or we could call that the row space of A. So let me write that down. So the column space of A transpose, and this is just general, let me write this generally. It doesn't just apply to this example."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equivalent to the subspace spanned by these rows, or we could call that the row space of A. So let me write that down. So the column space of A transpose, and this is just general, let me write this generally. It doesn't just apply to this example. So the column space of the transpose of any matrix, this is called the row space of A. And it's a very natural name because if A's got a bunch of rows, we could call them the transpose of some vector. So that's the first row, you got the second row, all the way to maybe the m-th row."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It doesn't just apply to this example. So the column space of the transpose of any matrix, this is called the row space of A. And it's a very natural name because if A's got a bunch of rows, we could call them the transpose of some vector. So that's the first row, you got the second row, all the way to maybe the m-th row. The m-th row, just like that. These are vector transposes. They're really just rows."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's the first row, you got the second row, all the way to maybe the m-th row. The m-th row, just like that. These are vector transposes. They're really just rows. If you imagine the space that's spanned by these vectors, by the different rows, that's essentially the column space of the transpose. Because when you transpose it, each of these guys become columns. So that's what the row space is."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They're really just rows. If you imagine the space that's spanned by these vectors, by the different rows, that's essentially the column space of the transpose. Because when you transpose it, each of these guys become columns. So that's what the row space is. Now, the null space of our transpose, let's write it like this. It was all of the vectors x that satisfied this equation. Equals the 0 vector right there."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's what the row space is. Now, the null space of our transpose, let's write it like this. It was all of the vectors x that satisfied this equation. Equals the 0 vector right there. Now, what happens if we take the transpose of both sides of this equation? So if we take the transpose of both sides of that equation? Well, we've learned from our transpose properties, this is equal to the reverse product of each of those transposes."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Equals the 0 vector right there. Now, what happens if we take the transpose of both sides of this equation? So if we take the transpose of both sides of that equation? Well, we've learned from our transpose properties, this is equal to the reverse product of each of those transposes. So this is going to be equal to, this is a vector, the vector x transpose. So now it's going to become, if this is a column vector before, now it's going to become a row vector. And then times A transpose transpose, and that's going to be equal to the transpose of the 0 vector."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we've learned from our transpose properties, this is equal to the reverse product of each of those transposes. So this is going to be equal to, this is a vector, the vector x transpose. So now it's going to become, if this is a column vector before, now it's going to become a row vector. And then times A transpose transpose, and that's going to be equal to the transpose of the 0 vector. Or we could just write this as, let's write it like this. We could write this as some matrix, well let me just write it like this, some column vector x times, what's the transpose of A transpose? Well, that's just equal to A."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then times A transpose transpose, and that's going to be equal to the transpose of the 0 vector. Or we could just write this as, let's write it like this. We could write this as some matrix, well let me just write it like this, some column vector x times, what's the transpose of A transpose? Well, that's just equal to A. So you take the transpose of this column vector, you now get a row vector. You could view it as a matrix if you want. If this was a member of Rn, this is now going to be an n by, this is an n by, or 1 by n matrix, if this was a member of Rn."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's just equal to A. So you take the transpose of this column vector, you now get a row vector. You could view it as a matrix if you want. If this was a member of Rn, this is now going to be an n by, this is an n by, or 1 by n matrix, if this was a member of Rn. And now it's essentially, we kind of switch the orders and we multiply it times the transpose of the transpose, we just get the matrix A. And we set that equal to the transpose of the 0 vector. Now this is interesting."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If this was a member of Rn, this is now going to be an n by, this is an n by, or 1 by n matrix, if this was a member of Rn. And now it's essentially, we kind of switch the orders and we multiply it times the transpose of the transpose, we just get the matrix A. And we set that equal to the transpose of the 0 vector. Now this is interesting. We now have it in terms of our original matrix A. Now what did the null space of our matrix A look like? The null space where all of the vectors x that satisfy this equation is equal to 0."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now this is interesting. We now have it in terms of our original matrix A. Now what did the null space of our matrix A look like? The null space where all of the vectors x that satisfy this equation is equal to 0. So the x was on the right. So the null space is all the x's that satisfy this. The null space of our transpose is all of the x's that satisfy this equation."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The null space where all of the vectors x that satisfy this equation is equal to 0. So the x was on the right. So the null space is all the x's that satisfy this. The null space of our transpose is all of the x's that satisfy this equation. And this is also called, so let me say the set of all of the x's such that A transpose times x is equal to 0. That is the null space of A transpose. That is equal to the null space of A transpose."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The null space of our transpose is all of the x's that satisfy this equation. And this is also called, so let me say the set of all of the x's such that A transpose times x is equal to 0. That is the null space of A transpose. That is equal to the null space of A transpose. Or we could also write this as the set of all of the x's such that the transpose of our x times A is equal to the transpose of the 0 vector. And we have another name for this. This is called the left null space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is equal to the null space of A transpose. Or we could also write this as the set of all of the x's such that the transpose of our x times A is equal to the transpose of the 0 vector. And we have another name for this. This is called the left null space. Why is it called the left null space? Because now we have x on our left. In just a regular null space, you have x on the right."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is called the left null space. Why is it called the left null space? Because now we have x on our left. In just a regular null space, you have x on the right. But now if you take the null space of the transpose using just our transpose properties, that's equivalent to this transpose vector right here. Actually let me write this as a transpose right there. This transpose vector multiplying A from the left-hand side."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In just a regular null space, you have x on the right. But now if you take the null space of the transpose using just our transpose properties, that's equivalent to this transpose vector right here. Actually let me write this as a transpose right there. This transpose vector multiplying A from the left-hand side. So all of the x's that satisfy this is the left null space. And it's going to be different than your null space. Because it was equal to, notice, your null space of A transpose was a span of this right here."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This transpose vector multiplying A from the left-hand side. So all of the x's that satisfy this is the left null space. And it's going to be different than your null space. Because it was equal to, notice, your null space of A transpose was a span of this right here. This is also the left null space of A. Now what was just the regular null space of A? The regular null space of A was essentially a plane in R3."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because it was equal to, notice, your null space of A transpose was a span of this right here. This is also the left null space of A. Now what was just the regular null space of A? The regular null space of A was essentially a plane in R3. That's the null space of A. The left null space of A is just a line in R2. Very different things."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The regular null space of A was essentially a plane in R3. That's the null space of A. The left null space of A is just a line in R2. Very different things. And if you go to the row space, what is the row space of A? The row space of A is a line in R3. Well what is the column space of A?"}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Very different things. And if you go to the row space, what is the row space of A? The row space of A is a line in R3. Well what is the column space of A? The column space of A right here, where did I have it? Well, this is the only linearly independent vector. It was essentially a line in R2."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well what is the column space of A? The column space of A right here, where did I have it? Well, this is the only linearly independent vector. It was essentially a line in R2. So they're all very different things. And we'll study a little bit more how they're all related. Now there's one thing I want to relate to you."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It was essentially a line in R2. So they're all very different things. And we'll study a little bit more how they're all related. Now there's one thing I want to relate to you. We figured out that the rank of this vector right here is 1. Because when you put it in reduced row echelon form, there was one pivot column. And the basis vectors are those associated with that pivot column."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now there's one thing I want to relate to you. We figured out that the rank of this vector right here is 1. Because when you put it in reduced row echelon form, there was one pivot column. And the basis vectors are those associated with that pivot column. And if you count your basis vectors, that's your dimension of your space. So the dimension of your column space is 1. And that's the same thing as your rank."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the basis vectors are those associated with that pivot column. And if you count your basis vectors, that's your dimension of your space. So the dimension of your column space is 1. And that's the same thing as your rank. Now what is the rank of A transpose? The rank of A transpose in the example, when you put it in reduced row echelon form, you got one linearly independent column vector. So the basis for our column space was also equal to 1."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that's the same thing as your rank. Now what is the rank of A transpose? The rank of A transpose in the example, when you put it in reduced row echelon form, you got one linearly independent column vector. So the basis for our column space was also equal to 1. And in general, that's always going to be the case. That the rank of A, which is the dimension of its column space, is equal to the rank of A transpose. And if you think about it, it makes a lot of sense."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the basis for our column space was also equal to 1. And in general, that's always going to be the case. That the rank of A, which is the dimension of its column space, is equal to the rank of A transpose. And if you think about it, it makes a lot of sense. To figure out the rank of A, you essentially just have to figure out how many pivot columns they have. Or another way to say it is how many pivot entries they have. Now, when you figure out the row space, when you want to find the rank of your transpose vector, you're essentially just saying, and I know this is maybe getting a little bit confusing."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if you think about it, it makes a lot of sense. To figure out the rank of A, you essentially just have to figure out how many pivot columns they have. Or another way to say it is how many pivot entries they have. Now, when you figure out the row space, when you want to find the rank of your transpose vector, you're essentially just saying, and I know this is maybe getting a little bit confusing. But when you want the rank of your transpose vector, you're saying, how many of these columns are linearly independent? Or which of these are linearly independent? And that's the same question as saying, how many of your rows up here are linearly independent?"}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, when you figure out the row space, when you want to find the rank of your transpose vector, you're essentially just saying, and I know this is maybe getting a little bit confusing. But when you want the rank of your transpose vector, you're saying, how many of these columns are linearly independent? Or which of these are linearly independent? And that's the same question as saying, how many of your rows up here are linearly independent? If you want to know how many columns in your transpose are linearly independent, that's equivalent to asking how many rows in your original matrix are linearly independent. And when you put this matrix in reduced row echelon form, everything in reduced row echelon form are just row operations. So they're just linear combinations of these things up here."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that's the same question as saying, how many of your rows up here are linearly independent? If you want to know how many columns in your transpose are linearly independent, that's equivalent to asking how many rows in your original matrix are linearly independent. And when you put this matrix in reduced row echelon form, everything in reduced row echelon form are just row operations. So they're just linear combinations of these things up here. Or you could go vice versa. Everything up here is just linear combinations of your matrix in reduced row echelon form. So if you only have one pivot entry, then this guy right here, by himself, or one pivot row, that guy by himself can represent a basis for your row space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So they're just linear combinations of these things up here. Or you could go vice versa. Everything up here is just linear combinations of your matrix in reduced row echelon form. So if you only have one pivot entry, then this guy right here, by himself, or one pivot row, that guy by himself can represent a basis for your row space. Or all of your rows can be represented by a linear combination of your pivot rows. And because of that, you just count that. You say, OK, there's one in this case."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you only have one pivot entry, then this guy right here, by himself, or one pivot row, that guy by himself can represent a basis for your row space. Or all of your rows can be represented by a linear combination of your pivot rows. And because of that, you just count that. You say, OK, there's one in this case. So the dimension of my row space is 1. And that's the same thing as the dimension of my column space's transpose. I know it's all of my transpose's column space."}, {"video_title": "Rowspace and left nullspace Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You say, OK, there's one in this case. So the dimension of my row space is 1. And that's the same thing as the dimension of my column space's transpose. I know it's all of my transpose's column space. I know it's getting all confusing. And it's late in the day for me as well. So that hopefully will convince you that the rank of our transpose is the same as the rank of our original matrix."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "We figured out its null space, its column space. We figured out the null space and column space of its transpose, which you can also call the left null space, and the row space, or what's essentially the space spanned by A's rows. But let's write it all in one place, because I realize it got a little disjointed. And see if we can visualize what all of these look like, especially relative to each other. So let me copy and paste my original matrix. And then let me scroll down here and paste it over here. Let me see if I can find our key takeaways from the last video."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And see if we can visualize what all of these look like, especially relative to each other. So let me copy and paste my original matrix. And then let me scroll down here and paste it over here. Let me see if I can find our key takeaways from the last video. So our column space right here of A was this thing right here, let me write this. This was our column space. It was the span of the R2 vector, 2, 4."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me see if I can find our key takeaways from the last video. So our column space right here of A was this thing right here, let me write this. This was our column space. It was the span of the R2 vector, 2, 4. We copy that. Bring it down. Paste."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It was the span of the R2 vector, 2, 4. We copy that. Bring it down. Paste. This was our column space. Let me write that. This is the column space of A was equal to that right there."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Paste. This was our column space. Let me write that. This is the column space of A was equal to that right there. And now what other things do we know? We know that the left null space was a span of 2, 1. Let me write that."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This is the column space of A was equal to that right there. And now what other things do we know? We know that the left null space was a span of 2, 1. Let me write that. So our left null space, or the null space of our transpose, either way, it was equal to the span of the R2 vector, 2, 1. Just like that. And then what was our null space?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that. So our left null space, or the null space of our transpose, either way, it was equal to the span of the R2 vector, 2, 1. Just like that. And then what was our null space? Our null space, we figured out in the last video. Here it is. It's the span of these two R3 vectors."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And then what was our null space? Our null space, we figured out in the last video. Here it is. It's the span of these two R3 vectors. Let me copy and paste that. So that's edit, copy. Let me go down here."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It's the span of these two R3 vectors. Let me copy and paste that. So that's edit, copy. Let me go down here. Let me paste it. So that was our null space right there. And then finally, what was our row space?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me go down here. Let me paste it. So that was our null space right there. And then finally, what was our row space? Or the column space of our transpose? The column space of our transpose. So the column space of our transpose was the span of this R3 vector right there."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, what was our row space? Or the column space of our transpose? The column space of our transpose. So the column space of our transpose was the span of this R3 vector right there. So it's this one right here. And so let me copy and paste it. Copy and scroll down."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So the column space of our transpose was the span of this R3 vector right there. So it's this one right here. And so let me copy and paste it. Copy and scroll down. And we can paste it. Just like that. OK, let's see if we can visualize this now, now that we have them all in one place."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Copy and scroll down. And we can paste it. Just like that. OK, let's see if we can visualize this now, now that we have them all in one place. So first of all, if we imagine a transformation, x that is equal to a times x, our transformation is going to be a mapping from what? x would be a member of R3. So R3 would be our domain."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "OK, let's see if we can visualize this now, now that we have them all in one place. So first of all, if we imagine a transformation, x that is equal to a times x, our transformation is going to be a mapping from what? x would be a member of R3. So R3 would be our domain. So it would be a mapping from R3. And then it would be a mapping to R2, because we have two rows here, right? You multiply a 2 by 3 matrix times a 3 by 1 vector, and you're going to get a 2 by 1 vector."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So R3 would be our domain. So it would be a mapping from R3. And then it would be a mapping to R2, because we have two rows here, right? You multiply a 2 by 3 matrix times a 3 by 1 vector, and you're going to get a 2 by 1 vector. So it's going to be a mapping to R2. So that's our codomain. So let's draw our domains and our codomains."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "You multiply a 2 by 3 matrix times a 3 by 1 vector, and you're going to get a 2 by 1 vector. So it's going to be a mapping to R2. So that's our codomain. So let's draw our domains and our codomains. So you have, well, I'll just write them very generally right here. So you could imagine R3 is our domain. And then our codomain is going to be R2, just like that."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let's draw our domains and our codomains. So you have, well, I'll just write them very generally right here. So you could imagine R3 is our domain. And then our codomain is going to be R2, just like that. Codomain. And our t is a mapping, or you could even imagine a is a mapping between any vector there and any vector there when you multiply them. Now, what is our column space of a?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And then our codomain is going to be R2, just like that. Codomain. And our t is a mapping, or you could even imagine a is a mapping between any vector there and any vector there when you multiply them. Now, what is our column space of a? Our column space of a is the span of the vector 2 minus 4. It's an R2 vector. This is a subspace of R2."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is our column space of a? Our column space of a is the span of the vector 2 minus 4. It's an R2 vector. This is a subspace of R2. We could write this. So let me write this. So our column space of a, these are just all of the vectors that are spanned by this."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This is a subspace of R2. We could write this. So let me write this. So our column space of a, these are just all of the vectors that are spanned by this. We figured out that these guys are just multiples of this first guy, or we could have done it the other way. We could have said this guy and that guy are multiples of that guy, either way. But the basis is just one of these vectors."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So our column space of a, these are just all of the vectors that are spanned by this. We figured out that these guys are just multiples of this first guy, or we could have done it the other way. We could have said this guy and that guy are multiples of that guy, either way. But the basis is just one of these vectors. We just had to have one of these vectors. And so it was equal to this right here. So the column space is a subset of R2."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But the basis is just one of these vectors. We just had to have one of these vectors. And so it was equal to this right here. So the column space is a subset of R2. And what else is a subset of R2? Well, our left null space is also a subset of R2. So let's graph them, actually."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So the column space is a subset of R2. And what else is a subset of R2? Well, our left null space is also a subset of R2. So let's graph them, actually. So I won't be too exact, but you can imagine. Let's see. If we draw the vector 2, 4, let me draw some axes here."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let's graph them, actually. So I won't be too exact, but you can imagine. Let's see. If we draw the vector 2, 4, let me draw some axes here. Let me scroll down a little bit. So if you have some vector, let me draw my, do this as neatly as possible. It's my vertical axis."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "If we draw the vector 2, 4, let me draw some axes here. Let me scroll down a little bit. So if you have some vector, let me draw my, do this as neatly as possible. It's my vertical axis. That is my horizontal axis. And then what does the span of our column space look like? What does the span of our column space look like?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It's my vertical axis. That is my horizontal axis. And then what does the span of our column space look like? What does the span of our column space look like? So you draw the vector 2 minus 4. So you're going to go out 1, 2, and then you're going to go down 1, 2, 3, 4. So that's what that vector looks like."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "What does the span of our column space look like? So you draw the vector 2 minus 4. So you're going to go out 1, 2, and then you're going to go down 1, 2, 3, 4. So that's what that vector looks like. And the span of this vector is essentially all of the multiples of this vector. You could say linear combinations of it, but you're taking a combination of just one vector. So it's just going to be all of the multiples of this vector."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So that's what that vector looks like. And the span of this vector is essentially all of the multiples of this vector. You could say linear combinations of it, but you're taking a combination of just one vector. So it's just going to be all of the multiples of this vector. So if I were to graph it, it would just be a line that is specified by all of the linear combinations of that vector. Right there. This right here is a graphical representation of the column space of A."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So it's just going to be all of the multiples of this vector. So if I were to graph it, it would just be a line that is specified by all of the linear combinations of that vector. Right there. This right here is a graphical representation of the column space of A. Now let's look at the left null space of A. Or you could imagine the null space of the transpose. They are the same thing."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This right here is a graphical representation of the column space of A. Now let's look at the left null space of A. Or you could imagine the null space of the transpose. They are the same thing. You saw why in the last video. What does this look like? So the left null space is a span of 2, 1."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "They are the same thing. You saw why in the last video. What does this look like? So the left null space is a span of 2, 1. So if you graph 2 and then you go up 1, this is a graph of 2, 1, it looks like this. Let me do it in a different color. So that's what the vector looks like."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So the left null space is a span of 2, 1. So if you graph 2 and then you go up 1, this is a graph of 2, 1, it looks like this. Let me do it in a different color. So that's what the vector looks like. The vector looks like that. But of course, we want the span of that vector. So it's going to be all of the combinations."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So that's what the vector looks like. The vector looks like that. But of course, we want the span of that vector. So it's going to be all of the combinations. Or all you can do when you combine one vector is just multiply it by a bunch of scalars. It's going to be all of the scalar multiples of that vector. So let me draw it like that."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be all of the combinations. Or all you can do when you combine one vector is just multiply it by a bunch of scalars. It's going to be all of the scalar multiples of that vector. So let me draw it like that. It's going to be like that. And the first thing you might notice, let me write this. This is our left null space of A, or the null space of our transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw it like that. It's going to be like that. And the first thing you might notice, let me write this. This is our left null space of A, or the null space of our transpose. This is equal to the left null space of A. This is equal to the left null space of A. And actually, since we're writing, we wrote this in terms of A transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This is our left null space of A, or the null space of our transpose. This is equal to the left null space of A. This is equal to the left null space of A. And actually, since we're writing, we wrote this in terms of A transpose. It's the null space of A transpose, which is the left null space of A. Let's write the column space of A also in terms of A transpose. This is equal to the row space of A transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And actually, since we're writing, we wrote this in terms of A transpose. It's the null space of A transpose, which is the left null space of A. Let's write the column space of A also in terms of A transpose. This is equal to the row space of A transpose. If you're looking at the columns of A, everything it spans, the columns of A are the same things as the rows of A transpose. But the first thing that you see when I just at least visually drew it like this is that these two spaces look to be orthogonal to each other. It looks like I drew it in R2."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to the row space of A transpose. If you're looking at the columns of A, everything it spans, the columns of A are the same things as the rows of A transpose. But the first thing that you see when I just at least visually drew it like this is that these two spaces look to be orthogonal to each other. It looks like I drew it in R2. It looks like there's a 90 degree angle there. And if we wanted to verify it, all we have to do is take the dot product. Well, any vector that is in our column space, so any vector, you could take an arbitrary vector that's in our column space, it's going to be equal to c times 2 minus 4."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It looks like I drew it in R2. It looks like there's a 90 degree angle there. And if we wanted to verify it, all we have to do is take the dot product. Well, any vector that is in our column space, so any vector, you could take an arbitrary vector that's in our column space, it's going to be equal to c times 2 minus 4. So let me write that down. So any vector here, so if v is a member, I want this stuff up here, I'll scroll down a little bit. So if v1 is a member of our column space, and that means that v1 is going to be equal to some scalar multiple times the spanning vector of our column space."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Well, any vector that is in our column space, so any vector, you could take an arbitrary vector that's in our column space, it's going to be equal to c times 2 minus 4. So let me write that down. So any vector here, so if v is a member, I want this stuff up here, I'll scroll down a little bit. So if v1 is a member of our column space, and that means that v1 is going to be equal to some scalar multiple times the spanning vector of our column space. So some scalar multiple of this. So we could say it's equal to c1 times 2 minus 4. That's some member of our column space."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So if v1 is a member of our column space, and that means that v1 is going to be equal to some scalar multiple times the spanning vector of our column space. So some scalar multiple of this. So we could say it's equal to c1 times 2 minus 4. That's some member of our column space. Now if we want some member of our left null space, let's write it here. So let's say that v2 is some member of our left null space. v2 is some member of our left null space, or the null space of the transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "That's some member of our column space. Now if we want some member of our left null space, let's write it here. So let's say that v2 is some member of our left null space. v2 is some member of our left null space, or the null space of the transpose. Now what does that mean? That means v2 is going to be equal to some scalar multiple of the spanning vector of our left null space, of 2, 1. So any vector that's in our column space can be represented this way."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "v2 is some member of our left null space, or the null space of the transpose. Now what does that mean? That means v2 is going to be equal to some scalar multiple of the spanning vector of our left null space, of 2, 1. So any vector that's in our column space can be represented this way. Any vector in our left null space can be represented this way. Now what happens if you take the dot product of these two characters? So let me do it down here."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So any vector that's in our column space can be represented this way. Any vector in our left null space can be represented this way. Now what happens if you take the dot product of these two characters? So let me do it down here. I want to save some space for what we're going to do in R3, but let me take the dot product of these two characters. So v1 dot v2 is equal to, I'll arbitrarily switch colors, c1 times 2 minus 4 dot c2 times 2, 1. And then the scalars, we've seen this before."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let me do it down here. I want to save some space for what we're going to do in R3, but let me take the dot product of these two characters. So v1 dot v2 is equal to, I'll arbitrarily switch colors, c1 times 2 minus 4 dot c2 times 2, 1. And then the scalars, we've seen this before. You can just say that this is the same thing as c1 c2 times the dot product of 2 minus 4 dot 2, 1. And then what is this equal to? This is going to be equal to c1 c2 times 2 times 2 is 4."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And then the scalars, we've seen this before. You can just say that this is the same thing as c1 c2 times the dot product of 2 minus 4 dot 2, 1. And then what is this equal to? This is going to be equal to c1 c2 times 2 times 2 is 4. And then plus minus 4 times 1 minus 4. Well, this is going to be equal to 0. So this whole expression is going to be equal to 0."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to c1 c2 times 2 times 2 is 4. And then plus minus 4 times 1 minus 4. Well, this is going to be equal to 0. So this whole expression is going to be equal to 0. And this was for any two vectors that are members of our column space and our left null space. They're orthogonal to each other. So every member of our column space is going to be orthogonal to every member of our left null space, or every member of the null space of our transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So this whole expression is going to be equal to 0. And this was for any two vectors that are members of our column space and our left null space. They're orthogonal to each other. So every member of our column space is going to be orthogonal to every member of our left null space, or every member of the null space of our transpose. And that was the case in this example. And it actually turns out this is always going to be the case. That your column space of a matrix, its orthogonal complement, is the left null space, or the null space of its transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So every member of our column space is going to be orthogonal to every member of our left null space, or every member of the null space of our transpose. And that was the case in this example. And it actually turns out this is always going to be the case. That your column space of a matrix, its orthogonal complement, is the left null space, or the null space of its transpose. I'll prove that probably in the next video. Either the next video or the video after that. But you can see it visually for this example."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "That your column space of a matrix, its orthogonal complement, is the left null space, or the null space of its transpose. I'll prove that probably in the next video. Either the next video or the video after that. But you can see it visually for this example. Now let's draw the other two characters that we're dealing with here. So we have our null space, which is a span of these two vectors in R3. It's a little bit more difficult to draw it."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But you can see it visually for this example. Now let's draw the other two characters that we're dealing with here. So we have our null space, which is a span of these two vectors in R3. It's a little bit more difficult to draw it. These two vectors in R3 right there. But what is the span of two vectors in R3? All of the linear combinations of two vectors in R3 is going to be a plane in R3."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It's a little bit more difficult to draw it. These two vectors in R3 right there. But what is the span of two vectors in R3? All of the linear combinations of two vectors in R3 is going to be a plane in R3. So I'll draw it in just very general terms right here. If we draw it in just very general terms. Let me see."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "All of the linear combinations of two vectors in R3 is going to be a plane in R3. So I'll draw it in just very general terms right here. If we draw it in just very general terms. Let me see. So let's say it's a plane in R3. It's a plane in R3 that looks like that. Maybe I'll fill in the plane a little bit."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me see. So let's say it's a plane in R3. It's a plane in R3 that looks like that. Maybe I'll fill in the plane a little bit. Give you some sense of what it looks like. This is the null space of A. And it's spanned by these two vectors."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I'll fill in the plane a little bit. Give you some sense of what it looks like. This is the null space of A. And it's spanned by these two vectors. You can imagine these two vectors look something like. I'm drawing it very general. But if you take any linear combinations of these two guys, you're going to get stuff, any vector that's along this plane that goes infinite directions."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And it's spanned by these two vectors. You can imagine these two vectors look something like. I'm drawing it very general. But if you take any linear combinations of these two guys, you're going to get stuff, any vector that's along this plane that goes infinite directions. And of course, the origin will be in these. All of these are valid subspaces. Now, what does the row space of A look like?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But if you take any linear combinations of these two guys, you're going to get stuff, any vector that's along this plane that goes infinite directions. And of course, the origin will be in these. All of these are valid subspaces. Now, what does the row space of A look like? Or you could say the column space of A transpose. Well, it's the span of this vector in R3. But let me show you something."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Now, what does the row space of A look like? Or you could say the column space of A transpose. Well, it's the span of this vector in R3. But let me show you something. Well, let's see something interesting about this vector in R3. How does it relate to these two vectors? Well, you may not see it immediately."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But let me show you something. Well, let's see something interesting about this vector in R3. How does it relate to these two vectors? Well, you may not see it immediately. Although if you kind of look at it closely, it might pop out at you that this guy is orthogonal to both of these guys. Notice, if you take the dot product of 2, minus 1, minus 3, and you dotted it with 1 half, 1, 0, what are you going to get? You're going to get 2 times 1 half, which is 1, plus minus 1 times 1, which is minus 1, plus minus 3 times 0, which is 0."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Well, you may not see it immediately. Although if you kind of look at it closely, it might pop out at you that this guy is orthogonal to both of these guys. Notice, if you take the dot product of 2, minus 1, minus 3, and you dotted it with 1 half, 1, 0, what are you going to get? You're going to get 2 times 1 half, which is 1, plus minus 1 times 1, which is minus 1, plus minus 3 times 0, which is 0. So that's when I dotted that guy with that guy right there. And then when I take the dot of this guy with that guy, what do you get? You get 3 halves, 0, and 1 dotted with, let me scroll down a little bit."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "You're going to get 2 times 1 half, which is 1, plus minus 1 times 1, which is minus 1, plus minus 3 times 0, which is 0. So that's when I dotted that guy with that guy right there. And then when I take the dot of this guy with that guy, what do you get? You get 3 halves, 0, and 1 dotted with, let me scroll down a little bit. Don't want to write too small. Dotted with 1, dotted with 2, minus 1, minus 3. In the row space of A, I wrote the spanning vector there."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "You get 3 halves, 0, and 1 dotted with, let me scroll down a little bit. Don't want to write too small. Dotted with 1, dotted with 2, minus 1, minus 3. In the row space of A, I wrote the spanning vector there. I wrote the spanning vector there this time. I probably shouldn't have switched the order. But here I'm dotting it with this guy, and then here I'm dotting it with this guy right there."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "In the row space of A, I wrote the spanning vector there. I wrote the spanning vector there this time. I probably shouldn't have switched the order. But here I'm dotting it with this guy, and then here I'm dotting it with this guy right there. So if you take it, 3 halves times 2 is equal to 3, plus 0 times minus 1 is 0, plus 1 times minus 3 is minus 3. So it's equal to 0. So the fact that this guy is orthogonal to both of that, both of these spanning vectors, it also means that it's orthogonal to any linear combination of those guys."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But here I'm dotting it with this guy, and then here I'm dotting it with this guy right there. So if you take it, 3 halves times 2 is equal to 3, plus 0 times minus 1 is 0, plus 1 times minus 3 is minus 3. So it's equal to 0. So the fact that this guy is orthogonal to both of that, both of these spanning vectors, it also means that it's orthogonal to any linear combination of those guys. Maybe it might be useful for you to see that. So let's say that, let's take some member of our null space. So let's say that v3 is a member of our null space."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So the fact that this guy is orthogonal to both of that, both of these spanning vectors, it also means that it's orthogonal to any linear combination of those guys. Maybe it might be useful for you to see that. So let's say that, let's take some member of our null space. So let's say that v3 is a member of our null space. That means it's a linear combination of that guy and that guy. Those are the two spanning vectors. I had written it up here."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that v3 is a member of our null space. That means it's a linear combination of that guy and that guy. Those are the two spanning vectors. I had written it up here. These are our two spanning vectors. I need the space down here, so let me scroll down a little bit, these are the two spanning vectors. So that means that v3 can be written as some linear combination of these two guys that I squared off in pink."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "I had written it up here. These are our two spanning vectors. I need the space down here, so let me scroll down a little bit, these are the two spanning vectors. So that means that v3 can be written as some linear combination of these two guys that I squared off in pink. So let me just write it as maybe a times 3 halves 0, 1, plus b times 2b times 1 half 1 0. Now, what happens if I take the dot product of v3 and I dot it with any member of my row space right here? So any member of my row space is going to be a multiple of this guy right here."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So that means that v3 can be written as some linear combination of these two guys that I squared off in pink. So let me just write it as maybe a times 3 halves 0, 1, plus b times 2b times 1 half 1 0. Now, what happens if I take the dot product of v3 and I dot it with any member of my row space right here? So any member of my row space is going to be a multiple of this guy right here. That is a spanning vector of my row space. So let me actually create that. So let me say that v4 is a member of my row space, which is the column space of the transpose of a."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So any member of my row space is going to be a multiple of this guy right here. That is a spanning vector of my row space. So let me actually create that. So let me say that v4 is a member of my row space, which is the column space of the transpose of a. And that means that v4 is equal to, let's say, some scaling vector. Well, as you see, you'll see a lot. Let me use d. Let's say it's d times my spanning vector."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So let me say that v4 is a member of my row space, which is the column space of the transpose of a. And that means that v4 is equal to, let's say, some scaling vector. Well, as you see, you'll see a lot. Let me use d. Let's say it's d times my spanning vector. d times 2 minus 1, 3. So what is v3, which is just any member of my null space, dotted with v4, which is any member of my row space? So what is this going to be equal to?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me use d. Let's say it's d times my spanning vector. d times 2 minus 1, 3. So what is v3, which is just any member of my null space, dotted with v4, which is any member of my row space? So what is this going to be equal to? This is going to be equal to this guy. So let me write it like this. a times 3 halves, 0, 1, plus b times 1 half, 1, 0, dotted with this guy, dot d times 2 minus 1, 3."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So what is this going to be equal to? This is going to be equal to this guy. So let me write it like this. a times 3 halves, 0, 1, plus b times 1 half, 1, 0, dotted with this guy, dot d times 2 minus 1, 3. What is this going to be equal to? Well, we know all of the properties of vector dot products. We can distribute it and then take the scalars out."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "a times 3 halves, 0, 1, plus b times 1 half, 1, 0, dotted with this guy, dot d times 2 minus 1, 3. What is this going to be equal to? Well, we know all of the properties of vector dot products. We can distribute it and then take the scalars out. So this is going to be equal to, I'll skip a few steps here, but it's going to be equal to ad times the dot product of 3 halves, 0, 1, dot 2 minus 1, 3. Just distribute it out to here. Plus bd times the dot product of 1 half, 1, 0, dotted with 2 minus 1, 3."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "We can distribute it and then take the scalars out. So this is going to be equal to, I'll skip a few steps here, but it's going to be equal to ad times the dot product of 3 halves, 0, 1, dot 2 minus 1, 3. Just distribute it out to here. Plus bd times the dot product of 1 half, 1, 0, dotted with 2 minus 1, 3. This is a dot product. I just distributed this term on these two terms right here. And we already know what these dot products are equal to."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "Plus bd times the dot product of 1 half, 1, 0, dotted with 2 minus 1, 3. This is a dot product. I just distributed this term on these two terms right here. And we already know what these dot products are equal to. We did it right here. This dot product right here is that dot product. I just switched the order, so this is equal to 0."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And we already know what these dot products are equal to. We did it right here. This dot product right here is that dot product. I just switched the order, so this is equal to 0. And this dot product is that dot product. So this is also equal to 0. So you take any member of your row space and you dot it with any member of your null space and you're going to get 0."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "I just switched the order, so this is equal to 0. And this dot product is that dot product. So this is also equal to 0. So you take any member of your row space and you dot it with any member of your null space and you're going to get 0. Or any member of your row space is orthogonal to any member of your null space. And I did all of that to help our visualization. So we just saw that any member of our row space, which is the span of this vector, is orthogonal to any member of our null space."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So you take any member of your row space and you dot it with any member of your null space and you're going to get 0. Or any member of your row space is orthogonal to any member of your null space. And I did all of that to help our visualization. So we just saw that any member of our row space, which is the span of this vector, is orthogonal to any member of our null space. So my row space, which is just going to be a line in R3 because it's just a multiple of a vector, is going to look like this. It's going to look like this. It's going to be a line and then it's going to maybe go behind it."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "So we just saw that any member of our row space, which is the span of this vector, is orthogonal to any member of our null space. So my row space, which is just going to be a line in R3 because it's just a multiple of a vector, is going to look like this. It's going to look like this. It's going to be a line and then it's going to maybe go behind it. You can't see it there. It's going to look like that. But it's going to be orthogonal."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be a line and then it's going to maybe go behind it. You can't see it there. It's going to look like that. But it's going to be orthogonal. So let me draw it like this. So this pink line right here in R3, that is our row space of A, which is equal to the column space of A transpose, because the rows of A are the same thing as the columns of A transpose and the row space is just the space spanned by your row vectors. And then this is the null space of A, which is a plane."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "But it's going to be orthogonal. So let me draw it like this. So this pink line right here in R3, that is our row space of A, which is equal to the column space of A transpose, because the rows of A are the same thing as the columns of A transpose and the row space is just the space spanned by your row vectors. And then this is the null space of A, which is a plane. It's spanned by two vectors in R3. Or we could also call that the left null space of A transpose. And I never used this term in the last video, but it's symmetric, right?"}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And then this is the null space of A, which is a plane. It's spanned by two vectors in R3. Or we could also call that the left null space of A transpose. And I never used this term in the last video, but it's symmetric, right? If the null space of A transpose is the left null space of A, then the null space of A is the left null space of A transpose, which is an interesting takeaway. Notice you have here the row space of A is orthogonal to the null space of A. And here you have the row space of A transpose is orthogonal to the null space of A transpose."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And I never used this term in the last video, but it's symmetric, right? If the null space of A transpose is the left null space of A, then the null space of A is the left null space of A transpose, which is an interesting takeaway. Notice you have here the row space of A is orthogonal to the null space of A. And here you have the row space of A transpose is orthogonal to the null space of A transpose. Or you could say the left null space of A is orthogonal to the column space of A. Or you could say the left null space of A transpose is orthogonal to the column space of A transpose. So these are just very interesting takeaways in general."}, {"video_title": "Visualizations of left nullspace and rowspace Linear Algebra Khan Academy.mp3", "Sentence": "And here you have the row space of A transpose is orthogonal to the null space of A transpose. Or you could say the left null space of A is orthogonal to the column space of A. Or you could say the left null space of A transpose is orthogonal to the column space of A transpose. So these are just very interesting takeaways in general. But this shows you, and just like I said here, that look, these happen to be orthogonal, these also happen to be orthogonal, and this isn't just some strange coincidence. In the next video or two, I'll show you that these are actually, that this space, this pink space, is the orthogonal complement of the null space right here, which means every vector in that is orthogonal to, or it represents all of the vectors that are orthogonal to the null space. And these two guys are orthogonal complements to each other."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And right from the get-go, we said, oh, the span of this matrix is just the span of the column vectors of it. And I just wrote it right there. But we weren't clear whether this was linearly independent. And if it's not linearly independent, it won't be a sufficient basis. And then we go off and we figure out the null space of a. We find out that the null space of a contains more than just the zero vector. It's just the span of these two vectors here."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if it's not linearly independent, it won't be a sufficient basis. And then we go off and we figure out the null space of a. We find out that the null space of a contains more than just the zero vector. It's just the span of these two vectors here. Which means that these columns are not linearly independent. We saw that several videos ago. And we use that information, that they're not linearly independent, to try to make them linearly independent by getting rid of the redundant ones."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just the span of these two vectors here. Which means that these columns are not linearly independent. We saw that several videos ago. And we use that information, that they're not linearly independent, to try to make them linearly independent by getting rid of the redundant ones. So we were able to get rid of this guy and this guy, because these were essentially the columns associated with the free variables. And we were able to do it using this little technique down here, where we were able to set one of these equal to zero, the other one equal to negative 1, and then solve for the pivot variables. And then we set the other one equal to zero and the other one equal to negative 1 and solved for the pivot variables."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we use that information, that they're not linearly independent, to try to make them linearly independent by getting rid of the redundant ones. So we were able to get rid of this guy and this guy, because these were essentially the columns associated with the free variables. And we were able to do it using this little technique down here, where we were able to set one of these equal to zero, the other one equal to negative 1, and then solve for the pivot variables. And then we set the other one equal to zero and the other one equal to negative 1 and solved for the pivot variables. And you could imagine that this is a kind of a generalizable process. You can always, if you have a bunch of free variables, you can set all of them but one to equal zero. And then all of the other one, and then that one that you didn't set to zero, set it to equal negative 1."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we set the other one equal to zero and the other one equal to negative 1 and solved for the pivot variables. And you could imagine that this is a kind of a generalizable process. You can always, if you have a bunch of free variables, you can set all of them but one to equal zero. And then all of the other one, and then that one that you didn't set to zero, set it to equal negative 1. And then you can express it as a sum of the pivot variables, where the pivot variables are a function of the free variables. So in general, and this would be kind of a quick way of doing it. We have to do it more slowly over here."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then all of the other one, and then that one that you didn't set to zero, set it to equal negative 1. And then you can express it as a sum of the pivot variables, where the pivot variables are a function of the free variables. So in general, and this would be kind of a quick way of doing it. We have to do it more slowly over here. If I have some matrix A, and I want to figure out the basis for the column space. The column space is just a span of these. But if I wanted a linearly independent basis, I need to figure out some set of these that are linearly independent."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have to do it more slowly over here. If I have some matrix A, and I want to figure out the basis for the column space. The column space is just a span of these. But if I wanted a linearly independent basis, I need to figure out some set of these that are linearly independent. What I can do is put this guy into reduced row echelon form. When I put them in reduced row echelon form, and I did that over here, this is the reduced row echelon form of A. I can look at the variables that are associated with the pivot entries. So this is x1."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if I wanted a linearly independent basis, I need to figure out some set of these that are linearly independent. What I can do is put this guy into reduced row echelon form. When I put them in reduced row echelon form, and I did that over here, this is the reduced row echelon form of A. I can look at the variables that are associated with the pivot entries. So this is x1. Let me write it this. Let me scroll up a little bit. This is associated with x1."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is x1. Let me write it this. Let me scroll up a little bit. This is associated with x1. When you multiply this times x1, you get this column times x1, this column times x2, this column times x3, this column times x4, like that. When you look at just regular A, when you look at just your matrix A, it's the same thing. If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3, and x4, like that."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is associated with x1. When you multiply this times x1, you get this column times x1, this column times x2, this column times x3, this column times x4, like that. When you look at just regular A, when you look at just your matrix A, it's the same thing. If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3, and x4, like that. What you can do is you put it in reduced row echelon form. You say, which columns have my pivot entries or associated with pivot variables? Say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3, and x4, like that. What you can do is you put it in reduced row echelon form. You say, which columns have my pivot entries or associated with pivot variables? Say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns. And so those first two columns would be a basis for the column space. And how did I get this? Am I just making up things on the fly?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns. And so those first two columns would be a basis for the column space. And how did I get this? Am I just making up things on the fly? Well, no. It all comes from the reality that you can always construct a situation where the vectors associated with the free variables, you can write them as linear combinations of the vectors associated with the pivot variables. And we did a special case of that last time."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Am I just making up things on the fly? Well, no. It all comes from the reality that you can always construct a situation where the vectors associated with the free variables, you can write them as linear combinations of the vectors associated with the pivot variables. And we did a special case of that last time. So a very quick and dirty way of doing it, I don't know if it's actually dirty, is just take your matrix, put it in reduced row echelon form, and you say, look, this column and this column are associated with my free variables. Therefore, this column and this column must be associated with my free variables. The solution sets are the same."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we did a special case of that last time. So a very quick and dirty way of doing it, I don't know if it's actually dirty, is just take your matrix, put it in reduced row echelon form, and you say, look, this column and this column are associated with my free variables. Therefore, this column and this column must be associated with my free variables. The solution sets are the same. To ax equal to 0, or the reduced row echelon form of ax is equal to 0. So they're the same. So if this column and this column are associated with free variables, so are this column and this column."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The solution sets are the same. To ax equal to 0, or the reduced row echelon form of ax is equal to 0. So they're the same. So if this column and this column are associated with free variables, so are this column and this column. Which means that they can be expressed just by judiciously selecting your values over free variables as linear combinations of the columns associated with the pivot variables, with the pivot entries, which are that column and that column. So this column and this column would be a basis for a. And we see that all the way down here."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if this column and this column are associated with free variables, so are this column and this column. Which means that they can be expressed just by judiciously selecting your values over free variables as linear combinations of the columns associated with the pivot variables, with the pivot entries, which are that column and that column. So this column and this column would be a basis for a. And we see that all the way down here. 1, 2, 3, and 1, 1, 4. We did a lot of work, and we got all the way there. And we said this is a basis for the column span of a."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we see that all the way down here. 1, 2, 3, and 1, 1, 4. We did a lot of work, and we got all the way there. And we said this is a basis for the column span of a. Now, doing all of that work, let's see if we can actually visualize what the column space of a looks like. I have a strange feeling I might have said column spans a couple of times, but the column space, what does it look like? Well, there's a couple of ways to think about what it looks like."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we said this is a basis for the column span of a. Now, doing all of that work, let's see if we can actually visualize what the column space of a looks like. I have a strange feeling I might have said column spans a couple of times, but the column space, what does it look like? Well, there's a couple of ways to think about what it looks like. One way is we can say, look, this span, that's a member of R3. That's a vector in R3, and this is a vector in R3. So we have two."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about what it looks like. One way is we can say, look, this span, that's a member of R3. That's a vector in R3, and this is a vector in R3. So we have two. If we imagine, let me draw my x, z, and normally it's drawn this is normally y, x, and z axes in R3 if I want to represent them as three-dimensional space. Then the vector 1, 2, 3 might look like this, where it's 1, 1, 2, 1, 2, 3. So we go out 1 down here, then up 3."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we have two. If we imagine, let me draw my x, z, and normally it's drawn this is normally y, x, and z axes in R3 if I want to represent them as three-dimensional space. Then the vector 1, 2, 3 might look like this, where it's 1, 1, 2, 1, 2, 3. So we go out 1 down here, then up 3. So the vector will look like that in its standard form. That's that one right there. And the vector 1, 1, 4 would be 1, 1, and go up 4."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we go out 1 down here, then up 3. So the vector will look like that in its standard form. That's that one right there. And the vector 1, 1, 4 would be 1, 1, and go up 4. So it might look something like this. It's hard to actually draw them very well in three dimensions, but you get the idea. But the column space is a span of these."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the vector 1, 1, 4 would be 1, 1, and go up 4. So it might look something like this. It's hard to actually draw them very well in three dimensions, but you get the idea. But the column space is a span of these. So all of the linear combinations of these two guys are going to create a plane that contains these two vectors, that if you just sum these guys up, multiple combinations, you can get any vector up there. If you just add them up, you'll get that vector right there. If you add this guy plus 2 times that, you'll get some vector up here."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But the column space is a span of these. So all of the linear combinations of these two guys are going to create a plane that contains these two vectors, that if you just sum these guys up, multiple combinations, you can get any vector up there. If you just add them up, you'll get that vector right there. If you add this guy plus 2 times that, you'll get some vector up here. So if you view them as position vectors, they'll essentially form a plane in R3. But let's see if we can get the equation for that plane. Well, how can we do that?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you add this guy plus 2 times that, you'll get some vector up here. So if you view them as position vectors, they'll essentially form a plane in R3. But let's see if we can get the equation for that plane. Well, how can we do that? Well, we know that we can figure out the equation of a plane based on the fact that a normal vector dotted with any let's say that this is some, let me write my normal vector like this, the normal vector dotted with any position vector specifying a position on the plane. So let me call that x minus any point on the plane or any vector position on the plane. So I could do that minus the vector 1, 2, 3 has to be equal to 0."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, how can we do that? Well, we know that we can figure out the equation of a plane based on the fact that a normal vector dotted with any let's say that this is some, let me write my normal vector like this, the normal vector dotted with any position vector specifying a position on the plane. So let me call that x minus any point on the plane or any vector position on the plane. So I could do that minus the vector 1, 2, 3 has to be equal to 0. And we can use this information to figure out the equation for this plane. But what is a normal vector? How can we find a normal vector to this plane?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could do that minus the vector 1, 2, 3 has to be equal to 0. And we can use this information to figure out the equation for this plane. But what is a normal vector? How can we find a normal vector to this plane? So this would be a vector. Let me see if I can draw this in a way that doesn't confuse the issue. The plane is like that."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "How can we find a normal vector to this plane? So this would be a vector. Let me see if I can draw this in a way that doesn't confuse the issue. The plane is like that. The normal vector would come out like that. So how can I create a normal vector? Well, we learned that you take the cross product of any two vectors in R3, and the cross product I've only defined so far in R3, and I will get a vector that's normal to both of those vectors."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The plane is like that. The normal vector would come out like that. So how can I create a normal vector? Well, we learned that you take the cross product of any two vectors in R3, and the cross product I've only defined so far in R3, and I will get a vector that's normal to both of those vectors. So let's take the cross product. Let me say, this is a nice way of thinking about it because it's really integrating everything that we've covered so far. So let me define my normal vector to be equal to 1, 2, 3 cross 1, 1, 4."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we learned that you take the cross product of any two vectors in R3, and the cross product I've only defined so far in R3, and I will get a vector that's normal to both of those vectors. So let's take the cross product. Let me say, this is a nice way of thinking about it because it's really integrating everything that we've covered so far. So let me define my normal vector to be equal to 1, 2, 3 cross 1, 1, 4. What does this equal? So my first term, I ignore that. I get 2 times 4 minus 3 times 1."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me define my normal vector to be equal to 1, 2, 3 cross 1, 1, 4. What does this equal? So my first term, I ignore that. I get 2 times 4 minus 3 times 1. 2 times 4 is 8. 2 times 4 minus 3 times 1. 8 minus 3."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I get 2 times 4 minus 3 times 1. 2 times 4 is 8. 2 times 4 minus 3 times 1. 8 minus 3. Then for my second row, I have 1 times 4, and my temptation is to do 1 times 4 minus 3 times 1. But you reverse it. You do 3 times 1, so it's 3, minus 1 times 4."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "8 minus 3. Then for my second row, I have 1 times 4, and my temptation is to do 1 times 4 minus 3 times 1. But you reverse it. You do 3 times 1, so it's 3, minus 1 times 4. We've done that multiple times. You might want to review the cross product video if that seemed strange. You ignore the middle row."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You do 3 times 1, so it's 3, minus 1 times 4. We've done that multiple times. You might want to review the cross product video if that seemed strange. You ignore the middle row. You normally would do 1 times 4 minus 3 times 1, but the middle row you switch. Or you keep switching as you go down. Well, we're only defining it for R3."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You ignore the middle row. You normally would do 1 times 4 minus 3 times 1, but the middle row you switch. Or you keep switching as you go down. Well, we're only defining it for R3. So instead, we do 3 times 1 minus 1 times 4. And then finally for the last row, we ignore it. We say 1 times 1, which is 1, minus 2 times 1, which is minus 2."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we're only defining it for R3. So instead, we do 3 times 1 minus 1 times 4. And then finally for the last row, we ignore it. We say 1 times 1, which is 1, minus 2 times 1, which is minus 2. And this is equal to the vector 5 minus 1 minus 1. Which by definition of the cross product, and I've shown this to you multiple times, is normal to both of these vectors. And so it'll be normal to any linear combination of these two vectors."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We say 1 times 1, which is 1, minus 2 times 1, which is minus 2. And this is equal to the vector 5 minus 1 minus 1. Which by definition of the cross product, and I've shown this to you multiple times, is normal to both of these vectors. And so it'll be normal to any linear combination of these two vectors. So now that we have our normal vector, we can define or we can figure out the traditional equation for the plane. So we now know that our normal vector, 5 minus 1 minus 1, that I got just by taking the cross product of our basis vectors, dot any vector in our plane. So let me just write any vector."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so it'll be normal to any linear combination of these two vectors. So now that we have our normal vector, we can define or we can figure out the traditional equation for the plane. So we now know that our normal vector, 5 minus 1 minus 1, that I got just by taking the cross product of our basis vectors, dot any vector in our plane. So let me just write any vector. Let me just write it x, y, z. So x, y, z, since that's how I defined my axes up here, this was the x-axis, x, y, and z. x, y, z minus, I just picked one of these. I could have picked either of them."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me just write any vector. Let me just write it x, y, z. So x, y, z, since that's how I defined my axes up here, this was the x-axis, x, y, and z. x, y, z minus, I just picked one of these. I could have picked either of them. Minus 1, 2, 3 has got to be equal to 0. So what's this? This is going to be equal to, let me write it a little smaller, a little neater."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could have picked either of them. Minus 1, 2, 3 has got to be equal to 0. So what's this? This is going to be equal to, let me write it a little smaller, a little neater. 5 minus 1 minus 1 dot, what's this guy going to be? x minus 1, y minus 2, and z minus 3 has got to be equal to 0. And what's the dot product?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to, let me write it a little smaller, a little neater. 5 minus 1 minus 1 dot, what's this guy going to be? x minus 1, y minus 2, and z minus 3 has got to be equal to 0. And what's the dot product? It's 5 times x minus 1. Or maybe I should say minus 1. So it's plus minus 1 times y minus 2 plus minus 1 times z minus 3 is equal to 0."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what's the dot product? It's 5 times x minus 1. Or maybe I should say minus 1. So it's plus minus 1 times y minus 2 plus minus 1 times z minus 3 is equal to 0. That's just the definition of our dot product. And if I simplify this, I get 5x minus 5 minus y plus 2 minus z plus 3 is equal to 0. You have 2 plus 3 is 5 minus 5, so those all cancel out."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's plus minus 1 times y minus 2 plus minus 1 times z minus 3 is equal to 0. That's just the definition of our dot product. And if I simplify this, I get 5x minus 5 minus y plus 2 minus z plus 3 is equal to 0. You have 2 plus 3 is 5 minus 5, so those all cancel out. Those are equal to 0. And we get 5x minus y minus z is equal to 0. And this plane in R3 is the column space of A."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You have 2 plus 3 is 5 minus 5, so those all cancel out. Those are equal to 0. And we get 5x minus y minus z is equal to 0. And this plane in R3 is the column space of A. So we've now shown you that it's truly a plane in A. And it actually makes sense that this plane intersects the origin. If you set x, y, and z equal 0, it satisfies this equation."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this plane in R3 is the column space of A. So we've now shown you that it's truly a plane in A. And it actually makes sense that this plane intersects the origin. If you set x, y, and z equal 0, it satisfies this equation. And that makes sense because we said that a column space of a matrix has to be a valid subspace. And a valid subspace has to contain the 0 vector. And in R3, that's the coordinate 0, 0, 0."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you set x, y, and z equal 0, it satisfies this equation. And that makes sense because we said that a column space of a matrix has to be a valid subspace. And a valid subspace has to contain the 0 vector. And in R3, that's the coordinate 0, 0, 0. Now what I want to do now is see if we can get at the same answer going a completely different way about the column, or approaching it in a completely different way. So let me get my original A, which I've forgotten. I've marked it up a good bit, but let me just copy and paste it."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And in R3, that's the coordinate 0, 0, 0. Now what I want to do now is see if we can get at the same answer going a completely different way about the column, or approaching it in a completely different way. So let me get my original A, which I've forgotten. I've marked it up a good bit, but let me just copy and paste it. So that's my original A right there. Let me copy it. Let me paste it."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I've marked it up a good bit, but let me just copy and paste it. So that's my original A right there. Let me copy it. Let me paste it. Nope, that's not what I wanted to do. I want to do it, let's see, so my original A, copied and pasted the wrong thing. Let me do it a little, don't want to waste your time."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me paste it. Nope, that's not what I wanted to do. I want to do it, let's see, so my original A, copied and pasted the wrong thing. Let me do it a little, don't want to waste your time. Edit, copy, edit, paste. There you go. Let me scroll this down, get to a point that's relatively clean."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it a little, don't want to waste your time. Edit, copy, edit, paste. There you go. Let me scroll this down, get to a point that's relatively clean. Bring my A down, I've used a lot of space. So here you go. This is my original A right there."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll this down, get to a point that's relatively clean. Bring my A down, I've used a lot of space. So here you go. This is my original A right there. My original A, and what I want to do is see if I can get this result completely different. I got this result by figuring out the basis of the column span, finding a normal vector by taking the cross product of our two basis vectors, and then using the dot product of the normal vector with the difference, this vector right here, where you take any vector on our plane minus one of our basis vectors, that's to find some vector in the plane. This is some vector in the plane."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is my original A right there. My original A, and what I want to do is see if I can get this result completely different. I got this result by figuring out the basis of the column span, finding a normal vector by taking the cross product of our two basis vectors, and then using the dot product of the normal vector with the difference, this vector right here, where you take any vector on our plane minus one of our basis vectors, that's to find some vector in the plane. This is some vector in the plane. So any vector in the plane dotted with my normal vector is going to be equal to zero. And actually, I should probably make a side note here, that the only reason I was able to say that the normal vector is a cross product of my two basis vectors is because I knew that these two basis vectors, not only do they specify some point on the plane, so let's say that this guy right here is this blue vector, not only does he specify some point on the plane right there, not only does he specify this point right there, but the vector lies completely on the plane. How did I know that?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is some vector in the plane. So any vector in the plane dotted with my normal vector is going to be equal to zero. And actually, I should probably make a side note here, that the only reason I was able to say that the normal vector is a cross product of my two basis vectors is because I knew that these two basis vectors, not only do they specify some point on the plane, so let's say that this guy right here is this blue vector, not only does he specify some point on the plane right there, not only does he specify this point right there, but the vector lies completely on the plane. How did I know that? Because I knew from the get-go that the 0, 0 vector is in my span, right? I knew that if I draw this guy in just standard position, the point 0, 0, 0 is in my span, and I know that its endpoint is in the span, so this whole vector has to be in my plane. And likewise, this whole vector has to be in my plane, so if I take the cross product, anything normal to these guys or any combination of these guys is going to be normal to the plane."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "How did I know that? Because I knew from the get-go that the 0, 0 vector is in my span, right? I knew that if I draw this guy in just standard position, the point 0, 0, 0 is in my span, and I know that its endpoint is in the span, so this whole vector has to be in my plane. And likewise, this whole vector has to be in my plane, so if I take the cross product, anything normal to these guys or any combination of these guys is going to be normal to the plane. We got this result right here. But let me take this right here and use our other definition of column span. Our other definition, or it's really an equivalent definition, is all of the valid solutions to Ax, let me write it this way, it's all of the valid solutions to Ax where x is a member of Rn."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And likewise, this whole vector has to be in my plane, so if I take the cross product, anything normal to these guys or any combination of these guys is going to be normal to the plane. We got this result right here. But let me take this right here and use our other definition of column span. Our other definition, or it's really an equivalent definition, is all of the valid solutions to Ax, let me write it this way, it's all of the valid solutions to Ax where x is a member of Rn. Or another way we could think of it is we could view it as all of the valid b's where Ax is equal to b and x is a member of Rn. These are equivalent statements. I'm just defining b here to be Ax."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Our other definition, or it's really an equivalent definition, is all of the valid solutions to Ax, let me write it this way, it's all of the valid solutions to Ax where x is a member of Rn. Or another way we could think of it is we could view it as all of the valid b's where Ax is equal to b and x is a member of Rn. These are equivalent statements. I'm just defining b here to be Ax. So these are equivalent statements. But let's run with this a little bit. So let's say that I define my b."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just defining b here to be Ax. So these are equivalent statements. But let's run with this a little bit. So let's say that I define my b. So b is going to be a vector in R3. All right, and we already have an intuition like that. So let's say that my b, when I take Ax, I get a b that's equal to x, y, z."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that I define my b. So b is going to be a vector in R3. All right, and we already have an intuition like that. So let's say that my b, when I take Ax, I get a b that's equal to x, y, z. And I want to figure out what x, y's, and z's can I get valid solutions for. So if I take my vector a right there, and then let me multiply it times, well, actually the best way to do it, if I'm just solving for, instead of going through, I think we're used to it right now. If I'm solving the equation Ax is equal to b, I can essentially just create the augmented matrix where I have the matrix a, and I can augment it with b, and put this in reduced row echelon form, and that'll essentially represent the solution set."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that my b, when I take Ax, I get a b that's equal to x, y, z. And I want to figure out what x, y's, and z's can I get valid solutions for. So if I take my vector a right there, and then let me multiply it times, well, actually the best way to do it, if I'm just solving for, instead of going through, I think we're used to it right now. If I'm solving the equation Ax is equal to b, I can essentially just create the augmented matrix where I have the matrix a, and I can augment it with b, and put this in reduced row echelon form, and that'll essentially represent the solution set. So let me do that. So if I just augment this matrix right here with b, so I write x, y, z, so this is a augmented with b. Let me put this in reduced row echelon form and find the solution set."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I'm solving the equation Ax is equal to b, I can essentially just create the augmented matrix where I have the matrix a, and I can augment it with b, and put this in reduced row echelon form, and that'll essentially represent the solution set. So let me do that. So if I just augment this matrix right here with b, so I write x, y, z, so this is a augmented with b. Let me put this in reduced row echelon form and find the solution set. These are the x, y's, and z's that define a valid b. So what do I get? So the first thing I want to do, and we've done this exercise before, let's keep my first row the same."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me put this in reduced row echelon form and find the solution set. These are the x, y's, and z's that define a valid b. So what do I get? So the first thing I want to do, and we've done this exercise before, let's keep my first row the same. 1, 1, 1, 1, and I get an x. And let's replace our second row with the second row minus the first row. Or let me say the second row minus 2 times the, actually let me do it this way."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing I want to do, and we've done this exercise before, let's keep my first row the same. 1, 1, 1, 1, and I get an x. And let's replace our second row with the second row minus the first row. Or let me say the second row minus 2 times the, actually let me do it this way. Let me replace the second row with 2 times the first row minus the second row. So 2 times the first row minus the second row, we're going to get a 2x minus y up there. And then 2 times 1 minus 2 is 0."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or let me say the second row minus 2 times the, actually let me do it this way. Let me replace the second row with 2 times the first row minus the second row. So 2 times the first row minus the second row, we're going to get a 2x minus y up there. And then 2 times 1 minus 2 is 0. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then 2 times 1 minus 2 is 0. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. Minus 2, minus 1. Fair enough. And now let me replace my third row with the third row minus 3 times the first row."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times 1 minus 3 is minus 1. Minus 2, minus 1. Fair enough. And now let me replace my third row with the third row minus 3 times the first row. So we're going to do the third row minus, let me do it this way, it's the third row minus 3 times the first row. So I'm just doing the b column first because I can remember what I did. The third row minus 3 times the first row."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now let me replace my third row with the third row minus 3 times the first row. So we're going to do the third row minus, let me do it this way, it's the third row minus 3 times the first row. So I'm just doing the b column first because I can remember what I did. The third row minus 3 times the first row. 3 minus 3 times 1 is 0. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The third row minus 3 times the first row. 3 minus 3 times 1 is 0. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And then 2 minus 3 times 1 is minus 1. Now, I could go all the way to reduced row echelon form, but something interesting is already happening. So let me just from the get-go, let me try to zero out this third row."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 3 times 1 is minus 2. And then 2 minus 3 times 1 is minus 1. Now, I could go all the way to reduced row echelon form, but something interesting is already happening. So let me just from the get-go, let me try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row, well I won't even write the first row, the second row is 1, 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me just from the get-go, let me try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row, well I won't even write the first row, the second row is 1, 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row, so you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row, so you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 is 0."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 is 0. And that's also 0. So we're only going to have a valid solution to Ax equals b if this guy right here is equal to 0. What happens if he's not equal to 0?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 is 0. And that's also 0. So we're only going to have a valid solution to Ax equals b if this guy right here is equal to 0. What happens if he's not equal to 0? Then we're going to have a bunch of 0's equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal 0, then I'll have no solution. If this guy equals 5, if I pick x, y's, and z's such that this expression is equal to 5, then Ax equals b will have no solution because I'll have 0 is equal to 5."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What happens if he's not equal to 0? Then we're going to have a bunch of 0's equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal 0, then I'll have no solution. If this guy equals 5, if I pick x, y's, and z's such that this expression is equal to 5, then Ax equals b will have no solution because I'll have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be in, to be a member of the column space of A, in order for it to be a valid vector that Ax can become or the product A times x can become for some x. So what is this equal to?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If this guy equals 5, if I pick x, y's, and z's such that this expression is equal to 5, then Ax equals b will have no solution because I'll have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be in, to be a member of the column space of A, in order for it to be a valid vector that Ax can become or the product A times x can become for some x. So what is this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is equal to 0, which is the exact same outcome we got when we figured out the basis vectors. We said, oh, you know what? The basis vectors, they have to be in the column space themselves by definition."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what is this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is equal to 0, which is the exact same outcome we got when we figured out the basis vectors. We said, oh, you know what? The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to 0. And then I got this equation."}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to 0. And then I got this equation. Or we could have done it the other way. We could have actually literally solved this equation, setting our b equal to this. And we said, what b's will give us a valid solution?"}, {"video_title": "Visualizing a column space as a plane in R3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I got this equation. Or we could have done it the other way. We could have actually literally solved this equation, setting our b equal to this. And we said, what b's will give us a valid solution? And our only valid solution will be obtained when this guy has to be equal to 0, because the rest of his row became 0. And when we set that equal to 0, we got the exact same equation. So hopefully you found this to be mildly satisfying, because we were able to tackle the same problem from two different directions and get the same result."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I showed you a method of how to do it. You literally put A in reduced row echelon form. So this matrix R is just the reduced row echelon form of A. And you look at its pivot columns. So this is a pivot column. It has a 1 and all 0s. This is a pivot column, 1 and all 0s."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you look at its pivot columns. So this is a pivot column. It has a 1 and all 0s. This is a pivot column, 1 and all 0s. And the 1 is the leading non-zero term in its row. And this is a pivot column. Let me circle them."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a pivot column, 1 and all 0s. And the 1 is the leading non-zero term in its row. And this is a pivot column. Let me circle them. These guys are pivot columns, and this guy is a pivot column right there. And you look at those in the reduced row echelon form of the matrix, and the corresponding columns in the original matrix will be your basis. So this guy, this guy, so the first, second, and fourth."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me circle them. These guys are pivot columns, and this guy is a pivot column right there. And you look at those in the reduced row echelon form of the matrix, and the corresponding columns in the original matrix will be your basis. So this guy, this guy, so the first, second, and fourth. So the first, second, and fourth columns. So if we call this, this is A1, this is A2, and let's call this A4. This would be A3, and this would be A5."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this guy, this guy, so the first, second, and fourth. So the first, second, and fourth columns. So if we call this, this is A1, this is A2, and let's call this A4. This would be A3, and this would be A5. So we could say that A1, A2, and A4 are a basis for the column span of A. And I didn't show you why two videos ago. I just said this is how you do it."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This would be A3, and this would be A5. So we could say that A1, A2, and A4 are a basis for the column span of A. And I didn't show you why two videos ago. I just said this is how you do it. You have to take it as a bit of an article of faith. Now, in order for these to be a basis, two things have to be true. They have to be linearly independent."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just said this is how you do it. You have to take it as a bit of an article of faith. Now, in order for these to be a basis, two things have to be true. They have to be linearly independent. And I showed you in the very last video, the second in our kind of series dealing with this vector, I showed you that by the fact that let's say that this guy is R1, this guy is R2, and this guy is R4, it's clear that these guys are linearly independent. They each have a 1 in a unique entry, and the rest of their entries are 0. So because all of the other, and this is true, we're looking at three pivot columns right now, but it's true of the, if we had n pivot columns, that each pivot column would have a 1 in a unique place, and all of the other pivot columns would have 0 in that entry."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They have to be linearly independent. And I showed you in the very last video, the second in our kind of series dealing with this vector, I showed you that by the fact that let's say that this guy is R1, this guy is R2, and this guy is R4, it's clear that these guys are linearly independent. They each have a 1 in a unique entry, and the rest of their entries are 0. So because all of the other, and this is true, we're looking at three pivot columns right now, but it's true of the, if we had n pivot columns, that each pivot column would have a 1 in a unique place, and all of the other pivot columns would have 0 in that entry. So there's no way that the other pivot columns, any linear combination of the other ones, could never add up to each of them. So these are definitely linearly independent. And I showed in the last video that if we know that these are linearly independent, we do know that they are, given that R has the same null space as A, we know that these guys have to be linearly independent."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So because all of the other, and this is true, we're looking at three pivot columns right now, but it's true of the, if we had n pivot columns, that each pivot column would have a 1 in a unique place, and all of the other pivot columns would have 0 in that entry. So there's no way that the other pivot columns, any linear combination of the other ones, could never add up to each of them. So these are definitely linearly independent. And I showed in the last video that if we know that these are linearly independent, we do know that they are, given that R has the same null space as A, we know that these guys have to be linearly independent. I did that in the very last video. Now the next requirement for a basis, we checked this one off, is to show that A1, A2, and An, that their span equals the column space of A. Now the column space of A is the span of all five of these vectors, so I have to throw A3 in there and A5."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I showed in the last video that if we know that these are linearly independent, we do know that they are, given that R has the same null space as A, we know that these guys have to be linearly independent. I did that in the very last video. Now the next requirement for a basis, we checked this one off, is to show that A1, A2, and An, that their span equals the column space of A. Now the column space of A is the span of all five of these vectors, so I have to throw A3 in there and A5. So to show that just these three vectors by themselves span our column space, we just have to show that, look, I can represent A3 and A5 as linear combinations of A1, A2, and A4. If I can do that, then I can say, well, you know, then these guys are redundant. And the span of A1, A2, A3, A4, and A5 doesn't need the A3 and the A5 terms."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now the column space of A is the span of all five of these vectors, so I have to throw A3 in there and A5. So to show that just these three vectors by themselves span our column space, we just have to show that, look, I can represent A3 and A5 as linear combinations of A1, A2, and A4. If I can do that, then I can say, well, you know, then these guys are redundant. And the span of A1, A2, A3, A4, and A5 doesn't need the A3 and the A5 terms. We can just reduce it to this, because these guys can be represented as linear combinations of the other three. These guys are redundant. And if we can get rid of them, if we can show that these guys can be represented as linear combinations of the other, then we can get rid of them, and then the span of these three guys would be the same as the span of these five guys, which is, of course, the definition of the column space of A."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the span of A1, A2, A3, A4, and A5 doesn't need the A3 and the A5 terms. We can just reduce it to this, because these guys can be represented as linear combinations of the other three. These guys are redundant. And if we can get rid of them, if we can show that these guys can be represented as linear combinations of the other, then we can get rid of them, and then the span of these three guys would be the same as the span of these five guys, which is, of course, the definition of the column space of A. So let's see if we can do that. So let's go back to our original, let me fill in each of these column vectors A1 through A5, and then each of these column vectors, let me label them R1, R2, R3, R4, and R5. Now let's explore the null spaces again, or not even the null spaces, let's just explore the equations."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if we can get rid of them, if we can show that these guys can be represented as linear combinations of the other, then we can get rid of them, and then the span of these three guys would be the same as the span of these five guys, which is, of course, the definition of the column space of A. So let's see if we can do that. So let's go back to our original, let me fill in each of these column vectors A1 through A5, and then each of these column vectors, let me label them R1, R2, R3, R4, and R5. Now let's explore the null spaces again, or not even the null spaces, let's just explore the equations. Ax is equal to, and let me write it this way, instead of x, let me write x1, x2, x3, x4, x5 is equal to 0. This is how we define the solution set of this, all the potential x1's through x5's, or all the potential vectors x right here, that represents our null space. And then also let's explore all of the R times x1, x2, x3, x4, x5's is equal to 0."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's explore the null spaces again, or not even the null spaces, let's just explore the equations. Ax is equal to, and let me write it this way, instead of x, let me write x1, x2, x3, x4, x5 is equal to 0. This is how we define the solution set of this, all the potential x1's through x5's, or all the potential vectors x right here, that represents our null space. And then also let's explore all of the R times x1, x2, x3, x4, x5's is equal to 0. This is the 0 vector, in which case we would have 4 entries in this particular case, it would be a member of Rm. So these equations can be rewritten. I can rewrite this as, what were the column vectors of A?"}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then also let's explore all of the R times x1, x2, x3, x4, x5's is equal to 0. This is the 0 vector, in which case we would have 4 entries in this particular case, it would be a member of Rm. So these equations can be rewritten. I can rewrite this as, what were the column vectors of A? They were A1, A2, through A5. So I can rewrite this as x1 times A1, plus x2 times A2, plus x3 times A3, plus x4 times A4, plus x5 times A5. Is equal to 0."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can rewrite this as, what were the column vectors of A? They were A1, A2, through A5. So I can rewrite this as x1 times A1, plus x2 times A2, plus x3 times A3, plus x4 times A4, plus x5 times A5. Is equal to 0. That's what we've learned, that was from our definition of matrix vector multiplication. And I just rewrote, this is just a bunch of column vectors A1 through A5. I drew it up here, A1 through A5, and I can just rewrite this equation like this."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Is equal to 0. That's what we've learned, that was from our definition of matrix vector multiplication. And I just rewrote, this is just a bunch of column vectors A1 through A5. I drew it up here, A1 through A5, and I can just rewrite this equation like this. Similarly, I can rewrite this equation as the vector R1 times x1, or x1 times R1, plus x2 times R2, plus x3 times R3, plus x4 times R4, plus x5 times R5, is equal to 0. Now we know that when we put this into reduced row echelon form, the x variables that are associated with the pivot columns are, so what are the x variables associated with the pivot columns? Well the pivot columns are R1, R2, and R4."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I drew it up here, A1 through A5, and I can just rewrite this equation like this. Similarly, I can rewrite this equation as the vector R1 times x1, or x1 times R1, plus x2 times R2, plus x3 times R3, plus x4 times R4, plus x5 times R5, is equal to 0. Now we know that when we put this into reduced row echelon form, the x variables that are associated with the pivot columns are, so what are the x variables associated with the pivot columns? Well the pivot columns are R1, R2, and R4. So it's R1, R2, and R4. The x variables associated with them, we can call them pivot variables. And then the ones that are not associated with our pivot columns are free variables."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well the pivot columns are R1, R2, and R4. So it's R1, R2, and R4. The x variables associated with them, we can call them pivot variables. And then the ones that are not associated with our pivot columns are free variables. So free variables in this case, so x3 and x5 are free variables. And that applies to A, this solution set, all of the sets, all of the vectors x that satisfy this equation also satisfy this equation and vice versa. They're the exact same null space, the exact same solution set."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then the ones that are not associated with our pivot columns are free variables. So free variables in this case, so x3 and x5 are free variables. And that applies to A, this solution set, all of the sets, all of the vectors x that satisfy this equation also satisfy this equation and vice versa. They're the exact same null space, the exact same solution set. So we can also call this x3 and this x5 as free variables. Now what does that mean? We've done multiple examples of this."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're the exact same null space, the exact same solution set. So we can also call this x3 and this x5 as free variables. Now what does that mean? We've done multiple examples of this. The free variables, you can set them to anything you want. So x3 in this case, x3 and x5, you can set it to any real number. You can set to anything you want."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We've done multiple examples of this. The free variables, you can set them to anything you want. So x3 in this case, x3 and x5, you can set it to any real number. You can set to anything you want. Set to anything. And then from this reduced row echelon form, we express the other pivot variables as functions of these guys. Maybe x1 is equal to A x3 plus B x5."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can set to anything you want. Set to anything. And then from this reduced row echelon form, we express the other pivot variables as functions of these guys. Maybe x1 is equal to A x3 plus B x5. Maybe x2 is equal to C x3 plus D x5. And maybe x4 is equal to E x3 plus F x5. That comes directly out of the literally multiplying this guy times this equals zero."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe x1 is equal to A x3 plus B x5. Maybe x2 is equal to C x3 plus D x5. And maybe x4 is equal to E x3 plus F x5. That comes directly out of the literally multiplying this guy times this equals zero. You would get a system of equations that you could solve for your pivot variables in terms of your free variables. Now given this, I want to show you that you can always construct one of your, in your original matrix. If you go to our original matrix, you can always construct one of the vectors that are associated with, we could call them the free columns."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That comes directly out of the literally multiplying this guy times this equals zero. You would get a system of equations that you could solve for your pivot variables in terms of your free variables. Now given this, I want to show you that you can always construct one of your, in your original matrix. If you go to our original matrix, you can always construct one of the vectors that are associated with, we could call them the free columns. You could always construct one of the free vectors using the linear combination of the ones that were associated with the pivot columns before. And how do I do that? Well let's say that I want to find some linear combination that gets me to this free column, that gets me to A3."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you go to our original matrix, you can always construct one of the vectors that are associated with, we could call them the free columns. You could always construct one of the free vectors using the linear combination of the ones that were associated with the pivot columns before. And how do I do that? Well let's say that I want to find some linear combination that gets me to this free column, that gets me to A3. So how could I do that? Let me rearrange this equation up here. So what do I get?"}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well let's say that I want to find some linear combination that gets me to this free column, that gets me to A3. So how could I do that? Let me rearrange this equation up here. So what do I get? If I subtract x2 A3 from both sides of this top equation, I get minus, sorry that's x3 A3. If I subtract x3 A3 from both sides of the equation, I get minus x3 A3 is equal to x1 A1 plus x2 A2 plus, I don't have the 3 there, plus x4 A4 plus x5, sorry x isn't a vector, x5 A5. All I did is I, this, I guess salmon colored statement here is just another rewriting of this equation right here."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what do I get? If I subtract x2 A3 from both sides of this top equation, I get minus, sorry that's x3 A3. If I subtract x3 A3 from both sides of the equation, I get minus x3 A3 is equal to x1 A1 plus x2 A2 plus, I don't have the 3 there, plus x4 A4 plus x5, sorry x isn't a vector, x5 A5. All I did is I, this, I guess salmon colored statement here is just another rewriting of this equation right here. And all I did is I subtracted this term right here, x3 A3 from both sides of the equation. Now x3 is a free variable, we can set it to anything we want and so is x5. So let's set x3 is equal to minus 1, so if we set x3 is equal to minus 1, then this term right here becomes a 1, right because that was minus x3."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I, this, I guess salmon colored statement here is just another rewriting of this equation right here. And all I did is I subtracted this term right here, x3 A3 from both sides of the equation. Now x3 is a free variable, we can set it to anything we want and so is x5. So let's set x3 is equal to minus 1, so if we set x3 is equal to minus 1, then this term right here becomes a 1, right because that was minus x3. And let's set x5 equal to 0. So if x5 is equal to 0, this term disappears and I did that because x5 was a free variable, I can set them to anything I want. Now I've written A3 as a linear combination of, I guess you could call it my potential basis vectors right now, or the vectors that were associated as A1, A2 and A4."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's set x3 is equal to minus 1, so if we set x3 is equal to minus 1, then this term right here becomes a 1, right because that was minus x3. And let's set x5 equal to 0. So if x5 is equal to 0, this term disappears and I did that because x5 was a free variable, I can set them to anything I want. Now I've written A3 as a linear combination of, I guess you could call it my potential basis vectors right now, or the vectors that were associated as A1, A2 and A4. They're the vectors in the original matrix that were associated with the pivot columns. Now, in order to show that I can always do this, we have to show that for this combination there's always some x1, x2 and x4 that satisfy this. Well of course there's always some x1, x2 that satisfy this, we just have to substitute our free variables, x3 is equal to minus 3 and x5 is equal to 0 into these equations that we get from our system when we did it with the reduced row echelon form."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I've written A3 as a linear combination of, I guess you could call it my potential basis vectors right now, or the vectors that were associated as A1, A2 and A4. They're the vectors in the original matrix that were associated with the pivot columns. Now, in order to show that I can always do this, we have to show that for this combination there's always some x1, x2 and x4 that satisfy this. Well of course there's always some x1, x2 that satisfy this, we just have to substitute our free variables, x3 is equal to minus 3 and x5 is equal to 0 into these equations that we get from our system when we did it with the reduced row echelon form. So you get, I don't know if I even have to go through, in this case you get x1 is equal to minus A plus 0, x2 is equal to minus C, so on and so forth. So you can always do that. You can always express the vectors that are associated with the non-pivot columns as linear combinations of the vectors that are associated with the pivot columns."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well of course there's always some x1, x2 that satisfy this, we just have to substitute our free variables, x3 is equal to minus 3 and x5 is equal to 0 into these equations that we get from our system when we did it with the reduced row echelon form. So you get, I don't know if I even have to go through, in this case you get x1 is equal to minus A plus 0, x2 is equal to minus C, so on and so forth. So you can always do that. You can always express the vectors that are associated with the non-pivot columns as linear combinations of the vectors that are associated with the pivot columns. What I just did for A3, you could just as easily have done for A5 by subtracting this term from both sides of the equation, setting x5 to negative 1 and setting x3 to 0 so that the 3 term disappears and you could run the same exact argument. So given that, I've hopefully shown you or at least helped you see or made you comfortable with the idea that the vectors, let me do them in a nice vibrant color, these magenta colored vectors here that are associated with the free columns, that are associated with the free columns or with the free variables, right, the free variables were x3 and x5, and that those were these columns right here, that they can always be expressed as linear combinations of the other columns because you just have to manipulate this equation, set the coefficient for whatever you're trying to find a linear combination for equal to minus 1 and set all the other free variables equal to 0 that you're not solving for, and then you can get a linear combination of the vectors that are associated with the pivot columns. So given that, we've shown you that look, these free vectors, and I'm using my terminology very loosely, but these ones that are associated with the non-pivot columns can be expressed as linear combinations of these guys."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can always express the vectors that are associated with the non-pivot columns as linear combinations of the vectors that are associated with the pivot columns. What I just did for A3, you could just as easily have done for A5 by subtracting this term from both sides of the equation, setting x5 to negative 1 and setting x3 to 0 so that the 3 term disappears and you could run the same exact argument. So given that, I've hopefully shown you or at least helped you see or made you comfortable with the idea that the vectors, let me do them in a nice vibrant color, these magenta colored vectors here that are associated with the free columns, that are associated with the free columns or with the free variables, right, the free variables were x3 and x5, and that those were these columns right here, that they can always be expressed as linear combinations of the other columns because you just have to manipulate this equation, set the coefficient for whatever you're trying to find a linear combination for equal to minus 1 and set all the other free variables equal to 0 that you're not solving for, and then you can get a linear combination of the vectors that are associated with the pivot columns. So given that, we've shown you that look, these free vectors, and I'm using my terminology very loosely, but these ones that are associated with the non-pivot columns can be expressed as linear combinations of these guys. So they're unnecessary. The span of this is equivalent to the span of this. The span of this is the column space of A, so the span of this is the column space of A."}, {"video_title": "Showing that the candidate basis does span C(A) Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So given that, we've shown you that look, these free vectors, and I'm using my terminology very loosely, but these ones that are associated with the non-pivot columns can be expressed as linear combinations of these guys. So they're unnecessary. The span of this is equivalent to the span of this. The span of this is the column space of A, so the span of this is the column space of A. So in the last video, I showed you that these guys are linearly independent, and now I've showed you that the span of these guys is the column space of A. So now you should be satisfied that these vectors that are associated, let me do it in a blue color, that that vector, that that column vector, this column vector, and this column vector that are associated with the pivot columns in the reduced row echelon form of the matrix do indeed represent a basis for the column space of A. Anyway, hopefully you didn't find that too convoluted."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "AD minus BC. That's the determinant right there. Now what if we were to multiply one of these rows by a scalar? Let's say we multiply it by K. So we have the situation. A, B. Let's multiply the second row times K. So we have K, C, and K, D. Now what's the determinant going to look like? We're going to have A times KD, or we could just write that as KAD, minus KC times B."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we multiply it by K. So we have the situation. A, B. Let's multiply the second row times K. So we have K, C, and K, D. Now what's the determinant going to look like? We're going to have A times KD, or we could just write that as KAD, minus KC times B. Or we could write that as KBC. If we factor out the K, we get that equals K times AD minus BC. Now you immediately see that this thing is the same thing as this thing."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're going to have A times KD, or we could just write that as KAD, minus KC times B. Or we could write that as KBC. If we factor out the K, we get that equals K times AD minus BC. Now you immediately see that this thing is the same thing as this thing. So this is equal to K times the determinant of A, B, C, D. So when you just multiply a row by a scalar, just one row, not the entire matrix. When you just multiply a row by some scalar, the resulting determinant will be the original determinant times that scalar. Now you might say, well, what happens if I multiply the whole matrix times that scalar?"}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now you immediately see that this thing is the same thing as this thing. So this is equal to K times the determinant of A, B, C, D. So when you just multiply a row by a scalar, just one row, not the entire matrix. When you just multiply a row by some scalar, the resulting determinant will be the original determinant times that scalar. Now you might say, well, what happens if I multiply the whole matrix times that scalar? Well, that's equivalent to multiplying by a scalar twice. If I say that, let's say I have the matrix A, and the matrix A is equal to A, B, C, D. If I were to think about the matrix KA, now I'm not just multiplying one row, I'm multiplying the whole matrix by a scalar. This is going to be equal to K lowercase a, KB, KC, and KD."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now you might say, well, what happens if I multiply the whole matrix times that scalar? Well, that's equivalent to multiplying by a scalar twice. If I say that, let's say I have the matrix A, and the matrix A is equal to A, B, C, D. If I were to think about the matrix KA, now I'm not just multiplying one row, I'm multiplying the whole matrix by a scalar. This is going to be equal to K lowercase a, KB, KC, and KD. And when you figure out its determinant, the determinant of K times A is going to be equal to the determinant of KA, KB, KC, and KD. And here you're immediately going to see you end up with K squared terms. You're going to have K squared times AD is going to be equal to K squared AD minus K squared BC."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to K lowercase a, KB, KC, and KD. And when you figure out its determinant, the determinant of K times A is going to be equal to the determinant of KA, KB, KC, and KD. And here you're immediately going to see you end up with K squared terms. You're going to have K squared times AD is going to be equal to K squared AD minus K squared BC. Or K squared times AD minus BC. Or K squared times the determinant of just A. So you have to be very careful."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're going to have K squared times AD is going to be equal to K squared AD minus K squared BC. Or K squared times AD minus BC. Or K squared times the determinant of just A. So you have to be very careful. And this is only for a 2 by 2 case. You'll find out if this was an N by N matrix, that this would have been K to the N. So the takeaway is the only way you can say that it's going to be a some scalar multiple times your original determinant is only if you multiply one row times that scalar multiple, not the whole matrix. Let's see how this extends to maybe a 3 by 3 case."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you have to be very careful. And this is only for a 2 by 2 case. You'll find out if this was an N by N matrix, that this would have been K to the N. So the takeaway is the only way you can say that it's going to be a some scalar multiple times your original determinant is only if you multiply one row times that scalar multiple, not the whole matrix. Let's see how this extends to maybe a 3 by 3 case. And you might say, hey, Sal, you just picked the second row, does it work with the first row? I'll leave that for you to determine. But it does."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's see how this extends to maybe a 3 by 3 case. And you might say, hey, Sal, you just picked the second row, does it work with the first row? I'll leave that for you to determine. But it does. It does work. It doesn't matter which row I multiplied it by. Let's take the 3 by 3 case."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But it does. It does work. It doesn't matter which row I multiplied it by. Let's take the 3 by 3 case. Let's say we have some matrix, let's call this A again. I'm redefining A. It's going to be A, B, C, D, E, F, G, H, I."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's take the 3 by 3 case. Let's say we have some matrix, let's call this A again. I'm redefining A. It's going to be A, B, C, D, E, F, G, H, I. And then if you take its determinant, let's just take its determinant. The determinant of A is going to be equal to, well, we could do it a couple of different ways, but I'll just pick some arbitrary row. Because that's the row that we're going to multiply by some scalar."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be A, B, C, D, E, F, G, H, I. And then if you take its determinant, let's just take its determinant. The determinant of A is going to be equal to, well, we could do it a couple of different ways, but I'll just pick some arbitrary row. Because that's the row that we're going to multiply by some scalar. So let's just take that row right there. So the determinant of A is going to be equal to, remember, the plus minus pattern. Remember, plus, minus, plus, minus, plus, minus, plus, minus, plus, that little checkerboard pattern."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because that's the row that we're going to multiply by some scalar. So let's just take that row right there. So the determinant of A is going to be equal to, remember, the plus minus pattern. Remember, plus, minus, plus, minus, plus, minus, plus, minus, plus, that little checkerboard pattern. So D is a minus right there. So it's going to be equal to minus D times the determinant of its submatrix. So you cross out that column and that row."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, plus, minus, plus, minus, plus, minus, plus, minus, plus, that little checkerboard pattern. So D is a minus right there. So it's going to be equal to minus D times the determinant of its submatrix. So you cross out that column and that row. It's B, C, H, I. And it's going to be plus E times its submatrix, A, C, G, I. And it's going to be minus F times, you get rid of that row, that column, D, E, G, H. The determinant of D, E, G, H. That's the determinant of this matrix A."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you cross out that column and that row. It's B, C, H, I. And it's going to be plus E times its submatrix, A, C, G, I. And it's going to be minus F times, you get rid of that row, that column, D, E, G, H. The determinant of D, E, G, H. That's the determinant of this matrix A. Now, what if we define some new matrix here? Let's call it A prime. Let me scroll down a little bit."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's going to be minus F times, you get rid of that row, that column, D, E, G, H. The determinant of D, E, G, H. That's the determinant of this matrix A. Now, what if we define some new matrix here? Let's call it A prime. Let me scroll down a little bit. Let me define A prime right here. A prime, I'm just going to multiply this row by a scalar. So it's going to be equal to A, B, C, K, D, K, E, and K, F. I'm not multiplying the whole matrix times the scalar."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll down a little bit. Let me define A prime right here. A prime, I'm just going to multiply this row by a scalar. So it's going to be equal to A, B, C, K, D, K, E, and K, F. I'm not multiplying the whole matrix times the scalar. I can't say that this is KA. I'm just multiplying one of its rows. And then I have G, H, and I."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be equal to A, B, C, K, D, K, E, and K, F. I'm not multiplying the whole matrix times the scalar. I can't say that this is KA. I'm just multiplying one of its rows. And then I have G, H, and I. So what's the determinant of A prime going to be? The determinant of A prime. And I put that prime there so it's different than A."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then I have G, H, and I. So what's the determinant of A prime going to be? The determinant of A prime. And I put that prime there so it's different than A. Or maybe it's, you know, but it's derived from A. I just multiplied one row of A times the scalar. Well, I can go along that same row that I did up here. I go along that same row."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I put that prime there so it's different than A. Or maybe it's, you know, but it's derived from A. I just multiplied one row of A times the scalar. Well, I can go along that same row that I did up here. I go along that same row. And the only difference is that instead of having a D, I now have a KD. Instead of an E, I now have a KE. So instead of a D, I'm going to have a KD there."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I go along that same row. And the only difference is that instead of having a D, I now have a KD. Instead of an E, I now have a KE. So instead of a D, I'm going to have a KD there. Instead of an E, I'm going to have a KE there. So it's going to be this exact same thing, but I can replace this guy, this guy, and this guy with them multiplied by K. So it's going to be equal to minus KD times the determinant of the submatrix, BCHI. I'm not even going to look over here because it's going to be the same thing as that one up there."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So instead of a D, I'm going to have a KD there. Instead of an E, I'm going to have a KE there. So it's going to be this exact same thing, but I can replace this guy, this guy, and this guy with them multiplied by K. So it's going to be equal to minus KD times the determinant of the submatrix, BCHI. I'm not even going to look over here because it's going to be the same thing as that one up there. Plus KE times the determinant of ACGI plus minus minus KF times the determinant of DEGH. And what is this equal to? This is equal to, if you just factor out the K, it's equal to K times this."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm not even going to look over here because it's going to be the same thing as that one up there. Plus KE times the determinant of ACGI plus minus minus KF times the determinant of DEGH. And what is this equal to? This is equal to, if you just factor out the K, it's equal to K times this. So it's equal to K times the determinant of A. So our result also worked for the 3 by 3 case. I just happened to pick the middle row, but I encourage you to pick other rows and to see what happens."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, if you just factor out the K, it's equal to K times this. So it's equal to K times the determinant of A. So our result also worked for the 3 by 3 case. I just happened to pick the middle row, but I encourage you to pick other rows and to see what happens. And so let's actually do it for the general case. Because I've just been giving you particular examples, and I like to show you the general proof when the general proof isn't too hairy. So let's say I have an n by n matrix."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I just happened to pick the middle row, but I encourage you to pick other rows and to see what happens. And so let's actually do it for the general case. Because I've just been giving you particular examples, and I like to show you the general proof when the general proof isn't too hairy. So let's say I have an n by n matrix. So let's say that I have a matrix A. Let's say that A is n by n. So it equals, you can write it like this. This is the first row, first column, A11, A12, all the way to A1n."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have an n by n matrix. So let's say that I have a matrix A. Let's say that A is n by n. So it equals, you can write it like this. This is the first row, first column, A11, A12, all the way to A1n. I'm going to pick some arbitrary row here that I'm going to end up multiplying by a scalar. So we could go down here. Let's say row AI."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the first row, first column, A11, A12, all the way to A1n. I'm going to pick some arbitrary row here that I'm going to end up multiplying by a scalar. So we could go down here. Let's say row AI. This is AI1, AI2, all the way to AIn. This is some row that I'm going to use to determine the determinant. Remember, we can go down any row to get the determinant."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say row AI. This is AI1, AI2, all the way to AIn. This is some row that I'm going to use to determine the determinant. Remember, we can go down any row to get the determinant. Then finally, you keep going, you get AN1, AN2, all the way to ANn. This is as general as you can get for an n by n matrix. Now let's figure out its determinant."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, we can go down any row to get the determinant. Then finally, you keep going, you get AN1, AN2, all the way to ANn. This is as general as you can get for an n by n matrix. Now let's figure out its determinant. So the determinant of A, and I'm just going to go down this row right there. So the determinant of A is equal to what? It's equal to, well, we have to remember the checkerboard pattern, and we don't know where we are in the checkerboard pattern."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's figure out its determinant. So the determinant of A, and I'm just going to go down this row right there. So the determinant of A is equal to what? It's equal to, well, we have to remember the checkerboard pattern, and we don't know where we are in the checkerboard pattern. Because I just picked an arbitrary general row here. But we can use the general formula that the sign is going to be determined by negative 1 to the i. I don't know if i is even or odd, so it's going to be i. Plus, for this term, 1 power."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to, well, we have to remember the checkerboard pattern, and we don't know where we are in the checkerboard pattern. Because I just picked an arbitrary general row here. But we can use the general formula that the sign is going to be determined by negative 1 to the i. I don't know if i is even or odd, so it's going to be i. Plus, for this term, 1 power. That's its sign. This is what gives us the checkerboard pattern. Let me make that clear."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Plus, for this term, 1 power. That's its sign. This is what gives us the checkerboard pattern. Let me make that clear. It looks complicated, but this is just the checkerboard. That's just that right there. Times this term right there."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me make that clear. It looks complicated, but this is just the checkerboard. That's just that right there. Times this term right there. So times ai, so the coefficient, ai1. And then times this guy's sub-matrix. And you remember the sub-matrix."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times this term right there. So times ai, so the coefficient, ai1. And then times this guy's sub-matrix. And you remember the sub-matrix. You get rid of this row, and this column is going to be everything that's left over. So times that sub-matrix of ai1. And then you're going to keep, and it's going to be plus, let me just keep doing it, plus negative 1 to the i plus 2 times ai2 times its sub-matrix, i2."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you remember the sub-matrix. You get rid of this row, and this column is going to be everything that's left over. So times that sub-matrix of ai1. And then you're going to keep, and it's going to be plus, let me just keep doing it, plus negative 1 to the i plus 2 times ai2 times its sub-matrix, i2. All the way, you just keep going, plus minus 1 to the i plus n times ai. You're in the nth column, and then its sub-matrix. This is going to be an n minus 1 by n minus 1 matrix."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you're going to keep, and it's going to be plus, let me just keep doing it, plus negative 1 to the i plus 2 times ai2 times its sub-matrix, i2. All the way, you just keep going, plus minus 1 to the i plus n times ai. You're in the nth column, and then its sub-matrix. This is going to be an n minus 1 by n minus 1 matrix. All of these are going to be in. Just like that. That's the determinant of a."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be an n minus 1 by n minus 1 matrix. All of these are going to be in. Just like that. That's the determinant of a. And we could actually rewrite it in sigma notation. That'll simplify things a little bit. So the determinant of a, we can rewrite it as the sum from j is equal to 1 to j, I'll write it explicitly here, j is equal to n of negative 1 to the i plus j times aij, and then each of the sub-matrices, aij."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the determinant of a. And we could actually rewrite it in sigma notation. That'll simplify things a little bit. So the determinant of a, we can rewrite it as the sum from j is equal to 1 to j, I'll write it explicitly here, j is equal to n of negative 1 to the i plus j times aij, and then each of the sub-matrices, aij. This thing right here is just another way of writing this thing I wrote up there. I'm just saying the sum. You just take j equal 1, put them in there, you get this term right there."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of a, we can rewrite it as the sum from j is equal to 1 to j, I'll write it explicitly here, j is equal to n of negative 1 to the i plus j times aij, and then each of the sub-matrices, aij. This thing right here is just another way of writing this thing I wrote up there. I'm just saying the sum. You just take j equal 1, put them in there, you get this term right there. You take j equal 2, you add it, you get this term right there. You keep doing it, you get j equal n, you get that term right there. So these two things are equivalent."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You just take j equal 1, put them in there, you get this term right there. You take j equal 2, you add it, you get this term right there. You keep doing it, you get j equal n, you get that term right there. So these two things are equivalent. So what happens if I have some new matrix? Let's say, let me copy and paste my current matrix. So let me copy and paste it."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these two things are equivalent. So what happens if I have some new matrix? Let's say, let me copy and paste my current matrix. So let me copy and paste it. Actually, let me copy and paste everything. That'll make things move quickly. I copied it, and now let me paste it, just like that."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me copy and paste it. Actually, let me copy and paste everything. That'll make things move quickly. I copied it, and now let me paste it, just like that. Let me define a new matrix. Let me define my new matrix, A prime. It's still an n by n matrix, but that row that I just happened to use to determine my determinant, I'm going to multiply it by a scalar k. So it's kAi, kA1, kAi2, kAin, just like that."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I copied it, and now let me paste it, just like that. Let me define a new matrix. Let me define my new matrix, A prime. It's still an n by n matrix, but that row that I just happened to use to determine my determinant, I'm going to multiply it by a scalar k. So it's kAi, kA1, kAi2, kAin, just like that. So what's the determinant of A prime? Well, we're just going to go down this row again. But now instead of just an Ai1, we have a kAi1."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's still an n by n matrix, but that row that I just happened to use to determine my determinant, I'm going to multiply it by a scalar k. So it's kAi, kA1, kAi2, kAin, just like that. So what's the determinant of A prime? Well, we're just going to go down this row again. But now instead of just an Ai1, we have a kAi1. Instead of an Ai2, we have a kAi2. Instead of a kAin, we have a kAin. So its determinant is just going to be this same thing, but instead of an Aij everywhere, we're going to have a kAij."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But now instead of just an Ai1, we have a kAi1. Instead of an Ai2, we have a kAi2. Instead of a kAin, we have a kAin. So its determinant is just going to be this same thing, but instead of an Aij everywhere, we're going to have a kAij. So this is going to be the determinant of A prime. And so this is equal to, we could just take out this constant right here. It has no i's or j's in it."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So its determinant is just going to be this same thing, but instead of an Aij everywhere, we're going to have a kAij. So this is going to be the determinant of A prime. And so this is equal to, we could just take out this constant right here. It has no i's or j's in it. So it has no j's in it in particular, so we can just take it out. So it's equal to k times the sum from j is equal to 1 to j is equal to n of minus 1 to the i plus j times Aij. And this is the coefficient, and then this is the submatrix for each of those coefficients."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It has no i's or j's in it. So it has no j's in it in particular, so we can just take it out. So it's equal to k times the sum from j is equal to 1 to j is equal to n of minus 1 to the i plus j times Aij. And this is the coefficient, and then this is the submatrix for each of those coefficients. Aij, that's a matrix right there, an n minus 1 by n minus 1 matrix. And then you immediately recognize, I think you saw where this was going. This right here is just the determinant of A."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is the coefficient, and then this is the submatrix for each of those coefficients. Aij, that's a matrix right there, an n minus 1 by n minus 1 matrix. And then you immediately recognize, I think you saw where this was going. This right here is just the determinant of A. So we get the result that the determinant of A prime is equal to k times the determinant of A. So we've just shown you in general, you have any n by n matrix, if you multiply only one row, not the whole matrix, only one row by some scalar multiple k, the resulting determinant will be your original determinant times k. Now, I touched on this in the original video. What happens if you multiply, what is the determinant of k times A?"}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This right here is just the determinant of A. So we get the result that the determinant of A prime is equal to k times the determinant of A. So we've just shown you in general, you have any n by n matrix, if you multiply only one row, not the whole matrix, only one row by some scalar multiple k, the resulting determinant will be your original determinant times k. Now, I touched on this in the original video. What happens if you multiply, what is the determinant of k times A? So now we're multiplying every row times k. Or another way to think about it is you're multiplying n rows times k. So you're doing this n times. So if you multiply k times itself n times, what do you get? You get k to the n. So this is going to be equal to k to the n times the determinant of A."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What happens if you multiply, what is the determinant of k times A? So now we're multiplying every row times k. Or another way to think about it is you're multiplying n rows times k. So you're doing this n times. So if you multiply k times itself n times, what do you get? You get k to the n. So this is going to be equal to k to the n times the determinant of A. If you just do it once, you get k times the determinant of A. Now, if you do a second row, you're going to get k times k times the determinant of A. You do a third row, you're going to get k to the third times the determinant of A."}, {"video_title": "Determinant when row multiplied by scalar Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get k to the n. So this is going to be equal to k to the n times the determinant of A. If you just do it once, you get k times the determinant of A. Now, if you do a second row, you're going to get k times k times the determinant of A. You do a third row, you're going to get k to the third times the determinant of A. The fourth row, k to the fourth times the determinant of A. If you do them all, all n rows, you're going to have k to the n times the determinant of A. Anyway, hopefully you found that interesting."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Let's say V is a subspace of Rn. And let's say the set B is a basis for V. So it's got a bunch of vectors in it. Let's say it's got V1, V2, all the way to Vk. So you can see we have k vectors. So V is a k-dimensional subspace. So that means that if I have some vector A that is a member of my subspace, that means that I can represent A as a linear combination of these characters right there. So I could write A as being equal to some constant times my first basis vector plus some other constant times my second basis vector."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So you can see we have k vectors. So V is a k-dimensional subspace. So that means that if I have some vector A that is a member of my subspace, that means that I can represent A as a linear combination of these characters right there. So I could write A as being equal to some constant times my first basis vector plus some other constant times my second basis vector. And then I could keep going all the way to a k-th constant times my k-th basis vector. Now, I've used the term coordinates fairly loosely in the past. And now we're going to have a more precise definition."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So I could write A as being equal to some constant times my first basis vector plus some other constant times my second basis vector. And then I could keep going all the way to a k-th constant times my k-th basis vector. Now, I've used the term coordinates fairly loosely in the past. And now we're going to have a more precise definition. I am going to call these constants here. So I'm going to call C1, C2, all the way through Ck. I'm going to call them the coordinates of A with respect to our basis B."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And now we're going to have a more precise definition. I am going to call these constants here. So I'm going to call C1, C2, all the way through Ck. I'm going to call them the coordinates of A with respect to our basis B. And we could also write it like this. We could also write, I have my vector A. But if I wanted to write my vector A in coordinates with respect to this basis set B, I would write it like this."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to call them the coordinates of A with respect to our basis B. And we could also write it like this. We could also write, I have my vector A. But if I wanted to write my vector A in coordinates with respect to this basis set B, I would write it like this. Put brackets around it and put the basis set right there. And this says I'm going to write this in coordinates with respect to this basis set. So I will then write it like this."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "But if I wanted to write my vector A in coordinates with respect to this basis set B, I would write it like this. Put brackets around it and put the basis set right there. And this says I'm going to write this in coordinates with respect to this basis set. So I will then write it like this. I'm just going to put these weights there, these constant terms on the linear combination that I have to get of my basis vectors to get A. So C1, C2, all the way to Ck. Now, there's one slightly interesting thing, or maybe very interesting thing, to point out here."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So I will then write it like this. I'm just going to put these weights there, these constant terms on the linear combination that I have to get of my basis vectors to get A. So C1, C2, all the way to Ck. Now, there's one slightly interesting thing, or maybe very interesting thing, to point out here. V is a basis of Rn. So anything in V is also going to be in Rn. But V has k vectors, has dimension k. And that k could be as high as n, but it might be something smaller."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Now, there's one slightly interesting thing, or maybe very interesting thing, to point out here. V is a basis of Rn. So anything in V is also going to be in Rn. But V has k vectors, has dimension k. And that k could be as high as n, but it might be something smaller. Maybe we have two vectors in R3, in which case V would be a plane in R3, but we can abstract that to further dimensions. But when you specify something that is in our subspace with respect to its basis, notice you only have to have that many, the dimension of your subspaces, you only have to have that many coordinates. So even though A is a member of Rn, I only had to give it k coordinates, because essentially you're giving it positions within, let's say if this was a plane, within the plane that is your subspace."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "But V has k vectors, has dimension k. And that k could be as high as n, but it might be something smaller. Maybe we have two vectors in R3, in which case V would be a plane in R3, but we can abstract that to further dimensions. But when you specify something that is in our subspace with respect to its basis, notice you only have to have that many, the dimension of your subspaces, you only have to have that many coordinates. So even though A is a member of Rn, I only had to give it k coordinates, because essentially you're giving it positions within, let's say if this was a plane, within the plane that is your subspace. Let me make this a little bit more concrete. Let me do some examples. So let's say we have some subspace."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So even though A is a member of Rn, I only had to give it k coordinates, because essentially you're giving it positions within, let's say if this was a plane, within the plane that is your subspace. Let me make this a little bit more concrete. Let me do some examples. So let's say we have some subspace. Let me clear this out. Let's say I have a couple of vectors. Let's say V1 is the vector 2, 1."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we have some subspace. Let me clear this out. Let's say I have a couple of vectors. Let's say V1 is the vector 2, 1. And let's say V2 is the vector 1, 2. Now you might immediately see that the basis, or the set, of V1 and V2, this is a basis for R2. This is a basis for R2, which means that any vector in R2 can be represented as a linear combination of these guys."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Let's say V1 is the vector 2, 1. And let's say V2 is the vector 1, 2. Now you might immediately see that the basis, or the set, of V1 and V2, this is a basis for R2. This is a basis for R2, which means that any vector in R2 can be represented as a linear combination of these guys. I could do a visual argument. Or we also know that, look, R2 is two dimensional, and we have two basis vectors right here, and they are linearly independent. You can verify that."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "This is a basis for R2, which means that any vector in R2 can be represented as a linear combination of these guys. I could do a visual argument. Or we also know that, look, R2 is two dimensional, and we have two basis vectors right here, and they are linearly independent. You can verify that. In fact, the easiest way to verify that is if you just take 2, 1, and 1, 2, and you put it in reduced row echelon form, you're going to get the 2 by 2 identity matrix. You're going to get 1, 0, 0, 1. And that lets you know that this guy and this guy are both basis vectors."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "You can verify that. In fact, the easiest way to verify that is if you just take 2, 1, and 1, 2, and you put it in reduced row echelon form, you're going to get the 2 by 2 identity matrix. You're going to get 1, 0, 0, 1. And that lets you know that this guy and this guy are both basis vectors. So that's all review. We've seen that before. But let's visualize these guys."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And that lets you know that this guy and this guy are both basis vectors. So that's all review. We've seen that before. But let's visualize these guys. So if I were to just graph it the way we normally graph these vectors, what does 2, 1 look like? Let me draw some axes here. Let me draw it."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "But let's visualize these guys. So if I were to just graph it the way we normally graph these vectors, what does 2, 1 look like? Let me draw some axes here. Let me draw it. Let me do it in a different color. Let's say that is my vertical axis, and this is my horizontal axis. And 2, 1 might look like this."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw it. Let me do it in a different color. Let's say that is my vertical axis, and this is my horizontal axis. And 2, 1 might look like this. So we're going to go out 1, 2, and then we're going to go up 1. So that is our vector 1 right there. That is 2, 1."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And 2, 1 might look like this. So we're going to go out 1, 2, and then we're going to go up 1. So that is our vector 1 right there. That is 2, 1. That's our vector 1. And then 1, 2 might look like, or it does look like this if I draw it in standard position. 1, then we go up 2."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "That is 2, 1. That's our vector 1. And then 1, 2 might look like, or it does look like this if I draw it in standard position. 1, then we go up 2. 1, 2 looks like this. So when we talk about coordinates with respect to this basis, let me pick some member of R2. So let's say, I'll engineer it so that I can easily find the linear combination."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "1, then we go up 2. 1, 2 looks like this. So when we talk about coordinates with respect to this basis, let me pick some member of R2. So let's say, I'll engineer it so that I can easily find the linear combination. Let me take 3 times v1 plus 2 times v2. What is that going to be equal to? That's going to be equal to the vector."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So let's say, I'll engineer it so that I can easily find the linear combination. Let me take 3 times v1 plus 2 times v2. What is that going to be equal to? That's going to be equal to the vector. So 3 times 2, which is 6, plus 2 times 1. So it's the vector 8. And then I have the vector 3 times v1 plus 2 times that."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be equal to the vector. So 3 times 2, which is 6, plus 2 times 1. So it's the vector 8. And then I have the vector 3 times v1 plus 2 times that. So 8, 7, right? 3 plus 2 times 2 is 4. 8, 7."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And then I have the vector 3 times v1 plus 2 times that. So 8, 7, right? 3 plus 2 times 2 is 4. 8, 7. So if we were to just graph 8, 7 in the traditional way, we would go 1, 2, 3, 4, 5, 6, 7, 8. And then we would go up 1, 2, 3, 4, 5, 6, 7. And we would have a vector."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "8, 7. So if we were to just graph 8, 7 in the traditional way, we would go 1, 2, 3, 4, 5, 6, 7, 8. And then we would go up 1, 2, 3, 4, 5, 6, 7. And we would have a vector. I'm not going to draw it out here, but it would specify that point right there. That would be this point. If we view these as coordinates, we would view that as the point 8, 7 right there."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And we would have a vector. I'm not going to draw it out here, but it would specify that point right there. That would be this point. If we view these as coordinates, we would view that as the point 8, 7 right there. Maybe I'll write it like that. That's the point 8, 7. If I wanted to draw this vector in standard position, I would draw a vector that ends right there."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "If we view these as coordinates, we would view that as the point 8, 7 right there. Maybe I'll write it like that. That's the point 8, 7. If I wanted to draw this vector in standard position, I would draw a vector that ends right there. Now, we have this basis here, this basis B, represented by these two vectors. This is v1 and v2. What we want to do is represent this guy."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "If I wanted to draw this vector in standard position, I would draw a vector that ends right there. Now, we have this basis here, this basis B, represented by these two vectors. This is v1 and v2. What we want to do is represent this guy. Let's say that I have this vector. Let me call this vector A. So vector A is equal to 8, 7."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "What we want to do is represent this guy. Let's say that I have this vector. Let me call this vector A. So vector A is equal to 8, 7. Now, we know that if we wanted to represent vector A as a linear combination of my basis vectors, it's going to be 3 times v1 plus 2 times v2. So just given what we just saw in the earlier part of this video, we can write that the vector A, with respect to the basis B, maybe I'll do it in the same color as the basis, is equal to these weights on the basis vectors. Is equal to 3 and 2."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So vector A is equal to 8, 7. Now, we know that if we wanted to represent vector A as a linear combination of my basis vectors, it's going to be 3 times v1 plus 2 times v2. So just given what we just saw in the earlier part of this video, we can write that the vector A, with respect to the basis B, maybe I'll do it in the same color as the basis, is equal to these weights on the basis vectors. Is equal to 3 and 2. Let's see if we can visually understand why this makes sense. We're saying that in some new coordinate system, this vector can be represented as 3, 2. And the way you think about the new coordinate system is, in this old coordinate system, we hashed out 1's in the horizontal axis, and that was our first coordinate."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Is equal to 3 and 2. Let's see if we can visually understand why this makes sense. We're saying that in some new coordinate system, this vector can be represented as 3, 2. And the way you think about the new coordinate system is, in this old coordinate system, we hashed out 1's in the horizontal axis, and that was our first coordinate. And we hashed out 1's in the vertical direction. That was our second coordinate. Now, in our new system, what's our first coordinate?"}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And the way you think about the new coordinate system is, in this old coordinate system, we hashed out 1's in the horizontal axis, and that was our first coordinate. And we hashed out 1's in the vertical direction. That was our second coordinate. Now, in our new system, what's our first coordinate? Our first coordinate is going to be multiples of v1. This is v1, or this is v1. So it's multiples of v1."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Now, in our new system, what's our first coordinate? Our first coordinate is going to be multiples of v1. This is v1, or this is v1. So it's multiples of v1. So that's 1 times v1. Then if we do 2 times v1, we're going to get over here. 2 times v1 would get us to 4, 2."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So it's multiples of v1. So that's 1 times v1. Then if we do 2 times v1, we're going to get over here. 2 times v1 would get us to 4, 2. 3 times v1 would get us to 6, 3. So let's see, 1, 2, 2, 3, 4, 5, 6, 7, 8. So 6 and then 3, just like that."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "2 times v1 would get us to 4, 2. 3 times v1 would get us to 6, 3. So let's see, 1, 2, 2, 3, 4, 5, 6, 7, 8. So 6 and then 3, just like that. And then 4 times v1 would get us to 8 and 4. So you can imagine that what I'm drawing here, this is kind of the axes, the first term axis generated by v1. So I could draw it, let me do it in this blue color, so you could imagine it like this."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So 6 and then 3, just like that. And then 4 times v1 would get us to 8 and 4. So you can imagine that what I'm drawing here, this is kind of the axes, the first term axis generated by v1. So I could draw it, let me do it in this blue color, so you could imagine it like this. This would be a straight line, just like that. And then the coordinate tells me how many v1's do I have. So I would hash off the coordinate system like this."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So I could draw it, let me do it in this blue color, so you could imagine it like this. This would be a straight line, just like that. And then the coordinate tells me how many v1's do I have. So I would hash off the coordinate system like this. Instead of doing increments of 1, I'm going to do increments of v1, just like that. As you go 9, 10, we're going to go up 1 more to 5, something like that. Now, the second coordinate tells you increments of v2."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So I would hash off the coordinate system like this. Instead of doing increments of 1, I'm going to do increments of v1, just like that. As you go 9, 10, we're going to go up 1 more to 5, something like that. Now, the second coordinate tells you increments of v2. So this is our first increment of v2, then our second increment, if we go to 4, it's going to be 4, 2, just like that, that's going to be 6 and 3, it's going to be just like that, then it's going to be 6 and 3, so it's going to look something like this. So if you want to think of it as a bit of a, well, you should be thinking of it as a coordinate system, you can have this new skewed graph paper, where any point, you can now specify it as going in the v1 direction by some amount, and then by going in the v2 direction by some amount, let me draw that as a bit of a graph paper. So I could draw another version of v2, just like that, just all the multiples of v2, I could shift them like that, I could do another one like this, I could do another, that one is a little bit not neat enough, I could do it like that, I could do, I think you're getting the idea, like let me make that a little bit neater, this might have been useful to do it with another tool, and then I could do all of the multiples of v1 like this, I'm doing a graph paper right here, so it would look something like this, it would look something like this, it would look something like this, and so you can imagine the skewed graph paper if I did it all over the place with this kind of green and this blue, so in our new coordinate system we're seeing 3, 2, so that means 3 times our first direction, which happens to be the v1 direction, it's no longer the horizontal direction, it's the v1 direction, so we go, we're going 1, 2, and then we go 3, like that, and then we're going to go 2 in the v2 direction, so we're going to go 1, 2 in the v2 direction, and so our point is going to be right there, right, you could imagine going like this, you go 3 in the v1 direction, and then you go 1, 2 in the v2 direction, you get to our point, or you could go kind of in your v2 direction and then your v1 direction, but either way you're going to get to your original point, so that vector, or the position specified by the vector 8, 7 could just as easily be specified in our new coordinate system by the coordinates 3, 2, because we're saying 3 times v1, and then plus 3 times v1, it takes us in this direction, we're going 3 notches in the v1 direction, and then we're going 2 notches in the v2 direction, and so that's why these are called coordinates."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Now, the second coordinate tells you increments of v2. So this is our first increment of v2, then our second increment, if we go to 4, it's going to be 4, 2, just like that, that's going to be 6 and 3, it's going to be just like that, then it's going to be 6 and 3, so it's going to look something like this. So if you want to think of it as a bit of a, well, you should be thinking of it as a coordinate system, you can have this new skewed graph paper, where any point, you can now specify it as going in the v1 direction by some amount, and then by going in the v2 direction by some amount, let me draw that as a bit of a graph paper. So I could draw another version of v2, just like that, just all the multiples of v2, I could shift them like that, I could do another one like this, I could do another, that one is a little bit not neat enough, I could do it like that, I could do, I think you're getting the idea, like let me make that a little bit neater, this might have been useful to do it with another tool, and then I could do all of the multiples of v1 like this, I'm doing a graph paper right here, so it would look something like this, it would look something like this, it would look something like this, and so you can imagine the skewed graph paper if I did it all over the place with this kind of green and this blue, so in our new coordinate system we're seeing 3, 2, so that means 3 times our first direction, which happens to be the v1 direction, it's no longer the horizontal direction, it's the v1 direction, so we go, we're going 1, 2, and then we go 3, like that, and then we're going to go 2 in the v2 direction, so we're going to go 1, 2 in the v2 direction, and so our point is going to be right there, right, you could imagine going like this, you go 3 in the v1 direction, and then you go 1, 2 in the v2 direction, you get to our point, or you could go kind of in your v2 direction and then your v1 direction, but either way you're going to get to your original point, so that vector, or the position specified by the vector 8, 7 could just as easily be specified in our new coordinate system by the coordinates 3, 2, because we're saying 3 times v1, and then plus 3 times v1, it takes us in this direction, we're going 3 notches in the v1 direction, and then we're going 2 notches in the v2 direction, and so that's why these are called coordinates. You're literally saying how many spaces in the v1 direction to go, and then how many spaces in the v2 direction to go. But this might, I guess, lead you to the obvious question, why have we been using the coordinates before? Like I might have been saying all along."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So I could draw another version of v2, just like that, just all the multiples of v2, I could shift them like that, I could do another one like this, I could do another, that one is a little bit not neat enough, I could do it like that, I could do, I think you're getting the idea, like let me make that a little bit neater, this might have been useful to do it with another tool, and then I could do all of the multiples of v1 like this, I'm doing a graph paper right here, so it would look something like this, it would look something like this, it would look something like this, and so you can imagine the skewed graph paper if I did it all over the place with this kind of green and this blue, so in our new coordinate system we're seeing 3, 2, so that means 3 times our first direction, which happens to be the v1 direction, it's no longer the horizontal direction, it's the v1 direction, so we go, we're going 1, 2, and then we go 3, like that, and then we're going to go 2 in the v2 direction, so we're going to go 1, 2 in the v2 direction, and so our point is going to be right there, right, you could imagine going like this, you go 3 in the v1 direction, and then you go 1, 2 in the v2 direction, you get to our point, or you could go kind of in your v2 direction and then your v1 direction, but either way you're going to get to your original point, so that vector, or the position specified by the vector 8, 7 could just as easily be specified in our new coordinate system by the coordinates 3, 2, because we're saying 3 times v1, and then plus 3 times v1, it takes us in this direction, we're going 3 notches in the v1 direction, and then we're going 2 notches in the v2 direction, and so that's why these are called coordinates. You're literally saying how many spaces in the v1 direction to go, and then how many spaces in the v2 direction to go. But this might, I guess, lead you to the obvious question, why have we been using the coordinates before? Like I might have been saying all along. Let's say I have some vector, I don't know, let's say I have some vector lowercase b that is equal to, I don't know, let's say it's equal to, I'll do it in R2 just because it's easy to visualize, let's say it's equal to 3 minus 1. And if we were to graph it, it would look something like this. We would go 1, 2, 3, and then we would go down 1."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Like I might have been saying all along. Let's say I have some vector, I don't know, let's say I have some vector lowercase b that is equal to, I don't know, let's say it's equal to, I'll do it in R2 just because it's easy to visualize, let's say it's equal to 3 minus 1. And if we were to graph it, it would look something like this. We would go 1, 2, 3, and then we would go down 1. So it would look something like this. It would specify this point. But why have we been calling 3 and negative 1 coordinates?"}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "We would go 1, 2, 3, and then we would go down 1. So it would look something like this. It would specify this point. But why have we been calling 3 and negative 1 coordinates? Why have we been calling 3 and minus 1 coordinates? We've been doing it well before we learned linear algebra, we called these coordinates all the way from when we first learned how to graph. Why are we calling those coordinates?"}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "But why have we been calling 3 and negative 1 coordinates? Why have we been calling 3 and minus 1 coordinates? We've been doing it well before we learned linear algebra, we called these coordinates all the way from when we first learned how to graph. Why are we calling those coordinates? Or how does this meaning of coordinates relate to these coordinates with respect to a basis? Well, these are coordinates with respect to basis. These are actually coordinates with respect to the standard basis."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Why are we calling those coordinates? Or how does this meaning of coordinates relate to these coordinates with respect to a basis? Well, these are coordinates with respect to basis. These are actually coordinates with respect to the standard basis. If you imagine, let's see, the standard basis in R2 looks like this. We could have E1, which is 1, 0. And we have E2, which is 0, 1."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "These are actually coordinates with respect to the standard basis. If you imagine, let's see, the standard basis in R2 looks like this. We could have E1, which is 1, 0. And we have E2, which is 0, 1. This is just the convention for the standard basis in R2. And so we could say, if we called, let's say, S is equal to the set of E1 and E2, then we say that S is the standard basis for R2. And it's the standard basis because these two guys are orthogonal."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And we have E2, which is 0, 1. This is just the convention for the standard basis in R2. And so we could say, if we called, let's say, S is equal to the set of E1 and E2, then we say that S is the standard basis for R2. And it's the standard basis because these two guys are orthogonal. This is 1 in the horizontal direction, this is 1 in the vertical direction. And any vector in R2, let's say I have some vector x, y in R2, it's going to be equal to x times E1 plus y times E2. So we could say that if you want to write some vector x, y, if you wanted to write it with respect to this standard basis right here, it's going to be equal to the coordinates by the definition that we did earlier in this video of the basis vectors right there, these weights on our E1's and E2's."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And it's the standard basis because these two guys are orthogonal. This is 1 in the horizontal direction, this is 1 in the vertical direction. And any vector in R2, let's say I have some vector x, y in R2, it's going to be equal to x times E1 plus y times E2. So we could say that if you want to write some vector x, y, if you wanted to write it with respect to this standard basis right here, it's going to be equal to the coordinates by the definition that we did earlier in this video of the basis vectors right there, these weights on our E1's and E2's. So it's going to be equal to, well, the weight there is x and the weight here is y. So these coordinates that we've been talking about from the get-go, these are definitely coordinates. They're consistent with our definition of coordinates in this video."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that if you want to write some vector x, y, if you wanted to write it with respect to this standard basis right here, it's going to be equal to the coordinates by the definition that we did earlier in this video of the basis vectors right there, these weights on our E1's and E2's. So it's going to be equal to, well, the weight there is x and the weight here is y. So these coordinates that we've been talking about from the get-go, these are definitely coordinates. They're consistent with our definition of coordinates in this video. But we can maybe be a little bit more precise. We can now call them the coordinates with respect to the standard basis. Or we could call them, we could call these right here, we could call these the standard coordinates."}, {"video_title": "Coordinates with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "They're consistent with our definition of coordinates in this video. But we can maybe be a little bit more precise. We can now call them the coordinates with respect to the standard basis. Or we could call them, we could call these right here, we could call these the standard coordinates. Standard coordinates. I just wanted to point this out. This might be almost trivially simple or a bit obvious."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I have two two-dimensional vectors right over here, vector A and vector B. And what I want to think about is how can we define, or what would be a reasonable way to define, the sum of vector A plus vector B? Well, one thing that might jump at your mind is, look, well, each of these are two-dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just the sum of the first two components of these two vectors? So why don't we just make it 6 plus negative 4?"}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just the sum of the first two components of these two vectors? So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two two-dimensional vectors. You add them together, you get another two two-dimensional vectors. If you think about it in terms of real coordinate spaces, both of these are members of R2."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So negative 2 plus 4 is also equal to 2. So we start with two two-dimensional vectors. You add them together, you get another two two-dimensional vectors. If you think about it in terms of real coordinate spaces, both of these are members of R2. I'll write this down here just so we get used to the notation. So vector A and vector B are both members of R2, which is just another way of saying that these are both two-tuples. They're both two-dimensional vectors right over here."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you think about it in terms of real coordinate spaces, both of these are members of R2. I'll write this down here just so we get used to the notation. So vector A and vector B are both members of R2, which is just another way of saying that these are both two-tuples. They're both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it. But how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it. But how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector A, we could visualize."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector A, we could visualize. This tells us how far this vector moves in each of these directions, horizontal direction and vertical direction. So if we put the, I guess you could say, the tail of the vector at the origin. Remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector A, we could visualize. This tells us how far this vector moves in each of these directions, horizontal direction and vertical direction. So if we put the, I guess you could say, the tail of the vector at the origin. Remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction, 1, 2, 3, 4, 5, 6, and then negative 2 in the vertical, so negative 2. So vector A could look like this. And once again, the important thing is the magnitude and the direction."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction, 1, 2, 3, 4, 5, 6, and then negative 2 in the vertical, so negative 2. So vector A could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector, and the direction is the direction that it's pointed in. And also, just to emphasize, I could have drawn vector A like that, or I could have put it over here. These are all equivalent vectors."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector, and the direction is the direction that it's pointed in. And also, just to emphasize, I could have drawn vector A like that, or I could have put it over here. These are all equivalent vectors. These are all equal to vector A. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector B. Vector B in the horizontal direction goes negative 4, 1, 2, 3, 4."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These are all equivalent vectors. These are all equal to vector A. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector B. Vector B in the horizontal direction goes negative 4, 1, 2, 3, 4. And in the vertical direction goes 4, 1, 2, 3, 4. So its tail, if we start at the origin, its head would be at negative 4, 4. So let me draw that just like that."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So with that in mind, let's also draw vector B. Vector B in the horizontal direction goes negative 4, 1, 2, 3, 4. And in the vertical direction goes 4, 1, 2, 3, 4. So its tail, if we start at the origin, its head would be at negative 4, 4. So let me draw that just like that. So that right over here is vector B. And once again, vector B, we could draw it like that, or we could draw it. So let me copy, let me paste it."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw that just like that. So that right over here is vector B. And once again, vector B, we could draw it like that, or we could draw it. So let me copy, let me paste it. So this would also be another way to draw vector B. This would also be another way to draw vector B. Once again, what I really care about is its magnitude and its direction."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me copy, let me paste it. So this would also be another way to draw vector B. This would also be another way to draw vector B. Once again, what I really care about is its magnitude and its direction. All of these green vectors have the same magnitude. They all have the same length, and they all have the same direction. So how does the way that I drew vector A and B gel with what its sum is?"}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Once again, what I really care about is its magnitude and its direction. All of these green vectors have the same magnitude. They all have the same length, and they all have the same direction. So how does the way that I drew vector A and B gel with what its sum is? So let me draw its sum like this. Let me draw its sum in this blue color. So the sum, based on this definition we just used, a vector addition, would be 2, 2."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So how does the way that I drew vector A and B gel with what its sum is? So let me draw its sum like this. Let me draw its sum in this blue color. So the sum, based on this definition we just used, a vector addition, would be 2, 2. So 2, 2. So it would look something like this. So how does this make sense, that the sum, that this purple vector plus this green vector is somehow going to be equal to this blue vector?"}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the sum, based on this definition we just used, a vector addition, would be 2, 2. So 2, 2. So it would look something like this. So how does this make sense, that the sum, that this purple vector plus this green vector is somehow going to be equal to this blue vector? I encourage you to pause the video and think about if that even makes sense. Well, one way to think about it is this first purple vector, it shifts us this much. It takes us from this point to that point."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So how does this make sense, that the sum, that this purple vector plus this green vector is somehow going to be equal to this blue vector? I encourage you to pause the video and think about if that even makes sense. Well, one way to think about it is this first purple vector, it shifts us this much. It takes us from this point to that point. And so if we were to add it, let's start at this point and put the green vector's tail right there and see where it ends up putting us. So the green vector, we already have a version. So once again, we start at the origin."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It takes us from this point to that point. And so if we were to add it, let's start at this point and put the green vector's tail right there and see where it ends up putting us. So the green vector, we already have a version. So once again, we start at the origin. The vector A takes us there. Now let's start over there with the green vector and see where green vector takes us. And this makes sense."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So once again, we start at the origin. The vector A takes us there. Now let's start over there with the green vector and see where green vector takes us. And this makes sense. Vector A plus vector B. Put the tail of vector B at the head of vector A. So if you were to start at the origin, vector A takes you there."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this makes sense. Vector A plus vector B. Put the tail of vector B at the head of vector A. So if you were to start at the origin, vector A takes you there. Then if you add on what vector B takes you, it takes you right over there. So relative to the origin, how much did you, I guess, you could say shift? And once again, vectors don't only apply to things like displacement."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you were to start at the origin, vector A takes you there. Then if you add on what vector B takes you, it takes you right over there. So relative to the origin, how much did you, I guess, you could say shift? And once again, vectors don't only apply to things like displacement. It can apply to velocity. It can apply to actual acceleration. It can apply to a whole series of things."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And once again, vectors don't only apply to things like displacement. It can apply to velocity. It can apply to actual acceleration. It can apply to a whole series of things. But when you visualize it this way, you see that it does make complete sense. This blue vector, the sum of the two, is what results where you start with vector A at that point right over there. Vector A takes you there."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It can apply to a whole series of things. But when you visualize it this way, you see that it does make complete sense. This blue vector, the sum of the two, is what results where you start with vector A at that point right over there. Vector A takes you there. Then you take vector B's tail, start over there, and it takes you to the tip of the sum. Now one question you might be having is, well, vector A plus vector B is this. But what is vector B plus vector A?"}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Vector A takes you there. Then you take vector B's tail, start over there, and it takes you to the tip of the sum. Now one question you might be having is, well, vector A plus vector B is this. But what is vector B plus vector A? Does this still work? Well, based on the definition we had, where you add the corresponding components, you're still going to get the same sum vector. So it should come out the same."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But what is vector B plus vector A? Does this still work? Well, based on the definition we had, where you add the corresponding components, you're still going to get the same sum vector. So it should come out the same. So this will just be negative 4 plus 6 is 2. 4 plus negative 2 is 2. But does that make visual sense?"}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it should come out the same. So this will just be negative 4 plus 6 is 2. 4 plus negative 2 is 2. But does that make visual sense? So if we start with vector B. So let's say you start right over here. Vector B takes you right over there."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But does that make visual sense? So if we start with vector B. So let's say you start right over here. Vector B takes you right over there. And then if you were to go there and you were to start with vector A. So let's do that. Actually, let me make this a little bit."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Vector B takes you right over there. And then if you were to go there and you were to start with vector A. So let's do that. Actually, let me make this a little bit. Actually, let me start with a new vector B. So let's say that that's our vector B right over there. And then if, actually, let me give us a place where I'll have some space to work with."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me make this a little bit. Actually, let me start with a new vector B. So let's say that that's our vector B right over there. And then if, actually, let me give us a place where I'll have some space to work with. So let's say that's my vector B right over there. And then let me get a copy of the vector A. That's a good one."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then if, actually, let me give us a place where I'll have some space to work with. So let's say that's my vector B right over there. And then let me get a copy of the vector A. That's a good one. So copy. And let me paste it. So I could put vector A's tail at the tip of vector B."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's a good one. So copy. And let me paste it. So I could put vector A's tail at the tip of vector B. And then it'll take me right over there. So if I start here, if I start right over here, vector B takes me there. And now I'm adding to that vector A, which starting here will take me there."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could put vector A's tail at the tip of vector B. And then it'll take me right over there. So if I start here, if I start right over here, vector B takes me there. And now I'm adding to that vector A, which starting here will take me there. And so from my original starting position, I have gone this far. Now what is this vector? Well, this is exactly the vector 2, 2, or another way of thinking about it."}, {"video_title": "Adding vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now I'm adding to that vector A, which starting here will take me there. And so from my original starting position, I have gone this far. Now what is this vector? Well, this is exactly the vector 2, 2, or another way of thinking about it. This vector shifts you 2 in the horizontal direction and 2 in the vertical direction. So either way, you're going to get the same result. And that should hopefully make visual or conceptual sense as well."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this was a neat idea because we're used to the length of things in two or three dimensional space, but it becomes very abstract when we get to n dimensions. If this has 100 components, at least for me, it's hard to visualize a 100 dimension vector. But we've actually defined its notion of length. And we saw that this is actually a scalar value. It's just a number. In this video, I want to attempt to define the notion of an angle between vectors. As you can see, we're building up this mathematics of vectors from the ground up."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we saw that this is actually a scalar value. It's just a number. In this video, I want to attempt to define the notion of an angle between vectors. As you can see, we're building up this mathematics of vectors from the ground up. And we can't just say, oh, I know what an angle is, because everything we know about angles and even lengths, it just applies to what we associate with two or three dimensional space. But the whole study of linear algebra is abstracting these ideas into multi-dimensional space. And I haven't even defined what dimension is yet, but I think you understand that idea to some degree already when people talk about one or two or three dimensions."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "As you can see, we're building up this mathematics of vectors from the ground up. And we can't just say, oh, I know what an angle is, because everything we know about angles and even lengths, it just applies to what we associate with two or three dimensional space. But the whole study of linear algebra is abstracting these ideas into multi-dimensional space. And I haven't even defined what dimension is yet, but I think you understand that idea to some degree already when people talk about one or two or three dimensions. So let's say I have two vectors. Vectors a and b. Vectors a and b. They're non-zero."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I haven't even defined what dimension is yet, but I think you understand that idea to some degree already when people talk about one or two or three dimensions. So let's say I have two vectors. Vectors a and b. Vectors a and b. They're non-zero. And they're members of Rn. And they're non-zero. And I don't have a notion of the angle between them yet."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're non-zero. And they're members of Rn. And they're non-zero. And I don't have a notion of the angle between them yet. But let me just draw them out. Let me just draw them as if I could draw them in two dimensions. So that would be vector a right there."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I don't have a notion of the angle between them yet. But let me just draw them out. Let me just draw them as if I could draw them in two dimensions. So that would be vector a right there. Maybe that's vector b right there. And then this vector right there would be the vector a minus b. And you can verify that just the way we've learned to add and subtract vectors, or this is heads to tails."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that would be vector a right there. Maybe that's vector b right there. And then this vector right there would be the vector a minus b. And you can verify that just the way we've learned to add and subtract vectors, or this is heads to tails. So b plus a minus b is, of course, going to be vector a. And that all just works out there. To help us define this notion of angle, let me construct another triangle that's going to look a lot like this one."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you can verify that just the way we've learned to add and subtract vectors, or this is heads to tails. So b plus a minus b is, of course, going to be vector a. And that all just works out there. To help us define this notion of angle, let me construct another triangle that's going to look a lot like this one. But remember, I'm just doing this for our simple minds to imagine it in two dimensions. But these aren't necessarily two-dimensional beasts. These each could have 100 components."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "To help us define this notion of angle, let me construct another triangle that's going to look a lot like this one. But remember, I'm just doing this for our simple minds to imagine it in two dimensions. But these aren't necessarily two-dimensional beasts. These each could have 100 components. But let me make another triangle that looks something similar. And I'm going to define the sides of the triangles to be the lengths of each of these vectors. Remember, the lengths of each of these vectors, I don't care how many components they are."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These each could have 100 components. But let me make another triangle that looks something similar. And I'm going to define the sides of the triangles to be the lengths of each of these vectors. Remember, the lengths of each of these vectors, I don't care how many components they are. They're just going to be your numbers. So the length of this side right here is just going to be the length of a. The length of this side right here is just going to be the length of vector a minus vector b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, the lengths of each of these vectors, I don't care how many components they are. They're just going to be your numbers. So the length of this side right here is just going to be the length of a. The length of this side right here is just going to be the length of vector a minus vector b. And the length of this side right here is going to be the length of vector b. Now, the first thing we want to make sure is that we can always construct a triangle like that. And so under what circumstances could we not construct a triangle like this?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The length of this side right here is just going to be the length of vector a minus vector b. And the length of this side right here is going to be the length of vector b. Now, the first thing we want to make sure is that we can always construct a triangle like that. And so under what circumstances could we not construct a triangle like this? Well, we wouldn't be able to construct a triangle like this if this side, if b, if the magnitude, so let me write this down. It's kind of a subtle point, but I want to make this very clear. In order to define an angle, I want to be comfortable that I can always make this construction."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so under what circumstances could we not construct a triangle like this? Well, we wouldn't be able to construct a triangle like this if this side, if b, if the magnitude, so let me write this down. It's kind of a subtle point, but I want to make this very clear. In order to define an angle, I want to be comfortable that I can always make this construction. And I need to make sure that, let me write reasons why I couldn't make this construction. Well, what if the magnitude of b was greater than, or the length of vector b was greater than the length of vector a plus the length of vector a minus b? In two dimensions, I could never draw a triangle like that then."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In order to define an angle, I want to be comfortable that I can always make this construction. And I need to make sure that, let me write reasons why I couldn't make this construction. Well, what if the magnitude of b was greater than, or the length of vector b was greater than the length of vector a plus the length of vector a minus b? In two dimensions, I could never draw a triangle like that then. Because you would have this length plus this length would be shorter than this thing right here, and so you could never construct it. And I could do it with all the sides. What if this length was larger than one of these two sides?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In two dimensions, I could never draw a triangle like that then. Because you would have this length plus this length would be shorter than this thing right here, and so you could never construct it. And I could do it with all the sides. What if this length was larger than one of these two sides? Or what if that length was larger than one of those two sides? I could just never draw a two-dimensional triangle that way. So what I'm going to do is I'm going to use the triangle, the vector triangle inequality to prove that each of these sides is less than or equal to the sum of the other sides."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What if this length was larger than one of these two sides? Or what if that length was larger than one of those two sides? I could just never draw a two-dimensional triangle that way. So what I'm going to do is I'm going to use the triangle, the vector triangle inequality to prove that each of these sides is less than or equal to the sum of the other sides. And I could do the same thing. Let me make the point clear. I could show that if a, for whatever reason, was greater than the other sides plus b, then I wouldn't be able to create a triangle."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what I'm going to do is I'm going to use the triangle, the vector triangle inequality to prove that each of these sides is less than or equal to the sum of the other sides. And I could do the same thing. Let me make the point clear. I could show that if a, for whatever reason, was greater than the other sides plus b, then I wouldn't be able to create a triangle. And the last one, of course, is if a minus b, for whatever reason, was greater than the other two sides, I just wouldn't be able to draw a triangle in a plus b. So I need to show that for any vectors that are any real vectors, non-zero real vectors, or members Rn, that none of these can ever happen. I need to prove that none of those can happen."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could show that if a, for whatever reason, was greater than the other sides plus b, then I wouldn't be able to create a triangle. And the last one, of course, is if a minus b, for whatever reason, was greater than the other two sides, I just wouldn't be able to draw a triangle in a plus b. So I need to show that for any vectors that are any real vectors, non-zero real vectors, or members Rn, that none of these can ever happen. I need to prove that none of those can happen. So what does the triangle inequality tell us? The triangle inequality tells us that if I have the sum of two vectors, if I take the length of the sum of two vectors, that that is always going to be less than, and these are non-zero vectors, this is always going to be less than or equal to the sum of each of their individual lengths. So let's see if we can apply that to this triangle right here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I need to prove that none of those can happen. So what does the triangle inequality tell us? The triangle inequality tells us that if I have the sum of two vectors, if I take the length of the sum of two vectors, that that is always going to be less than, and these are non-zero vectors, this is always going to be less than or equal to the sum of each of their individual lengths. So let's see if we can apply that to this triangle right here. So what is the magnitude, the length of a? Well, I can rewrite vector a. What is vector a equal to?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can apply that to this triangle right here. So what is the magnitude, the length of a? Well, I can rewrite vector a. What is vector a equal to? Vector a is equal to vector b plus vector a minus b. This is just, I mean, I'm just rewriting the vector here. I'm just rewriting a here as a sum of the other two vectors."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is vector a equal to? Vector a is equal to vector b plus vector a minus b. This is just, I mean, I'm just rewriting the vector here. I'm just rewriting a here as a sum of the other two vectors. Nothing fancy there. I haven't used the triangle inequality or anything. I've just used my definition of vector addition."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just rewriting a here as a sum of the other two vectors. Nothing fancy there. I haven't used the triangle inequality or anything. I've just used my definition of vector addition. But here now, if I put little parentheses here, now I can apply the triangle inequality. I say, well, you know what? This is going to be, by the triangle inequality which we proved, it's going to be less than or equal to the lengths of each of these vectors."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I've just used my definition of vector addition. But here now, if I put little parentheses here, now I can apply the triangle inequality. I say, well, you know what? This is going to be, by the triangle inequality which we proved, it's going to be less than or equal to the lengths of each of these vectors. Vector b plus the length of vector a minus b. So we know that the length of a is less than the sum of that one and that one. So we don't have to worry about this being our problem."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be, by the triangle inequality which we proved, it's going to be less than or equal to the lengths of each of these vectors. Vector b plus the length of vector a minus b. So we know that the length of a is less than the sum of that one and that one. So we don't have to worry about this being our problem. We know that that is not true. Now let's look at b. So is there any way that I can rewrite b as a sum of two other vectors?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we don't have to worry about this being our problem. We know that that is not true. Now let's look at b. So is there any way that I can rewrite b as a sum of two other vectors? Well, sure. I can write it as a sum of a plus, let me put it this way, if that vector right there is a minus b, the same vector in the reverse direction is going to be the vector b minus a. So a plus the vector b minus a, that's the same thing as b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So is there any way that I can rewrite b as a sum of two other vectors? Well, sure. I can write it as a sum of a plus, let me put it this way, if that vector right there is a minus b, the same vector in the reverse direction is going to be the vector b minus a. So a plus the vector b minus a, that's the same thing as b. And you can see it right here. The a's would cancel out and you're just left with a b there. Now by the triangle inequality, we know that this is less than or equal to the length of vector a plus the length of vector b minus a."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So a plus the vector b minus a, that's the same thing as b. And you can see it right here. The a's would cancel out and you're just left with a b there. Now by the triangle inequality, we know that this is less than or equal to the length of vector a plus the length of vector b minus a. Now you're saying, hey Sal, you're dealing with b minus a. This is the length of a minus b. And I could leave this for you to prove it based on our definition of vector lengths."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now by the triangle inequality, we know that this is less than or equal to the length of vector a plus the length of vector b minus a. Now you're saying, hey Sal, you're dealing with b minus a. This is the length of a minus b. And I could leave this for you to prove it based on our definition of vector lengths. But the length of b minus a is equal to minus 1 times a minus b. And I'll leave it to you to say that look, these lengths are equal because essentially, well, I could leave that, but I think you can take that based on just the visual depiction of them, that they're the exact same vectors just in different directions. And I have to be careful with length because it's not just in two dimensions, but I think you get the idea."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I could leave this for you to prove it based on our definition of vector lengths. But the length of b minus a is equal to minus 1 times a minus b. And I'll leave it to you to say that look, these lengths are equal because essentially, well, I could leave that, but I think you can take that based on just the visual depiction of them, that they're the exact same vectors just in different directions. And I have to be careful with length because it's not just in two dimensions, but I think you get the idea. And I think I'll leave that for you to prove that these lengths are the same thing. So we know that b is less than the length of those two things, so we don't have to worry about that one right there. Finally, a minus b, can we write that, the magnitude or the length of vector a minus b?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I have to be careful with length because it's not just in two dimensions, but I think you get the idea. And I think I'll leave that for you to prove that these lengths are the same thing. So we know that b is less than the length of those two things, so we don't have to worry about that one right there. Finally, a minus b, can we write that, the magnitude or the length of vector a minus b? Well, I can write that as vector a plus vector minus b. You could say if we just put a minus b right there and go in the other directions, we could say minus b, which would be in that direction, plus a would give us our vector a minus b. And that's obvious."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Finally, a minus b, can we write that, the magnitude or the length of vector a minus b? Well, I can write that as vector a plus vector minus b. You could say if we just put a minus b right there and go in the other directions, we could say minus b, which would be in that direction, plus a would give us our vector a minus b. And that's obvious. Actually, I don't even have to go there. That's obvious from this. I just kind of put the negative in the parentheses."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's obvious. Actually, I don't even have to go there. That's obvious from this. I just kind of put the negative in the parentheses. Well, the triangle inequality, and this might seem a little mundane to you, but it really shows us that we can always define a regular planar triangle based on these vectors in this way. Tells us this is less than or equal to the length of our vector a plus the length of minus b. And I just said, and you could prove it to yourself, that this is the same thing as the length of b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just kind of put the negative in the parentheses. Well, the triangle inequality, and this might seem a little mundane to you, but it really shows us that we can always define a regular planar triangle based on these vectors in this way. Tells us this is less than or equal to the length of our vector a plus the length of minus b. And I just said, and you could prove it to yourself, that this is the same thing as the length of b. So we just saw that this is definitely less than those two. This is definitely less than those two, and that is definitely less than those two. So we don't have, none of the reasons that would keep us from constructing a triangle are valid."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I just said, and you could prove it to yourself, that this is the same thing as the length of b. So we just saw that this is definitely less than those two. This is definitely less than those two, and that is definitely less than those two. So we don't have, none of the reasons that would keep us from constructing a triangle are valid. So we can always construct a triangle in this way from any arbitrary non-zero vectors in Rn. We can always construct this. Now, to define an angle, let me redraw it down here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we don't have, none of the reasons that would keep us from constructing a triangle are valid. So we can always construct a triangle in this way from any arbitrary non-zero vectors in Rn. We can always construct this. Now, to define an angle, let me redraw it down here. Let me redraw the vectors, maybe a little bit bigger. That's vector a. I'll draw it slightly. This is vector b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, to define an angle, let me redraw it down here. Let me redraw the vectors, maybe a little bit bigger. That's vector a. I'll draw it slightly. This is vector b. And then, let me just draw it this way. This is the vector right there. That's the vector a minus b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is vector b. And then, let me just draw it this way. This is the vector right there. That's the vector a minus b. And we said we're going to define a corresponding regular run-of-the-mill vanilla triangle whose lengths are defined by the lengths of the vectors, by the vector length. So this is the length of b, that side. This is the length of a minus b."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's the vector a minus b. And we said we're going to define a corresponding regular run-of-the-mill vanilla triangle whose lengths are defined by the lengths of the vectors, by the vector length. So this is the length of b, that side. This is the length of a minus b. And then, this is the length of a. Now, now that I know that I can always construct a triangle like this, I can attempt to define, or actually, I will define my definition of an angle between two vectors. So we know what an angle means in this context."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is the length of a minus b. And then, this is the length of a. Now, now that I know that I can always construct a triangle like this, I can attempt to define, or actually, I will define my definition of an angle between two vectors. So we know what an angle means in this context. This is just a regular run-of-the-mill geometric triangle. Now, my definition of an angle between two vectors, I'm going to say, so this is what I'm trying to define. This is what I'm going to define."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we know what an angle means in this context. This is just a regular run-of-the-mill geometric triangle. Now, my definition of an angle between two vectors, I'm going to say, so this is what I'm trying to define. This is what I'm going to define. Because these can have arbitrary number of components, so it's hard to visualize. But I'm going to define this angle as the corresponding angle in a regular run-of-the-mill triangle where the sides of the run-of-the-mill triangle are the two vectors, and then the opposite side is the subtraction, is the length of the difference between the two vectors. This is just the definition."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is what I'm going to define. Because these can have arbitrary number of components, so it's hard to visualize. But I'm going to define this angle as the corresponding angle in a regular run-of-the-mill triangle where the sides of the run-of-the-mill triangle are the two vectors, and then the opposite side is the subtraction, is the length of the difference between the two vectors. This is just the definition. I'm just saying that these two things, I'm defining this, the angle between two vectors in Rn that could have an arbitrary number of components, I'm defining this angle to be the same as this angle, the angle between the two sides, the two lengths of those vectors, in just a regular run-of-the-mill triangle. Now, what can I do with this? Well, can we find a relationship between all of these things right here?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is just the definition. I'm just saying that these two things, I'm defining this, the angle between two vectors in Rn that could have an arbitrary number of components, I'm defining this angle to be the same as this angle, the angle between the two sides, the two lengths of those vectors, in just a regular run-of-the-mill triangle. Now, what can I do with this? Well, can we find a relationship between all of these things right here? Well, sure. If you remember from your trigonometry class, and if you don't, I've proved it in their playlist, you have the law of cosines. And I'll do it with an arbitrary triangle right here just because I don't want to confuse you."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, can we find a relationship between all of these things right here? Well, sure. If you remember from your trigonometry class, and if you don't, I've proved it in their playlist, you have the law of cosines. And I'll do it with an arbitrary triangle right here just because I don't want to confuse you. So if this is side a, b, and c, and this is theta, the law of cosines tells us that c squared is equal to a squared plus b squared minus 2ab cosine of theta. I always think of it as kind of a broader Pythagorean theorem because this thing does not have to be a right angle, it accounts for all angles. If this becomes a right angle, then this term disappears and you're just left with the Pythagorean theorem."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I'll do it with an arbitrary triangle right here just because I don't want to confuse you. So if this is side a, b, and c, and this is theta, the law of cosines tells us that c squared is equal to a squared plus b squared minus 2ab cosine of theta. I always think of it as kind of a broader Pythagorean theorem because this thing does not have to be a right angle, it accounts for all angles. If this becomes a right angle, then this term disappears and you're just left with the Pythagorean theorem. But we've proven this. This applies to just regular run-of-the-mill triangles. And lucky for us, we have a regular run-of-the-mill triangle here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If this becomes a right angle, then this term disappears and you're just left with the Pythagorean theorem. But we've proven this. This applies to just regular run-of-the-mill triangles. And lucky for us, we have a regular run-of-the-mill triangle here. So let's apply the law of cosines to this triangle right here. So we would get, and the way I drew it, they correspond, the length of this side squared, so that means the length of a minus b squared, the length of vector a minus vector b, that's just the length of that side. So I'm just squaring that side."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And lucky for us, we have a regular run-of-the-mill triangle here. So let's apply the law of cosines to this triangle right here. So we would get, and the way I drew it, they correspond, the length of this side squared, so that means the length of a minus b squared, the length of vector a minus vector b, that's just the length of that side. So I'm just squaring that side. It equals the length of vector b squared plus the length of vector a squared minus 2 times the length of, I'll just write 2 times length of vector a times the length of vector b times the cosine of this angle right here, times the cosine of that angle. And I'm defining this angle between these two vectors to be the same as this angle right there. So if we know this angle, by definition, we know that angle right there."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just squaring that side. It equals the length of vector b squared plus the length of vector a squared minus 2 times the length of, I'll just write 2 times length of vector a times the length of vector b times the cosine of this angle right here, times the cosine of that angle. And I'm defining this angle between these two vectors to be the same as this angle right there. So if we know this angle, by definition, we know that angle right there. Well, we know that the square of our lengths of a vector, when we use our vector definition of length, that this is just the same thing as a vector dotted with itself. So that's a minus b dot a minus b. And it's all going to be equal to this whole stuff on the right-hand side, but let me simplify the left-hand side of this equation."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if we know this angle, by definition, we know that angle right there. Well, we know that the square of our lengths of a vector, when we use our vector definition of length, that this is just the same thing as a vector dotted with itself. So that's a minus b dot a minus b. And it's all going to be equal to this whole stuff on the right-hand side, but let me simplify the left-hand side of this equation. a minus b dot a minus b, this is the same thing as a dot a, those two terms, minus a dot b. And then I have minus b dot a. That's those two terms right there."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And it's all going to be equal to this whole stuff on the right-hand side, but let me simplify the left-hand side of this equation. a minus b dot a minus b, this is the same thing as a dot a, those two terms, minus a dot b. And then I have minus b dot a. That's those two terms right there. And then you have the minus b dot minus b. That's the same thing as a plus b dot b. Remember, this is just a simplification of the left-hand side."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's those two terms right there. And then you have the minus b dot minus b. That's the same thing as a plus b dot b. Remember, this is just a simplification of the left-hand side. And I can rewrite this. a dot a, we know that's just the length of a squared. a dot b and b dot a are the same thing."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, this is just a simplification of the left-hand side. And I can rewrite this. a dot a, we know that's just the length of a squared. a dot b and b dot a are the same thing. So we have two of these. So this right here, this term right there, we'll simplify to minus 2 times a dot b. And then finally, b dot b, we know that that's just the length of b squared."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "a dot b and b dot a are the same thing. So we have two of these. So this right here, this term right there, we'll simplify to minus 2 times a dot b. And then finally, b dot b, we know that that's just the length of b squared. Now I just simplified, or maybe I just expanded, that's a better word, when you go from one term here to three terms, you can't say you simplified it. But I expanded just the left-hand side. And so this has to be equal to the right-hand side by the law of cosines."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, b dot b, we know that that's just the length of b squared. Now I just simplified, or maybe I just expanded, that's a better word, when you go from one term here to three terms, you can't say you simplified it. But I expanded just the left-hand side. And so this has to be equal to the right-hand side by the law of cosines. So that is equal to, I almost feel like instead of rewriting it, let me just copy and paste it. What did I just do? Copy, and then edit, copy, and paste."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so this has to be equal to the right-hand side by the law of cosines. So that is equal to, I almost feel like instead of rewriting it, let me just copy and paste it. What did I just do? Copy, and then edit, copy, and paste. There you go. I don't know if that was worth it, but maybe I saved a little bit of time. So that is equal to that right there."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Copy, and then edit, copy, and paste. There you go. I don't know if that was worth it, but maybe I saved a little bit of time. So that is equal to that right there. And then we can simplify. We have a length of a squared here, length of a squared there, subtract it from both sides. Length of b squared here, length of b squared there, subtract it from both sides."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that is equal to that right there. And then we can simplify. We have a length of a squared here, length of a squared there, subtract it from both sides. Length of b squared here, length of b squared there, subtract it from both sides. And then, what can we do? We can divide both sides by minus 2, because everything else has disappeared. And so that term and that term will both become 1's."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Length of b squared here, length of b squared there, subtract it from both sides. And then, what can we do? We can divide both sides by minus 2, because everything else has disappeared. And so that term and that term will both become 1's. And all we're left with is the vector a dot the vector b. And this is interesting, because all of a sudden we're getting a relationship between the dot products of two vectors. We've kind of gone away from their definition by lengths, but that the dot product of two vectors is equal to the product of their lengths, their vector lengths, and they can have an arbitrary number of components, times the cosine of the angle between them."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so that term and that term will both become 1's. And all we're left with is the vector a dot the vector b. And this is interesting, because all of a sudden we're getting a relationship between the dot products of two vectors. We've kind of gone away from their definition by lengths, but that the dot product of two vectors is equal to the product of their lengths, their vector lengths, and they can have an arbitrary number of components, times the cosine of the angle between them. Remember, this theta, I said, this is the same as when you draw this kind of analogous regular triangle, but I'm defining the angle between them to be the same as that. So I can say that this is the angle between them. And obviously, the idea of between two vectors, it's hard to visualize if you go beyond three dimensions, but now we have it at least mathematically defined, angle between them."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We've kind of gone away from their definition by lengths, but that the dot product of two vectors is equal to the product of their lengths, their vector lengths, and they can have an arbitrary number of components, times the cosine of the angle between them. Remember, this theta, I said, this is the same as when you draw this kind of analogous regular triangle, but I'm defining the angle between them to be the same as that. So I can say that this is the angle between them. And obviously, the idea of between two vectors, it's hard to visualize if you go beyond three dimensions, but now we have it at least mathematically defined, angle between them. So if you give me two vectors, we can now, using this formula that we proved using this definition up here, we can now calculate the angle between any two vectors using this right here. And just to make it clear, what happens if a is a, and maybe it's not clear from that definition, so I'll make it clear here, that by definition, if a is equal to some scalar multiple of b, where c is greater than 0, we'll define theta to be equal to 0. And if c is less than 0, so a is collinear but goes in the exact opposite direction, we'll define theta to be equal to 180 degrees."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And obviously, the idea of between two vectors, it's hard to visualize if you go beyond three dimensions, but now we have it at least mathematically defined, angle between them. So if you give me two vectors, we can now, using this formula that we proved using this definition up here, we can now calculate the angle between any two vectors using this right here. And just to make it clear, what happens if a is a, and maybe it's not clear from that definition, so I'll make it clear here, that by definition, if a is equal to some scalar multiple of b, where c is greater than 0, we'll define theta to be equal to 0. And if c is less than 0, so a is collinear but goes in the exact opposite direction, we'll define theta to be equal to 180 degrees. And that's consistent with what we understand about just two-dimensional vectors. If they're collinear and the scalar multiples are the same, that means a looks something like that, and b looks something like that. So we say, that's a 0 angle."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if c is less than 0, so a is collinear but goes in the exact opposite direction, we'll define theta to be equal to 180 degrees. And that's consistent with what we understand about just two-dimensional vectors. If they're collinear and the scalar multiples are the same, that means a looks something like that, and b looks something like that. So we say, that's a 0 angle. And if they go the other way, if a looks something like, this is a case where a is just going in the other direction from b, a goes like that and b goes like that, we define the angle between them to be 180 degrees. But everything else is pretty well defined by the triangle example. I had to make the special case of these because it's not clear you really get a triangle in these cases, because then the triangle kind of disappears."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we say, that's a 0 angle. And if they go the other way, if a looks something like, this is a case where a is just going in the other direction from b, a goes like that and b goes like that, we define the angle between them to be 180 degrees. But everything else is pretty well defined by the triangle example. I had to make the special case of these because it's not clear you really get a triangle in these cases, because then the triangle kind of disappears. It flattens out if a and b are on top of each other, or if they're going in the exact opposite direction. So that's why I wanted to make a little bit of a side note right there. Now, using this definition of the angle between the vectors, we can now define the idea of perpendicular vectors."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I had to make the special case of these because it's not clear you really get a triangle in these cases, because then the triangle kind of disappears. It flattens out if a and b are on top of each other, or if they're going in the exact opposite direction. So that's why I wanted to make a little bit of a side note right there. Now, using this definition of the angle between the vectors, we can now define the idea of perpendicular vectors. So we can now say perpendicular vectors, this is another definition, and this won't be earth shattering, but it kind of is because we've generalized this to vectors that have an arbitrary number of components. We're defining perpendicular to mean the theta between, so perpendicular, two vectors a and b are perpendicular if the angle between them is 90 degrees. And we can define that."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, using this definition of the angle between the vectors, we can now define the idea of perpendicular vectors. So we can now say perpendicular vectors, this is another definition, and this won't be earth shattering, but it kind of is because we've generalized this to vectors that have an arbitrary number of components. We're defining perpendicular to mean the theta between, so perpendicular, two vectors a and b are perpendicular if the angle between them is 90 degrees. And we can define that. We can take two vectors, dot them, take their dot product, figure out their two lengths, and then you could figure out their angle between them. And if it's 90 degrees, you can say that they are perpendicular angles. And I want to be very clear here that this is actually not defined for the zero vector right here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we can define that. We can take two vectors, dot them, take their dot product, figure out their two lengths, and then you could figure out their angle between them. And if it's 90 degrees, you can say that they are perpendicular angles. And I want to be very clear here that this is actually not defined for the zero vector right here. So this situation right here, not defined for the zero vector, because if you have the zero vector, then this quantity right here is going to be zero, and then this quantity right here is going to be zero, and there's no clear definition for your angle. If this is zero right here, you'd get zero is equal to 0 times cosine of theta. And so if you wanted to solve for theta, you'd get cosine of theta is equal to 0 over 0, which is undefined."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want to be very clear here that this is actually not defined for the zero vector right here. So this situation right here, not defined for the zero vector, because if you have the zero vector, then this quantity right here is going to be zero, and then this quantity right here is going to be zero, and there's no clear definition for your angle. If this is zero right here, you'd get zero is equal to 0 times cosine of theta. And so if you wanted to solve for theta, you'd get cosine of theta is equal to 0 over 0, which is undefined. But what we can do is create a slightly more general word than the word perpendicular. So you have to have a defined angle to even talk about perpendicular. If the angle between two vectors is 90 degrees, we're saying by definition those two vectors are perpendicular."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so if you wanted to solve for theta, you'd get cosine of theta is equal to 0 over 0, which is undefined. But what we can do is create a slightly more general word than the word perpendicular. So you have to have a defined angle to even talk about perpendicular. If the angle between two vectors is 90 degrees, we're saying by definition those two vectors are perpendicular. But what if we made the statement, and if you look at them, if the angle between two vectors is 90 degrees, what does that mean? So let's say that theta is 90 degrees. Let me draw a line here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If the angle between two vectors is 90 degrees, we're saying by definition those two vectors are perpendicular. But what if we made the statement, and if you look at them, if the angle between two vectors is 90 degrees, what does that mean? So let's say that theta is 90 degrees. Let me draw a line here. Let's say that theta is 90 degrees. Theta is equal to 90 degrees. What does this formula tell us?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw a line here. Let's say that theta is 90 degrees. Theta is equal to 90 degrees. What does this formula tell us? It tells us that a dot b is equal to the length of a times the length of b times cosine of 90 degrees. Well, what's cosine of 90 degrees? It's 0."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What does this formula tell us? It tells us that a dot b is equal to the length of a times the length of b times cosine of 90 degrees. Well, what's cosine of 90 degrees? It's 0. You can review your unit circle if that doesn't make a lot of sense, but that is equal to 0. So this whole term is going to be equal to 0. So if theta is equal to 90 degrees, then a dot b is equal to 0, and so this is another interesting takeaway."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's 0. You can review your unit circle if that doesn't make a lot of sense, but that is equal to 0. So this whole term is going to be equal to 0. So if theta is equal to 90 degrees, then a dot b is equal to 0, and so this is another interesting takeaway. If a and b are perpendicular, then their dot product is going to be equal to 0. Now, if their dot product is equal to 0, can we necessarily say that they are perpendicular? Well, what if a or b is the 0 vector?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if theta is equal to 90 degrees, then a dot b is equal to 0, and so this is another interesting takeaway. If a and b are perpendicular, then their dot product is going to be equal to 0. Now, if their dot product is equal to 0, can we necessarily say that they are perpendicular? Well, what if a or b is the 0 vector? Let me call z for 0 vector. I mean, or I could just draw the 0 vector dot anything is always going to be equal to 0. So does that mean that the 0 vector is perpendicular to everything?"}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, what if a or b is the 0 vector? Let me call z for 0 vector. I mean, or I could just draw the 0 vector dot anything is always going to be equal to 0. So does that mean that the 0 vector is perpendicular to everything? Well, no. Because the 0 vector, I said, we have to have the notion of an angle between things in order to use the word perpendicular. So we can't use the 0 vector."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So does that mean that the 0 vector is perpendicular to everything? Well, no. Because the 0 vector, I said, we have to have the notion of an angle between things in order to use the word perpendicular. So we can't use the 0 vector. We can't say just because two vectors dot products are equal to 0 that they are perpendicular, and that's because the 0 vector would mess that up, because the 0 vector is not defined. But if we say, and we have been saying, that a and b are non-zero, if they are non-zero vectors, then we can say that if a and b are non-zero and their dot product is equal to 0, then a and b are perpendicular. So now it goes both ways."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we can't use the 0 vector. We can't say just because two vectors dot products are equal to 0 that they are perpendicular, and that's because the 0 vector would mess that up, because the 0 vector is not defined. But if we say, and we have been saying, that a and b are non-zero, if they are non-zero vectors, then we can say that if a and b are non-zero and their dot product is equal to 0, then a and b are perpendicular. So now it goes both ways. But what if we just have this condition right here? What if we just have the condition that a dot b is equal to 0? It seems like that's kind of just a simple, pure condition."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now it goes both ways. But what if we just have this condition right here? What if we just have the condition that a dot b is equal to 0? It seems like that's kind of just a simple, pure condition. And we can write a word for that. And these words are often used synonymously, but hopefully you understand the distinction now. We can say that if two vectors dot product is equal to 0, we will call them orthogonal."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It seems like that's kind of just a simple, pure condition. And we can write a word for that. And these words are often used synonymously, but hopefully you understand the distinction now. We can say that if two vectors dot product is equal to 0, we will call them orthogonal. Orthogonal. As I always say, spelling isn't my best subject. But this is kind of a neat idea."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can say that if two vectors dot product is equal to 0, we will call them orthogonal. Orthogonal. As I always say, spelling isn't my best subject. But this is kind of a neat idea. This tells us that all perpendicular vectors are orthogonal. And it also tells us that the 0 vector is orthogonal to everything else, to everything, even to itself. The 0 dot 0 vector, you still get 0."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But this is kind of a neat idea. This tells us that all perpendicular vectors are orthogonal. And it also tells us that the 0 vector is orthogonal to everything else, to everything, even to itself. The 0 dot 0 vector, you still get 0. So by definition, it's orthogonal. So for the first time, probably in your mathematical career, you're seeing that the words, every time you first got exposed to the words perpendicular and orthogonal in geometry or maybe in physics or wherever else, they were always kind of the same words. But now I'm introducing a nice little distinction here."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The 0 dot 0 vector, you still get 0. So by definition, it's orthogonal. So for the first time, probably in your mathematical career, you're seeing that the words, every time you first got exposed to the words perpendicular and orthogonal in geometry or maybe in physics or wherever else, they were always kind of the same words. But now I'm introducing a nice little distinction here. And you can kind of be a little smart aleck with teachers. Oh, it's perpendicular only if neither of them are a 0 vector. Otherwise, if they're dot product is 0, you can only say that they're orthogonal."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But now I'm introducing a nice little distinction here. And you can kind of be a little smart aleck with teachers. Oh, it's perpendicular only if neither of them are a 0 vector. Otherwise, if they're dot product is 0, you can only say that they're orthogonal. But if they're non-zero, you can say that they're orthogonal and perpendicular. But anyway, I thought that I would introduce this little distinction for you in case you have someone who likes to trip you up with words. But it also, I think, highlights that we are building a mathematics from the ground up."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Otherwise, if they're dot product is 0, you can only say that they're orthogonal. But if they're non-zero, you can say that they're orthogonal and perpendicular. But anyway, I thought that I would introduce this little distinction for you in case you have someone who likes to trip you up with words. But it also, I think, highlights that we are building a mathematics from the ground up. And we have to be careful about the words we use. And we have to be very precise about our definitions. Because if we're not precise about our definitions and we build up a bunch of mathematics on top of this and do a bunch of proofs, one day we might scratch our heads and reach some type of weird ambiguity."}, {"video_title": "Defining the angle between vectors Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But it also, I think, highlights that we are building a mathematics from the ground up. And we have to be careful about the words we use. And we have to be very precise about our definitions. Because if we're not precise about our definitions and we build up a bunch of mathematics on top of this and do a bunch of proofs, one day we might scratch our heads and reach some type of weird ambiguity. And it might have all came out of the fact that we weren't precise enough in defining what some of these terms mean. Well, anyway, hopefully you found this useful. We can now take the angle, or we can now determine the angle between vectors with an arbitrary number of components."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're talking about the distance between this plane, between this plane, and some plane that contains these two lines. So in order to talk realistically about distances between planes, those planes will have to be parallel, because if they're not parallel, if they intersect with each other, the distance is clearly 0. And they're telling us here that the distance is the square root of 6. So we have a situation so that the planes can't intersect, they must be parallel. So you have this plane up here, you have this plane up here, we could call this, the equation here is AX minus 2Y plus Z is equal to D. And then you're going to have another plane that's going to be parallel to it, maybe it looks something like this. You have the other plane that is parallel, and it's going to contain both of these lines. So maybe it has this line, so this line is in, I'll say it in green, maybe this line looks something like this, it's on that blue plane, and then this line, maybe in magenta, is also going to be on the blue plane."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we have a situation so that the planes can't intersect, they must be parallel. So you have this plane up here, you have this plane up here, we could call this, the equation here is AX minus 2Y plus Z is equal to D. And then you're going to have another plane that's going to be parallel to it, maybe it looks something like this. You have the other plane that is parallel, and it's going to contain both of these lines. So maybe it has this line, so this line is in, I'll say it in green, maybe this line looks something like this, it's on that blue plane, and then this line, maybe in magenta, is also going to be on the blue plane. So how can we figure out the distances? Well, a good starting point would be to try to figure out the equation for this blue plane here. And since these planes are parallel, this equation should look very much like this orange equation, at least on the left-hand side, it might just have a different D value, and that's because it has the exact same inclination."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So maybe it has this line, so this line is in, I'll say it in green, maybe this line looks something like this, it's on that blue plane, and then this line, maybe in magenta, is also going to be on the blue plane. So how can we figure out the distances? Well, a good starting point would be to try to figure out the equation for this blue plane here. And since these planes are parallel, this equation should look very much like this orange equation, at least on the left-hand side, it might just have a different D value, and that's because it has the exact same inclination. And then once we figure out the equation for this plane over here, then we could actually probably figure out what A is, and then we could find some point on the blue plane, and then use our knowledge of finding the distance between points and planes to figure out the actual distance from any point to this orange plane. So let's figure out the equation of this blue plane first. And a good place to start is just to try to figure out two vectors on this blue plane, then we could take the cross product of those two vectors to find out a normal to the blue plane, and then use that information to actually figure out the equation for the blue plane."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And since these planes are parallel, this equation should look very much like this orange equation, at least on the left-hand side, it might just have a different D value, and that's because it has the exact same inclination. And then once we figure out the equation for this plane over here, then we could actually probably figure out what A is, and then we could find some point on the blue plane, and then use our knowledge of finding the distance between points and planes to figure out the actual distance from any point to this orange plane. So let's figure out the equation of this blue plane first. And a good place to start is just to try to figure out two vectors on this blue plane, then we could take the cross product of those two vectors to find out a normal to the blue plane, and then use that information to actually figure out the equation for the blue plane. So let's figure out first some points that sit on the blue plane. So on this green line right over here, you have, see if I want all of these to be equal to 0, you'd have the point x is equal to 1, y is equal to 2, z is equal to 3. So you'd have the point 1, 1, 2, 3."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And a good place to start is just to try to figure out two vectors on this blue plane, then we could take the cross product of those two vectors to find out a normal to the blue plane, and then use that information to actually figure out the equation for the blue plane. So let's figure out first some points that sit on the blue plane. So on this green line right over here, you have, see if I want all of these to be equal to 0, you'd have the point x is equal to 1, y is equal to 2, z is equal to 3. So you'd have the point 1, 1, 2, 3. That definitely sits on the blue plane. Let's come up with another point. Let's see, if I want all of these to be equal to 1, I could make this, if I want this to evaluate to 2, I would have the point 3."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you'd have the point 1, 1, 2, 3. That definitely sits on the blue plane. Let's come up with another point. Let's see, if I want all of these to be equal to 1, I could make this, if I want this to evaluate to 2, I would have the point 3. I would have the point, if I want this to evaluate to 1, I would want 5 minus 2 over 3, so 3, 5, and then I would want this to be 7. 7 minus 3 over 4 is also 1. So that's another point."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, if I want all of these to be equal to 1, I could make this, if I want this to evaluate to 2, I would have the point 3. I would have the point, if I want this to evaluate to 1, I would want 5 minus 2 over 3, so 3, 5, and then I would want this to be 7. 7 minus 3 over 4 is also 1. So that's another point. Actually, both of these points sit on this line right over here, 1, 2, 3, and then 3, 5, and 7. And then let's do the same thing for this. Let's find two points on this plane."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's another point. Actually, both of these points sit on this line right over here, 1, 2, 3, and then 3, 5, and 7. And then let's do the same thing for this. Let's find two points on this plane. Actually, we just have to find one point on this plane, because if you have three points, that's enough to figure out two different vectors, vectors that aren't scalar multiples of each other, which would be enough to figure out the normal to this plane. So let's just figure out one more point over here, and one more point would be, if we want all of these three to be equal to 0, it would be the point 2, 3, 4. 2, 2, 3, 4, because this would be 0, 0, 0."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's find two points on this plane. Actually, we just have to find one point on this plane, because if you have three points, that's enough to figure out two different vectors, vectors that aren't scalar multiples of each other, which would be enough to figure out the normal to this plane. So let's just figure out one more point over here, and one more point would be, if we want all of these three to be equal to 0, it would be the point 2, 3, 4. 2, 2, 3, 4, because this would be 0, 0, 0. So that's also sitting on the plane, and that sits on the magenta line right over there. So let's use these three points to figure out two vectors on the plane that aren't multiples of each other, then we can take their cross product to actually figure out a normal vector to the blue plane. So let's say the first vector, a, that sits on this plane, let's say it's the difference in the position vectors that specify these two points, and we know that'll be on the plane."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2, 2, 3, 4, because this would be 0, 0, 0. So that's also sitting on the plane, and that sits on the magenta line right over there. So let's use these three points to figure out two vectors on the plane that aren't multiples of each other, then we can take their cross product to actually figure out a normal vector to the blue plane. So let's say the first vector, a, that sits on this plane, let's say it's the difference in the position vectors that specify these two points, and we know that'll be on the plane. And so that will be 3 minus 1 is 2i, plus 5 minus 2 is 3j, plus 7 minus 3 is 4k. So vector a is actually going to sit on this green line, because both of these points are on this line, so it's going to sit on that line. If we put it on the plane, or if we were to start it at one of those points, it'll sit on that line."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say the first vector, a, that sits on this plane, let's say it's the difference in the position vectors that specify these two points, and we know that'll be on the plane. And so that will be 3 minus 1 is 2i, plus 5 minus 2 is 3j, plus 7 minus 3 is 4k. So vector a is actually going to sit on this green line, because both of these points are on this line, so it's going to sit on that line. If we put it on the plane, or if we were to start it at one of those points, it'll sit on that line. And then we could do another vector, and it's essentially going between a point on the green line and a point on the purple line, but that's definitely going to be a vector on our blue plane. And let's go between these two points. That looks pretty straightforward, so let's call vector b, let me do another color so we don't confuse ourselves with that purple, let's call vector b, let's call that, let's see, 2 minus 1 is i, 3 minus 2 is j, and then 4 minus 3, that's just 1k."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we put it on the plane, or if we were to start it at one of those points, it'll sit on that line. And then we could do another vector, and it's essentially going between a point on the green line and a point on the purple line, but that's definitely going to be a vector on our blue plane. And let's go between these two points. That looks pretty straightforward, so let's call vector b, let me do another color so we don't confuse ourselves with that purple, let's call vector b, let's call that, let's see, 2 minus 1 is i, 3 minus 2 is j, and then 4 minus 3, that's just 1k. So this also, this vector right here is also sitting on the plane. So if I take the cross product of a and b, I am going to get a vector that is perpendicular to the plane, or a normal vector to the plane. So let's do that."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That looks pretty straightforward, so let's call vector b, let me do another color so we don't confuse ourselves with that purple, let's call vector b, let's call that, let's see, 2 minus 1 is i, 3 minus 2 is j, and then 4 minus 3, that's just 1k. So this also, this vector right here is also sitting on the plane. So if I take the cross product of a and b, I am going to get a vector that is perpendicular to the plane, or a normal vector to the plane. So let's do that. So let's find what a cross b is. a cross b is equal to, and this is how I find it easiest, I just write i, j, k, this is really the definition of the cross product, or I guess one of them, and we write our first vector. We have 2, 3, 4, and then we have our second vector, which is just 1, 1, 1, and then this is going to be equal to, first we'll look at the i component, so cross that row, that column out, 3 times 1 minus 1 times 4, so that's just 3 minus 4, so it's negative i, and then minus, we're going to have the j, so let me write a minus here, minus, we just swap signs, we have positive, negative, positive, so j, get rid of that column, that row, 2 times 1, which is 2 minus 1 times 4, so that's minus 4, is negative 2, so we could write a negative 2 here, but the negatives cancel out, so it becomes plus 2j, and then finally for the k, get rid of that row, that column, 2 times 1 is 2 minus 1 times 3, is 2 minus 3, which is negative 1, so it's negative k. So this right here is a normal vector."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So let's find what a cross b is. a cross b is equal to, and this is how I find it easiest, I just write i, j, k, this is really the definition of the cross product, or I guess one of them, and we write our first vector. We have 2, 3, 4, and then we have our second vector, which is just 1, 1, 1, and then this is going to be equal to, first we'll look at the i component, so cross that row, that column out, 3 times 1 minus 1 times 4, so that's just 3 minus 4, so it's negative i, and then minus, we're going to have the j, so let me write a minus here, minus, we just swap signs, we have positive, negative, positive, so j, get rid of that column, that row, 2 times 1, which is 2 minus 1 times 4, so that's minus 4, is negative 2, so we could write a negative 2 here, but the negatives cancel out, so it becomes plus 2j, and then finally for the k, get rid of that row, that column, 2 times 1 is 2 minus 1 times 3, is 2 minus 3, which is negative 1, so it's negative k. So this right here is a normal vector. This right here is a normal vector to the plane. So if we want to find the equation for that plane, we've done it multiple times, we just have to take the dot product of that normal vector, and any arbitrary vector on that that we can specify with an arbitrary x, y, and z, and we've done this multiple times in multiple videos, if this is any point x, y, z that sits on the plane, then the vector, let me draw the vector, let's say we go to this point right over here, so this vector right over here is going to be, let me draw it the other way actually, so this vector right over here, let's say we're going between this point and x, y, z, this vector right here is going to be x minus 3i plus y minus 5j plus z minus 7k. That's what this vector is, it sits on the plane, assuming x, y, and z sit on the plane."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have 2, 3, 4, and then we have our second vector, which is just 1, 1, 1, and then this is going to be equal to, first we'll look at the i component, so cross that row, that column out, 3 times 1 minus 1 times 4, so that's just 3 minus 4, so it's negative i, and then minus, we're going to have the j, so let me write a minus here, minus, we just swap signs, we have positive, negative, positive, so j, get rid of that column, that row, 2 times 1, which is 2 minus 1 times 4, so that's minus 4, is negative 2, so we could write a negative 2 here, but the negatives cancel out, so it becomes plus 2j, and then finally for the k, get rid of that row, that column, 2 times 1 is 2 minus 1 times 3, is 2 minus 3, which is negative 1, so it's negative k. So this right here is a normal vector. This right here is a normal vector to the plane. So if we want to find the equation for that plane, we've done it multiple times, we just have to take the dot product of that normal vector, and any arbitrary vector on that that we can specify with an arbitrary x, y, and z, and we've done this multiple times in multiple videos, if this is any point x, y, z that sits on the plane, then the vector, let me draw the vector, let's say we go to this point right over here, so this vector right over here is going to be, let me draw it the other way actually, so this vector right over here, let's say we're going between this point and x, y, z, this vector right here is going to be x minus 3i plus y minus 5j plus z minus 7k. That's what this vector is, it sits on the plane, assuming x, y, and z sit on the plane. So if we take the dot product of this and the normal vector, that has got to be equal to 0 because it sits on the plane, and then we'll have our equation. So let's take n dot that over there, so n dot x minus 3i plus y minus 5j plus z minus 7k. If any of this is confusing to you, I've gone into a little bit more depth in previous videos, especially in the linear algebra playlist, where I talk about constructing the equation of a plane given a point on the plane and the normal vector, and even how do you find that normal vector."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's what this vector is, it sits on the plane, assuming x, y, and z sit on the plane. So if we take the dot product of this and the normal vector, that has got to be equal to 0 because it sits on the plane, and then we'll have our equation. So let's take n dot that over there, so n dot x minus 3i plus y minus 5j plus z minus 7k. If any of this is confusing to you, I've gone into a little bit more depth in previous videos, especially in the linear algebra playlist, where I talk about constructing the equation of a plane given a point on the plane and the normal vector, and even how do you find that normal vector. So you might want to watch those if you want some review there. But these are going to be equal to 0, so when you take the dot product, n, our normal vector is this, so we just take the x term, which is negative 1, times this x term right over here, so negative 1 times this is just 3 minus x, and then plus this y component times this y component, so it's 2 times this, so it's plus 2y minus 10, and then finally the z component, negative 1 times this, so this is plus 7 minus z is equal to 0, and what do we get? So we have our negative x plus 2y minus z, and then is equal to, let's subtract 3 from both sides, so if we take it out there, it'll be minus 3."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If any of this is confusing to you, I've gone into a little bit more depth in previous videos, especially in the linear algebra playlist, where I talk about constructing the equation of a plane given a point on the plane and the normal vector, and even how do you find that normal vector. So you might want to watch those if you want some review there. But these are going to be equal to 0, so when you take the dot product, n, our normal vector is this, so we just take the x term, which is negative 1, times this x term right over here, so negative 1 times this is just 3 minus x, and then plus this y component times this y component, so it's 2 times this, so it's plus 2y minus 10, and then finally the z component, negative 1 times this, so this is plus 7 minus z is equal to 0, and what do we get? So we have our negative x plus 2y minus z, and then is equal to, let's subtract 3 from both sides, so if we take it out there, it'll be minus 3. If we add 10 to both sides, so then you have a plus 10 over here, and then we subtract 7 from both sides, this becomes a minus 7, so then on the right-hand side, negative 3 plus 10 minus 7, well, that's just going to be 0. That's just going to be 0, and just like that, we have the equation for this blue plane over here, the plane that contains these two lines. Now remember what we said at the beginning of the video."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we have our negative x plus 2y minus z, and then is equal to, let's subtract 3 from both sides, so if we take it out there, it'll be minus 3. If we add 10 to both sides, so then you have a plus 10 over here, and then we subtract 7 from both sides, this becomes a minus 7, so then on the right-hand side, negative 3 plus 10 minus 7, well, that's just going to be 0. That's just going to be 0, and just like that, we have the equation for this blue plane over here, the plane that contains these two lines. Now remember what we said at the beginning of the video. These two planes are parallel, so the ratio of the coefficients on the x terms, the y term, and the z term has got to be the same, and so this one has a positive 2, that has a negative 2. This is just to simplify it so it looks very similar to each other. Let's multiply this equation right here, both sides, by negative 1, and then we're going to get x minus 2y plus z is equal to 0."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now remember what we said at the beginning of the video. These two planes are parallel, so the ratio of the coefficients on the x terms, the y term, and the z term has got to be the same, and so this one has a positive 2, that has a negative 2. This is just to simplify it so it looks very similar to each other. Let's multiply this equation right here, both sides, by negative 1, and then we're going to get x minus 2y plus z is equal to 0. This is a completely valid, another alternate way of expressing the same plane, and what I like about this is it looks very similar to this, at least the ratios of the x, y's, and z's. Negative 2y, negative 2y, 1z, 1z, and remember the ratios have to be the same. Here we have a 1 to 1 ratio between the z coefficient and the z coefficient, the y coefficient and the y coefficient, so it's also going to be for the x coefficient."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's multiply this equation right here, both sides, by negative 1, and then we're going to get x minus 2y plus z is equal to 0. This is a completely valid, another alternate way of expressing the same plane, and what I like about this is it looks very similar to this, at least the ratios of the x, y's, and z's. Negative 2y, negative 2y, 1z, 1z, and remember the ratios have to be the same. Here we have a 1 to 1 ratio between the z coefficient and the z coefficient, the y coefficient and the y coefficient, so it's also going to be for the x coefficient. Here we know if this is going to be parallel to the blue plane, we know that a has got to be equal to 1. This is x minus 2y plus z is equal to d. Now let's figure out the actual distance. Let's actually figure out the actual distance between these two planes."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Here we have a 1 to 1 ratio between the z coefficient and the z coefficient, the y coefficient and the y coefficient, so it's also going to be for the x coefficient. Here we know if this is going to be parallel to the blue plane, we know that a has got to be equal to 1. This is x minus 2y plus z is equal to d. Now let's figure out the actual distance. Let's actually figure out the actual distance between these two planes. What we can do is we can take a point on this blue plane, and we have several examples of points on the blue plane, and find the distance between that point and this plane over here. Actually, I just finished doing some videos on how do you find the distance between a point and a plane. I'm just going to use that formula."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's actually figure out the actual distance between these two planes. What we can do is we can take a point on this blue plane, and we have several examples of points on the blue plane, and find the distance between that point and this plane over here. Actually, I just finished doing some videos on how do you find the distance between a point and a plane. I'm just going to use that formula. If you want it to be proved, go watch that video. It's actually a pretty interesting proof, I think. The distance between, let's say, this point, 1, 2, 3, and this plane over here, so this distance right here is going to be in the direction of the normal, the distance is going to be, and I want to be careful not using that, so I'll actually write out the word."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just going to use that formula. If you want it to be proved, go watch that video. It's actually a pretty interesting proof, I think. The distance between, let's say, this point, 1, 2, 3, and this plane over here, so this distance right here is going to be in the direction of the normal, the distance is going to be, and I want to be careful not using that, so I'll actually write out the word. Don't want to overload variables. The distance is going to be, you literally just evaluate this, let me do it this way, you literally put in this point for the x, y, and z, and then you subtract the d in the numerator, and we saw that as the formula for finding the distance. So it's literally going to be 1, I'm actually using this point right over here, it's going to be 1, 1, because we just have 1x, so it's just going to be 1 minus 2 times 2, 1 minus 4, that's 2 times 2, plus 3, minus d, well here, the d is just d, so we're just going to write minus d, just like that, all of that over what is essentially the magnitude of the normal vector, and we saw in several videos, that's just the square of the coefficients on each of these terms right here, and taking the sum of those, taking the square root, so it's going to be equal to 1 squared plus negative 2 squared, which is 4, plus 1 squared, which is 1."}, {"video_title": "Distance between planes Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The distance between, let's say, this point, 1, 2, 3, and this plane over here, so this distance right here is going to be in the direction of the normal, the distance is going to be, and I want to be careful not using that, so I'll actually write out the word. Don't want to overload variables. The distance is going to be, you literally just evaluate this, let me do it this way, you literally put in this point for the x, y, and z, and then you subtract the d in the numerator, and we saw that as the formula for finding the distance. So it's literally going to be 1, I'm actually using this point right over here, it's going to be 1, 1, because we just have 1x, so it's just going to be 1 minus 2 times 2, 1 minus 4, that's 2 times 2, plus 3, minus d, well here, the d is just d, so we're just going to write minus d, just like that, all of that over what is essentially the magnitude of the normal vector, and we saw in several videos, that's just the square of the coefficients on each of these terms right here, and taking the sum of those, taking the square root, so it's going to be equal to 1 squared plus negative 2 squared, which is 4, plus 1 squared, which is 1. So this is going to simplify to, the distance is equal to, 1 minus 4 plus 3 is 0, so in the numerator we have negative d, all of that over, the square root of 1 plus 4 plus 1, so all over the square root of 6. So they say the distance, if the distance between the plane Ax minus 2i plus z is equal, between this plane and this plane over here, is square root of 6, so they're saying the distance is equal to, is equal to the square root of 6, that's what this information right over here is, maybe I should do that in another color, so this distance right, that's not another color, the distance between the two planes is going to be the square root of 6, and so then if we solve for d, multiply both sides of this equation times the square root of 6, you get 6 is equal to negative d, or d is equal to negative 6. Now what they care about is the absolute value of d, or the absolute distance, so that's actually, this would be kind of the signed distance, it kind of specifies whether we're above or below the plane, since we're below the plane, we got a negative number, I just happened to draw it right, if we were above the plane, we would get a positive number, so this distance is negative 6, the absolute value of it, the absolute value of d, which is the same thing as the absolute value of negative 6, is equal to 6, so take any point, any point on this blue plane, and you look for the closest point on the orange plane, and they will be, they will be exactly 6 apart, anyway, hopefully you found that interesting."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So a function literally is, so I'll write it like this, a function really is just a relation between the members of one set and the members of the other set. So let's say I have some set x. And for every member of that set x, I'm going to relate that member or associate that member with another member of a set y. So if I imagine that that is my set x and that this is my set y right there, and y doesn't have to be smaller, that's just the way I drew it. The function is just a relation that if I just take a member of my set x, so if I just take a member of it, let's say that's the member that I'm taking it, we're visualizing it as a point. This function will say, OK, you gave me a member of x, then I will give you a member of y associated with that member of x. So the function will say, you give me that, then I will map it to that member right there."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I imagine that that is my set x and that this is my set y right there, and y doesn't have to be smaller, that's just the way I drew it. The function is just a relation that if I just take a member of my set x, so if I just take a member of it, let's say that's the member that I'm taking it, we're visualizing it as a point. This function will say, OK, you gave me a member of x, then I will give you a member of y associated with that member of x. So the function will say, you give me that, then I will map it to that member right there. I'll use the word map it. And that really just means relating it to or associating with another member of y. And if you'd give me some other point right here, I'll relate it to another member of y. I'll relate it to another member of y right there."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the function will say, you give me that, then I will map it to that member right there. I'll use the word map it. And that really just means relating it to or associating with another member of y. And if you'd give me some other point right here, I'll relate it to another member of y. I'll relate it to another member of y right there. I might even relate it to the same member of y. And so this notation just says this is a mapping from one set x, and I'm speaking in very general terms, to another set y. And so you're probably saying, hey, Sal, this is very abstract."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if you'd give me some other point right here, I'll relate it to another member of y. I'll relate it to another member of y right there. I might even relate it to the same member of y. And so this notation just says this is a mapping from one set x, and I'm speaking in very general terms, to another set y. And so you're probably saying, hey, Sal, this is very abstract. How does this relate to the functions that I've seen in the past? Well, let me just write down a function. You've probably seen a lot in the past."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so you're probably saying, hey, Sal, this is very abstract. How does this relate to the functions that I've seen in the past? Well, let me just write down a function. You've probably seen a lot in the past. You've seen people write, or you've dealt with f of x is equal to x squared. How would we write this in this notation? Well, this is a function, assuming that it's kind of the traditional way that you see it."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You've probably seen a lot in the past. You've seen people write, or you've dealt with f of x is equal to x squared. How would we write this in this notation? Well, this is a function, assuming that it's kind of the traditional way that you see it. This function, and actually, let me just write it with the f, I was going to write it with the g of x, just so that this doesn't always have to be an f. But I think you get that idea. In this case, f is a mapping from real numbers, right? The real numbers are everything that I can put in here."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is a function, assuming that it's kind of the traditional way that you see it. This function, and actually, let me just write it with the f, I was going to write it with the g of x, just so that this doesn't always have to be an f. But I think you get that idea. In this case, f is a mapping from real numbers, right? The real numbers are everything that I can put in here. And actually, this is part of the function definition. I could constrain this to just be integers, or just be even numbers, or just be even integers. But this is part of the function definition."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The real numbers are everything that I can put in here. And actually, this is part of the function definition. I could constrain this to just be integers, or just be even numbers, or just be even integers. But this is part of the function definition. I'm defining the function to be a mapping from real numbers. I'm saying you can put any real number here, and it's going to map to real numbers. So in this case, if x is real numbers, it's going to map to itself, which is completely legitimate."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this is part of the function definition. I'm defining the function to be a mapping from real numbers. I'm saying you can put any real number here, and it's going to map to real numbers. So in this case, if x is real numbers, it's going to map to itself, which is completely legitimate. So if this is the real numbers, and obviously the real numbers would go off in every direction forever, but if this is the real numbers, this function mapping is just taking every point in r and mapping it to another point in r, it's taking every point and associating with it its perfect square. And I want to make a very subtle notation. Or at least in my mind, the first time that I got exposed to functions, I was thinking, you give me an x and I square it, and I'm giving you the square of x."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, if x is real numbers, it's going to map to itself, which is completely legitimate. So if this is the real numbers, and obviously the real numbers would go off in every direction forever, but if this is the real numbers, this function mapping is just taking every point in r and mapping it to another point in r, it's taking every point and associating with it its perfect square. And I want to make a very subtle notation. Or at least in my mind, the first time that I got exposed to functions, I was thinking, you give me an x and I square it, and I'm giving you the square of x. And that's true. You are doing that. But at least the way my brain worked, I kind of thought of it as I was changing my x into another number."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or at least in my mind, the first time that I got exposed to functions, I was thinking, you give me an x and I square it, and I'm giving you the square of x. And that's true. You are doing that. But at least the way my brain worked, I kind of thought of it as I was changing my x into another number. And you can maybe view it that way, and that might actually be the best way to view it. But the mathematical definition I'm introducing here is more that I'm associating. I'm associating x with x squared."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But at least the way my brain worked, I kind of thought of it as I was changing my x into another number. And you can maybe view it that way, and that might actually be the best way to view it. But the mathematical definition I'm introducing here is more that I'm associating. I'm associating x with x squared. And this is actually another way, this is another function notation of writing this exact same thing. These two statements right here, this statement and this statement are identical. This statement you've probably never seen before, but I kind of like it because it kind of shows the mapping or the association more."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm associating x with x squared. And this is actually another way, this is another function notation of writing this exact same thing. These two statements right here, this statement and this statement are identical. This statement you've probably never seen before, but I kind of like it because it kind of shows the mapping or the association more. While this association, I kind of think that, look, you're putting an x into a little meat grinder or some machine that's going to ground up the x or square the x or do whatever it needs to do to the x. This notation to me implies the actual mapping. You give me an x, and I'm going to associate another number in real numbers called x squared."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This statement you've probably never seen before, but I kind of like it because it kind of shows the mapping or the association more. While this association, I kind of think that, look, you're putting an x into a little meat grinder or some machine that's going to ground up the x or square the x or do whatever it needs to do to the x. This notation to me implies the actual mapping. You give me an x, and I'm going to associate another number in real numbers called x squared. So it's going to be just another point. And just as a little bit of terminology, and I think you've seen this terminology before, the set that you are mapping from is called the domain. And it's part of the function definition."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You give me an x, and I'm going to associate another number in real numbers called x squared. So it's going to be just another point. And just as a little bit of terminology, and I think you've seen this terminology before, the set that you are mapping from is called the domain. And it's part of the function definition. I as a function creator have to tell you that, look, every valid input here has to be a set of real numbers. Now, the set that I'm mapping to, this is called the codomain. And I guess the obvious question that you're probably asking is, hey, Sal, when I learned all of this function stuff in Algebra 2 or whenever you first learned it, you're like, we never use this codomain word."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's part of the function definition. I as a function creator have to tell you that, look, every valid input here has to be a set of real numbers. Now, the set that I'm mapping to, this is called the codomain. And I guess the obvious question that you're probably asking is, hey, Sal, when I learned all of this function stuff in Algebra 2 or whenever you first learned it, you're like, we never use this codomain word. And actually, I don't think it has a hyphen in it. But we never use that codomain word. We have this idea of range."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I guess the obvious question that you're probably asking is, hey, Sal, when I learned all of this function stuff in Algebra 2 or whenever you first learned it, you're like, we never use this codomain word. And actually, I don't think it has a hyphen in it. But we never use that codomain word. We have this idea of range. I learned the word range when I was in ninth or tenth grade. How does this codomain relate to range? And it's a very subtle notation."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have this idea of range. I learned the word range when I was in ninth or tenth grade. How does this codomain relate to range? And it's a very subtle notation. So the codomain is a set that you're mapping to. In this example, this is the codomain. Codomain."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's a very subtle notation. So the codomain is a set that you're mapping to. In this example, this is the codomain. Codomain. In this example, the real numbers are the domain and the codomain. So the question is, how does the range relate to this? So the codomain is the set that can be possibly mapped to."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Codomain. In this example, the real numbers are the domain and the codomain. So the question is, how does the range relate to this? So the codomain is the set that can be possibly mapped to. You're not necessarily mapping to every point in the codomain. I'm just saying that this function is generally mapping from members of this set to that set. The range is the subset."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the codomain is the set that can be possibly mapped to. You're not necessarily mapping to every point in the codomain. I'm just saying that this function is generally mapping from members of this set to that set. The range is the subset. Let me write it this way. It could be equal to the codomain. It's some subset."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The range is the subset. Let me write it this way. It could be equal to the codomain. It's some subset. A set is a subset of itself. Every member of a set is also a member of itself. So it's a subset of itself."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's some subset. A set is a subset of itself. Every member of a set is also a member of itself. So it's a subset of itself. So range is a subset of the codomain. I keep adding a hyphen there. Or dash."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's a subset of itself. So range is a subset of the codomain. I keep adding a hyphen there. Or dash. Is a subset of the codomain, which the function actually maps to. So let me give you an example. Let's say I define the function g. And it is a mapping from the set of real numbers."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or dash. Is a subset of the codomain, which the function actually maps to. So let me give you an example. Let's say I define the function g. And it is a mapping from the set of real numbers. Well, let me say it's a mapping from R2 to R. So I'm essentially taking two tuples and I'm mapping it to R. And I will define g. I'll write it a couple of different ways. So now I'm going to take g of, let's say, two values. So I could say x, y."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I define the function g. And it is a mapping from the set of real numbers. Well, let me say it's a mapping from R2 to R. So I'm essentially taking two tuples and I'm mapping it to R. And I will define g. I'll write it a couple of different ways. So now I'm going to take g of, let's say, two values. So I could say x, y. Or I could say x1, x2. Let me do it that way. g of x1, x2 is equal to, well, let's say it's always equal to 2."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I could say x, y. Or I could say x1, x2. Let me do it that way. g of x1, x2 is equal to, well, let's say it's always equal to 2. It just always equals to 2. It's a mapping from R2 to R, but this just always equals 2. So what is our, and actually let me write it the other notation, just because you probably haven't seen this much."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "g of x1, x2 is equal to, well, let's say it's always equal to 2. It just always equals to 2. It's a mapping from R2 to R, but this just always equals 2. So what is our, and actually let me write it the other notation, just because you probably haven't seen this much. But I could write g of, g maps any points x1 and x2 to the point 2. This makes the mapping a little bit clearer. But just to get the notation right, what is our domain?"}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is our, and actually let me write it the other notation, just because you probably haven't seen this much. But I could write g of, g maps any points x1 and x2 to the point 2. This makes the mapping a little bit clearer. But just to get the notation right, what is our domain? What's the real numbers? That was part of my function definition. I said we're mapping from R2."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But just to get the notation right, what is our domain? What's the real numbers? That was part of my function definition. I said we're mapping from R2. So my domain is R2. And I should actually make that with that little line there. Now what is my codomain?"}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I said we're mapping from R2. So my domain is R2. And I should actually make that with that little line there. Now what is my codomain? My codomain. Well, my codomain is the set that I am potentially mapping to. And it's part of the function definition."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what is my codomain? My codomain. Well, my codomain is the set that I am potentially mapping to. And it's part of the function definition. This, by definition, is the codomain. So my codomain is R. Now, what is the range of my function? The range is the set of values that the function actually maps to."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's part of the function definition. This, by definition, is the codomain. So my codomain is R. Now, what is the range of my function? The range is the set of values that the function actually maps to. In this case, we always map to the value 2. So the range is actually just the value 2. And if we were to visualize this, R2 is actually, I would draw it as a blurb."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The range is the set of values that the function actually maps to. In this case, we always map to the value 2. So the range is actually just the value 2. And if we were to visualize this, R2 is actually, I would draw it as a blurb. I would draw it as the entire Cartesian space. But I'm just giving you kind of an abstract notion. That's R2."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if we were to visualize this, R2 is actually, I would draw it as a blurb. I would draw it as the entire Cartesian space. But I'm just giving you kind of an abstract notion. That's R2. If I really had to draw R, I would draw it as some type of a number line. Actually, let me do it that way, just for fun. You don't normally see it written that way."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's R2. If I really had to draw R, I would draw it as some type of a number line. Actually, let me do it that way, just for fun. You don't normally see it written that way. But I could just draw R like, that's R2. And I could just draw R as some straight line. So this is the set R. I could draw it like that as well."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You don't normally see it written that way. But I could just draw R like, that's R2. And I could just draw R as some straight line. So this is the set R. I could draw it like that as well. But let's just say it's a set R. And my function g essentially maps any point over here to exactly the point 2. 2 is just one little point in R. And my function g takes any point in R2, any coordinate, this could be the point 3, minus 5, whatever it is, and it always maps it to the point 2 in R. So if I take that point, it maps it to the point 2. That's what g always does."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the set R. I could draw it like that as well. But let's just say it's a set R. And my function g essentially maps any point over here to exactly the point 2. 2 is just one little point in R. And my function g takes any point in R2, any coordinate, this could be the point 3, minus 5, whatever it is, and it always maps it to the point 2 in R. So if I take that point, it maps it to the point 2. That's what g always does. You could say it's all of the real numbers, but its range is really just 2. Now, if I write the example, if I say that, let me do another example that might be interesting. If I just write h is a function that goes from R2 to R3, and I'll be a little careful here, h goes from R2 to R3, and I'll write here that h of x1, x2 is equal to, so now I'm mapping, I'm going to a higher dimension space."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what g always does. You could say it's all of the real numbers, but its range is really just 2. Now, if I write the example, if I say that, let me do another example that might be interesting. If I just write h is a function that goes from R2 to R3, and I'll be a little careful here, h goes from R2 to R3, and I'll write here that h of x1, x2 is equal to, so now I'm mapping, I'm going to a higher dimension space. So I'm going to say that that is going to be equal to, let's say my first coordinate, I could say an R3, or my first component R3 is x1 plus x2. Let's say my second coordinate is x2 minus x1. And let's say my third coordinate is x2 times x1."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I just write h is a function that goes from R2 to R3, and I'll be a little careful here, h goes from R2 to R3, and I'll write here that h of x1, x2 is equal to, so now I'm mapping, I'm going to a higher dimension space. So I'm going to say that that is going to be equal to, let's say my first coordinate, I could say an R3, or my first component R3 is x1 plus x2. Let's say my second coordinate is x2 minus x1. And let's say my third coordinate is x2 times x1. Now, what is my domain and my range and my co-domain? So my domain by definition is this right there. My co-domain by definition is R3."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say my third coordinate is x2 times x1. Now, what is my domain and my range and my co-domain? So my domain by definition is this right there. My co-domain by definition is R3. And notice, I'm going from a space that has two dimensions to a space that has three dimensions, or three components, but I can always associate some point in x1, x2 with some point in my R3 there. And now a slightly trickier question here is, what is the range? Can I always associate every point, maybe this wasn't the best example because it's not simple enough, but can I associate every point in R3?"}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "My co-domain by definition is R3. And notice, I'm going from a space that has two dimensions to a space that has three dimensions, or three components, but I can always associate some point in x1, x2 with some point in my R3 there. And now a slightly trickier question here is, what is the range? Can I always associate every point, maybe this wasn't the best example because it's not simple enough, but can I associate every point in R3? So this is my co-domain. My domain was R2. Now, and my function goes from R2 to R3, so that's h. And so my range, as you can see, it's not like every coordinate you can express in this way in some way."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Can I always associate every point, maybe this wasn't the best example because it's not simple enough, but can I associate every point in R3? So this is my co-domain. My domain was R2. Now, and my function goes from R2 to R3, so that's h. And so my range, as you can see, it's not like every coordinate you can express in this way in some way. Let me give you an example. For example, clearly the term, I mean I could put some x1's and x2's here and figure it out. Let's do that."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, and my function goes from R2 to R3, so that's h. And so my range, as you can see, it's not like every coordinate you can express in this way in some way. Let me give you an example. For example, clearly the term, I mean I could put some x1's and x2's here and figure it out. Let's do that. Let's take our h of, and let me use my other notation. Let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2, 3. And then my function tells me that this will map to the point in R3, this will map to, I add the two terms, so 2 plus 3, so it's 5."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's do that. Let's take our h of, and let me use my other notation. Let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2, 3. And then my function tells me that this will map to the point in R3, this will map to, I add the two terms, so 2 plus 3, so it's 5. I find the difference between x2 and x1, so 3 minus 2 is 1. And then I multiply the two, 6. So clearly this will be in the range."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then my function tells me that this will map to the point in R3, this will map to, I add the two terms, so 2 plus 3, so it's 5. I find the difference between x2 and x1, so 3 minus 2 is 1. And then I multiply the two, 6. So clearly this will be in the range. This is a member of the range. I shouldn't write like that, I should write it like this. The member of the range."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So clearly this will be in the range. This is a member of the range. I shouldn't write like that, I should write it like this. The member of the range. So for example, the point 2, 3, which might be right there, will be mapped to the three-dimensional point. It's kind of just drawn as a two-dimensional blurb right there, but I think you get the idea. Would be mapped to the three-dimensional point 5, 1, 6."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The member of the range. So for example, the point 2, 3, which might be right there, will be mapped to the three-dimensional point. It's kind of just drawn as a two-dimensional blurb right there, but I think you get the idea. Would be mapped to the three-dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say I have the point, let me do it in a different color. Let me say I have the point there."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Would be mapped to the three-dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say I have the point, let me do it in a different color. Let me say I have the point there. Let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me say I have the point there. Let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain. It's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range?"}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's definitely a member of the codomain. It's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? Well, if I take, this 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know if 5 is the sum and 1 is the difference, we're dealing with 2 and 3. There's no way that you can get the product of those numbers to be equal to 0."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But is this in our range? Well, if I take, this 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know if 5 is the sum and 1 is the difference, we're dealing with 2 and 3. There's no way that you can get the product of those numbers to be equal to 0. So this guy is not in the range. So the range would be the subset of all of these points in R3. So there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "There's no way that you can get the product of those numbers to be equal to 0. So this guy is not in the range. So the range would be the subset of all of these points in R3. So there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range. Now I want to introduce you to one more piece of terminology when it comes to functions. These functions up here, this function that mapped from points in R2 to R, so it mapped, its codomain was R. This function up here is probably the most common function you see in mathematics. This is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range. Now I want to introduce you to one more piece of terminology when it comes to functions. These functions up here, this function that mapped from points in R2 to R, so it mapped, its codomain was R. This function up here is probably the most common function you see in mathematics. This is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it. But if they mapped to kind of a one-dimensional space, we call them a scalar-valued function or a real-valued function. Which is pretty much all of the functions that you've probably dealt with up to this point in your mathematical career, unless you've kind of taken some vector calculus or whatever not. Now, the functions that map to spaces or subspaces that have more than one dimension."}, {"video_title": "A more formal understanding of functions Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it. But if they mapped to kind of a one-dimensional space, we call them a scalar-valued function or a real-valued function. Which is pretty much all of the functions that you've probably dealt with up to this point in your mathematical career, unless you've kind of taken some vector calculus or whatever not. Now, the functions that map to spaces or subspaces that have more than one dimension. So if you map to R or any subset of R, you have a real valued function or a scalar-valued function. If you map to Rn, where n is greater than 1, so if you map to R2, R3, R4, R100, you're then dealing with a vector valued function. So this last function that I defined over here, h is a vector-valued function."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have a subspace V that is equal to all of the vectors, all the vectors, let me write it this way, all of the x1, x2, x3's, so all the vectors like this, that satisfy x1 plus x2 plus x3 is equal to 0. So if you think about it, this is just a plane in R3. So this subspace is a plane in R3. And I'm interested in finding the transformation matrix for the projection of any vector x in R3 onto V. So how can we do that? So we could do it like we did in the last video. We can find the basis for this subspace right there. And that's not too hard to do."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And I'm interested in finding the transformation matrix for the projection of any vector x in R3 onto V. So how can we do that? So we could do it like we did in the last video. We can find the basis for this subspace right there. And that's not too hard to do. We could say x1, if we assume that, let's say that x2 and x3 are kind of free variables, then we could say x1 is equal to minus x2 minus x3. And then let's just, so that we can write it in kind of our parametric form, or if we can write our solution set as the combination of basis vectors, we can say x2 is equal to, let's say it's equal to some arbitrary constant c2. And let's say that x3 is equal to some arbitrary constant c3."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And that's not too hard to do. We could say x1, if we assume that, let's say that x2 and x3 are kind of free variables, then we could say x1 is equal to minus x2 minus x3. And then let's just, so that we can write it in kind of our parametric form, or if we can write our solution set as the combination of basis vectors, we can say x2 is equal to, let's say it's equal to some arbitrary constant c2. And let's say that x3 is equal to some arbitrary constant c3. Then we can say that V, we can rewrite V, we can say that V is, I'll do it here, V is equal to the set of all x1's, x2's, and x3's that are equal to c2 times, so x1 is equal to minus, let me rewrite this with the c2's. This is equal to c2, this is equal to c3, so x1 is equal to minus c2 minus c3. So x1 is equal to minus 1 times c2 plus c3 times what?"}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that x3 is equal to some arbitrary constant c3. Then we can say that V, we can rewrite V, we can say that V is, I'll do it here, V is equal to the set of all x1's, x2's, and x3's that are equal to c2 times, so x1 is equal to minus, let me rewrite this with the c2's. This is equal to c2, this is equal to c3, so x1 is equal to minus c2 minus c3. So x1 is equal to minus 1 times c2 plus c3 times what? Plus c3 times minus 1. And then what is x2 equal to? x2 is just equal to c2."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So x1 is equal to minus 1 times c2 plus c3 times what? Plus c3 times minus 1. And then what is x2 equal to? x2 is just equal to c2. So it's 1 times c2 plus 0 times c3. x3 is just equal to c3, so it's 0 times c2 plus 1 times c3. And so this is another way of defining our subspace."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "x2 is just equal to c2. So it's 1 times c2 plus 0 times c3. x3 is just equal to c3, so it's 0 times c2 plus 1 times c3. And so this is another way of defining our subspace. All of the vectors that satisfy this is equal to this definition here. It's all the vectors whose components satisfy or that lie in this plane, whose entries lie in that plane. And that's for any real numbers right there."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And so this is another way of defining our subspace. All of the vectors that satisfy this is equal to this definition here. It's all the vectors whose components satisfy or that lie in this plane, whose entries lie in that plane. And that's for any real numbers right there. Or another way of writing that, another way of writing this is that v is equal to the span of the vectors minus 1, 1, and 0, and the vector minus 1, 0, and 1. Just like that. And we know that these are actually a basis for v because they're linearly independent."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And that's for any real numbers right there. Or another way of writing that, another way of writing this is that v is equal to the span of the vectors minus 1, 1, and 0, and the vector minus 1, 0, and 1. Just like that. And we know that these are actually a basis for v because they're linearly independent. There's no way I can take linear combinations of this guy and make the second entry be 1 here. And likewise, there's no way I can take linear combinations of this guy and make this third entry equal a 1 here. So these are also a basis for v. So given that, just using the technique we did before, we could set some matrix A equal to minus 1, 1, 0, and then minus 1, 0, and 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And we know that these are actually a basis for v because they're linearly independent. There's no way I can take linear combinations of this guy and make the second entry be 1 here. And likewise, there's no way I can take linear combinations of this guy and make this third entry equal a 1 here. So these are also a basis for v. So given that, just using the technique we did before, we could set some matrix A equal to minus 1, 1, 0, and then minus 1, 0, and 1. And then we can figure out that the projection of any vector x in R3 onto v is going to be equal to, and we saw this, it's going to be equal to A times the inverse of A transpose A, all of that times A transpose and all of that times x. And you can do it. You have A here."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So these are also a basis for v. So given that, just using the technique we did before, we could set some matrix A equal to minus 1, 1, 0, and then minus 1, 0, and 1. And then we can figure out that the projection of any vector x in R3 onto v is going to be equal to, and we saw this, it's going to be equal to A times the inverse of A transpose A, all of that times A transpose and all of that times x. And you can do it. You have A here. You can figure out what the transpose of A is. Very easy. You can take out A transpose A."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "You have A here. You can figure out what the transpose of A is. Very easy. You can take out A transpose A. Then you can invert it. And it'll be very similar to what we did in the last video. It'll be a little less work because this is a 3 by 2 matrix instead of a 4 by 2 matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "You can take out A transpose A. Then you can invert it. And it'll be very similar to what we did in the last video. It'll be a little less work because this is a 3 by 2 matrix instead of a 4 by 2 matrix. But you saw, it is actually a lot of work. It's very hairy and you might make some careless mistakes. So let's figure out if there's another way that we can come up with this matrix right here."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "It'll be a little less work because this is a 3 by 2 matrix instead of a 4 by 2 matrix. But you saw, it is actually a lot of work. It's very hairy and you might make some careless mistakes. So let's figure out if there's another way that we can come up with this matrix right here. Now, we know that if x is a member of R3, that x can be represented as a combination of some vector v that is in our subspace plus some vector w that is in the orthogonal complement to the subspace, where v is a member of our subspace and w is a member of the orthogonal complement of our subspace. Now, by definition, that right there is the projection of x onto v, and this is the projection of x onto the orthogonal complement of v. So we can write that x is equal to the projection onto v of x plus the projection onto v's orthogonal complement or the orthogonal complement of v of x. So this is by definition that any member of R3 can be represented this way."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's figure out if there's another way that we can come up with this matrix right here. Now, we know that if x is a member of R3, that x can be represented as a combination of some vector v that is in our subspace plus some vector w that is in the orthogonal complement to the subspace, where v is a member of our subspace and w is a member of the orthogonal complement of our subspace. Now, by definition, that right there is the projection of x onto v, and this is the projection of x onto the orthogonal complement of v. So we can write that x is equal to the projection onto v of x plus the projection onto v's orthogonal complement or the orthogonal complement of v of x. So this is by definition that any member of R3 can be represented this way. Now, if we want to write this as matrix vector products, and two videos ago I showed you that these are linear transformations. So let me write that here. So they're linear transformations, so they can be written as matrix vector products."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this is by definition that any member of R3 can be represented this way. Now, if we want to write this as matrix vector products, and two videos ago I showed you that these are linear transformations. So let me write that here. So they're linear transformations, so they can be written as matrix vector products. You see that right there. Let me define this matrix, I don't know, let me call this matrix T, and let me call, let me do another, let me do a letter, let me do b. And let's say that the projection onto the orthogonal complement of v of x, let's say that that's equal to some other vector, sorry, some other matrix c times x."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So they're linear transformations, so they can be written as matrix vector products. You see that right there. Let me define this matrix, I don't know, let me call this matrix T, and let me call, let me do another, let me do a letter, let me do b. And let's say that the projection onto the orthogonal complement of v of x, let's say that that's equal to some other vector, sorry, some other matrix c times x. We know this is a linear transformation, so it can be represented as some matrix c times x. So what are these going to be equal to? Well, x, if I want to write it as a linear transformation of x, I could just write it as the 3 by 3 identity matrix times x."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that the projection onto the orthogonal complement of v of x, let's say that that's equal to some other vector, sorry, some other matrix c times x. We know this is a linear transformation, so it can be represented as some matrix c times x. So what are these going to be equal to? Well, x, if I want to write it as a linear transformation of x, I could just write it as the 3 by 3 identity matrix times x. That's the same thing as x. That's going to be equal to the projection of x onto v. Well, that's just the same thing as b times x. That's the same thing as b times x."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well, x, if I want to write it as a linear transformation of x, I could just write it as the 3 by 3 identity matrix times x. That's the same thing as x. That's going to be equal to the projection of x onto v. Well, that's just the same thing as b times x. That's the same thing as b times x. And then plus the projection of x onto v's orthogonal complement. Well, that's just c times x. And if you want to factor out the x on this side, we know that the matrix vector products exhibit the distributive property."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "That's the same thing as b times x. And then plus the projection of x onto v's orthogonal complement. Well, that's just c times x. And if you want to factor out the x on this side, we know that the matrix vector products exhibit the distributive property. So we could write that the identity matrix times x is equal to b plus c times x. Or another way to view this equation is that this matrix must be equal to these two matrices. So we get that the identity matrix in R3 is equal to the projection matrix onto v plus the projection matrix onto v's orthogonal complement."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And if you want to factor out the x on this side, we know that the matrix vector products exhibit the distributive property. So we could write that the identity matrix times x is equal to b plus c times x. Or another way to view this equation is that this matrix must be equal to these two matrices. So we get that the identity matrix in R3 is equal to the projection matrix onto v plus the projection matrix onto v's orthogonal complement. So if we're trying to, remember, the whole point of this problem is to figure out this thing right here, is to solve for b. And we know a technique for doing it. You take A transpose, you can do this whole thing, but that might be pretty hairy."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we get that the identity matrix in R3 is equal to the projection matrix onto v plus the projection matrix onto v's orthogonal complement. So if we're trying to, remember, the whole point of this problem is to figure out this thing right here, is to solve for b. And we know a technique for doing it. You take A transpose, you can do this whole thing, but that might be pretty hairy. But maybe it's easy to find this guy. Maybe, I don't know. It actually turns out in this video, this one will be easy."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "You take A transpose, you can do this whole thing, but that might be pretty hairy. But maybe it's easy to find this guy. Maybe, I don't know. It actually turns out in this video, this one will be easy. So if it's easy to find this guy, we can just solve for b. If we subtract c from both sides, we get that b is equal to, is equal to i, is equal to the identity matrix minus the transformation matrix for the transformation onto v's orthogonal complement. So let's see what this is."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "It actually turns out in this video, this one will be easy. So if it's easy to find this guy, we can just solve for b. If we subtract c from both sides, we get that b is equal to, is equal to i, is equal to the identity matrix minus the transformation matrix for the transformation onto v's orthogonal complement. So let's see what this is. Let's see if we can figure out what c is right there. So let's go back to our original. So remember, let me rewrite the problem actually."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's see what this is. Let's see if we can figure out what c is right there. So let's go back to our original. So remember, let me rewrite the problem actually. Remember that v was equal to, essentially, is equal to all of the x1's, x2's, x3's that satisfy x1 plus x2 plus x3 is equal to 0. Or another way to say it is all the x1's, x2's, and x3's that satisfy the equation 1, 1, 1 times x1, x2, x3 is equal to the 0 vector. Or in this case, it'll just be 0."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So remember, let me rewrite the problem actually. Remember that v was equal to, essentially, is equal to all of the x1's, x2's, x3's that satisfy x1 plus x2 plus x3 is equal to 0. Or another way to say it is all the x1's, x2's, and x3's that satisfy the equation 1, 1, 1 times x1, x2, x3 is equal to the 0 vector. Or in this case, it'll just be 0. We could write the 0 vector like that, just like that. So 1 times x1 plus 1 times x2 plus 1 times x3 is going to equal the 0 vector. This is another way to write v. Now, all of the x's that satisfy this right here, what is that?"}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Or in this case, it'll just be 0. We could write the 0 vector like that, just like that. So 1 times x1 plus 1 times x2 plus 1 times x3 is going to equal the 0 vector. This is another way to write v. Now, all of the x's that satisfy this right here, what is that? This is saying that v is equal to the null space of this matrix right there. The null space of this matrix is all of the vectors that satisfy this equation. So v is equal to the null space of 1, 1, 1, just like that."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is another way to write v. Now, all of the x's that satisfy this right here, what is that? This is saying that v is equal to the null space of this matrix right there. The null space of this matrix is all of the vectors that satisfy this equation. So v is equal to the null space of 1, 1, 1, just like that. Up here, we kind of figured out v in kind of the traditional way. We figured out that v is the span of these things, but now we know that's the same thing as the null space of 1, 1, 1. These two statements are equivalent."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So v is equal to the null space of 1, 1, 1, just like that. Up here, we kind of figured out v in kind of the traditional way. We figured out that v is the span of these things, but now we know that's the same thing as the null space of 1, 1, 1. These two statements are equivalent. Now, we at least had a hunch that maybe we could figure out straight up this b here by doing all of this a transpose and by doing all of this silliness here. But our hunch is maybe if we can figure out the transformation matrix for the orthogonal complement of v right there, that we can just apply this kind of, that we can just solve for b given that the identity matrix minus this guy is going to be equal to b. So let's see if we can figure out the projection matrix, if we can figure out the transformation matrix for the orthogonal projection, for x onto the orthogonal projection of v. So if this is v, what is v complement?"}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "These two statements are equivalent. Now, we at least had a hunch that maybe we could figure out straight up this b here by doing all of this a transpose and by doing all of this silliness here. But our hunch is maybe if we can figure out the transformation matrix for the orthogonal complement of v right there, that we can just apply this kind of, that we can just solve for b given that the identity matrix minus this guy is going to be equal to b. So let's see if we can figure out the projection matrix, if we can figure out the transformation matrix for the orthogonal projection, for x onto the orthogonal projection of v. So if this is v, what is v complement? v complement is going to be equal to the orthogonal complement, or v perp is going to be equal to the orthogonal complement of the null space of this matrix right here, which is equal to what? Remember, the null space, it's orthogonal complement. A null space's orthogonal complement is equivalent to the row space, or the column space of a transpose."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can figure out the projection matrix, if we can figure out the transformation matrix for the orthogonal projection, for x onto the orthogonal projection of v. So if this is v, what is v complement? v complement is going to be equal to the orthogonal complement, or v perp is going to be equal to the orthogonal complement of the null space of this matrix right here, which is equal to what? Remember, the null space, it's orthogonal complement. A null space's orthogonal complement is equivalent to the row space, or the column space of a transpose. We saw that multiple times. Or you could say the orthogonal complement of the row space is the null space. We've seen this many, many times before."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "A null space's orthogonal complement is equivalent to the row space, or the column space of a transpose. We saw that multiple times. Or you could say the orthogonal complement of the row space is the null space. We've seen this many, many times before. So the orthogonal complement of this guy is going to be the column space of his transpose. So the column space of the transpose of this guy. So it's 1, 1, 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "We've seen this many, many times before. So the orthogonal complement of this guy is going to be the column space of his transpose. So the column space of the transpose of this guy. So it's 1, 1, 1. Just like that. Or we can write that v's orthogonal complement is equal to the span of 1, 1, 1. The column space of this matrix, we only have one column in it."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So it's 1, 1, 1. Just like that. Or we can write that v's orthogonal complement is equal to the span of 1, 1, 1. The column space of this matrix, we only have one column in it. So its column space is going to be the span of that one column. So just to visualize what we're doing here, that original equation for v that satisfies that, that's just going to be some plane in R3. That's going to be some plane in R3."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "The column space of this matrix, we only have one column in it. So its column space is going to be the span of that one column. So just to visualize what we're doing here, that original equation for v that satisfies that, that's just going to be some plane in R3. That's going to be some plane in R3. That is v right there. And now we just figured out what v's orthogonal complement is. It's going to be a line in R3."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be some plane in R3. That is v right there. And now we just figured out what v's orthogonal complement is. It's going to be a line in R3. It's going to be all of the linear combinations of this guy. So it's going to be some line in R3. I haven't drawn it."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be a line in R3. It's going to be all of the linear combinations of this guy. So it's going to be some line in R3. I haven't drawn it. This is going to be tilted more, and so is this. But it's going to be some line. So this is the orthogonal complement of v. So let's see if we can figure out."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "I haven't drawn it. This is going to be tilted more, and so is this. But it's going to be some line. So this is the orthogonal complement of v. So let's see if we can figure out. So remember, the projection, let me do it this way. So this is the basis for v's orthogonal complement. So let's construct some matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this is the orthogonal complement of v. So let's see if we can figure out. So remember, the projection, let me do it this way. So this is the basis for v's orthogonal complement. So let's construct some matrix. I don't know, let me use a new letter that I haven't used before. Let me construct some matrix D whose columns are the basis vectors for the orthogonal complement of v. Well, there's only one basis vector, so it's going to be that. And we learned in the last video and the video before that, that the projection of any vector in R3 onto v's orthogonal complement is going to be equal to D times D transpose D inverse times D transpose times x."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's construct some matrix. I don't know, let me use a new letter that I haven't used before. Let me construct some matrix D whose columns are the basis vectors for the orthogonal complement of v. Well, there's only one basis vector, so it's going to be that. And we learned in the last video and the video before that, that the projection of any vector in R3 onto v's orthogonal complement is going to be equal to D times D transpose D inverse times D transpose times x. Or another way to view it is that this thing right here is the transformation matrix for this projection. That is the transformation matrix. So let's see if this is easier to solve, this thing, than this business up here where we had a 3 by 2 matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And we learned in the last video and the video before that, that the projection of any vector in R3 onto v's orthogonal complement is going to be equal to D times D transpose D inverse times D transpose times x. Or another way to view it is that this thing right here is the transformation matrix for this projection. That is the transformation matrix. So let's see if this is easier to solve, this thing, than this business up here where we had a 3 by 2 matrix. That was the whole motivation for doing this problem. To figure out the projection matrix for v subspace, we'd have to do this with a 3 by 2 matrix. It seemed pretty difficult."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if this is easier to solve, this thing, than this business up here where we had a 3 by 2 matrix. That was the whole motivation for doing this problem. To figure out the projection matrix for v subspace, we'd have to do this with a 3 by 2 matrix. It seemed pretty difficult. Instead, let's find the projection matrix to get to the projection onto v's orthogonal complement, which is this. So what is D transpose? So D transpose is just going to be equal to 1, 1, 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "It seemed pretty difficult. Instead, let's find the projection matrix to get to the projection onto v's orthogonal complement, which is this. So what is D transpose? So D transpose is just going to be equal to 1, 1, 1. What is D transpose times D? Well, that's D transpose. This is D, just like that."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So D transpose is just going to be equal to 1, 1, 1. What is D transpose times D? Well, that's D transpose. This is D, just like that. So what is this going to be equal to? This is just the dot product of that and that, right? 1 times 1 plus 1 times 1 plus 1 times 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is D, just like that. So what is this going to be equal to? This is just the dot product of that and that, right? 1 times 1 plus 1 times 1 plus 1 times 1. It equals 3. So this thing right here is equal to a 1 by 1 matrix, 3. So let's write down."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 times 1 plus 1 times 1 plus 1 times 1. It equals 3. So this thing right here is equal to a 1 by 1 matrix, 3. So let's write down. So this is equal to D, which is this matrix, 1, 1, 1, times D transpose D inverse. So D transpose D is just a 1 by 1 matrix. We're going to have to invert it."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's write down. So this is equal to D, which is this matrix, 1, 1, 1, times D transpose D inverse. So D transpose D is just a 1 by 1 matrix. We're going to have to invert it. Actually, I've never defined the inverse of a 1 by 1 matrix for you just now, so this is mildly exciting. Times D transpose. So D transpose looks like this."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "We're going to have to invert it. Actually, I've never defined the inverse of a 1 by 1 matrix for you just now, so this is mildly exciting. Times D transpose. So D transpose looks like this. 1, 1, 1. And then all of that's times x. But this is the transformation matrix right there."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So D transpose looks like this. 1, 1, 1. And then all of that's times x. But this is the transformation matrix right there. Now, what is the inverse of a 1 by 1 matrix? Now, you just have to remember. You just have to remember that A inverse times A is equal to the identity matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "But this is the transformation matrix right there. Now, what is the inverse of a 1 by 1 matrix? Now, you just have to remember. You just have to remember that A inverse times A is equal to the identity matrix. If we're dealing with a 1 by 1 matrix, then I'm just trying to figure out what matrix times 3 is going to be equal to the 1 by 1 identity matrix. So if I say, let's say, you know, I don't know. Let's say that 3 inverse times 3 has to be equal to the identity matrix, 1 by 1 identity matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "You just have to remember that A inverse times A is equal to the identity matrix. If we're dealing with a 1 by 1 matrix, then I'm just trying to figure out what matrix times 3 is going to be equal to the 1 by 1 identity matrix. So if I say, let's say, you know, I don't know. Let's say that 3 inverse times 3 has to be equal to the identity matrix, 1 by 1 identity matrix. Well, the only matrix that's going to make this work out, to get this entry, I have to take this guy's entry times that guy's entry, is going to be this guy right here. The inverse of this 1 by 1 matrix has to be the matrix 1 third. 1 third times 3 is equal to 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that 3 inverse times 3 has to be equal to the identity matrix, 1 by 1 identity matrix. Well, the only matrix that's going to make this work out, to get this entry, I have to take this guy's entry times that guy's entry, is going to be this guy right here. The inverse of this 1 by 1 matrix has to be the matrix 1 third. 1 third times 3 is equal to 1. This is almost trivially simple, but this is the inverse, that right there is the inverse matrix for the 1 by 1 matrix 3. So this right here is just 1 third. And we could actually just take that out."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 third times 3 is equal to 1. This is almost trivially simple, but this is the inverse, that right there is the inverse matrix for the 1 by 1 matrix 3. So this right here is just 1 third. And we could actually just take that out. It's a 1 by 1 matrix, which is essentially equivalent to a scalar. So this is going to be equal to, let me just draw a line here, this thing is equal to 1 third. I don't want to confuse you."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And we could actually just take that out. It's a 1 by 1 matrix, which is essentially equivalent to a scalar. So this is going to be equal to, let me just draw a line here, this thing is equal to 1 third. I don't want to confuse you. Let me rewrite it. So we get the projection of any vector in R3 onto the orthogonal complement of v is equal to 1 third times the vector 1, 1, 1 times, sorry, or yeah, it is a vector, or the matrix 1, 1, 1 times that matrix transpose, 1, 1, 1. And then all of that times x."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to confuse you. Let me rewrite it. So we get the projection of any vector in R3 onto the orthogonal complement of v is equal to 1 third times the vector 1, 1, 1 times, sorry, or yeah, it is a vector, or the matrix 1, 1, 1 times that matrix transpose, 1, 1, 1. And then all of that times x. And you can see, this is a lot simpler than when we did it with, if we had to do all of this business, if we did all of this business with this matrix. That's a harder matrix to deal with. This 1, 1, 1 matrix is very easy."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then all of that times x. And you can see, this is a lot simpler than when we did it with, if we had to do all of this business, if we did all of this business with this matrix. That's a harder matrix to deal with. This 1, 1, 1 matrix is very easy. Now what is this going to be equal to? This is going to be equal to 1 third times, we have a 3 by 1 times a 1 by 3 matrix. So it's going to result in a 3 by 3 matrix."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "This 1, 1, 1 matrix is very easy. Now what is this going to be equal to? This is going to be equal to 1 third times, we have a 3 by 1 times a 1 by 3 matrix. So it's going to result in a 3 by 3 matrix. And what do we get? The first entry is going to be 1 times 1, which is 1. Second entry is going to be 1 times 1, which is 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to result in a 3 by 3 matrix. And what do we get? The first entry is going to be 1 times 1, which is 1. Second entry is going to be 1 times 1, which is 1. Third entry is going to be 1 times 1, which is 1. I think you see the pattern. This guy, the second row, first column, 1 times 1 is 1."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Second entry is going to be 1 times 1, which is 1. Third entry is going to be 1 times 1, which is 1. I think you see the pattern. This guy, the second row, first column, 1 times 1 is 1. So this is just going to be a 3 by 3 matrix of 1's. So just like that, we were able to get, that was a pretty straightforward situation, we were able to get the projection matrix for any vector in R3 onto v's orthogonal complement. Now, we know that this thing right here is our original c that we said."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "This guy, the second row, first column, 1 times 1 is 1. So this is just going to be a 3 by 3 matrix of 1's. So just like that, we were able to get, that was a pretty straightforward situation, we were able to get the projection matrix for any vector in R3 onto v's orthogonal complement. Now, we know that this thing right here is our original c that we said. And we said that the identity matrix, we wrote it up here. Let me refer back to what I wrote way up here. We said, look, the identity matrix is equal to the transformation matrix for the projection onto v, plus the transformation matrix for the projection onto v's orthogonal complement."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now, we know that this thing right here is our original c that we said. And we said that the identity matrix, we wrote it up here. Let me refer back to what I wrote way up here. We said, look, the identity matrix is equal to the transformation matrix for the projection onto v, plus the transformation matrix for the projection onto v's orthogonal complement. Or we could write that the transformation matrix for the projection onto v is equal to the identity matrix minus the transformation matrix for the projection onto v's orthogonal complement. So we can write, so B is our transformation matrix onto. So if we say that the projection onto V of x is equal to B times x, we know that B is equal to the 3 by 3 identity matrix minus C. And this is C right there."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "We said, look, the identity matrix is equal to the transformation matrix for the projection onto v, plus the transformation matrix for the projection onto v's orthogonal complement. Or we could write that the transformation matrix for the projection onto v is equal to the identity matrix minus the transformation matrix for the projection onto v's orthogonal complement. So we can write, so B is our transformation matrix onto. So if we say that the projection onto V of x is equal to B times x, we know that B is equal to the 3 by 3 identity matrix minus C. And this is C right there. So B is equal to the identity matrix. So that's just 1, 0, 0, 0, 1, 0, 0, 0, 1 minus C. Minus 1 third times 1, 1, 1, 1, 1, 1, 1, 1, 1, 1. Just like that."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "So if we say that the projection onto V of x is equal to B times x, we know that B is equal to the 3 by 3 identity matrix minus C. And this is C right there. So B is equal to the identity matrix. So that's just 1, 0, 0, 0, 1, 0, 0, 0, 1 minus C. Minus 1 third times 1, 1, 1, 1, 1, 1, 1, 1, 1, 1. Just like that. What is this going to be equal to? This is going to be equal to, let's see, in our heads multiply this out. All of these entries are going to be 1 third essentially if we multiply this out like that."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. What is this going to be equal to? This is going to be equal to, let's see, in our heads multiply this out. All of these entries are going to be 1 third essentially if we multiply this out like that. So if we have 1 minus 1 third, I could write it out like that's 1 third, 1 third, 1 third. Everything is 1 third. 1 third, 1 third, 1 third, 1 third, 1 third, 1 third."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "All of these entries are going to be 1 third essentially if we multiply this out like that. So if we have 1 minus 1 third, I could write it out like that's 1 third, 1 third, 1 third. Everything is 1 third. 1 third, 1 third, 1 third, 1 third, 1 third, 1 third. And this just becomes a 1. So 1 minus 1 third is 2 thirds. And all of the 1's minus 1 third are going to be 2 thirds."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 third, 1 third, 1 third, 1 third, 1 third, 1 third. And this just becomes a 1. So 1 minus 1 third is 2 thirds. And all of the 1's minus 1 third are going to be 2 thirds. We could just go down the diagonal. And then the 0's minus 1 third are going to be minus 1 third. Minus 1 third, minus 1 third, minus 1 third."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "And all of the 1's minus 1 third are going to be 2 thirds. We could just go down the diagonal. And then the 0's minus 1 third are going to be minus 1 third. Minus 1 third, minus 1 third, minus 1 third. You have minus 1 third, minus 1 third, and minus 1 third. And just like that, we've been able to figure out our projection, our transformation matrix for the projection of any vector x onto v. By essentially finding this guy first, by finding the transformation matrix for the projection of any x onto v's orthogonal complement. Anyway, I thought that was pretty neat."}, {"video_title": "Another example of a projection matrix Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 third, minus 1 third, minus 1 third. You have minus 1 third, minus 1 third, and minus 1 third. And just like that, we've been able to figure out our projection, our transformation matrix for the projection of any vector x onto v. By essentially finding this guy first, by finding the transformation matrix for the projection of any x onto v's orthogonal complement. Anyway, I thought that was pretty neat. And you could rewrite this. You could rewrite this as being equal to 1 third times 2, 2, 2, 2's along the diagonals. And then you have minus 1's everywhere else."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We've seen in several videos that the column space of a matrix is pretty straightforward to find. In this situation, the column space of A is just equal to all of the linear combinations of the column vectors of A. So it's equal to, oh, another way of saying all of the linear combinations is just the span of each of these column vectors. So we call this one right here A1, this is A2, A3, A4, this is A5, then the column space of A is just equal to the span of A1, A2, A3, A4, and A5. Fair enough. But a more interesting question is whether these guys form a basis for the column space. Or even more interesting, what is the basis for the column space of A?"}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we call this one right here A1, this is A2, A3, A4, this is A5, then the column space of A is just equal to the span of A1, A2, A3, A4, and A5. Fair enough. But a more interesting question is whether these guys form a basis for the column space. Or even more interesting, what is the basis for the column space of A? And in this video, I'm going to show you a method for determining the basis. And along the way, we'll get an intuition for maybe why it works. And if I have time, actually I probably won't have time in this video, in the next video I'll prove to you why it works."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or even more interesting, what is the basis for the column space of A? And in this video, I'm going to show you a method for determining the basis. And along the way, we'll get an intuition for maybe why it works. And if I have time, actually I probably won't have time in this video, in the next video I'll prove to you why it works. So what you do to figure out the basis for the column space of A, remember the basis just means that vectors span CA. Clearly these vectors span our column space. I mean the span of these vectors is the column space."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if I have time, actually I probably won't have time in this video, in the next video I'll prove to you why it works. So what you do to figure out the basis for the column space of A, remember the basis just means that vectors span CA. Clearly these vectors span our column space. I mean the span of these vectors is the column space. But in order to be a basis, the vectors also have to be linearly independent. And we don't know whether these guys or what subset of these guys are linearly independent. So what you do, and I'm just really going to describe the process here as opposed to the proof, is you put this guy in reduced row echelon form."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I mean the span of these vectors is the column space. But in order to be a basis, the vectors also have to be linearly independent. And we don't know whether these guys or what subset of these guys are linearly independent. So what you do, and I'm just really going to describe the process here as opposed to the proof, is you put this guy in reduced row echelon form. So let's do that, let's keep our first row the same. Let me do it actually in the right side right here. So let's keep the first row the same, 1, 0, minus 1, 0, 4."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what you do, and I'm just really going to describe the process here as opposed to the proof, is you put this guy in reduced row echelon form. So let's do that, let's keep our first row the same. Let me do it actually in the right side right here. So let's keep the first row the same, 1, 0, minus 1, 0, 4. And then let's replace our second row with the second row minus 2 times the first row. So then our second row, 2 minus 2 times 1 is 0. 1 minus 2 times 0 is 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's keep the first row the same, 1, 0, minus 1, 0, 4. And then let's replace our second row with the second row minus 2 times the first row. So then our second row, 2 minus 2 times 1 is 0. 1 minus 2 times 0 is 1. 0 minus 2 times negative 1. So that's 0 plus 2. 0 minus 2 times 0 is just 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 times 0 is 1. 0 minus 2 times negative 1. So that's 0 plus 2. 0 minus 2 times 0 is just 0. And then 9 minus 2 times 4 is 1. Fair enough. Now we want to zero out this guy."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 2 times 0 is just 0. And then 9 minus 2 times 4 is 1. Fair enough. Now we want to zero out this guy. Well it seems like a pretty straightforward way. Just add, replace this row with this row plus the first row, so minus 1 plus 1 is 0. 2 plus 0 is 2."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now we want to zero out this guy. Well it seems like a pretty straightforward way. Just add, replace this row with this row plus the first row, so minus 1 plus 1 is 0. 2 plus 0 is 2. 5 minus 1 is 4. 1 plus 0 is 1. Minus 5 plus 4 is minus 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 plus 0 is 2. 5 minus 1 is 4. 1 plus 0 is 1. Minus 5 plus 4 is minus 1. And then finally we got this guy right here, and then maybe we can just, in order to zero him out, let's replace him with him minus the first row. So 1 minus 1 is 0. Minus 1 minus 0 is minus 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 5 plus 4 is minus 1. And then finally we got this guy right here, and then maybe we can just, in order to zero him out, let's replace him with him minus the first row. So 1 minus 1 is 0. Minus 1 minus 0 is minus 1. Minus 3 minus negative 1. That's minus 3 plus 1, so that's minus 2. Minus 2 minus 0 is minus 2."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 minus 0 is minus 1. Minus 3 minus negative 1. That's minus 3 plus 1, so that's minus 2. Minus 2 minus 0 is minus 2. And then 9 minus 4 is 5. So we did one round. We got our first pivot column going."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 minus 0 is minus 2. And then 9 minus 4 is 5. So we did one round. We got our first pivot column going. Now let's do another round of row operations. Well we want to zero all of these guys out. Luckily this is already 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We got our first pivot column going. Now let's do another round of row operations. Well we want to zero all of these guys out. Luckily this is already 0. So we don't have to change our first row or our second row. So we get 1, 0, minus 1, 0, 4. Our second row becomes 0, 1, 2, 0, 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Luckily this is already 0. So we don't have to change our first row or our second row. So we get 1, 0, minus 1, 0, 4. Our second row becomes 0, 1, 2, 0, 1. And now let us see if we can eliminate this guy right here. And let's do it by replacing our blue row, our third row, with the third row minus 2 times the second row. So 0 minus 2 times 0 is 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Our second row becomes 0, 1, 2, 0, 1. And now let us see if we can eliminate this guy right here. And let's do it by replacing our blue row, our third row, with the third row minus 2 times the second row. So 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 4 minus 2 times 2 is 0. 1 minus 2 times 0 is 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 4 minus 2 times 2 is 0. 1 minus 2 times 0 is 1. Minus 1 minus 2 times 1 is minus 3. All right. Now this last guy, we want to eliminate him."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 times 0 is 1. Minus 1 minus 2 times 1 is minus 3. All right. Now this last guy, we want to eliminate him. And we want to turn this into a 0. Let's replace this fourth row with the fourth row plus the second row. So 0 plus 0 is 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now this last guy, we want to eliminate him. And we want to turn this into a 0. Let's replace this fourth row with the fourth row plus the second row. So 0 plus 0 is 0. Minus 1 plus minus 1 is 0. Minus 2 plus minus 2 is 0. Minus 2 plus 0 is minus 2."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 0 plus 0 is 0. Minus 1 plus minus 1 is 0. Minus 2 plus minus 2 is 0. Minus 2 plus 0 is minus 2. And then 5 plus 1 is 6. We're getting close. So let's see."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 plus 0 is minus 2. And then 5 plus 1 is 6. We're getting close. So let's see. Let's look at our pivot entries. We have this is a pivot entry. That's a pivot entry."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see. Let's look at our pivot entries. We have this is a pivot entry. That's a pivot entry. And this is not a pivot entry. And let's see. Because it's following, obviously, another."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's a pivot entry. And this is not a pivot entry. And let's see. Because it's following, obviously, another. This guy is a pivot entry right here. Or will be. We need to 0 this minus 2 out."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because it's following, obviously, another. This guy is a pivot entry right here. Or will be. We need to 0 this minus 2 out. So let's, I think we'll be done. So let me write my first row just the way it is. Because everything above it is 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We need to 0 this minus 2 out. So let's, I think we'll be done. So let me write my first row just the way it is. Because everything above it is 0. So we don't have to worry about it. So my first row I can just write as 1, 0, minus 1, 0, 4. I can write my second row 0, 1, 2, 0, 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because everything above it is 0. So we don't have to worry about it. So my first row I can just write as 1, 0, minus 1, 0, 4. I can write my second row 0, 1, 2, 0, 1. I can write my third row as 0, 0, 0, 1, minus 3. And now let's replace my fourth row. Let's replace it with it plus 2 times the second row."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can write my second row 0, 1, 2, 0, 1. I can write my third row as 0, 0, 0, 1, minus 3. And now let's replace my fourth row. Let's replace it with it plus 2 times the second row. So 0 plus 2 times 0, minus 2 plus 2 times 1 is just 0. 6 plus 2 times minus 3, that's 6 minus 6. That's just 0."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's replace it with it plus 2 times the second row. So 0 plus 2 times 0, minus 2 plus 2 times 1 is just 0. 6 plus 2 times minus 3, that's 6 minus 6. That's just 0. And there we have, we've actually put our matrix in reduced row echelon form. So let me put little brackets around it. It's not so bad if you just kind of go and just do the manipulations."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's just 0. And there we have, we've actually put our matrix in reduced row echelon form. So let me put little brackets around it. It's not so bad if you just kind of go and just do the manipulations. And sometimes when you kind of get a headache thinking about doing something like this, but this wasn't too bad. So this is the reduced, let me just say, the reduced row echelon form of A. Let me just call that matrix R. So this is matrix R right there."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's not so bad if you just kind of go and just do the manipulations. And sometimes when you kind of get a headache thinking about doing something like this, but this wasn't too bad. So this is the reduced, let me just say, the reduced row echelon form of A. Let me just call that matrix R. So this is matrix R right there. Now, what do we see about matrix R? Well, it has three pivot entries, or three pivot columns. Let me square them out or circle them out."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me just call that matrix R. So this is matrix R right there. Now, what do we see about matrix R? Well, it has three pivot entries, or three pivot columns. Let me square them out or circle them out. Column 1 is a pivot column. Column 2 is a pivot column. And column 3 is a pivot column."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me square them out or circle them out. Column 1 is a pivot column. Column 2 is a pivot column. And column 3 is a pivot column. Now, and we've done this in previous videos, well, there's two things that you can see. These three columns, they're clearly linearly independent. How do we know that?"}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And column 3 is a pivot column. Now, and we've done this in previous videos, well, there's two things that you can see. These three columns, they're clearly linearly independent. How do we know that? Well, this guy's got a 1 where, and that's just with respect to each other. If we just took a set of, let's call this R1, R2, and this would be R3, this would be R4 right here. It's clear that the set R1, R2, and R4, it's clear that this is linearly independent."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "How do we know that? Well, this guy's got a 1 where, and that's just with respect to each other. If we just took a set of, let's call this R1, R2, and this would be R3, this would be R4 right here. It's clear that the set R1, R2, and R4, it's clear that this is linearly independent. And you say, why is that? Well, look, R1 has got a 1 here, while the other two have a 0 in that entry. And this is by definition of pivot entries."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's clear that the set R1, R2, and R4, it's clear that this is linearly independent. And you say, why is that? Well, look, R1 has got a 1 here, while the other two have a 0 in that entry. And this is by definition of pivot entries. Pivot entries have 0s, or pivot columns have 0s everywhere except for where they have a 1. So if for any pivot column, it will be the only pivot column that has 0s there, or it will be the only pivot column that has a 1 there. So there's no way that you can add up combinations of these guys to get a 1."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is by definition of pivot entries. Pivot entries have 0s, or pivot columns have 0s everywhere except for where they have a 1. So if for any pivot column, it will be the only pivot column that has 0s there, or it will be the only pivot column that has a 1 there. So there's no way that you can add up combinations of these guys to get a 1. You could say 100 times 0 minus 3 times 0, you're just going to get a bunch of 0s. So no combination of these two guys is going to be equal to that guy. By the same reasoning, no combination of that and that is going to equal this."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So there's no way that you can add up combinations of these guys to get a 1. You could say 100 times 0 minus 3 times 0, you're just going to get a bunch of 0s. So no combination of these two guys is going to be equal to that guy. By the same reasoning, no combination of that and that is going to equal this. This is by definition of a pivot entry. When you put it in reduced row echelon form, it's very clear that any pivot column will be the only one to have 1 in that place, so it's very clear that these guys are linearly independent. Now it turns out, and I haven't proven it to you, that the corresponding columns in A, this is R1, but in A, before we put it in reduced row echelon form, that these guys right here, so A1, A2, and A4, are also linearly independent."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "By the same reasoning, no combination of that and that is going to equal this. This is by definition of a pivot entry. When you put it in reduced row echelon form, it's very clear that any pivot column will be the only one to have 1 in that place, so it's very clear that these guys are linearly independent. Now it turns out, and I haven't proven it to you, that the corresponding columns in A, this is R1, but in A, before we put it in reduced row echelon form, that these guys right here, so A1, A2, and A4, are also linearly independent. So A1, A2, and A4, so if I write it like this, A1, A2, and A4, let me write it in set notation, these guys are also linearly independent, which I haven't proven, but I think you can kind of get a sense that these row operations really don't change the sense of the matrix. And I'll do a better explanation of this, but I really just want you to understand how to develop a basis for the column space. So they're linear independent, and so the next question is do they span our column space?"}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now it turns out, and I haven't proven it to you, that the corresponding columns in A, this is R1, but in A, before we put it in reduced row echelon form, that these guys right here, so A1, A2, and A4, are also linearly independent. So A1, A2, and A4, so if I write it like this, A1, A2, and A4, let me write it in set notation, these guys are also linearly independent, which I haven't proven, but I think you can kind of get a sense that these row operations really don't change the sense of the matrix. And I'll do a better explanation of this, but I really just want you to understand how to develop a basis for the column space. So they're linear independent, and so the next question is do they span our column space? And in order for them to span, obviously all of these 5 vectors, if you have all of them, that's going to span your column space by definition. But if we can show, and I'm not going to show it in this video, but it turns out that you can always represent the free, the non-pivot columns, so you can always represent the non-pivot columns as linear combinations of the pivot columns. And we've kind of touched on that in previous videos where we find the solution for the null space and all of that."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So they're linear independent, and so the next question is do they span our column space? And in order for them to span, obviously all of these 5 vectors, if you have all of them, that's going to span your column space by definition. But if we can show, and I'm not going to show it in this video, but it turns out that you can always represent the free, the non-pivot columns, so you can always represent the non-pivot columns as linear combinations of the pivot columns. And we've kind of touched on that in previous videos where we find the solution for the null space and all of that. So these guys can definitely be represented as linear combinations of these guys. I haven't shown you that, but if you take that as kind of on faith, then you don't need that column, that column to span. If you did, then, or I guess a better way to think of it, you don't need them to span, although they are part of the span, because if you needed this guy, you can just construct him with linear combinations of these guys."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we've kind of touched on that in previous videos where we find the solution for the null space and all of that. So these guys can definitely be represented as linear combinations of these guys. I haven't shown you that, but if you take that as kind of on faith, then you don't need that column, that column to span. If you did, then, or I guess a better way to think of it, you don't need them to span, although they are part of the span, because if you needed this guy, you can just construct him with linear combinations of these guys. So if you wanted to figure out a basis for the column space of A, you literally just take A into reduced row echelon form, you look at the pivot entries in the reduced row echelon form of A, and that's those three, and then you look at the corresponding columns to those pivot columns in your original A. And those form the basis, because any linear combination of them, or linear combinations of them, can be used to construct the non-pivot columns, and they're linearly independent. So I haven't shown you that."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you did, then, or I guess a better way to think of it, you don't need them to span, although they are part of the span, because if you needed this guy, you can just construct him with linear combinations of these guys. So if you wanted to figure out a basis for the column space of A, you literally just take A into reduced row echelon form, you look at the pivot entries in the reduced row echelon form of A, and that's those three, and then you look at the corresponding columns to those pivot columns in your original A. And those form the basis, because any linear combination of them, or linear combinations of them, can be used to construct the non-pivot columns, and they're linearly independent. So I haven't shown you that. But for this case, if you want to know the basis, it's just these three vectors, sorry, it's just A1, A2, and A4 right there, and now we can answer another question. So let me say, so the basis, so A1, A2, and A4 form a basis basis for the column space of A, because you can construct the other two guys with linear combinations of our basis vectors, and they're also linearly independent. Now, and the next question is, what is the dimension of the basis, or what is the dimension, not the dimension of the basis, what is the dimension of the column space of A?"}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I haven't shown you that. But for this case, if you want to know the basis, it's just these three vectors, sorry, it's just A1, A2, and A4 right there, and now we can answer another question. So let me say, so the basis, so A1, A2, and A4 form a basis basis for the column space of A, because you can construct the other two guys with linear combinations of our basis vectors, and they're also linearly independent. Now, and the next question is, what is the dimension of the basis, or what is the dimension, not the dimension of the basis, what is the dimension of the column space of A? Well, the dimension is just the number of vectors in any basis for the column space, and all basis have the same number of vectors for any given subspace. So we have one, two, three vectors. So the dimension of our column space is equal to 3."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, and the next question is, what is the dimension of the basis, or what is the dimension, not the dimension of the basis, what is the dimension of the column space of A? Well, the dimension is just the number of vectors in any basis for the column space, and all basis have the same number of vectors for any given subspace. So we have one, two, three vectors. So the dimension of our column space is equal to 3. And the dimension of a column space actually has a specific term for it, and that's called the rank. So the rank of A, which is the exact same thing as the dimension of the column space, it is equal to 3. And another way to think about it, it's the number, the rank of A is the number of linearly independent column vectors that you have that can span your entire column space, or the number of linearly independent column vectors that can be used to construct all of the other column vectors."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the dimension of our column space is equal to 3. And the dimension of a column space actually has a specific term for it, and that's called the rank. So the rank of A, which is the exact same thing as the dimension of the column space, it is equal to 3. And another way to think about it, it's the number, the rank of A is the number of linearly independent column vectors that you have that can span your entire column space, or the number of linearly independent column vectors that can be used to construct all of the other column vectors. But hopefully this didn't confuse you too much, because the idea is very simple. Take A, put it into reduced row echelon form, see which row columns are pivot columns, the corresponding columns are going to be basis for your column space. If you want to know the rank for your matrix, you can just count them."}, {"video_title": "Dimension of the column space or rank Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And another way to think about it, it's the number, the rank of A is the number of linearly independent column vectors that you have that can span your entire column space, or the number of linearly independent column vectors that can be used to construct all of the other column vectors. But hopefully this didn't confuse you too much, because the idea is very simple. Take A, put it into reduced row echelon form, see which row columns are pivot columns, the corresponding columns are going to be basis for your column space. If you want to know the rank for your matrix, you can just count them. Or even if you don't want to count those, you could literally just count the number of pivot columns you have in your reduced row echelon form. So that's how you do it. In the next video, I'll explain why this worked."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And I think we'll appreciate that it's a good bit more difficult, just because the math becomes a little hairier. So lambda is an eigenvalue of A, by definition, if and only if, I'll write it like this, if and only if A times some non-zero vector v is equal to lambda times that non-zero vector v. Let me write that. For some non-zero, I could call it eigenvector v, but I'll just call it for some non-zero vector v, or some non-zero v. Now this is true if and only if this leads to, I'll write it like this, this is true if and only if, and this is a bit of a review, but I like to review it just because when you do this 10 years from now, I don't want you to remember the formula. I want you to just remember the logic of how we got to it. So this is true if and only if, let's just subtract Av from both sides, the zero vector is equal to lambda. Instead of writing lambda times v, I'm going to write lambda times the identity matrix times v. This is the same thing, identity matrix times v is just v, minus Av. I just subtracted Av from both sides, rewrote v as the identity matrix times v. Well, this is only true if and only if the zero vector is equal to lambda times the identity matrix minus A times v. I just factored the vector v out from the right-hand side of both of these guys, and I'm just left with some matrix times v. Well, this is only true, let me rewrite this over here."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I want you to just remember the logic of how we got to it. So this is true if and only if, let's just subtract Av from both sides, the zero vector is equal to lambda. Instead of writing lambda times v, I'm going to write lambda times the identity matrix times v. This is the same thing, identity matrix times v is just v, minus Av. I just subtracted Av from both sides, rewrote v as the identity matrix times v. Well, this is only true if and only if the zero vector is equal to lambda times the identity matrix minus A times v. I just factored the vector v out from the right-hand side of both of these guys, and I'm just left with some matrix times v. Well, this is only true, let me rewrite this over here. This equation, just in a form you might recognize it, lambda times the identity matrix times A. This is just some matrix. This matrix times v has got to be equal to 0 for some non-zero vector v. That means that the null space of this matrix has got to be non-trivial."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I just subtracted Av from both sides, rewrote v as the identity matrix times v. Well, this is only true if and only if the zero vector is equal to lambda times the identity matrix minus A times v. I just factored the vector v out from the right-hand side of both of these guys, and I'm just left with some matrix times v. Well, this is only true, let me rewrite this over here. This equation, just in a form you might recognize it, lambda times the identity matrix times A. This is just some matrix. This matrix times v has got to be equal to 0 for some non-zero vector v. That means that the null space of this matrix has got to be non-trivial. Or another way to think about it is that its columns are not linearly independent. Or another way to think about it is it's not invertible, or it has a determinant of 0. So lambda is the eigenvalue of A if and only if each of these steps are true."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This matrix times v has got to be equal to 0 for some non-zero vector v. That means that the null space of this matrix has got to be non-trivial. Or another way to think about it is that its columns are not linearly independent. Or another way to think about it is it's not invertible, or it has a determinant of 0. So lambda is the eigenvalue of A if and only if each of these steps are true. And this is true if and only if, for some non-zero vector, the determinant of lambda times the identity matrix minus A is equal to 0. And that was our takeaway. I think it was two videos ago or three videos ago."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So lambda is the eigenvalue of A if and only if each of these steps are true. And this is true if and only if, for some non-zero vector, the determinant of lambda times the identity matrix minus A is equal to 0. And that was our takeaway. I think it was two videos ago or three videos ago. But let's apply it now to this 3 by 3 matrix A. So we're going to use the 3 by 3 identity matrix. So we want to concern ourselves, let's see, lambda times the identity matrix is just going to be times the 3 by 3 identity matrix is just going to be, let me write this, this is lambda times the identity matrix in R3."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I think it was two videos ago or three videos ago. But let's apply it now to this 3 by 3 matrix A. So we're going to use the 3 by 3 identity matrix. So we want to concern ourselves, let's see, lambda times the identity matrix is just going to be times the 3 by 3 identity matrix is just going to be, let me write this, this is lambda times the identity matrix in R3. So it's just going to be lambda, lambda, lambda. And everything else is going to be 0's. The identity matrix had 1's across here, so that's the only thing that becomes non-zero when you multiply it by lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we want to concern ourselves, let's see, lambda times the identity matrix is just going to be times the 3 by 3 identity matrix is just going to be, let me write this, this is lambda times the identity matrix in R3. So it's just going to be lambda, lambda, lambda. And everything else is going to be 0's. The identity matrix had 1's across here, so that's the only thing that becomes non-zero when you multiply it by lambda. Everything else was a 0. So that's the identity matrix times lambda. Now what is, so lambda times the identity matrix minus A is going to be equal to, it's actually pretty straightforward to find."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The identity matrix had 1's across here, so that's the only thing that becomes non-zero when you multiply it by lambda. Everything else was a 0. So that's the identity matrix times lambda. Now what is, so lambda times the identity matrix minus A is going to be equal to, it's actually pretty straightforward to find. Everything along the diagonal is going to be lambda minus, let's just do it, lambda minus minus 1, I'll do the diagonals here. Lambda minus minus 1 is lambda plus 1. And then 0 minus 2, I'll do that in a different color."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now what is, so lambda times the identity matrix minus A is going to be equal to, it's actually pretty straightforward to find. Everything along the diagonal is going to be lambda minus, let's just do it, lambda minus minus 1, I'll do the diagonals here. Lambda minus minus 1 is lambda plus 1. And then 0 minus 2, I'll do that in a different color. 0 minus 2 is minus 2. 0 minus 2 is minus 2. 0 minus 2 is minus 2."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then 0 minus 2, I'll do that in a different color. 0 minus 2 is minus 2. 0 minus 2 is minus 2. 0 minus 2 is minus 2. Let's do this one. 0 minus 2 is minus 2. 0 plus, or minus minus 1 is 0 plus 1, which is 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 2 is minus 2. Let's do this one. 0 minus 2 is minus 2. 0 plus, or minus minus 1 is 0 plus 1, which is 1. And then let's just do this one. 0 minus minus 1, that's 1. And let me finish up the diagonal."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "0 plus, or minus minus 1 is 0 plus 1, which is 1. And then let's just do this one. 0 minus minus 1, that's 1. And let me finish up the diagonal. And then you have lambda minus 2. And then you have lambda minus 2. So lambda is an eigenvalue of A if and only if the determinant of this matrix right here is equal to 0."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And let me finish up the diagonal. And then you have lambda minus 2. And then you have lambda minus 2. So lambda is an eigenvalue of A if and only if the determinant of this matrix right here is equal to 0. So let's figure out its determinant. And the easiest way, at least in my head to do this, is to use the rule of Saris. So let's use the rule of Saris to find this determinant."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So lambda is an eigenvalue of A if and only if the determinant of this matrix right here is equal to 0. So let's figure out its determinant. And the easiest way, at least in my head to do this, is to use the rule of Saris. So let's use the rule of Saris to find this determinant. So I just rewrite these rows right there. I could just copy and paste them, really. So I just take those two rows, and then let me paste them, put them right there."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's use the rule of Saris to find this determinant. So I just rewrite these rows right there. I could just copy and paste them, really. So I just take those two rows, and then let me paste them, put them right there. It's a little bit too close to this guy, but I think you get the idea. And now the rule of Saris, I just take this product plus this product plus this product. And then I subtract out this product times this product times this product."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I just take those two rows, and then let me paste them, put them right there. It's a little bit too close to this guy, but I think you get the idea. And now the rule of Saris, I just take this product plus this product plus this product. And then I subtract out this product times this product times this product. We'll do that next. So this product is lambda plus 1 times lambda minus 2 times lambda minus 2. That's that one there."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then I subtract out this product times this product times this product. We'll do that next. So this product is lambda plus 1 times lambda minus 2 times lambda minus 2. That's that one there. And then plus, you see, minus 2 times minus 2. That's plus 4. And then we have minus 2 times minus 2 plus 4 times 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's that one there. And then plus, you see, minus 2 times minus 2. That's plus 4. And then we have minus 2 times minus 2 plus 4 times 1. So that is plus 4 again. And then we do minus this column times this column. Or minus this column minus this column."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then we have minus 2 times minus 2 plus 4 times 1. So that is plus 4 again. And then we do minus this column times this column. Or minus this column minus this column. And then, or I shouldn't say column, but diagonal, really. So we say minus, you see, minus 2 times minus 2. Let me write this."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or minus this column minus this column. And then, or I shouldn't say column, but diagonal, really. So we say minus, you see, minus 2 times minus 2. Let me write this. Minus minus 2 times minus 2, which is 4 times lambda minus 2, that was this diagonal. And then we have minus, what is this going to be? It's going to be minus 1 times lambda plus 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this. Minus minus 2 times minus 2, which is 4 times lambda minus 2, that was this diagonal. And then we have minus, what is this going to be? It's going to be minus 1 times lambda plus 1. So minus lambda plus 1. And then you go down this diagonal. See, minus 2 times minus 2 is 4."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be minus 1 times lambda plus 1. So minus lambda plus 1. And then you go down this diagonal. See, minus 2 times minus 2 is 4. So it's going to be 4 times lambda minus 2. And we're subtracting. So minus 4 times lambda minus 2."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "See, minus 2 times minus 2 is 4. So it's going to be 4 times lambda minus 2. And we're subtracting. So minus 4 times lambda minus 2. And let's see if we can simplify this. So this blue stuff over here. Let's see, these guys right here become an 8."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So minus 4 times lambda minus 2. And let's see if we can simplify this. So this blue stuff over here. Let's see, these guys right here become an 8. And then this becomes lambda plus 1 times, if I multiply these two guys out, lambda squared minus 4 lambda. Right? Minus 2 lambda, then minus 2 lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, these guys right here become an 8. And then this becomes lambda plus 1 times, if I multiply these two guys out, lambda squared minus 4 lambda. Right? Minus 2 lambda, then minus 2 lambda. So minus 4 lambda plus 4. And then I have this plus 8 here. And then I have minus 4 times lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 lambda, then minus 2 lambda. So minus 4 lambda plus 4. And then I have this plus 8 here. And then I have minus 4 times lambda. Let me just multiply everything out. So I have minus 4 lambda plus 8 minus lambda minus 1 minus 4 lambda plus 8. And then let me simplify this up a little bit."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then I have minus 4 times lambda. Let me just multiply everything out. So I have minus 4 lambda plus 8 minus lambda minus 1 minus 4 lambda plus 8. And then let me simplify this up a little bit. So this guy over here, let's see, constant terms. I have an 8. I have a minus 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then let me simplify this up a little bit. So this guy over here, let's see, constant terms. I have an 8. I have a minus 1. I have an 8. And I have an 8. So that's 24 minus 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I have a minus 1. I have an 8. And I have an 8. So that's 24 minus 1. So that is a 23. And then the lambda terms. I have a minus 4 lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that's 24 minus 1. So that is a 23. And then the lambda terms. I have a minus 4 lambda. I have a minus lambda. I have a minus 4 lambda. So that's minus 8 minus 1."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I have a minus 4 lambda. I have a minus lambda. I have a minus 4 lambda. So that's minus 8 minus 1. So I have minus 9 lambda. Minus 9 lambda plus 23. And now I have to simplify this out."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that's minus 8 minus 1. So I have minus 9 lambda. Minus 9 lambda plus 23. And now I have to simplify this out. So first I can take lambda and multiply it times this whole guy right there. So it's going to be lambda cubed minus 4 lambda squared plus 4 lambda. And then I can take this 1 and multiply it times that guy."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And now I have to simplify this out. So first I can take lambda and multiply it times this whole guy right there. So it's going to be lambda cubed minus 4 lambda squared plus 4 lambda. And then I can take this 1 and multiply it times that guy. So plus lambda squared minus 4 lambda plus 4. And now, of course, we have these terms over here. So we're going to have to simplify it again."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then I can take this 1 and multiply it times that guy. So plus lambda squared minus 4 lambda plus 4. And now, of course, we have these terms over here. So we're going to have to simplify it again. So what are all of our constant terms? We have a 23 and we have a plus 4. So we have a 27, a plus 27."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to have to simplify it again. So what are all of our constant terms? We have a 23 and we have a plus 4. So we have a 27, a plus 27. And then what are all of our lambda terms? We have a minus 9 lambda. And then we have a plus 4 lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we have a 27, a plus 27. And then what are all of our lambda terms? We have a minus 9 lambda. And then we have a plus 4 lambda. And then we have a minus 4 lambda. So these two cancel out. So I just have a minus 9 lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then we have a plus 4 lambda. And then we have a minus 4 lambda. So these two cancel out. So I just have a minus 9 lambda. And then what are my lambda squared terms? I have a plus lambda squared and I have a minus 4 lambda squared. So if you add those two, it's going to be minus 3 lambda squared."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I just have a minus 9 lambda. And then what are my lambda squared terms? I have a plus lambda squared and I have a minus 4 lambda squared. So if you add those two, it's going to be minus 3 lambda squared. Minus 3 lambda squared. And then finally I have only one lambda cubed term, that right there. So this is the characteristic polynomial for our matrix."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if you add those two, it's going to be minus 3 lambda squared. Minus 3 lambda squared. And then finally I have only one lambda cubed term, that right there. So this is the characteristic polynomial for our matrix. So this is the characteristic polynomial. And this represents the determinant for any lambda. The determinant of this matrix for any lambda."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is the characteristic polynomial for our matrix. So this is the characteristic polynomial. And this represents the determinant for any lambda. The determinant of this matrix for any lambda. And we said that this has to be equal to 0 if and only if lambda is truly an eigenvalue. So we're going to have to set this equal to 0. And unlucky or lucky for us, there is no real trivial, there's no quadratic."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The determinant of this matrix for any lambda. And we said that this has to be equal to 0 if and only if lambda is truly an eigenvalue. So we're going to have to set this equal to 0. And unlucky or lucky for us, there is no real trivial, there's no quadratic. Well, there is actually for a trivial, but it's very complicated and so it's usually a waste of time. So we're going to have to do kind of the art of factoring a quadratic polynomial. I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class, or really in an algebra class generally, it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And unlucky or lucky for us, there is no real trivial, there's no quadratic. Well, there is actually for a trivial, but it's very complicated and so it's usually a waste of time. So we're going to have to do kind of the art of factoring a quadratic polynomial. I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class, or really in an algebra class generally, it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions. And if you are dealing with integer solutions, then your roots are going to be factors of this term right here. Especially if you have a 1 coefficient out here. So your potential roots in this case, what are the factors of 27?"}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class, or really in an algebra class generally, it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions. And if you are dealing with integer solutions, then your roots are going to be factors of this term right here. Especially if you have a 1 coefficient out here. So your potential roots in this case, what are the factors of 27? So 1, 3, 9, and 27. So all of these are potential roots. So we could just try them out."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So your potential roots in this case, what are the factors of 27? So 1, 3, 9, and 27. So all of these are potential roots. So we could just try them out. 1 cubed is 1 minus 3. Let me try 1. If we try 1, it's 1 minus 3 minus 9 plus 27."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we could just try them out. 1 cubed is 1 minus 3. Let me try 1. If we try 1, it's 1 minus 3 minus 9 plus 27. That does not equal 0. I don't even feel like it's minus 2, minus 9 is minus 11, plus it's 16, that does not equal 0. So 1 is not a root."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "If we try 1, it's 1 minus 3 minus 9 plus 27. That does not equal 0. I don't even feel like it's minus 2, minus 9 is minus 11, plus it's 16, that does not equal 0. So 1 is not a root. If we try 3, we get 3 cubed, which is 27, minus 3 times 3 squared, which is minus 3 times 3, which is minus 27, minus 9 times 3, which is minus 27, plus 27. That does equal 0. So lucky for us, on our second try, we were able to find 1 0 for this."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 1 is not a root. If we try 3, we get 3 cubed, which is 27, minus 3 times 3 squared, which is minus 3 times 3, which is minus 27, minus 9 times 3, which is minus 27, plus 27. That does equal 0. So lucky for us, on our second try, we were able to find 1 0 for this. So if 3 is a 0, that means that x minus 3 is one of the factors of this. So that means that this is going to be x minus 3 times something else, or I should say lambda minus 3. So let's see what the other root is."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So lucky for us, on our second try, we were able to find 1 0 for this. So if 3 is a 0, that means that x minus 3 is one of the factors of this. So that means that this is going to be x minus 3 times something else, or I should say lambda minus 3. So let's see what the other root is. So if I take lambda minus 3 and I divide it into this guy up here, into lambda cubed minus 3 lambda squared minus 9 lambda plus 27, what do I get? See, lambda goes into lambda cubed lambda squared times. Lambda squared times that, lambda squared times lambda, is lambda cubed."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's see what the other root is. So if I take lambda minus 3 and I divide it into this guy up here, into lambda cubed minus 3 lambda squared minus 9 lambda plus 27, what do I get? See, lambda goes into lambda cubed lambda squared times. Lambda squared times that, lambda squared times lambda, is lambda cubed. Lambda squared times minus 3 is minus 3 lambda squared. You subtract these guys, you get a 0. You get 0."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Lambda squared times that, lambda squared times lambda, is lambda cubed. Lambda squared times minus 3 is minus 3 lambda squared. You subtract these guys, you get a 0. You get 0. And then we can put here, well, we could do it either way, we could put it down the minus 9, we could bring down everything, really. So now you have minus 9 lambda plus 27. You can almost imagine we just subtracted this from this whole thing up here and we're just left with these terms right here."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You get 0. And then we can put here, well, we could do it either way, we could put it down the minus 9, we could bring down everything, really. So now you have minus 9 lambda plus 27. You can almost imagine we just subtracted this from this whole thing up here and we're just left with these terms right here. And so lambda minus 3 goes into this. Well, lambda minus 3 goes into 9 lambda minus 9 times. So I'll just write a minus 9 here."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You can almost imagine we just subtracted this from this whole thing up here and we're just left with these terms right here. And so lambda minus 3 goes into this. Well, lambda minus 3 goes into 9 lambda minus 9 times. So I'll just write a minus 9 here. Minus 9 times lambda minus 3 is minus 9 lambda plus 27. So it went in very nicely. So you get to 0."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I'll just write a minus 9 here. Minus 9 times lambda minus 3 is minus 9 lambda plus 27. So it went in very nicely. So you get to 0. So this, our characteristic polynomial has simplified to lambda minus 3 times lambda squared minus 9. And of course, we're going to have to set this equal to 0 if lambda is truly an eigenvalue of our matrix. And this is very easy to factor."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you get to 0. So this, our characteristic polynomial has simplified to lambda minus 3 times lambda squared minus 9. And of course, we're going to have to set this equal to 0 if lambda is truly an eigenvalue of our matrix. And this is very easy to factor. So this becomes lambda minus 3 times lambda squared minus 9 is just lambda plus 3 times lambda minus 3. And all of that equals 0. And these roots, we already know one of them."}, {"video_title": "Eigenvalues of a 3x3 matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And this is very easy to factor. So this becomes lambda minus 3 times lambda squared minus 9 is just lambda plus 3 times lambda minus 3. And all of that equals 0. And these roots, we already know one of them. We know that 3 is a root. And actually, this tells us that 3 is a root as well. So the possible eigenvalues of our matrix A, our 3 by 3 matrix A that we had way up there, this matrix A right there, the possible eigenvalues are lambda is equal to 3 or lambda is equal to minus 3."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What I want to do is figure out whether matrix products exhibit the distributive property. So let's test out A times B plus C. And of course, these are all matrices. So B, just to make things clear, the matrix B could be represented as just a bunch of column vectors. B1, B2, all the way to Bn. And the matrix C can also be represented as just a bunch of column vectors. So could be the matrix A, but I don't have to draw that just yet. So the matrix C could just be represented as a bunch of column vectors."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "B1, B2, all the way to Bn. And the matrix C can also be represented as just a bunch of column vectors. So could be the matrix A, but I don't have to draw that just yet. So the matrix C could just be represented as a bunch of column vectors. C1, C2, all the way to Cn. Maybe I should have drawn this taller. These are column vectors, so they actually have some verticality to them."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the matrix C could just be represented as a bunch of column vectors. C1, C2, all the way to Cn. Maybe I should have drawn this taller. These are column vectors, so they actually have some verticality to them. I think you've seen that multiple times. So what is A times B plus C? Well, let's figure out what B plus C is."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are column vectors, so they actually have some verticality to them. I think you've seen that multiple times. So what is A times B plus C? Well, let's figure out what B plus C is. This is equal to A times B plus C. When you add B plus C, the definition of matrix addition is you just add the corresponding columns, which essentially boils down to adding the corresponding entries. So that is going to be equal to the first column is going to be equal to B1 plus C1. The second column is going to be B2 plus C2."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, let's figure out what B plus C is. This is equal to A times B plus C. When you add B plus C, the definition of matrix addition is you just add the corresponding columns, which essentially boils down to adding the corresponding entries. So that is going to be equal to the first column is going to be equal to B1 plus C1. The second column is going to be B2 plus C2. And you're going to go all the way to the nth column is going to be Bn plus Cn. Now, by our definition of matrix-matrix products, this is this product right here is going to be equal to the matrix where we take the matrix A and multiply it by each of the column vectors of this matrix here of B plus C. Which, as you can imagine, these are both m by n. In fact, they both have to have the same dimensions for this addition to be well-defined. So this is going to be an m by n matrix."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The second column is going to be B2 plus C2. And you're going to go all the way to the nth column is going to be Bn plus Cn. Now, by our definition of matrix-matrix products, this is this product right here is going to be equal to the matrix where we take the matrix A and multiply it by each of the column vectors of this matrix here of B plus C. Which, as you can imagine, these are both m by n. In fact, they both have to have the same dimensions for this addition to be well-defined. So this is going to be an m by n matrix. I already told you that A is a k by m. And we know this is well-defined because A has the same number of columns as B plus C has of rows. So this is well-defined. And this is going to be equal to A times the column vector B1 plus C1."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be an m by n matrix. I already told you that A is a k by m. And we know this is well-defined because A has the same number of columns as B plus C has of rows. So this is well-defined. And this is going to be equal to A times the column vector B1 plus C1. The second column is going to be A times the column vector B2 plus C2. Running out of space. It's going to be all the way to A times the column vector Bn plus Cn."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is going to be equal to A times the column vector B1 plus C1. The second column is going to be A times the column vector B2 plus C2. Running out of space. It's going to be all the way to A times the column vector Bn plus Cn. This is our definition of matrix-matrix products. You just take the first matrix and you multiply it times each of the column vectors of the second matrix. And we can say that because we've already defined matrix-vector products."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be all the way to A times the column vector Bn plus Cn. This is our definition of matrix-matrix products. You just take the first matrix and you multiply it times each of the column vectors of the second matrix. And we can say that because we've already defined matrix-vector products. So what is this thing on the right equal to? I'll keep switching colors. This is equal to, we know that matrix-vector products exhibit the distributive property."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we can say that because we've already defined matrix-vector products. So what is this thing on the right equal to? I'll keep switching colors. This is equal to, we know that matrix-vector products exhibit the distributive property. I don't even remember when I did that video, but we've assumed it for a while. It's a very trivial thing to prove. So each of these columns are going to be equal to, let me write it this way."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, we know that matrix-vector products exhibit the distributive property. I don't even remember when I did that video, but we've assumed it for a while. It's a very trivial thing to prove. So each of these columns are going to be equal to, let me write it this way. This guy right here can be rewritten. The first column is going to be A times the column vector B1 plus A times the column vector C1. This term right there is the same thing as that term right there."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So each of these columns are going to be equal to, let me write it this way. This guy right here can be rewritten. The first column is going to be A times the column vector B1 plus A times the column vector C1. This term right there is the same thing as that term right there. The next one is going to be AB2 plus matrix A times the vector C2. And then the nth column is going to be the matrix, so we have to keep going. And then the nth column is going to be the matrix A times the column vector Bn plus matrix A times the column vector Cn, just like that."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This term right there is the same thing as that term right there. The next one is going to be AB2 plus matrix A times the vector C2. And then the nth column is going to be the matrix, so we have to keep going. And then the nth column is going to be the matrix A times the column vector Bn plus matrix A times the column vector Cn, just like that. Now we can write this matrix as the sum of two different matrices. So what is this going to be equal to? This is equal to, let me see, I'll just write it right here."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the nth column is going to be the matrix A times the column vector Bn plus matrix A times the column vector Cn, just like that. Now we can write this matrix as the sum of two different matrices. So what is this going to be equal to? This is equal to, let me see, I'll just write it right here. This is equal to AB1 as the first column, AB2 as the second column, all the way to ABn as the third column. So that's these terms right there. And if I were to add to that the matrix A times vector C1, A times the column vector C2, these are just the different columns of this matrix."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, let me see, I'll just write it right here. This is equal to AB1 as the first column, AB2 as the second column, all the way to ABn as the third column. So that's these terms right there. And if I were to add to that the matrix A times vector C1, A times the column vector C2, these are just the different columns of this matrix. And we just then have the matrix A times the column vector Cn. These represent these terms. So clearly if I add these two matrices, I just add the corresponding column vectors, and I'll get this matrix up here."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if I were to add to that the matrix A times vector C1, A times the column vector C2, these are just the different columns of this matrix. And we just then have the matrix A times the column vector Cn. These represent these terms. So clearly if I add these two matrices, I just add the corresponding column vectors, and I'll get this matrix up here. But what is this equal to? This right here, by definition, this is the matrix A times the matrix B. The definition of matrix products is you take the first matrix and multiply it times the column vectors of the second matrix."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So clearly if I add these two matrices, I just add the corresponding column vectors, and I'll get this matrix up here. But what is this equal to? This right here, by definition, this is the matrix A times the matrix B. The definition of matrix products is you take the first matrix and multiply it times the column vectors of the second matrix. And by the same argument, I guess you could say, this is equivalent to A times C. And all of this, remember, we just had a bunch of equal signs, is equal to A times B plus C. So now we can say definitively that as long as the products are well-defined and the sums are well-defined, so they all have to have the correct dimensions, that A times B plus C is equal to AB plus AC. So matrix products do exhibit the distributive property, at least in this direction. And I say that because, remember, matrix products are not commutative."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The definition of matrix products is you take the first matrix and multiply it times the column vectors of the second matrix. And by the same argument, I guess you could say, this is equivalent to A times C. And all of this, remember, we just had a bunch of equal signs, is equal to A times B plus C. So now we can say definitively that as long as the products are well-defined and the sums are well-defined, so they all have to have the correct dimensions, that A times B plus C is equal to AB plus AC. So matrix products do exhibit the distributive property, at least in this direction. And I say that because, remember, matrix products are not commutative. So we don't know necessarily that B plus C times A is equivalent to that. In fact, most of the times, these two things are not equivalent. So we don't know quite yet that if we reversed this, whether it's still going to exhibit the distributive property."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I say that because, remember, matrix products are not commutative. So we don't know necessarily that B plus C times A is equivalent to that. In fact, most of the times, these two things are not equivalent. So we don't know quite yet that if we reversed this, whether it's still going to exhibit the distributive property. So let's try to do that. And I'll do a little bit quicker, because I think you know the general argument here. So let's take B plus C times A."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we don't know quite yet that if we reversed this, whether it's still going to exhibit the distributive property. So let's try to do that. And I'll do a little bit quicker, because I think you know the general argument here. So let's take B plus C times A. And I'll just write A as its column vectors. A1, A2, all the way to A has m columns, if I remember correctly. A has m columns."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's take B plus C times A. And I'll just write A as its column vectors. A1, A2, all the way to A has m columns, if I remember correctly. A has m columns. So all the way to A m. And by the definition of matrix products, this is going to be equal to the matrix. B plus C is just a matrix. We can represent it as a sum of two matrix, but this is just a matrix."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A has m columns. So all the way to A m. And by the definition of matrix products, this is going to be equal to the matrix. B plus C is just a matrix. We can represent it as a sum of two matrix, but this is just a matrix. So it's B plus C times each of the column vectors of A. So it's going to be equal to B plus C times A1, B plus C times A2, all the way to B plus C times A n. And I once again, it was many videos ago, that I think we showed that matrix vector products are distributive. So we can just distribute this vector along these two matrices."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can represent it as a sum of two matrix, but this is just a matrix. So it's B plus C times each of the column vectors of A. So it's going to be equal to B plus C times A1, B plus C times A2, all the way to B plus C times A n. And I once again, it was many videos ago, that I think we showed that matrix vector products are distributive. So we can just distribute this vector along these two matrices. And if I haven't proven it yet, it's actually a very straightforward proof to do. So we could say that this is equal to B A1 plus C A2. That's the first column."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can just distribute this vector along these two matrices. And if I haven't proven it yet, it's actually a very straightforward proof to do. So we could say that this is equal to B A1 plus C A2. That's the first column. The second column is B times A2 plus C times A2, all the way to B times A n plus C times A n. And then what is this equal to? Well, I think we can skip. Well, I'll write it out."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the first column. The second column is B times A2 plus C times A2, all the way to B times A n plus C times A n. And then what is this equal to? Well, I think we can skip. Well, I'll write it out. This is equal to B times A. This is A1. Oh, this is an A1 right here."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, I'll write it out. This is equal to B times A. This is A1. Oh, this is an A1 right here. B times A1, and then B times A2, all the way to B A n, plus the vector C times A1, C times A2, all the way to C times A n. This guy represents these terms right there. This guy represents the first terms in each of these column vectors. And this, by the definition of matrix products, is just equivalent to B A."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Oh, this is an A1 right here. B times A1, and then B times A2, all the way to B A n, plus the vector C times A1, C times A2, all the way to C times A n. This guy represents these terms right there. This guy represents the first terms in each of these column vectors. And this, by the definition of matrix products, is just equivalent to B A. And then this is just equivalent to C A. So now we've seen that the distributive property works both ways with matrix vector products. That B plus C times A is equal to B A plus C A."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this, by the definition of matrix products, is just equivalent to B A. And then this is just equivalent to C A. So now we've seen that the distributive property works both ways with matrix vector products. That B plus C times A is equal to B A plus C A. And that A times B plus C is equal to A B plus A C. Now the one thing that you have to be careful of is that these two things are not equivalent to each other. We just figured out that this guy is equal to B A plus C A. So the distribution works in both ways."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That B plus C times A is equal to B A plus C A. And that A times B plus C is equal to A B plus A C. Now the one thing that you have to be careful of is that these two things are not equivalent to each other. We just figured out that this guy is equal to B A plus C A. So the distribution works in both ways. But when you're dealing with matrices, it's very important to keep your order. So that this is going to be, you have the A second here. So it's B A plus C A."}, {"video_title": "Distributive property of matrix products Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the distribution works in both ways. But when you're dealing with matrices, it's very important to keep your order. So that this is going to be, you have the A second here. So it's B A plus C A. You can't say that this is equal to A B plus A C. You can't just switch these up, because we've shown multiple times, or we've talked about it multiple times, that matrix products are not commutative. You can't just switch the order of the products. But we've at least shown in this video that the distributive property works both ways."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But, to be frank, it is somewhat mundane, but I'm doing it for two reasons. One is, this is the type of thing that's often asked of you when you take a linear algebra class. But more importantly, it gives you the appreciation that we really are kind of building up a mathematics of vectors from the ground up. You really can't assume anything. You really need to prove everything for yourself. So the first thing I want to prove is that the dot product, that when you take the vector dot product, if I take v dot w, that it's commutative. That the order that I take the dot product doesn't matter."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You really can't assume anything. You really need to prove everything for yourself. So the first thing I want to prove is that the dot product, that when you take the vector dot product, if I take v dot w, that it's commutative. That the order that I take the dot product doesn't matter. I want to prove to myself that that is equal to w dot v. And so how do we do that? Well, and this is the general pattern for a lot of these vector proofs. Write out the vectors."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That the order that I take the dot product doesn't matter. I want to prove to myself that that is equal to w dot v. And so how do we do that? Well, and this is the general pattern for a lot of these vector proofs. Write out the vectors. So v will look like v1, v2, all the way down to vn. Let's say that this is equal to v. And let's say that w is equal to w1, w2, all the way down to wn. So what does v dot w equal?"}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Write out the vectors. So v will look like v1, v2, all the way down to vn. Let's say that this is equal to v. And let's say that w is equal to w1, w2, all the way down to wn. So what does v dot w equal? v dot w is equal to, I'll switch colors here, v1 times w1, whoops, v1w1 plus v2w2 plus all the way to vnwn. Fair enough. Now what does w dot v equal?"}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what does v dot w equal? v dot w is equal to, I'll switch colors here, v1 times w1, whoops, v1w1 plus v2w2 plus all the way to vnwn. Fair enough. Now what does w dot v equal? Well, w dot v, when I made the definition, you just multiply the products, but I'll just do it in the order that they gave it to us. So it equals w1v1 plus w2v2 plus all the way to wnvn. Now, these are clearly equal to each other, because if you just match up the first term with the first term, those are clearly equal to each other."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what does w dot v equal? Well, w dot v, when I made the definition, you just multiply the products, but I'll just do it in the order that they gave it to us. So it equals w1v1 plus w2v2 plus all the way to wnvn. Now, these are clearly equal to each other, because if you just match up the first term with the first term, those are clearly equal to each other. So v1w1 is equal to w1v1, and I can say this now, because now we're just dealing with regular numbers. Here we were dealing with vectors, and we were taking this weird type of multiplication called the dot product, but now I can definitely say that these are equal, because this is just regular multiplication, and this is just a commutative property. Commutative property of multiplication."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, these are clearly equal to each other, because if you just match up the first term with the first term, those are clearly equal to each other. So v1w1 is equal to w1v1, and I can say this now, because now we're just dealing with regular numbers. Here we were dealing with vectors, and we were taking this weird type of multiplication called the dot product, but now I can definitely say that these are equal, because this is just regular multiplication, and this is just a commutative property. Commutative property of multiplication. We learned this in, I don't know, when you learned this, in second or third grade. So you know that those are equal, and by the same argument, you know that these two are equal, and you can just rewrite each of these terms just by switching that around. That's just from basic multiplication of scalar numbers, of just regular real numbers."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Commutative property of multiplication. We learned this in, I don't know, when you learned this, in second or third grade. So you know that those are equal, and by the same argument, you know that these two are equal, and you can just rewrite each of these terms just by switching that around. That's just from basic multiplication of scalar numbers, of just regular real numbers. So that's what tells us that these two things are equal, or these two things are equal. So we've proven to ourselves that order doesn't matter when you take the dot product. Now the next thing we could take a look at is whether the dot product exhibits the distributive property."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's just from basic multiplication of scalar numbers, of just regular real numbers. So that's what tells us that these two things are equal, or these two things are equal. So we've proven to ourselves that order doesn't matter when you take the dot product. Now the next thing we could take a look at is whether the dot product exhibits the distributive property. So let me just define another vector x here. Another vector x, and you can imagine how I'm going to define it. x1, x2, all the way down to xn."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now the next thing we could take a look at is whether the dot product exhibits the distributive property. So let me just define another vector x here. Another vector x, and you can imagine how I'm going to define it. x1, x2, all the way down to xn. Now, what I want to see, if the dot product deals with the distributive property the way I would expect it to, then if I were to add v plus w, and then multiply that by x, and first of all, it shouldn't matter what order I do that with. I just showed it here. I could do x dot this thing."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x1, x2, all the way down to xn. Now, what I want to see, if the dot product deals with the distributive property the way I would expect it to, then if I were to add v plus w, and then multiply that by x, and first of all, it shouldn't matter what order I do that with. I just showed it here. I could do x dot this thing. It shouldn't matter, because I just showed you it's commutative. But if the distribution works, then this should be the same thing as v dot x plus w dot x. If these were just numbers, this was just regular multiplication, and this was multiplied by each of the terms, and that's what I'm showing here."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could do x dot this thing. It shouldn't matter, because I just showed you it's commutative. But if the distribution works, then this should be the same thing as v dot x plus w dot x. If these were just numbers, this was just regular multiplication, and this was multiplied by each of the terms, and that's what I'm showing here. So let's see if this is true for the dot product. So what is v plus w? v plus w is equal to, we just add up each of their corresponding terms, v1 plus w1, v2 plus w2, all the way down to vn plus wn."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If these were just numbers, this was just regular multiplication, and this was multiplied by each of the terms, and that's what I'm showing here. So let's see if this is true for the dot product. So what is v plus w? v plus w is equal to, we just add up each of their corresponding terms, v1 plus w1, v2 plus w2, all the way down to vn plus wn. That's that right there. And then when we dot that, with x1, x2, all the way down to xn, what do we get? Well, we get v1 plus w1 times x1, plus v2 plus w2 times x2, plus all the way to vn plus wn times xn."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "v plus w is equal to, we just add up each of their corresponding terms, v1 plus w1, v2 plus w2, all the way down to vn plus wn. That's that right there. And then when we dot that, with x1, x2, all the way down to xn, what do we get? Well, we get v1 plus w1 times x1, plus v2 plus w2 times x2, plus all the way to vn plus wn times xn. I just took the dot product of these two, I just multiplied corresponding components, and then added them all up. That was the dot product. This is v plus w dot x."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we get v1 plus w1 times x1, plus v2 plus w2 times x2, plus all the way to vn plus wn times xn. I just took the dot product of these two, I just multiplied corresponding components, and then added them all up. That was the dot product. This is v plus w dot x. Let me write that down. This is v plus w dot x. Now, let's work on these things up here."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is v plus w dot x. Let me write that down. This is v plus w dot x. Now, let's work on these things up here. What is v dot, let me write it over here. What is v dot x? v dot x, we've seen this before, this is just v1, x1, no vectors now, these are just their actual components, plus v2, x2, all the way to vn, xn."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's work on these things up here. What is v dot, let me write it over here. What is v dot x? v dot x, we've seen this before, this is just v1, x1, no vectors now, these are just their actual components, plus v2, x2, all the way to vn, xn. What is w dot x? w dot x is equal to w1, x1, plus w2, x2, all the way to wn, xn. What do you get when you add these two things?"}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "v dot x, we've seen this before, this is just v1, x1, no vectors now, these are just their actual components, plus v2, x2, all the way to vn, xn. What is w dot x? w dot x is equal to w1, x1, plus w2, x2, all the way to wn, xn. What do you get when you add these two things? Notice, I'm just adding, here I'm adding two scalar quantities. That's a scalar, that's a scalar. We're not doing vector addition anymore."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What do you get when you add these two things? Notice, I'm just adding, here I'm adding two scalar quantities. That's a scalar, that's a scalar. We're not doing vector addition anymore. This is a scalar quantity, and this is a scalar quantity. What do I get when I add them? v dot x plus w dot x is equal to v1, x1, plus w1, x1, plus v2, x2, plus w2, x2, all the way to vn, xn, plus wn, xn."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We're not doing vector addition anymore. This is a scalar quantity, and this is a scalar quantity. What do I get when I add them? v dot x plus w dot x is equal to v1, x1, plus w1, x1, plus v2, x2, plus w2, x2, all the way to vn, xn, plus wn, xn. I know, it's very monotonous, but you can immediately see we're just dealing with regular numbers here. We can take the x's out, and what do you get? This is equal to, let me write it here, this is equal to, we can just take the x out, factor the x out, v1 plus w1, x1, plus v2, plus w2, x2, all the way to vn, plus wn, xn, which we see this is the same thing as this thing right here."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "v dot x plus w dot x is equal to v1, x1, plus w1, x1, plus v2, x2, plus w2, x2, all the way to vn, xn, plus wn, xn. I know, it's very monotonous, but you can immediately see we're just dealing with regular numbers here. We can take the x's out, and what do you get? This is equal to, let me write it here, this is equal to, we can just take the x out, factor the x out, v1 plus w1, x1, plus v2, plus w2, x2, all the way to vn, plus wn, xn, which we see this is the same thing as this thing right here. We just showed that this expression right here is the same thing as that expression, or that distribution, the distributive property, seems to or does apply the way we would expect to the dot product. I know this is so mundane, why are we doing this? I'm doing this to show you that we're building things up."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, let me write it here, this is equal to, we can just take the x out, factor the x out, v1 plus w1, x1, plus v2, plus w2, x2, all the way to vn, plus wn, xn, which we see this is the same thing as this thing right here. We just showed that this expression right here is the same thing as that expression, or that distribution, the distributive property, seems to or does apply the way we would expect to the dot product. I know this is so mundane, why are we doing this? I'm doing this to show you that we're building things up. We couldn't just assume this, but the proof is pretty straightforward. In general, I didn't do these proofs when I did it for vector addition and scalar multiplication, and I really should have, but you can prove the commutativity of it, or for the scalar multiplication, you could prove that distribution works for it, doing a proof exactly the same way as this. A lot of math books or linear algebra books just leave these as exercises to the student because it's mundane, so they didn't think it was worth their paper."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm doing this to show you that we're building things up. We couldn't just assume this, but the proof is pretty straightforward. In general, I didn't do these proofs when I did it for vector addition and scalar multiplication, and I really should have, but you can prove the commutativity of it, or for the scalar multiplication, you could prove that distribution works for it, doing a proof exactly the same way as this. A lot of math books or linear algebra books just leave these as exercises to the student because it's mundane, so they didn't think it was worth their paper. Let me just show you the last property. Let me just say associativity, the associative property. That doesn't matter."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A lot of math books or linear algebra books just leave these as exercises to the student because it's mundane, so they didn't think it was worth their paper. Let me just show you the last property. Let me just say associativity, the associative property. That doesn't matter. Let me show you. If I take some scalar, and I multiply it times v, some vector v, and then I take the dot product of that with w, if this is associative, the way multiplication in our everyday world normally works, this should be equal to, and it's still a question mark because I haven't proven it to you, it should be equal to c times v dot w. Let's figure it out. What's c times the vector v?"}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That doesn't matter. Let me show you. If I take some scalar, and I multiply it times v, some vector v, and then I take the dot product of that with w, if this is associative, the way multiplication in our everyday world normally works, this should be equal to, and it's still a question mark because I haven't proven it to you, it should be equal to c times v dot w. Let's figure it out. What's c times the vector v? c times the vector v is c times v1, c times v2, all the way down to c times vn, and then the vector w, we already know what that is, so dot w is equal to what? It's equal to this times the first term of w, so cv1w1, plus this times the second term of w, cv2w2, all the way to cvnwn. Fair enough."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What's c times the vector v? c times the vector v is c times v1, c times v2, all the way down to c times vn, and then the vector w, we already know what that is, so dot w is equal to what? It's equal to this times the first term of w, so cv1w1, plus this times the second term of w, cv2w2, all the way to cvnwn. Fair enough. That's what this side is equal to. Now let's do this side. What is v dot w?"}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. That's what this side is equal to. Now let's do this side. What is v dot w? I'll write it here. v dot w, we've done this multiple times, this is just v1w1 plus v2w2, all the way to vnwn. I'm getting tired of doing this, and you're probably tired of watching it, but it's good to go through the exercises."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is v dot w? I'll write it here. v dot w, we've done this multiple times, this is just v1w1 plus v2w2, all the way to vnwn. I'm getting tired of doing this, and you're probably tired of watching it, but it's good to go through the exercises. If someone asks you to do this, now you'll be able to do this. Now what is c times this? If I multiply some scalar times this, that's the same thing as multiplying some scalar times that."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm getting tired of doing this, and you're probably tired of watching it, but it's good to go through the exercises. If someone asks you to do this, now you'll be able to do this. Now what is c times this? If I multiply some scalar times this, that's the same thing as multiplying some scalar times that. I'm just multiplying a scalar times a big, this is just the regular distributive property of just numbers, of just regular real numbers. This is going to be equal to cv1w1, plus cv2w2, plus all the way to cvnwn. We see that this is equal to this, because this is equal to this."}, {"video_title": "Proving vector dot product properties Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply some scalar times this, that's the same thing as multiplying some scalar times that. I'm just multiplying a scalar times a big, this is just the regular distributive property of just numbers, of just regular real numbers. This is going to be equal to cv1w1, plus cv2w2, plus all the way to cvnwn. We see that this is equal to this, because this is equal to this. Now the hardest part of this, I remember when I first took linear algebra, I found when the professor would assign, prove this, I would have trouble doing it because it almost seemed so ridiculously obvious. Obviously, if you just look at the components of them, it just turns into multiplying of each individual component and adding them up, and those are associative, so that's obviously, so what's there to prove? It only took me a little while, they just wanted me to write that down."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think you're pretty familiar with the idea of matrix vector products. What I want to do in this video is show you that taking a product of a vector with a matrix is equivalent to a transformation. It's actually a linear transformation. Let's say we have some matrix A. And let's say that its columns are V1, their column vector is V2, all the way to Vn. So this guy has n columns, and let's say he has m rows, so it's an m by n matrix. And let's say I define some transformation."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have some matrix A. And let's say that its columns are V1, their column vector is V2, all the way to Vn. So this guy has n columns, and let's say he has m rows, so it's an m by n matrix. And let's say I define some transformation. Let's say my transformation goes from Rn to Rm. This is the domain. I can take any vector in Rn, and it will map it to some vector in Rm."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say I define some transformation. Let's say my transformation goes from Rn to Rm. This is the domain. I can take any vector in Rn, and it will map it to some vector in Rm. And I define my transformation, so T of x, where this is some vector in Rn, is equal to A, this is this A, let me write it in this color right here, it should be bolded. I kind of get careless sometimes with the bolding. That big bold A times the vector x."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I can take any vector in Rn, and it will map it to some vector in Rm. And I define my transformation, so T of x, where this is some vector in Rn, is equal to A, this is this A, let me write it in this color right here, it should be bolded. I kind of get careless sometimes with the bolding. That big bold A times the vector x. So the first thing you might say, Sal, gee, this transformation looks very odd relative to how we've been defining transformations or functions so far. So the first thing we have to just feel comfortable with is the idea that this is a transformation. So what are we doing?"}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "That big bold A times the vector x. So the first thing you might say, Sal, gee, this transformation looks very odd relative to how we've been defining transformations or functions so far. So the first thing we have to just feel comfortable with is the idea that this is a transformation. So what are we doing? We're taking something from Rn, and then what does Ax produce? If we write Ax like this, if this is x, where it's x1, x2, it's going to have n terms, because it's in Rn. This can be rewritten as x1 times v1 plus x2 times v2, all the way to xn times vn."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what are we doing? We're taking something from Rn, and then what does Ax produce? If we write Ax like this, if this is x, where it's x1, x2, it's going to have n terms, because it's in Rn. This can be rewritten as x1 times v1 plus x2 times v2, all the way to xn times vn. So it's going to be a sum of a bunch of these column vectors, and each of these column vectors, v1, v2, all the way to vn, what set are they members of? This is an m by n matrix, so the matrix has m rows, or each of these column vectors will have m entries. So all of these guys are members of Rm."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This can be rewritten as x1 times v1 plus x2 times v2, all the way to xn times vn. So it's going to be a sum of a bunch of these column vectors, and each of these column vectors, v1, v2, all the way to vn, what set are they members of? This is an m by n matrix, so the matrix has m rows, or each of these column vectors will have m entries. So all of these guys are members of Rm. So if I just take a linear combination of all of these guys, I'm going to get another member of Rm. So this guy right here is going to be a member of Rm, another vector. So clearly, by multiplying my vector x times a, I'm mapping, I'm creating a mapping from Rn, and let me pick another color, to Rm."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So all of these guys are members of Rm. So if I just take a linear combination of all of these guys, I'm going to get another member of Rm. So this guy right here is going to be a member of Rm, another vector. So clearly, by multiplying my vector x times a, I'm mapping, I'm creating a mapping from Rn, and let me pick another color, to Rm. And I'm saying it in very general terms. Maybe n is 3, maybe m is 5, who knows? But I'm saying it in very general terms."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So clearly, by multiplying my vector x times a, I'm mapping, I'm creating a mapping from Rn, and let me pick another color, to Rm. And I'm saying it in very general terms. Maybe n is 3, maybe m is 5, who knows? But I'm saying it in very general terms. And so if this is a particular instance, a particular member of set Rn, so it's that vector, our transformation, or our function, is going to map it to this guy right here, and this guy will be a member of Rm, and we could call him Ax. Or maybe if we said Ax equals b, we could call him the vector b, whatever. But this is our transformation mapping."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "But I'm saying it in very general terms. And so if this is a particular instance, a particular member of set Rn, so it's that vector, our transformation, or our function, is going to map it to this guy right here, and this guy will be a member of Rm, and we could call him Ax. Or maybe if we said Ax equals b, we could call him the vector b, whatever. But this is our transformation mapping. So this does fit our kind of definition, or our terminology for a function or a transformation, as a mapping from one set to another. But it still might not be satisfying, because everything we saw before looked kind of like this. If we had a transformation, I would write it like the transformation of x1 and x2 and xn is equal to, and I'd write m terms here in commas."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this is our transformation mapping. So this does fit our kind of definition, or our terminology for a function or a transformation, as a mapping from one set to another. But it still might not be satisfying, because everything we saw before looked kind of like this. If we had a transformation, I would write it like the transformation of x1 and x2 and xn is equal to, and I'd write m terms here in commas. And you're like, how does this relate to that? And to do that, I'll do a specific example. So let's say that I have the matrix."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we had a transformation, I would write it like the transformation of x1 and x2 and xn is equal to, and I'd write m terms here in commas. And you're like, how does this relate to that? And to do that, I'll do a specific example. So let's say that I have the matrix. Let me do a different letter. Let's say I have my matrix B, and it is a fairly simple matrix. Let's say 2, minus 1, 3, and 4."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that I have the matrix. Let me do a different letter. Let's say I have my matrix B, and it is a fairly simple matrix. Let's say 2, minus 1, 3, and 4. And I define some transformation, so I define some transformation T, and it goes from R2 to R2. And I define T, T of some vector x, is equal to this matrix B times that vector x. Now what would that equal?"}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say 2, minus 1, 3, and 4. And I define some transformation, so I define some transformation T, and it goes from R2 to R2. And I define T, T of some vector x, is equal to this matrix B times that vector x. Now what would that equal? Well, the matrix is right there. It's, let me write it in purple, 2, minus 1, 3, and 4, times x, x1, x2. And so what does this equal?"}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what would that equal? Well, the matrix is right there. It's, let me write it in purple, 2, minus 1, 3, and 4, times x, x1, x2. And so what does this equal? Well, this equals another vector. It equals a vector in the codomain R2, where the first term is 2 times x1. I'm just doing the definition of matrix-vector multiplication."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so what does this equal? Well, this equals another vector. It equals a vector in the codomain R2, where the first term is 2 times x1. I'm just doing the definition of matrix-vector multiplication. 2 times x1, plus minus 1 times x2, or minus x2. That's that row times our vector. And then the second row times that vector, we get 3 times x1, plus 4 times x2."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just doing the definition of matrix-vector multiplication. 2 times x1, plus minus 1 times x2, or minus x2. That's that row times our vector. And then the second row times that vector, we get 3 times x1, plus 4 times x2. So this is what we might be more familiar with, and I could rewrite this transformation as T of x1, x2 is equal to 2x1 minus x2, comma, let me scroll over a little bit, comma, 3x1 plus 4x2. So hopefully you're satisfied now that a matrix multiplication, it isn't some new exotic form of transformation, that they really are just another way that this statement right here is just another way of writing this exact transformation right here. Now the next question you might ask, and I already told you the answer to this at the beginning of the video, is, is multiplication by a matrix always going to be a linear transformation?"}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the second row times that vector, we get 3 times x1, plus 4 times x2. So this is what we might be more familiar with, and I could rewrite this transformation as T of x1, x2 is equal to 2x1 minus x2, comma, let me scroll over a little bit, comma, 3x1 plus 4x2. So hopefully you're satisfied now that a matrix multiplication, it isn't some new exotic form of transformation, that they really are just another way that this statement right here is just another way of writing this exact transformation right here. Now the next question you might ask, and I already told you the answer to this at the beginning of the video, is, is multiplication by a matrix always going to be a linear transformation? And what are the two constraints for being a linear transformation? We know that the transformation of two vectors, A plus B, the sum of two vectors, should be equal to the sum of their transformations. Transformation of A plus the transformation of B."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now the next question you might ask, and I already told you the answer to this at the beginning of the video, is, is multiplication by a matrix always going to be a linear transformation? And what are the two constraints for being a linear transformation? We know that the transformation of two vectors, A plus B, the sum of two vectors, should be equal to the sum of their transformations. Transformation of A plus the transformation of B. And then the other requirement is that the transformation of a scaled version of a vector should be equal to a scaled version of the transformation. These are our two requirements for being a linear transformation. So let's see if matrix multiplication applies there."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Transformation of A plus the transformation of B. And then the other requirement is that the transformation of a scaled version of a vector should be equal to a scaled version of the transformation. These are our two requirements for being a linear transformation. So let's see if matrix multiplication applies there. And I've touched on this in the past, and I've even told you that you should prove it, and I've already assumed you know it, but I'll prove it to you here because I'm tired of telling you that you should prove it. I should do it at least once. So let's see, matrix multiplication."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if matrix multiplication applies there. And I've touched on this in the past, and I've even told you that you should prove it, and I've already assumed you know it, but I'll prove it to you here because I'm tired of telling you that you should prove it. I should do it at least once. So let's see, matrix multiplication. If I multiply a matrix A times some vector X, we know that, let's say, let me write it this way. We know that this is equivalent to, I said our matrix, we could just write it, let's say this is an M by N matrix. We can write any matrix as just a series of column vectors."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see, matrix multiplication. If I multiply a matrix A times some vector X, we know that, let's say, let me write it this way. We know that this is equivalent to, I said our matrix, we could just write it, let's say this is an M by N matrix. We can write any matrix as just a series of column vectors. So this guy could have N column vectors. So let's say it's V1, V2, all the way to VN column vectors. And each of these guys are going to have M components, times X1, X2, all the way down to XN."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can write any matrix as just a series of column vectors. So this guy could have N column vectors. So let's say it's V1, V2, all the way to VN column vectors. And each of these guys are going to have M components, times X1, X2, all the way down to XN. And we've seen this multiple, multiple times before. This, by the definition of matrix vector multiplication, is equal to X1 times V1, that times that, this scalar times that vector, plus X2 times V2, all the way to plus XN times VN. This was by definition of a matrix vector multiplication."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And each of these guys are going to have M components, times X1, X2, all the way down to XN. And we've seen this multiple, multiple times before. This, by the definition of matrix vector multiplication, is equal to X1 times V1, that times that, this scalar times that vector, plus X2 times V2, all the way to plus XN times VN. This was by definition of a matrix vector multiplication. And of course, this is going to, and I did this at the top of the video, this is going to have, right here, this vector is going to be a member of RM. It's going to have M components. So what happens if I take some matrix A, some M by N matrix A, and I multiply it times the sum of two vectors A plus B."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This was by definition of a matrix vector multiplication. And of course, this is going to, and I did this at the top of the video, this is going to have, right here, this vector is going to be a member of RM. It's going to have M components. So what happens if I take some matrix A, some M by N matrix A, and I multiply it times the sum of two vectors A plus B. Where, so I could rewrite this as this thing right here, so my matrix A times the sum of A plus B, the first term will just be A1 plus V1, second term is A2 plus B2, all the way down to AN plus BN. This is the same thing as this. I'm not saying A of A plus B, I'm saying A times, maybe I should put a dot right there."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what happens if I take some matrix A, some M by N matrix A, and I multiply it times the sum of two vectors A plus B. Where, so I could rewrite this as this thing right here, so my matrix A times the sum of A plus B, the first term will just be A1 plus V1, second term is A2 plus B2, all the way down to AN plus BN. This is the same thing as this. I'm not saying A of A plus B, I'm saying A times, maybe I should put a dot right there. I'm multiplying the matrix, it's not, maybe I want to be careful with my notation, this is the matrix vector multiplication, it's not some type of new matrix dot product. But this is the same thing as this multiplication right here, and based on what I just told you up here, I've seen multiple, multiple times, this is the same thing as A1 plus B1 times the first column in A, which is that vector right there, this A is the same as this A, so times V1, plus A2 plus B2 times V2, all the way to plus AN plus BN times VN. All I did, each XI term here is just being replaced by AI plus BI term, so each X1 here is replaced by A1 plus B1 here, so this is equivalent to this, and then from the fact that we know that vector products times scalars exhibit the distributive property, we can say that this is equal to A1 times V1, let me actually write all of the A1 terms, let me write this, A1 times V1 plus B1 times V1, plus A2 times V2 plus B2 times V2, all the way to plus AN times VN plus BN times VN, and then if we just reassociate this, if we just group all of the A's together, all of the A terms together, we get A1 plus, let me write it this way, A1 times V1 plus A2 times V2, plus all the way AN times VN, I just grabbed all the A terms, we get that, plus all the B terms, all the B terms I'll do in this color, all the B terms are like that, so plus B1 times V1 plus B2 times V2, all the way to plus BN times VN, that's that guy right there, is equivalent to this statement up here, I just regrouped everything, which is of course equivalent to that statement over there."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm not saying A of A plus B, I'm saying A times, maybe I should put a dot right there. I'm multiplying the matrix, it's not, maybe I want to be careful with my notation, this is the matrix vector multiplication, it's not some type of new matrix dot product. But this is the same thing as this multiplication right here, and based on what I just told you up here, I've seen multiple, multiple times, this is the same thing as A1 plus B1 times the first column in A, which is that vector right there, this A is the same as this A, so times V1, plus A2 plus B2 times V2, all the way to plus AN plus BN times VN. All I did, each XI term here is just being replaced by AI plus BI term, so each X1 here is replaced by A1 plus B1 here, so this is equivalent to this, and then from the fact that we know that vector products times scalars exhibit the distributive property, we can say that this is equal to A1 times V1, let me actually write all of the A1 terms, let me write this, A1 times V1 plus B1 times V1, plus A2 times V2 plus B2 times V2, all the way to plus AN times VN plus BN times VN, and then if we just reassociate this, if we just group all of the A's together, all of the A terms together, we get A1 plus, let me write it this way, A1 times V1 plus A2 times V2, plus all the way AN times VN, I just grabbed all the A terms, we get that, plus all the B terms, all the B terms I'll do in this color, all the B terms are like that, so plus B1 times V1 plus B2 times V2, all the way to plus BN times VN, that's that guy right there, is equivalent to this statement up here, I just regrouped everything, which is of course equivalent to that statement over there. But what's this equal to? This is equal to my vector, these columns, remember the column for the matrix capital A, so this is equal to the matrix capital A times A1, A2, all the way down to AN, which was our vector A, and what's this equal to? This is equal to plus these V1's, these are the columns for A, so it's equal to the matrix A times my vector B, V1, V2, all the way down to BN, this is my vector B."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "All I did, each XI term here is just being replaced by AI plus BI term, so each X1 here is replaced by A1 plus B1 here, so this is equivalent to this, and then from the fact that we know that vector products times scalars exhibit the distributive property, we can say that this is equal to A1 times V1, let me actually write all of the A1 terms, let me write this, A1 times V1 plus B1 times V1, plus A2 times V2 plus B2 times V2, all the way to plus AN times VN plus BN times VN, and then if we just reassociate this, if we just group all of the A's together, all of the A terms together, we get A1 plus, let me write it this way, A1 times V1 plus A2 times V2, plus all the way AN times VN, I just grabbed all the A terms, we get that, plus all the B terms, all the B terms I'll do in this color, all the B terms are like that, so plus B1 times V1 plus B2 times V2, all the way to plus BN times VN, that's that guy right there, is equivalent to this statement up here, I just regrouped everything, which is of course equivalent to that statement over there. But what's this equal to? This is equal to my vector, these columns, remember the column for the matrix capital A, so this is equal to the matrix capital A times A1, A2, all the way down to AN, which was our vector A, and what's this equal to? This is equal to plus these V1's, these are the columns for A, so it's equal to the matrix A times my vector B, V1, V2, all the way down to BN, this is my vector B. So we just saw that, we just showed you that if I add my two vectors, A and B, and then multiply it by the matrix, it's completely equivalent to multiplying each of the vectors times the matrix first, and then adding them up. So we've satisfied, and this is for any M by N matrix, so we've now satisfied this first condition right there, and then what about the second condition? This one's even more straightforward to understand."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to plus these V1's, these are the columns for A, so it's equal to the matrix A times my vector B, V1, V2, all the way down to BN, this is my vector B. So we just saw that, we just showed you that if I add my two vectors, A and B, and then multiply it by the matrix, it's completely equivalent to multiplying each of the vectors times the matrix first, and then adding them up. So we've satisfied, and this is for any M by N matrix, so we've now satisfied this first condition right there, and then what about the second condition? This one's even more straightforward to understand. C times A1, so let me write it this way, if I were to write, the vector A times, sorry, the matrix capital A times the vector lowercase a, let me do it this way, times the vector C lowercase a, so I'm multiplying my vector times the scalar first, is equal to, I can write my big matrix A, I've already labeled its columns, V1, V2, all the way to VN, that's my matrix A, and then what does CA look like? CA, you just multiply that scalar times each of the terms of A, so it's CA1, CA2, all the way down to CAN, and what does this equal? We know this, we've seen this show multiple times before right there."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This one's even more straightforward to understand. C times A1, so let me write it this way, if I were to write, the vector A times, sorry, the matrix capital A times the vector lowercase a, let me do it this way, times the vector C lowercase a, so I'm multiplying my vector times the scalar first, is equal to, I can write my big matrix A, I've already labeled its columns, V1, V2, all the way to VN, that's my matrix A, and then what does CA look like? CA, you just multiply that scalar times each of the terms of A, so it's CA1, CA2, all the way down to CAN, and what does this equal? We know this, we've seen this show multiple times before right there. So it just equals, this equals, I'll write a little bit lower, equals CA1 times this column vector, times V1, plus CA2 times V2, times this guy, all the way to plus CAN times VN, times the vector VN. And if you just factor this C out, once again, scalar multiplication times vectors exhibits the distributive property. I believe I've done a video on that, but it's very easy to prove."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know this, we've seen this show multiple times before right there. So it just equals, this equals, I'll write a little bit lower, equals CA1 times this column vector, times V1, plus CA2 times V2, times this guy, all the way to plus CAN times VN, times the vector VN. And if you just factor this C out, once again, scalar multiplication times vectors exhibits the distributive property. I believe I've done a video on that, but it's very easy to prove. So this will be equal to C times, I'll just stay in one color right now, A1V1 plus A2V2, plus all the way to ANVN, and what is this thing equal to? Well that's just our matrix A, that's just our matrix A times our vector, or our matrix uppercase A, maybe I'm overloading the letter A, my matrix uppercase A times my vector lowercase A, times my lowercase A, right? Where the lowercase A is just this thing right here, A1, A2, and so forth."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I believe I've done a video on that, but it's very easy to prove. So this will be equal to C times, I'll just stay in one color right now, A1V1 plus A2V2, plus all the way to ANVN, and what is this thing equal to? Well that's just our matrix A, that's just our matrix A times our vector, or our matrix uppercase A, maybe I'm overloading the letter A, my matrix uppercase A times my vector lowercase A, times my lowercase A, right? Where the lowercase A is just this thing right here, A1, A2, and so forth. This thing up here was the same thing as that. So I just showed you that if I take my matrix and multiply it times some vector that was multiplied by a scalar first, that's equivalent to first multiplying the matrix times the vector, and then multiplying by the scalar. So we've shown you that matrix times vector products or matrix vector products satisfy this condition of linear transformations and this condition."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Where the lowercase A is just this thing right here, A1, A2, and so forth. This thing up here was the same thing as that. So I just showed you that if I take my matrix and multiply it times some vector that was multiplied by a scalar first, that's equivalent to first multiplying the matrix times the vector, and then multiplying by the scalar. So we've shown you that matrix times vector products or matrix vector products satisfy this condition of linear transformations and this condition. So the big takeaway right here is matrix multiplication. And this is an important takeaway. Matrix multiplication or matrix products with vectors is always a linear transformation."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we've shown you that matrix times vector products or matrix vector products satisfy this condition of linear transformations and this condition. So the big takeaway right here is matrix multiplication. And this is an important takeaway. Matrix multiplication or matrix products with vectors is always a linear transformation. And this is a bit of a side note. In the next video I'm going to show you that any linear transformation, this is incredibly powerful, can be represented by a matrix product or by any transformation on any vector can be equivalently written as a product of that vector with a matrix. It has huge repercussions."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Matrix multiplication or matrix products with vectors is always a linear transformation. And this is a bit of a side note. In the next video I'm going to show you that any linear transformation, this is incredibly powerful, can be represented by a matrix product or by any transformation on any vector can be equivalently written as a product of that vector with a matrix. It has huge repercussions. And just as a side note, kind of tying this back to your everyday life, you probably have your Xbox or your Sony PlayStation, and you have these 3D graphic programs where you're running around and shooting at things. And the way that the software renders those programs where you can see things from every different angle, you have a cube, and then if you move this way a little bit the cube will look more like this, and it gets rotated, and you move up and down. These are all transformations of matrices, and we'll do this in more detail."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "It has huge repercussions. And just as a side note, kind of tying this back to your everyday life, you probably have your Xbox or your Sony PlayStation, and you have these 3D graphic programs where you're running around and shooting at things. And the way that the software renders those programs where you can see things from every different angle, you have a cube, and then if you move this way a little bit the cube will look more like this, and it gets rotated, and you move up and down. These are all transformations of matrices, and we'll do this in more detail. These are all transformations of vectors or the positions of vectors, and I'll do that in a lot more detail. And all of that is really just matrix multiplications. So all of these things that you're doing in your fancy 3D games on your Xbox or your PlayStation, they're all just matrix multiplications."}, {"video_title": "Matrix vector products as linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are all transformations of matrices, and we'll do this in more detail. These are all transformations of vectors or the positions of vectors, and I'll do that in a lot more detail. And all of that is really just matrix multiplications. So all of these things that you're doing in your fancy 3D games on your Xbox or your PlayStation, they're all just matrix multiplications. I'm going to prove that to you in the next video. And so when you have these graphics cards or these graphics engines, all they are, and this is kind of a, you know, we're jumping away from the theoretical, but all these graphics processors are are hardwired matrix multipliers. If I have just a generalized, some type of CPU, I have to in software write how to multiply matrices."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "What I want to do in this video, and actually the next few videos, is to show you how to essentially design linear transformations to do things to vectors that you want them to do. So we already know that if I have some linear transformation, t, and it's a mapping from Rn to Rm, that we can represent t, what t does to any vector in x, or the mapping of x in Rn to Rm, we can represent it as some matrix times the vector x, where this would be an m by n matrix, and we know that we can always construct this matrix, that any linear transformation can be represented by a matrix this way, and we can represent it by taking our identity matrix, you've seen that before, by taking the identity matrix with n rows and n columns, so it literally just looks like this. So it's a 1, and then it has n minus 1 zeros all the way down, then it's a 0, 1, and then everything else is zeros all the way down, and so essentially you just have 1's down as diagonal, it's an n by n matrix, all of these are zeros, just like that. You take your identity matrix and you perform the transformation on each of its columns, and we call each of these columns the standard basis for Rn, so this is column E1, this is column E2, and it has n columns, En. And each of these columns are, of course, members of Rn, because this is an n rows and n column matrix, and we know that A, our matrix A, can be represented as the transformation being operated on each of these columns, so the transformation on E1, and the transformation on E2, so forth, and so on, all the way to the transformation to En. And this is a really useful thing to know, because it's very easy to operate any transformation on each of these basis vectors that only have a 1 kind of in its corresponding dimension, or with respect to the corresponding variable, and everything else is 0. So let's, you know, all of this is review."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "You take your identity matrix and you perform the transformation on each of its columns, and we call each of these columns the standard basis for Rn, so this is column E1, this is column E2, and it has n columns, En. And each of these columns are, of course, members of Rn, because this is an n rows and n column matrix, and we know that A, our matrix A, can be represented as the transformation being operated on each of these columns, so the transformation on E1, and the transformation on E2, so forth, and so on, all the way to the transformation to En. And this is a really useful thing to know, because it's very easy to operate any transformation on each of these basis vectors that only have a 1 kind of in its corresponding dimension, or with respect to the corresponding variable, and everything else is 0. So let's, you know, all of this is review. Let's actually use this information to construct some interesting transformations. So let's start with some set in Rn, and actually, everything I'm going to do is going to be in R2, but you can extend a lot of this into just general dimensions, but we're dealing with R2 right here, obviously. It's only two dimensions right here."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So let's, you know, all of this is review. Let's actually use this information to construct some interesting transformations. So let's start with some set in Rn, and actually, everything I'm going to do is going to be in R2, but you can extend a lot of this into just general dimensions, but we're dealing with R2 right here, obviously. It's only two dimensions right here. Let's say we have a triangle formed by the points. Let's say the first point is right here. Let's say it's the point 3, 2, and then you have the point, let's say that your next point in your triangle is the point, well, let's just make it the point minus 3, 2."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "It's only two dimensions right here. Let's say we have a triangle formed by the points. Let's say the first point is right here. Let's say it's the point 3, 2, and then you have the point, let's say that your next point in your triangle is the point, well, let's just make it the point minus 3, 2. I shouldn't have written that as a fraction. I don't know why I did that. 3, 2, then you have the point minus 3, 2, and that's this point right here, minus 3, 2."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's the point 3, 2, and then you have the point, let's say that your next point in your triangle is the point, well, let's just make it the point minus 3, 2. I shouldn't have written that as a fraction. I don't know why I did that. 3, 2, then you have the point minus 3, 2, and that's this point right here, minus 3, 2. And then let's say, just for fun, let's say you have the point, let's say we have the point, or the vector, the position vector, right? 3, minus 2, which is right here. Now, each of these are position vectors, and I can draw them."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "3, 2, then you have the point minus 3, 2, and that's this point right here, minus 3, 2. And then let's say, just for fun, let's say you have the point, let's say we have the point, or the vector, the position vector, right? 3, minus 2, which is right here. Now, each of these are position vectors, and I can draw them. I could draw this 3, 2 as in the standard position by kind of drawing an arrow like that. I could do the minus 3, 2 in its standard position like that. And 3, minus 2, I could draw like that."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Now, each of these are position vectors, and I can draw them. I could draw this 3, 2 as in the standard position by kind of drawing an arrow like that. I could do the minus 3, 2 in its standard position like that. And 3, minus 2, I could draw like that. But more than the actual position vectors, I'm more concerned with the positions that they specify. And we know that if we take the set of all of the positions, or all of the position vectors that specify the triangle that's essentially formed by connecting these dots, the transformation of this set is essentially, you take the transformation of each of these endpoints, and then you connect the dots in the same order. And we saw that several videos ago."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And 3, minus 2, I could draw like that. But more than the actual position vectors, I'm more concerned with the positions that they specify. And we know that if we take the set of all of the positions, or all of the position vectors that specify the triangle that's essentially formed by connecting these dots, the transformation of this set is essentially, you take the transformation of each of these endpoints, and then you connect the dots in the same order. And we saw that several videos ago. But let's actually design a transformation here. So let's say we want to, let's just write down in words what we want to do with whatever we start in our domain. Let's say we want to reflect around the x-axis, reflect around the, well actually, let's reflect around the y-axis."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And we saw that several videos ago. But let's actually design a transformation here. So let's say we want to, let's just write down in words what we want to do with whatever we start in our domain. Let's say we want to reflect around the x-axis, reflect around the, well actually, let's reflect around the y-axis. So we essentially want to flip it over. We want to flip it over that way, so I'm kind of envisioning something that'll look something like that when we flip it over. So we're going to reflect it around the y-axis."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we want to reflect around the x-axis, reflect around the, well actually, let's reflect around the y-axis. So we essentially want to flip it over. We want to flip it over that way, so I'm kind of envisioning something that'll look something like that when we flip it over. So we're going to reflect it around the y-axis. And let's say we want to stretch in y direction by 2. In y direction times 2. So what I envision, we're going to flip it over like this, what I just drew here."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to reflect it around the y-axis. And let's say we want to stretch in y direction by 2. In y direction times 2. So what I envision, we're going to flip it over like this, what I just drew here. And then we want to stretch it. So we're going to first flip it, that's kind of step one. And then step two is we're going to stretch it."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So what I envision, we're going to flip it over like this, what I just drew here. And then we want to stretch it. So we're going to first flip it, that's kind of step one. And then step two is we're going to stretch it. So instead of looking like this, it'll be twice as tall, so it'll look like this. Without necessarily stretching the x. So how can we do that?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And then step two is we're going to stretch it. So instead of looking like this, it'll be twice as tall, so it'll look like this. Without necessarily stretching the x. So how can we do that? So the first idea of reflecting around the y-axis. So what we want is this point that was a minus 3 in the x coordinate right there, we want this point to have its same y coordinate. We want it to still have a 2 there."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So how can we do that? So the first idea of reflecting around the y-axis. So what we want is this point that was a minus 3 in the x coordinate right there, we want this point to have its same y coordinate. We want it to still have a 2 there. And I'm calling the second coordinate here our y coordinate. I could call that our x2 coordinate, but we're used to dealing with the y coordinate when we graph things. So I'll just keep calling it the y coordinate."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "We want it to still have a 2 there. And I'm calling the second coordinate here our y coordinate. I could call that our x2 coordinate, but we're used to dealing with the y coordinate when we graph things. So I'll just keep calling it the y coordinate. But what we want is this negative 3 to turn to a positive 3. Because we want this point to end up over here. And we want this positive 3 here to end up becoming a negative 3 over here."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So I'll just keep calling it the y coordinate. But what we want is this negative 3 to turn to a positive 3. Because we want this point to end up over here. And we want this positive 3 here to end up becoming a negative 3 over here. And we want this positive 3 for the x coordinate to end up as a negative 3 over there. So you can imagine all we're doing is we're flipping the sign. This reflection around y, this is just equivalent to flipping the sign."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And we want this positive 3 here to end up becoming a negative 3 over here. And we want this positive 3 for the x coordinate to end up as a negative 3 over there. So you can imagine all we're doing is we're flipping the sign. This reflection around y, this is just equivalent to flipping the sign. Flipping the sign of the x coordinate. So this statement right here is equivalent to minus 1 times the x coordinate. So let me call that, or let's call that times x1."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "This reflection around y, this is just equivalent to flipping the sign. Flipping the sign of the x coordinate. So this statement right here is equivalent to minus 1 times the x coordinate. So let me call that, or let's call that times x1. Because this is x1. And then stretching in the y direction. So what does that mean?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So let me call that, or let's call that times x1. Because this is x1. And then stretching in the y direction. So what does that mean? That means that whatever height we have here, so this next step here is whatever height we have here, I want it to be 2 times as much. So right here, this coordinate is 3, 2. If I didn't do this first step first, I would want to make it 3, 4."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So what does that mean? That means that whatever height we have here, so this next step here is whatever height we have here, I want it to be 2 times as much. So right here, this coordinate is 3, 2. If I didn't do this first step first, I would want to make it 3, 4. I want to make it 2 times the y coordinate. So the next thing I want to do is I want to do 2 times, well, I could either call it, well, let me just call it the y coordinate. It's a little bit different convention than I've been using, but I'm just calling these vectors, instead of calling them x1 and x2, I'm saying that my vectors in R2, the first term I'm calling the x term or the x entry, and the second term I'm calling the y entry."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "If I didn't do this first step first, I would want to make it 3, 4. I want to make it 2 times the y coordinate. So the next thing I want to do is I want to do 2 times, well, I could either call it, well, let me just call it the y coordinate. It's a little bit different convention than I've been using, but I'm just calling these vectors, instead of calling them x1 and x2, I'm saying that my vectors in R2, the first term I'm calling the x term or the x entry, and the second term I'm calling the y entry. But it's the same idea that we've been doing before. I'm just switching to this notation because we're used to thinking of this as the y-axis as opposed to the x1 and the x2 axis. So how do we construct this transformation?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "It's a little bit different convention than I've been using, but I'm just calling these vectors, instead of calling them x1 and x2, I'm saying that my vectors in R2, the first term I'm calling the x term or the x entry, and the second term I'm calling the y entry. But it's the same idea that we've been doing before. I'm just switching to this notation because we're used to thinking of this as the y-axis as opposed to the x1 and the x2 axis. So how do we construct this transformation? I mean, I can write it down in kind of transformation words. I could say, I could define my transformation as T of some vector x, or let me write it this way, T of some vector xy is going to be equal to, I want to take minus 1 times the x, so I'm going to minus the x, and then I'm going to multiply 2 times the y. So that's how I could just write it in transformation language, and that's pretty straightforward."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So how do we construct this transformation? I mean, I can write it down in kind of transformation words. I could say, I could define my transformation as T of some vector x, or let me write it this way, T of some vector xy is going to be equal to, I want to take minus 1 times the x, so I'm going to minus the x, and then I'm going to multiply 2 times the y. So that's how I could just write it in transformation language, and that's pretty straightforward. But how would I actually construct a matrix for this? So what you do is you just take, we're dealing in R2, so you start off with the identity matrix in R2, which is just 1, 0, 0, 1, and you apply this transformation to each of the columns of this identity matrix. So if you apply the transformation to this first column, what do you get?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So that's how I could just write it in transformation language, and that's pretty straightforward. But how would I actually construct a matrix for this? So what you do is you just take, we're dealing in R2, so you start off with the identity matrix in R2, which is just 1, 0, 0, 1, and you apply this transformation to each of the columns of this identity matrix. So if you apply the transformation to this first column, what do you get? So what we're going to do is we're going to create a new matrix A, and say that is equal to the transformation of, let me write it like this, transformation of 1, 0. That's going to be our new column. We're just going to transform this column."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So if you apply the transformation to this first column, what do you get? So what we're going to do is we're going to create a new matrix A, and say that is equal to the transformation of, let me write it like this, transformation of 1, 0. That's going to be our new column. We're just going to transform this column. And then the second column is going to be the transformation of that column. So it's transformation of 0, 1, just like that. So what are these equal to?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "We're just going to transform this column. And then the second column is going to be the transformation of that column. So it's transformation of 0, 1, just like that. So what are these equal to? The transformation of 1, 0. So let me write it down here in green. So A is equal to, what's the transformation of 1, 0, where x is 1, where we just take the minus of the x term, so we get minus 1, and then 2 times the y term, so 2 times 0 is just 0."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So what are these equal to? The transformation of 1, 0. So let me write it down here in green. So A is equal to, what's the transformation of 1, 0, where x is 1, where we just take the minus of the x term, so we get minus 1, and then 2 times the y term, so 2 times 0 is just 0. Now do the second term. The minus of the 0 term is just minus 0, so that just stays as 0. And then you multiply 2 times the y term."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So A is equal to, what's the transformation of 1, 0, where x is 1, where we just take the minus of the x term, so we get minus 1, and then 2 times the y term, so 2 times 0 is just 0. Now do the second term. The minus of the 0 term is just minus 0, so that just stays as 0. And then you multiply 2 times the y term. So 2 times y is going to be equal to 2 times 1, so it's equal to 2. So now we can describe this transformation. So now we could say the transformation of some vector x, y, we can describe it as a matrix vector product."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And then you multiply 2 times the y term. So 2 times y is going to be equal to 2 times 1, so it's equal to 2. So now we can describe this transformation. So now we could say the transformation of some vector x, y, we can describe it as a matrix vector product. It is equal to minus 1, 0, 0, 2 times our vector, times x, y. And let's apply it to verify that it works, to verify that our matrix works. So this first point, and I'll try to do it color coded, let's do this first point right here."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So now we could say the transformation of some vector x, y, we can describe it as a matrix vector product. It is equal to minus 1, 0, 0, 2 times our vector, times x, y. And let's apply it to verify that it works, to verify that our matrix works. So this first point, and I'll try to do it color coded, let's do this first point right here. This is minus 3, 2. So that's minus 3, 2. So what is minus 3, 2?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So this first point, and I'll try to do it color coded, let's do this first point right here. This is minus 3, 2. So that's minus 3, 2. So what is minus 3, 2? I'll do it right over here. I could just look at that. So what is minus 1, 0, 0, 2 times minus 3, 2?"}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So what is minus 3, 2? I'll do it right over here. I could just look at that. So what is minus 1, 0, 0, 2 times minus 3, 2? Well, this is just a straight up matrix vector product. Minus 1 times minus 3 is positive 3, plus 0 times 2. So plus 0."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So what is minus 1, 0, 0, 2 times minus 3, 2? Well, this is just a straight up matrix vector product. Minus 1 times minus 3 is positive 3, plus 0 times 2. So plus 0. So this is 3. And then 0 times minus 3 is 0, plus 2 times 2. So it's 3, 4."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So plus 0. So this is 3. And then 0 times minus 3 is 0, plus 2 times 2. So it's 3, 4. So that point right there will now become the point 3, 4. It now becomes that point right there. Let's look at this point right here, the point 3, 2."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So it's 3, 4. So that point right there will now become the point 3, 4. It now becomes that point right there. Let's look at this point right here, the point 3, 2. So let's take our transformation matrix, minus 1, 0, 0, 2 times 3, 2. This is equal to minus 1 times 3 is minus 3 plus 0 times 2. So it's just minus 3."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Let's look at this point right here, the point 3, 2. So let's take our transformation matrix, minus 1, 0, 0, 2 times 3, 2. This is equal to minus 1 times 3 is minus 3 plus 0 times 2. So it's just minus 3. You have 0 times 3, which is 0, plus 2 times 2, which is 4. So you get that point. So this point, by our transformation T, becomes minus 3, 4."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So it's just minus 3. You have 0 times 3, which is 0, plus 2 times 2, which is 4. So you get that point. So this point, by our transformation T, becomes minus 3, 4. So minus 3, 4. And I kind of switch in my terminology. I say it becomes, or you could say it's mapped to, if you want to use the language that I use when I introduce the ideas of functions and transformation."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So this point, by our transformation T, becomes minus 3, 4. So minus 3, 4. And I kind of switch in my terminology. I say it becomes, or you could say it's mapped to, if you want to use the language that I use when I introduce the ideas of functions and transformation. This point is mapped to this point in R2. And then finally, let's look at this point right here. Apply our transformation matrix that we've engineered."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "I say it becomes, or you could say it's mapped to, if you want to use the language that I use when I introduce the ideas of functions and transformation. This point is mapped to this point in R2. And then finally, let's look at this point right here. Apply our transformation matrix that we've engineered. Let's multiply minus 1, 0, 0, 2 times this point right here, which is 3 minus 2, which is equal to minus 1 times 3 is minus 3. And then 0 times minus 2 is just 0. So this becomes minus 3."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Apply our transformation matrix that we've engineered. Let's multiply minus 1, 0, 0, 2 times this point right here, which is 3 minus 2, which is equal to minus 1 times 3 is minus 3. And then 0 times minus 2 is just 0. So this becomes minus 3. And then 0 times 3 is 0. 2 times minus 2 is minus 4. So minus 3, minus 4."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So this becomes minus 3. And then 0 times 3 is 0. 2 times minus 2 is minus 4. So minus 3, minus 4. So this point right here becomes minus 3, minus 4. Becomes that point right there. And we know that the set in R2 that connects these dots by the same transformation will be mapped to the set in R3 that connects these dots."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So minus 3, minus 4. So this point right here becomes minus 3, minus 4. Becomes that point right there. And we know that the set in R2 that connects these dots by the same transformation will be mapped to the set in R3 that connects these dots. We've seen that already. I think that was three videos ago. So the image of this set that I've drawn here, this triangle is just a set of points specified by a set of vectors."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And we know that the set in R2 that connects these dots by the same transformation will be mapped to the set in R3 that connects these dots. We've seen that already. I think that was three videos ago. So the image of this set that I've drawn here, this triangle is just a set of points specified by a set of vectors. The image of that set of position vectors specifies these points right here. Specifies the points that I'm drawing right here. Let me see if I'm doing it right."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "So the image of this set that I've drawn here, this triangle is just a set of points specified by a set of vectors. The image of that set of position vectors specifies these points right here. Specifies the points that I'm drawing right here. Let me see if I'm doing it right. There you go. Just like that. And lo and behold, it has done what we wanted to do."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Let me see if I'm doing it right. There you go. Just like that. And lo and behold, it has done what we wanted to do. We flipped it over so that we got kind of this side onto the other side like that. And then we stretched it. And we stretched it in the y direction."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And lo and behold, it has done what we wanted to do. We flipped it over so that we got kind of this side onto the other side like that. And then we stretched it. And we stretched it in the y direction. And we see that it has stretched by a factor of 2. We flipped it first, and then we stretched it by a factor of 2. And in general, any of these operations can be performed."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And we stretched it in the y direction. And we see that it has stretched by a factor of 2. We flipped it first, and then we stretched it by a factor of 2. And in general, any of these operations can be performed. I mean, you can always go back to the basics. You can always say, look, I can write my transformation in this type of form. Then I can just apply that to my basis vectors or the columns in my identity matrix."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And in general, any of these operations can be performed. I mean, you can always go back to the basics. You can always say, look, I can write my transformation in this type of form. Then I can just apply that to my basis vectors or the columns in my identity matrix. But a general theme is any of these transformations that literally just scale in either the x or y direction, or essentially just, well, you could say scale. They can either shrink or expand in the x or y direction, or flip in the x or y direction, creating a reflection. These are going to be diagonal matrices."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Then I can just apply that to my basis vectors or the columns in my identity matrix. But a general theme is any of these transformations that literally just scale in either the x or y direction, or essentially just, well, you could say scale. They can either shrink or expand in the x or y direction, or flip in the x or y direction, creating a reflection. These are going to be diagonal matrices. Diagonal matrices. And why are they diagonal matrices? Because they only have non-zero terms along their diagonal."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "These are going to be diagonal matrices. Diagonal matrices. And why are they diagonal matrices? Because they only have non-zero terms along their diagonal. This is a 2 by 2 case. If I did a 3 by 3, it would be 0's everywhere except along the diagonal. And it makes a lot of sense, because this first term is essentially what you're doing to the x1 term."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "Because they only have non-zero terms along their diagonal. This is a 2 by 2 case. If I did a 3 by 3, it would be 0's everywhere except along the diagonal. And it makes a lot of sense, because this first term is essentially what you're doing to the x1 term. The second term is what you're doing to the x2 term. If you had, or the y term in our example, if I had multiple terms, if this was a 3 by 3, that would be what I would do to the third dimension. Then the next term would be what I would do to the fourth dimension."}, {"video_title": "Linear transformation examples Scaling and reflections Linear Algebra Khan Academy.mp3", "Sentence": "And it makes a lot of sense, because this first term is essentially what you're doing to the x1 term. The second term is what you're doing to the x2 term. If you had, or the y term in our example, if I had multiple terms, if this was a 3 by 3, that would be what I would do to the third dimension. Then the next term would be what I would do to the fourth dimension. So you could expand this idea to an arbitrary Rn. Anyway, this whole point of this video is to kind of introduce you to this idea of creating custom transformations. And I think you're already starting to realize this could be very useful if you wanted to, especially in computer programming, if you're going to do some graphics or create some type of multidimensional games."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if there are other useful reasons to have an orthonormal basis. So we already know, let's say I have some subspace V. Let's say V is a subspace of Rn. And let's say we have B, which is an orthonormal basis. B is equal to V1, V2, all the way to Vk. And it is an orthonormal basis for V, which is just a fancy way of saying that all of these vectors have length 1, and they're all orthogonal with respect to each other. Now, we've seen many times before that if I have just any member of Rn, so let's say that I have some vector x that is a member of Rn, then x can be represented as a sum of a member of V, as some vector V that is in our subspace, and some vector W that is in the orthogonal complement of our subspace. Let me write that down."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "B is equal to V1, V2, all the way to Vk. And it is an orthonormal basis for V, which is just a fancy way of saying that all of these vectors have length 1, and they're all orthogonal with respect to each other. Now, we've seen many times before that if I have just any member of Rn, so let's say that I have some vector x that is a member of Rn, then x can be represented as a sum of a member of V, as some vector V that is in our subspace, and some vector W that is in the orthogonal complement of our subspace. Let me write that down. Where V is a member of my subspace, and W is a member of my subspace's orthogonal complement. We saw this when I was doing the whole set of videos on orthogonal complements. Now, what is this thing right here?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. Where V is a member of my subspace, and W is a member of my subspace's orthogonal complement. We saw this when I was doing the whole set of videos on orthogonal complements. Now, what is this thing right here? What is this thing right there? By definition, that is the projection of x onto V. This would be the projection of x onto V's orthogonal complement. And we know in the past that this is not an easy thing to find, that if I set up some matrix A that has my basis vectors as the columns."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this thing right here? What is this thing right there? By definition, that is the projection of x onto V. This would be the projection of x onto V's orthogonal complement. And we know in the past that this is not an easy thing to find, that if I set up some matrix A that has my basis vectors as the columns. So if I set up some matrix A that looks like this, V1, V2, all the way to Vk, we learned before that if we wanted to figure out, have a kind of a general way of figuring out what the projection is, we learned that the projection of any vector x onto V is equal to A times A transpose A inverse times A times x. And this was a pain to figure out. That is a pain to figure out."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And we know in the past that this is not an easy thing to find, that if I set up some matrix A that has my basis vectors as the columns. So if I set up some matrix A that looks like this, V1, V2, all the way to Vk, we learned before that if we wanted to figure out, have a kind of a general way of figuring out what the projection is, we learned that the projection of any vector x onto V is equal to A times A transpose A inverse times A times x. And this was a pain to figure out. That is a pain to figure out. But let's see if the assumption that these guys are orthonormal, or that this is an orthonormal set, in any way simplifies this. So the first thing we can do is just explore this a little bit. This vector V, this is a member of our subspace, which means it can be represented as a linear combination of my basis vectors."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "That is a pain to figure out. But let's see if the assumption that these guys are orthonormal, or that this is an orthonormal set, in any way simplifies this. So the first thing we can do is just explore this a little bit. This vector V, this is a member of our subspace, which means it can be represented as a linear combination of my basis vectors. So I can write x is equal to, instead of V, I can write C1 times V1 plus C2 times V2, all the way to plus Ck times Vk. This is the same thing as just any or some unique member of my subspace V. So that's V right there. And you can also view this as the projection of x onto the subspace V. So x can be represented as some member of V and then some member of V's orthogonal complement, plus W right there."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This vector V, this is a member of our subspace, which means it can be represented as a linear combination of my basis vectors. So I can write x is equal to, instead of V, I can write C1 times V1 plus C2 times V2, all the way to plus Ck times Vk. This is the same thing as just any or some unique member of my subspace V. So that's V right there. And you can also view this as the projection of x onto the subspace V. So x can be represented as some member of V and then some member of V's orthogonal complement, plus W right there. Now, what happens if we take both sides of this equation, if we dot it with some one of these guys, with let's say Vi, let's dot both sides of this equation with Vi. So if I take Vi dot x, where Vi is the ith basis vector up here, the ith basis vector and the basis for my subspace V, what am I going to get? This is going to be C1 times Vi times V1 plus C2 times Vi times V2, plus you're going to keep going."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And you can also view this as the projection of x onto the subspace V. So x can be represented as some member of V and then some member of V's orthogonal complement, plus W right there. Now, what happens if we take both sides of this equation, if we dot it with some one of these guys, with let's say Vi, let's dot both sides of this equation with Vi. So if I take Vi dot x, where Vi is the ith basis vector up here, the ith basis vector and the basis for my subspace V, what am I going to get? This is going to be C1 times Vi times V1 plus C2 times Vi times V2, plus you're going to keep going. Eventually you'll get to the ith term, which will be Ci times Vi dotted with Vi. And then, assuming that i isn't 1, 2, or k, eventually you'll get to Ck times Vi dotted with Vk. We just saw this in the last video."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be C1 times Vi times V1 plus C2 times Vi times V2, plus you're going to keep going. Eventually you'll get to the ith term, which will be Ci times Vi dotted with Vi. And then, assuming that i isn't 1, 2, or k, eventually you'll get to Ck times Vi dotted with Vk. We just saw this in the last video. I'm just dotting both sides. But we also have this W term. So then we're going to plus Vi dot W. Now, just to clarify things, in the last video we assumed that x was inside of the subspace, so that x could be represented with coordinates here."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "We just saw this in the last video. I'm just dotting both sides. But we also have this W term. So then we're going to plus Vi dot W. Now, just to clarify things, in the last video we assumed that x was inside of the subspace, so that x could be represented with coordinates here. Now x can be any member of Rn, and we're just looking at the projection of x. And because it's any member, it's going to be some combination of these guys plus some member of V's orthogonal complement. Now when I take the dot product of one of my basis vectors, the ith basis vector with both sides of the equation, this side is just that."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So then we're going to plus Vi dot W. Now, just to clarify things, in the last video we assumed that x was inside of the subspace, so that x could be represented with coordinates here. Now x can be any member of Rn, and we're just looking at the projection of x. And because it's any member, it's going to be some combination of these guys plus some member of V's orthogonal complement. Now when I take the dot product of one of my basis vectors, the ith basis vector with both sides of the equation, this side is just that. But on the right side, something very similar happens to what we saw in the last video. What is Vi dot V1? Well, they're different members of this orthonormal set, so they're orthogonal."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now when I take the dot product of one of my basis vectors, the ith basis vector with both sides of the equation, this side is just that. But on the right side, something very similar happens to what we saw in the last video. What is Vi dot V1? Well, they're different members of this orthonormal set, so they're orthogonal. So that's going to be 0. Vi dot V2, that's 0, assuming Vi doesn't equal 2. Vi dot Vi is 1."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Well, they're different members of this orthonormal set, so they're orthogonal. So that's going to be 0. Vi dot V2, that's 0, assuming Vi doesn't equal 2. Vi dot Vi is 1. So this term is just going to be Ci. Vi dot Vk, that's also 0. Doesn't matter what our constant is."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Vi dot Vi is 1. So this term is just going to be Ci. Vi dot Vk, that's also 0. Doesn't matter what our constant is. 0 times anything is 0. And then what is Vi dot W? Well, by definition, W is a member of our orthogonal complement to V, which means that it's orthogonal to every member of V. Well, this is a member of V, so these two guys are orthogonal."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Doesn't matter what our constant is. 0 times anything is 0. And then what is Vi dot W? Well, by definition, W is a member of our orthogonal complement to V, which means that it's orthogonal to every member of V. Well, this is a member of V, so these two guys are orthogonal. So that is also equal to 0. And just like that, you get Ci is equal to Vi times x. Just like that."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Well, by definition, W is a member of our orthogonal complement to V, which means that it's orthogonal to every member of V. Well, this is a member of V, so these two guys are orthogonal. So that is also equal to 0. And just like that, you get Ci is equal to Vi times x. Just like that. So what does this do? This is kind of a very similar result that we got last time. But remember, we're not assuming that x is a member of V. In that case, then the Ci's would be the coordinates for x."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. So what does this do? This is kind of a very similar result that we got last time. But remember, we're not assuming that x is a member of V. In that case, then the Ci's would be the coordinates for x. In this case, we're looking for the projection of x onto V, or the member of V that is kind of x's component in V, or that represents x's projection onto V. So if we now want to find the projection of x onto V, it's equal to these Ci's times the respective basis vectors. But now we know what the Ci's are. They're that basis vector times your vector x."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But remember, we're not assuming that x is a member of V. In that case, then the Ci's would be the coordinates for x. In this case, we're looking for the projection of x onto V, or the member of V that is kind of x's component in V, or that represents x's projection onto V. So if we now want to find the projection of x onto V, it's equal to these Ci's times the respective basis vectors. But now we know what the Ci's are. They're that basis vector times your vector x. So just like that, we get a pretty simple way of figuring out the projection onto a subspace with an orthonormal basis. So let's see. C1 is just going to be V1 dot x."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "They're that basis vector times your vector x. So just like that, we get a pretty simple way of figuring out the projection onto a subspace with an orthonormal basis. So let's see. C1 is just going to be V1 dot x. That's C1. And then we're going to multiply that times the vector V1. That's a vector 2."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "C1 is just going to be V1 dot x. That's C1. And then we're going to multiply that times the vector V1. That's a vector 2. And then the next coefficient on V2 is going to be V2 dot x times the vector V2. And then you're going to go all the way to plus Vk dot x times Vk. And I don't know if you remember what we did when we took the projection of x onto some line."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "That's a vector 2. And then the next coefficient on V2 is going to be V2 dot x times the vector V2. And then you're going to go all the way to plus Vk dot x times Vk. And I don't know if you remember what we did when we took the projection of x onto some line. When we were taking the projection of x onto some line where L is equal to the span of some unit vector, where this had a length 1, for t is any real number. That's just a line. Some of the span of some unit vector."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And I don't know if you remember what we did when we took the projection of x onto some line. When we were taking the projection of x onto some line where L is equal to the span of some unit vector, where this had a length 1, for t is any real number. That's just a line. Some of the span of some unit vector. We assume this has length 1. Then the projection onto a line just simplified to the formula x dot u times the vector u. This was a projection onto a line."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Some of the span of some unit vector. We assume this has length 1. Then the projection onto a line just simplified to the formula x dot u times the vector u. This was a projection onto a line. Notice when we're dealing with an orthonormal basis for a subspace, when you take a projection of any vector in Rn onto that subspace, it's essentially you're just finding the projection onto the line spanned by each of these vectors. x dot V1 times the vector V1. x times V1 times the vector V1."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This was a projection onto a line. Notice when we're dealing with an orthonormal basis for a subspace, when you take a projection of any vector in Rn onto that subspace, it's essentially you're just finding the projection onto the line spanned by each of these vectors. x dot V1 times the vector V1. x times V1 times the vector V1. You're taking x's projections onto the line spanned by each of these guys. That's all it is. But clearly this is a much, much simpler way of finding a projection than going through this mess of saying a times the inverse of a transpose a times x."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "x times V1 times the vector V1. You're taking x's projections onto the line spanned by each of these guys. That's all it is. But clearly this is a much, much simpler way of finding a projection than going through this mess of saying a times the inverse of a transpose a times x. This is clearly a lot easier. But you might say, OK, this is easier. But you told me that a projection is a linear transformation."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But clearly this is a much, much simpler way of finding a projection than going through this mess of saying a times the inverse of a transpose a times x. This is clearly a lot easier. But you might say, OK, this is easier. But you told me that a projection is a linear transformation. You've told me it's a linear transformation, so I want to figure out the matrix here. So let's see if being orthonormal in any way simplifies this. So we could always just figure out for any particular x, we can just apply the dot product with each of the basis vectors."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But you told me that a projection is a linear transformation. You've told me it's a linear transformation, so I want to figure out the matrix here. So let's see if being orthonormal in any way simplifies this. So we could always just figure out for any particular x, we can just apply the dot product with each of the basis vectors. Those will be the coefficients. And then apply those coefficients times the basis vectors, add them up, and you know your projection. But some of us might actually want the transformation matrix, so let's figure out what it is."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So we could always just figure out for any particular x, we can just apply the dot product with each of the basis vectors. Those will be the coefficients. And then apply those coefficients times the basis vectors, add them up, and you know your projection. But some of us might actually want the transformation matrix, so let's figure out what it is. So let me just rewrite what we already know. We already know that the projection onto any subspace V of x is equal to a times a transpose a inverse times a times x. And where a's column vectors are just the basis vectors."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But some of us might actually want the transformation matrix, so let's figure out what it is. So let me just rewrite what we already know. We already know that the projection onto any subspace V of x is equal to a times a transpose a inverse times a times x. And where a's column vectors are just the basis vectors. V1, V2, all the way to Vk. Now, let's see if the assumption that these guys are an orthonormal basis, let's see if this simplifies it at all. Let's take the case in particular of a transpose a. a transpose a is going to be equal to what?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And where a's column vectors are just the basis vectors. V1, V2, all the way to Vk. Now, let's see if the assumption that these guys are an orthonormal basis, let's see if this simplifies it at all. Let's take the case in particular of a transpose a. a transpose a is going to be equal to what? It's going to be equal to a transpose. Let's think about this. These guys are members of Rn, so this is going to be an n by k matrix."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let's take the case in particular of a transpose a. a transpose a is going to be equal to what? It's going to be equal to a transpose. Let's think about this. These guys are members of Rn, so this is going to be an n by k matrix. So this is n by k. This guy right here is k by n times an n by k. We're going to have a k by k product. k by n times n by k is going to be k by k. a transpose a is going to be k by k. And what is a transpose equal to? Well, each of these columns are going to become rows."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "These guys are members of Rn, so this is going to be an n by k matrix. So this is n by k. This guy right here is k by n times an n by k. We're going to have a k by k product. k by n times n by k is going to be k by k. a transpose a is going to be k by k. And what is a transpose equal to? Well, each of these columns are going to become rows. So the first row here is going to be V1 transpose. The second column here is going to be V2 transpose. And you're going to go all the way down."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Well, each of these columns are going to become rows. So the first row here is going to be V1 transpose. The second column here is going to be V2 transpose. And you're going to go all the way down. The k-th column there is going to be Vk transpose. Just like that. And then a is, of course, this thing right there."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And you're going to go all the way down. The k-th column there is going to be Vk transpose. Just like that. And then a is, of course, this thing right there. So a looks like this. You have V1 at, you have V2 like that. And then you keep going and you have Vk, just like that."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then a is, of course, this thing right there. So a looks like this. You have V1 at, you have V2 like that. And then you keep going and you have Vk, just like that. What's going to happen when we take this product? Let's do a couple of rows right here. So when I take this product, I'm going to get a k by k matrix, let me write it big so I can explain it reasonably."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then you keep going and you have Vk, just like that. What's going to happen when we take this product? Let's do a couple of rows right here. So when I take this product, I'm going to get a k by k matrix, let me write it big so I can explain it reasonably. So what's the first row, first column going to be? It's going to be this row dotted with this column. Or V1 dot V1."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So when I take this product, I'm going to get a k by k matrix, let me write it big so I can explain it reasonably. So what's the first row, first column going to be? It's going to be this row dotted with this column. Or V1 dot V1. Well, V1 dot V1, that's nice. That's just 1. And then what's the second row, second column?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Or V1 dot V1. Well, V1 dot V1, that's nice. That's just 1. And then what's the second row, second column? Well, that's going to be V2. You're going to get your row from this guy and your column from that guy. This row dotted with that column."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then what's the second row, second column? Well, that's going to be V2. You're going to get your row from this guy and your column from that guy. This row dotted with that column. So V2 dot V2. So that's nice. That'll be a 1."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This row dotted with that column. So V2 dot V2. So that's nice. That'll be a 1. And in general, if you're finding the AII or you're finding anything along the diagonal, you're going to take the, let's say, the i-th row with the i-th column. So you're just going to have 1's that go all the way down the diagonal. Now what about everything else?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "That'll be a 1. And in general, if you're finding the AII or you're finding anything along the diagonal, you're going to take the, let's say, the i-th row with the i-th column. So you're just going to have 1's that go all the way down the diagonal. Now what about everything else? Let's say that you're looking for this entry right here, which is the first row, second column. This guy right here is going to be the dot product of V2. It's going to be the dot product of this row."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now what about everything else? Let's say that you're looking for this entry right here, which is the first row, second column. This guy right here is going to be the dot product of V2. It's going to be the dot product of this row. Sorry, the dot product of V1. It's going to be the dot product of this row with this column right there. So this is going to be V1 dot V2."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be the dot product of this row. Sorry, the dot product of V1. It's going to be the dot product of this row with this column right there. So this is going to be V1 dot V2. But these two guys are orthogonal. So what's that going to be equal to? It's going to be equal to 0."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be V1 dot V2. But these two guys are orthogonal. So what's that going to be equal to? It's going to be equal to 0. This one right here is going to be V1 dot V3. Well, that's going to be 0. V1 dot anything other than V1 is going to be 0."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be equal to 0. This one right here is going to be V1 dot V3. Well, that's going to be 0. V1 dot anything other than V1 is going to be 0. Similarly, everything here in the second row, it's going to be V2, the first column in the second row is going to be V2 dot V1, which is clearly 0. Then you have V2 dot V2, which is 1. And then V2 dot all the rest of the stuff is going to be 0."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "V1 dot anything other than V1 is going to be 0. Similarly, everything here in the second row, it's going to be V2, the first column in the second row is going to be V2 dot V1, which is clearly 0. Then you have V2 dot V2, which is 1. And then V2 dot all the rest of the stuff is going to be 0. They're all orthogonal with respect to each other. And so everything else is, if your row and your column is not the same, well, if your row and your column is the same, you're going to be dotting the same vector. So you're going to be getting 1, because all their lengths are 1."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then V2 dot all the rest of the stuff is going to be 0. They're all orthogonal with respect to each other. And so everything else is, if your row and your column is not the same, well, if your row and your column is the same, you're going to be dotting the same vector. So you're going to be getting 1, because all their lengths are 1. But if your row and column are not the same, you're going to be taking the dot product of two different members of your orthonormal basis. And they're all orthogonal, so you're just going to get a bunch of 0's. Now, what is this?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to be getting 1, because all their lengths are 1. But if your row and column are not the same, you're going to be taking the dot product of two different members of your orthonormal basis. And they're all orthogonal, so you're just going to get a bunch of 0's. Now, what is this? You have 0's everywhere, with 1's down the diagonal. It's a k by k matrix. This is the identity matrix in Rk."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this? You have 0's everywhere, with 1's down the diagonal. It's a k by k matrix. This is the identity matrix in Rk. So that simply, if you assume, so normally this was our definition, or this is our way of finding our transformation matrix for the projection of x onto some subspace. But that simply, if we assume an orthonormal basis, then A transpose A becomes the k by k identity matrix. And so what's the inverse of the identity matrix?"}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This is the identity matrix in Rk. So that simply, if you assume, so normally this was our definition, or this is our way of finding our transformation matrix for the projection of x onto some subspace. But that simply, if we assume an orthonormal basis, then A transpose A becomes the k by k identity matrix. And so what's the inverse of the identity matrix? So A transpose A inverse becomes the inverse of the k by k identity matrix, which is just the k by k identity matrix. So this simplifies to the projection onto v of our vector x simplifies to A times the inverse of the identity matrix, which is just the identity matrix. So it's just A times ik times A transpose."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And so what's the inverse of the identity matrix? So A transpose A inverse becomes the inverse of the k by k identity matrix, which is just the k by k identity matrix. So this simplifies to the projection onto v of our vector x simplifies to A times the inverse of the identity matrix, which is just the identity matrix. So it's just A times ik times A transpose. I always forget that second A transpose right there. Times x. And we could just ignore this."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So it's just A times ik times A transpose. I always forget that second A transpose right there. Times x. And we could just ignore this. That does nothing to it. So it's just equal to A times A transpose times x. Which is a huge simplification."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And we could just ignore this. That does nothing to it. So it's just equal to A times A transpose times x. Which is a huge simplification. I still have to do a matrix-matrix product. But finding the transpose of a matrix is pretty straightforward. You just switch the rows and the column."}, {"video_title": "Projections onto subspaces with orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Which is a huge simplification. I still have to do a matrix-matrix product. But finding the transpose of a matrix is pretty straightforward. You just switch the rows and the column. Finding the inverse, first multiplying the transpose times A, that's a lot of work. But then it's a huge amount of work to find the inverse of this thing. But now since we assumed that these columns here are all, they form an orthonormal set, this just gets reduced to the identity matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is our definition right here, AD minus BC. And then we were able to broaden that a bit by creating a definition for the determinant of a 3 by 3 matrix, and we did that right here. Where we essentially said the determinant is equal to each of these terms, you could call these maybe the coefficient terms, times the determinant of the matrix, you can kind of view it as a sub-matrix produced when you get rid of each of these guys, column and row. So when you got rid of this guy's column and row, you're left with this matrix. So we said this guy times the determinant of this. And then we kept switching signs, minus this guy times the determinant if you remove his column and his row. So it was left with these terms right there to get that determinant."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So when you got rid of this guy's column and row, you're left with this matrix. So we said this guy times the determinant of this. And then we kept switching signs, minus this guy times the determinant if you remove his column and his row. So it was left with these terms right there to get that determinant. Then finally, you switch signs again. So plus this guy times the determinant of the 2 by 2 matrix if you get rid of this row and this column. So it's this thing right here, which was this matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it was left with these terms right there to get that determinant. Then finally, you switch signs again. So plus this guy times the determinant of the 2 by 2 matrix if you get rid of this row and this column. So it's this thing right here, which was this matrix. Now let's see if we can extend this to a general n by n matrix. So let's write out our n by n matrix right over here. Do it in blue."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's this thing right here, which was this matrix. Now let's see if we can extend this to a general n by n matrix. So let's write out our n by n matrix right over here. Do it in blue. So let's say I have some matrix A that is an n by n matrix. So it's going to look like this. This would be A1 1, that would be A1 2, and it would go all the way to, you're going to have n columns, A1 n. And when you go down, this is going to be your second row, A2 1, it's going to go all the way down to A n 1, because you have n rows as well."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Do it in blue. So let's say I have some matrix A that is an n by n matrix. So it's going to look like this. This would be A1 1, that would be A1 2, and it would go all the way to, you're going to have n columns, A1 n. And when you go down, this is going to be your second row, A2 1, it's going to go all the way down to A n 1, because you have n rows as well. And then if you go down the diagonal all the way, this right here would be A n n. So there is my n by n matrix. Now, before I define how to find the determinant of this, let me make another definition. Let me define, so this is my matrix A."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This would be A1 1, that would be A1 2, and it would go all the way to, you're going to have n columns, A1 n. And when you go down, this is going to be your second row, A2 1, it's going to go all the way down to A n 1, because you have n rows as well. And then if you go down the diagonal all the way, this right here would be A n n. So there is my n by n matrix. Now, before I define how to find the determinant of this, let me make another definition. Let me define, so this is my matrix A. Let me define a sub matrix, Aij, to be equal to, this is n by n, right? So this is going to be an n minus 1 by n minus 1 matrix. So if this is 7 by 7, the sub matrix is going to be 6 by 6, 1 less in each direction."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me define, so this is my matrix A. Let me define a sub matrix, Aij, to be equal to, this is n by n, right? So this is going to be an n minus 1 by n minus 1 matrix. So if this is 7 by 7, the sub matrix is going to be 6 by 6, 1 less in each direction. So this is going to be the n minus 1 by n minus 1 matrix you get if you essentially ignore, or if you take away, maybe I should say take away. I'm going to say ignore, I like the word ignore. If you ignore the i-th row, this right here is the row, the i-th row, and the j-th column of A."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this is 7 by 7, the sub matrix is going to be 6 by 6, 1 less in each direction. So this is going to be the n minus 1 by n minus 1 matrix you get if you essentially ignore, or if you take away, maybe I should say take away. I'm going to say ignore, I like the word ignore. If you ignore the i-th row, this right here is the row, the i-th row, and the j-th column of A. So for example, let's go back to our 3 by 3 right here. This thing could be denoted, based on that definition, we could have called this, this was A11, this term right here. We could denote the matrix when you get rid of the first column and the first row, or the first row and the first column, we could call this thing right here, we could call that big matrix A11."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you ignore the i-th row, this right here is the row, the i-th row, and the j-th column of A. So for example, let's go back to our 3 by 3 right here. This thing could be denoted, based on that definition, we could have called this, this was A11, this term right here. We could denote the matrix when you get rid of the first column and the first row, or the first row and the first column, we could call this thing right here, we could call that big matrix A11. So this was big matrix A11. This is big matrix A21. Or actually, this matrix was called C, so this would be C11."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could denote the matrix when you get rid of the first column and the first row, or the first row and the first column, we could call this thing right here, we could call that big matrix A11. So this was big matrix A11. This is big matrix A21. Or actually, this matrix was called C, so this would be C11. This would, right there, C11. We could call this one, this would be matrix C12. Why is that?"}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or actually, this matrix was called C, so this would be C11. This would, right there, C11. We could call this one, this would be matrix C12. Why is that? Because if you get rid of the first row, let me get rid of the first row, the first term is your row. If you get rid of the first row and the second column, this is the matrix that's left over. 2, 3, 4, 1."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Why is that? Because if you get rid of the first row, let me get rid of the first row, the first term is your row. If you get rid of the first row and the second column, this is the matrix that's left over. 2, 3, 4, 1. So this is this guy and this guy. 2, 3, 4, 1. So this is the sub-matrix C, because this is the big matrix C, but this one is C12."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2, 3, 4, 1. So this is this guy and this guy. 2, 3, 4, 1. So this is the sub-matrix C, because this is the big matrix C, but this one is C12. I know it's a little bit messy there. C12. So that's all we mean by the sub-matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the sub-matrix C, because this is the big matrix C, but this one is C12. I know it's a little bit messy there. C12. So that's all we mean by the sub-matrix. Very similar to what we did in the 3 by 3 case. You essentially get rid of, so if you want to find out this guy's sub-matrix, you would call that A11, and you would literally cross out the first row and the first column, and everything left over here would be that sub-matrix. Now, with that out of the way, we can create a definition."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's all we mean by the sub-matrix. Very similar to what we did in the 3 by 3 case. You essentially get rid of, so if you want to find out this guy's sub-matrix, you would call that A11, and you would literally cross out the first row and the first column, and everything left over here would be that sub-matrix. Now, with that out of the way, we can create a definition. And it might seem a little circular to you at first. And on some level, it is. We're going to define the determinant of A to be equal to, and this is interesting."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, with that out of the way, we can create a definition. And it might seem a little circular to you at first. And on some level, it is. We're going to define the determinant of A to be equal to, and this is interesting. It's actually a recursive definition. I'll talk about that in a second. It's equal to, we start with a plus."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're going to define the determinant of A to be equal to, and this is interesting. It's actually a recursive definition. I'll talk about that in a second. It's equal to, we start with a plus. It's equal to A11 times the sub-matrix. If you remove this guy's row and column. So that, by definition, is just A, big capital A, 11's determinant."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to, we start with a plus. It's equal to A11 times the sub-matrix. If you remove this guy's row and column. So that, by definition, is just A, big capital A, 11's determinant. So that's exactly what we did. Let me write that a little bit neater. So times the determinant of its sub-matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that, by definition, is just A, big capital A, 11's determinant. So that's exactly what we did. Let me write that a little bit neater. So times the determinant of its sub-matrix. So the determinant of A11. So you take A11, you get rid of its column and its row, or its row and its column, and everything else, you find the determinant of that. Actually, let me write it in terms of, let me write it this way."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So times the determinant of its sub-matrix. So the determinant of A11. So you take A11, you get rid of its column and its row, or its row and its column, and everything else, you find the determinant of that. Actually, let me write it in terms of, let me write it this way. A11 times the determinant of the sub-matrix A11. And then we switch sides. We're just going to go along this row."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me write it in terms of, let me write it this way. A11 times the determinant of the sub-matrix A11. And then we switch sides. We're just going to go along this row. We're just going to go along this row. And then you do minus A12 times the determinant of its sub-matrix, which we'll just call A12. We would get rid of this row and this column, and everything left would be this matrix A12."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're just going to go along this row. We're just going to go along this row. And then you do minus A12 times the determinant of its sub-matrix, which we'll just call A12. We would get rid of this row and this column, and everything left would be this matrix A12. We want to find its determinant. And then we'll take the next guy over here, who'd be A13. So we switch sides, we went minus."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We would get rid of this row and this column, and everything left would be this matrix A12. We want to find its determinant. And then we'll take the next guy over here, who'd be A13. So we switch sides, we went minus. Now you go plus. So A13 times the determinant of its sub-matrix. So if this is n by n, these are each going to be n minus 1 by n minus 1."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we switch sides, we went minus. Now you go plus. So A13 times the determinant of its sub-matrix. So if this is n by n, these are each going to be n minus 1 by n minus 1. So the determinant of A13. And you're just going to keep doing that, keep switching sides. So there's going to be a minus and then a plus."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this is n by n, these are each going to be n minus 1 by n minus 1. So the determinant of A13. And you're just going to keep doing that, keep switching sides. So there's going to be a minus and then a plus. And you keep going all the way. And then, I don't know, it depends on whether An, whether we're dealing with an odd number or an even number. If we're dealing with an even number, this is going to be a minus sign."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So there's going to be a minus and then a plus. And you keep going all the way. And then, I don't know, it depends on whether An, whether we're dealing with an odd number or an even number. If we're dealing with an even number, this is going to be a minus sign. If it's an odd number, it's going to be a plus sign. But you get the idea. It's either going to be a plus or a minus."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we're dealing with an even number, this is going to be a minus sign. If it's an odd number, it's going to be a plus sign. But you get the idea. It's either going to be a plus or a minus. Not just if it's odd, this is going to be a plus. If it's an even n, it's going to be a minus. All the way to A1n, the nth column, times its sub-matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's either going to be a plus or a minus. Not just if it's odd, this is going to be a plus. If it's an even n, it's going to be a minus. All the way to A1n, the nth column, times its sub-matrix. A1n. With that sub-matrix, you get rid of the first row and the nth column, and it's going to be everything that's left in between. And you immediately might say, Sal, what kind of a definition is this?"}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All the way to A1n, the nth column, times its sub-matrix. A1n. With that sub-matrix, you get rid of the first row and the nth column, and it's going to be everything that's left in between. And you immediately might say, Sal, what kind of a definition is this? You've defined a determinant for an arbitrary n by n matrix in terms of another definition of a determinant. How does this work? And the reason why this works is because the determinants that you use in the definition are determinants of a smaller matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you immediately might say, Sal, what kind of a definition is this? You've defined a determinant for an arbitrary n by n matrix in terms of another definition of a determinant. How does this work? And the reason why this works is because the determinants that you use in the definition are determinants of a smaller matrix. So this is a determinant of an n minus 1 by n minus 1 matrix. And you're saying, hey, Sal, that still doesn't make any sense, because we don't know how to find the determinant of an n minus 1 by n minus 1 matrix. Well, you apply this definition again, and then it's going to be in terms of n minus 2 by n minus 2 matrices."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the reason why this works is because the determinants that you use in the definition are determinants of a smaller matrix. So this is a determinant of an n minus 1 by n minus 1 matrix. And you're saying, hey, Sal, that still doesn't make any sense, because we don't know how to find the determinant of an n minus 1 by n minus 1 matrix. Well, you apply this definition again, and then it's going to be in terms of n minus 2 by n minus 2 matrices. And you're like, how do you do that? Well, you keep doing it, and you're going to get all the way down to a 2 by 2 matrix. You're going to get all the way down to a 2 by 2 matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, you apply this definition again, and then it's going to be in terms of n minus 2 by n minus 2 matrices. And you're like, how do you do that? Well, you keep doing it, and you're going to get all the way down to a 2 by 2 matrix. You're going to get all the way down to a 2 by 2 matrix. And that one we defined well. We defined the determinant of a 2 by 2 matrix not in terms of a determinant. We just defined it in terms of a times d minus b times c. And you can see, I mean, we could just go down to the 3 by 3, but the 2 by 2 is really the most fundamental definition."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're going to get all the way down to a 2 by 2 matrix. And that one we defined well. We defined the determinant of a 2 by 2 matrix not in terms of a determinant. We just defined it in terms of a times d minus b times c. And you can see, I mean, we could just go down to the 3 by 3, but the 2 by 2 is really the most fundamental definition. And you could see that the definition of a 3 by 3 determinant is essentially a special case of the general case for an n by n. We take this guy, and we multiply him times the determinant of his sub-matrix right there. Then we take this guy, where we switch signs, we have a minus, and we multiply him times the determinant of his sub-matrix, which is that right there. And then you do a plus."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just defined it in terms of a times d minus b times c. And you can see, I mean, we could just go down to the 3 by 3, but the 2 by 2 is really the most fundamental definition. And you could see that the definition of a 3 by 3 determinant is essentially a special case of the general case for an n by n. We take this guy, and we multiply him times the determinant of his sub-matrix right there. Then we take this guy, where we switch signs, we have a minus, and we multiply him times the determinant of his sub-matrix, which is that right there. And then you do a plus. You switch signs, and then you multiply this guy times the determinant of his sub-matrix, which is that right there. So this is a general case of what I just defined. But we know it's never that satisfying to deal in the abstract or the generalities."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you do a plus. You switch signs, and then you multiply this guy times the determinant of his sub-matrix, which is that right there. So this is a general case of what I just defined. But we know it's never that satisfying to deal in the abstract or the generalities. We want to do a specific case. And actually, before I do that, let me just introduce a term to you. This is called a recursive formula."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But we know it's never that satisfying to deal in the abstract or the generalities. We want to do a specific case. And actually, before I do that, let me just introduce a term to you. This is called a recursive formula. And if you become a computer science major, you'll see this a lot, but a recursive function or recursive formula is defined in terms of itself. But the things that you use in the definition use a slightly simpler version of it. And as you keep going through, you keep recursing through it, you get simpler and simpler versions of it until you get to some type of base case."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is called a recursive formula. And if you become a computer science major, you'll see this a lot, but a recursive function or recursive formula is defined in terms of itself. But the things that you use in the definition use a slightly simpler version of it. And as you keep going through, you keep recursing through it, you get simpler and simpler versions of it until you get to some type of base case. In this case, our base case is the case of a 2 by 2 matrix. You keep doing this, and eventually you'll get to a determinant of a 2 by 2 matrix. And we know how to find those."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And as you keep going through, you keep recursing through it, you get simpler and simpler versions of it until you get to some type of base case. In this case, our base case is the case of a 2 by 2 matrix. You keep doing this, and eventually you'll get to a determinant of a 2 by 2 matrix. And we know how to find those. So this is a recursive definition. But let's actually apply it, because I think that's what actually makes things concrete. So let's take a, this is going to be computationally intensive, but I think if we focus, we can get there."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we know how to find those. So this is a recursive definition. But let's actually apply it, because I think that's what actually makes things concrete. So let's take a, this is going to be computationally intensive, but I think if we focus, we can get there. So I'm going to have a 4 by 4 matrix. 1, 2, 3, 4, 1, throw some 0's in there to make the computation a little bit simpler. 0, 1, 2, 3, and then 2, 3, 0, 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's take a, this is going to be computationally intensive, but I think if we focus, we can get there. So I'm going to have a 4 by 4 matrix. 1, 2, 3, 4, 1, throw some 0's in there to make the computation a little bit simpler. 0, 1, 2, 3, and then 2, 3, 0, 0. So let's figure out this determinant. Let's figure out this determinant right there. This is the determinant of the matrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0, 1, 2, 3, and then 2, 3, 0, 0. So let's figure out this determinant. Let's figure out this determinant right there. This is the determinant of the matrix. If I put some brackets there, that would have been the matrix, but let's find the determinant of this matrix. So this is going to be equal to, by our definition, it's going to be equal to 1 times the determinant of this matrix right here. If you get rid of this row and this column, it's going to be 1 times the determinant of 0, 2, 0, 1, 2, 3, 3, 0, 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the determinant of the matrix. If I put some brackets there, that would have been the matrix, but let's find the determinant of this matrix. So this is going to be equal to, by our definition, it's going to be equal to 1 times the determinant of this matrix right here. If you get rid of this row and this column, it's going to be 1 times the determinant of 0, 2, 0, 1, 2, 3, 3, 0, 0. That's just this guy right here. This matrix right there. Then I'm going to have a 2, but I'm going to switch signs."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you get rid of this row and this column, it's going to be 1 times the determinant of 0, 2, 0, 1, 2, 3, 3, 0, 0. That's just this guy right here. This matrix right there. Then I'm going to have a 2, but I'm going to switch signs. So it's minus 2. Minus 2 times the determinant, if I get rid of that row and this column, so it's 1, 2, 0. I'm ignoring the 0 because it's in the same column as the 2."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then I'm going to have a 2, but I'm going to switch signs. So it's minus 2. Minus 2 times the determinant, if I get rid of that row and this column, so it's 1, 2, 0. I'm ignoring the 0 because it's in the same column as the 2. 1, 2, 0. So 1, 2, 0. 0, 2, 3."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm ignoring the 0 because it's in the same column as the 2. 1, 2, 0. So 1, 2, 0. 0, 2, 3. 0, 2, 3. And then 2, 0, 0. And then I switch signs again."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0, 2, 3. 0, 2, 3. And then 2, 0, 0. And then I switch signs again. It was a minus, so now I go back to plus. So I do that guy. So plus 3 times the determinant of his submatrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then I switch signs again. It was a minus, so now I go back to plus. So I do that guy. So plus 3 times the determinant of his submatrix. Get rid of that row, get rid of that column. I get a 1, 0, 0. I get a 0, 1, 3."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So plus 3 times the determinant of his submatrix. Get rid of that row, get rid of that column. I get a 1, 0, 0. I get a 0, 1, 3. I skip this column every time. Then I get a 2, 3, 0. I get a 2, 3, 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I get a 0, 1, 3. I skip this column every time. Then I get a 2, 3, 0. I get a 2, 3, 0. Just like that. We're almost done. One more in this column."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I get a 2, 3, 0. Just like that. We're almost done. One more in this column. Let me switch to another color. I haven't used the blue in this yet. So then I'm going to do a minus 4."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "One more in this column. Let me switch to another color. I haven't used the blue in this yet. So then I'm going to do a minus 4. Remember, it's plus, minus, plus, minus 4 times the determinant of its submatrix. That's going to be that right there. So it's 1, 0, 2."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So then I'm going to do a minus 4. Remember, it's plus, minus, plus, minus 4 times the determinant of its submatrix. That's going to be that right there. So it's 1, 0, 2. 0, 1, 2. 2, 3, 0. Just like that."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's 1, 0, 2. 0, 1, 2. 2, 3, 0. Just like that. And now we're down to the 3 by 3 case. We could use the definition of the 3 by 3, but we could just keep applying this recursive definition. We could keep applying this recursive definition here."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. And now we're down to the 3 by 3 case. We could use the definition of the 3 by 3, but we could just keep applying this recursive definition. We could keep applying this recursive definition here. So this is going to be equal to, let me write it here. So 1 times, what's this determinant? This determinant is going to be 0 times the determinant of that submatrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could keep applying this recursive definition here. So this is going to be equal to, let me write it here. So 1 times, what's this determinant? This determinant is going to be 0 times the determinant of that submatrix. 2, 3, 0, 0. That was this one right here. And then we have minus 2, minus this 2."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This determinant is going to be 0 times the determinant of that submatrix. 2, 3, 0, 0. That was this one right here. And then we have minus 2, minus this 2. Remember, we switch signs. Plus, minus, plus. So minus 2 times its submatrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we have minus 2, minus this 2. Remember, we switch signs. Plus, minus, plus. So minus 2 times its submatrix. It's 1, 3, 3, 0. 1, 3, 3, 0. And then finally, plus 0 times its submatrix, which is this thing right here."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So minus 2 times its submatrix. It's 1, 3, 3, 0. 1, 3, 3, 0. And then finally, plus 0 times its submatrix, which is this thing right here. 1, 2, 3, 0. 1, 2, 3, 0. Just like that."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, plus 0 times its submatrix, which is this thing right here. 1, 2, 3, 0. 1, 2, 3, 0. Just like that. And then we have this next guy right here. As you can see, this can get a little bit tedious, but we'll keep our spirits up. So minus 2 times 1 times its submatrix."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. And then we have this next guy right here. As you can see, this can get a little bit tedious, but we'll keep our spirits up. So minus 2 times 1 times its submatrix. So that's this guy right here. Times the determinant of its submatrix. 2, 3, 0, 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So minus 2 times 1 times its submatrix. So that's this guy right here. Times the determinant of its submatrix. 2, 3, 0, 0. Then minus 2 times, get rid of that row, that column, 0, 3, 2, 0. 0, 3, 2, 0. And then plus 0 times 0, 2, 2, 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2, 3, 0, 0. Then minus 2 times, get rid of that row, that column, 0, 3, 2, 0. 0, 3, 2, 0. And then plus 0 times 0, 2, 2, 0. That's this one right there. Halfway there, at least for now. And then we get this next one."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then plus 0 times 0, 2, 2, 0. That's this one right there. Halfway there, at least for now. And then we get this next one. So we have a plus 3. Bring out our parentheses. And then we're going to have 1 times its sub, I guess call it subdeterminant."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we get this next one. So we have a plus 3. Bring out our parentheses. And then we're going to have 1 times its sub, I guess call it subdeterminant. So 1 times the determinant 1, 3, 3, 0. You get rid of this guy's row and column. You get this guy right there."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to have 1 times its sub, I guess call it subdeterminant. So 1 times the determinant 1, 3, 3, 0. You get rid of this guy's row and column. You get this guy right there. And then minus 0, minus 0, get rid of this row and column, times 0, 3, 2, 0. And then you have plus 0 times its subdeterminant, 0, 1, 2, 3. 3 4ths of the way there."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get this guy right there. And then minus 0, minus 0, get rid of this row and column, times 0, 3, 2, 0. And then you have plus 0 times its subdeterminant, 0, 1, 2, 3. 3 4ths of the way there. One last term. Let's hope we haven't made any careless mistakes. Minus 4 times 1 times 1, 2, 3, 0, right there."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "3 4ths of the way there. One last term. Let's hope we haven't made any careless mistakes. Minus 4 times 1 times 1, 2, 3, 0, right there. 1, 2, 3, 0. Minus 0 times, get rid of those two guys, 0, 2, 2, 0. And then plus 2 times 0, 1, 2, 3."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 4 times 1 times 1, 2, 3, 0, right there. 1, 2, 3, 0. Minus 0 times, get rid of those two guys, 0, 2, 2, 0. And then plus 2 times 0, 1, 2, 3. Plus 2, get rid of these guys, 0, 1, 2, 3. Now we've defined our, or we've calculated, or we've defined our determinant of this matrix in terms of just a bunch of 2 by 2 matrices. So hopefully you saw in this example that their recursion worked out."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then plus 2 times 0, 1, 2, 3. Plus 2, get rid of these guys, 0, 1, 2, 3. Now we've defined our, or we've calculated, or we've defined our determinant of this matrix in terms of just a bunch of 2 by 2 matrices. So hopefully you saw in this example that their recursion worked out. So let's actually find what this number is equal to. A determinant is always just going to be a number. So let me get a nice vibrant color."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So hopefully you saw in this example that their recursion worked out. So let's actually find what this number is equal to. A determinant is always just going to be a number. So let me get a nice vibrant color. So this is a 0 times, I don't care, 0 times anything is going to be 0. 0 times anything is going to be 0. 0 times anything is going to be 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me get a nice vibrant color. So this is a 0 times, I don't care, 0 times anything is going to be 0. 0 times anything is going to be 0. 0 times anything is going to be 0. 0 times anything is going to be 0. Just simplifying it. These guys are 0, because it's 0 times that."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0 times anything is going to be 0. 0 times anything is going to be 0. Just simplifying it. These guys are 0, because it's 0 times that. 0 times this is going to be equal to 0. So what are we left with? This is going to be equal to 1 times, this is all we have left here is a minus 2, times, and what does this determine, it's 1 times 0, which is 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These guys are 0, because it's 0 times that. 0 times this is going to be equal to 0. So what are we left with? This is going to be equal to 1 times, this is all we have left here is a minus 2, times, and what does this determine, it's 1 times 0, which is 0. It's 0, let me write this, this is going to be 1 times 0 is 0, minus 3 times 3, 0 minus 9. So minus 9. This right here is just minus 9."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to 1 times, this is all we have left here is a minus 2, times, and what does this determine, it's 1 times 0, which is 0. It's 0, let me write this, this is going to be 1 times 0 is 0, minus 3 times 3, 0 minus 9. So minus 9. This right here is just minus 9. So minus 2 times minus 9. That's our first thing. I'll simplify it in a second."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This right here is just minus 9. So minus 2 times minus 9. That's our first thing. I'll simplify it in a second. Now let's do this term right here. So it's minus 2 times. Now what's this determine?"}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll simplify it in a second. Now let's do this term right here. So it's minus 2 times. Now what's this determine? 2 times 0 minus 0 times 3. That's 0 minus 0. So this is 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what's this determine? 2 times 0 minus 0 times 3. That's 0 minus 0. So this is 0. That guy became 0. So we can ignore that term. This term right here is 0 times 0, which is 0, minus 2 times 3, so it's minus 6."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is 0. That guy became 0. So we can ignore that term. This term right here is 0 times 0, which is 0, minus 2 times 3, so it's minus 6. So it's minus 2 times, so this is a minus 6 right here. You have a minus 2 times a minus 6, so that's a plus 12. So I'll just write a plus 12 here."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This term right here is 0 times 0, which is 0, minus 2 times 3, so it's minus 6. So it's minus 2 times, so this is a minus 6 right here. You have a minus 2 times a minus 6, so that's a plus 12. So I'll just write a plus 12 here. This minus 2 is that minus 2 right there. And then we have a plus 3. We have this plus 3."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'll just write a plus 12 here. This minus 2 is that minus 2 right there. And then we have a plus 3. We have this plus 3. And then this first term is 1 times 0, which is 0, minus, let me make a parentheses here, 1 times 0, which is 0, minus 3 times 3, which is minus 9, times 1. So it's minus 9. Everything else was a 0."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have this plus 3. And then this first term is 1 times 0, which is 0, minus, let me make a parentheses here, 1 times 0, which is 0, minus 3 times 3, which is minus 9, times 1. So it's minus 9. Everything else was a 0. We're in the home stretch. We have a minus 4. Let's see, this is 1 times 0, which is 0, minus 3 times 2, so minus 6."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Everything else was a 0. We're in the home stretch. We have a minus 4. Let's see, this is 1 times 0, which is 0, minus 3 times 2, so minus 6. So this is minus 6 right here. This is 0. And then we have this guy right here."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, this is 1 times 0, which is 0, minus 3 times 2, so minus 6. So this is minus 6 right here. This is 0. And then we have this guy right here. So we have 0 times 3, which is 0, minus 2 times 1. So that's minus 2. And then you have a minus 2 times a plus 2 is minus 4."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we have this guy right here. So we have 0 times 3, which is 0, minus 2 times 1. So that's minus 2. And then you have a minus 2 times a plus 2 is minus 4. So now we just have to make sure we do our arithmetic properly. This is 1 times plus 18, so this is 18. Minus 2 times minus 9."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have a minus 2 times a plus 2 is minus 4. So now we just have to make sure we do our arithmetic properly. This is 1 times plus 18, so this is 18. Minus 2 times minus 9. This right here is minus 24. This right here is minus 27. And then this right here, let's see, this is minus 10 right here."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 times minus 9. This right here is minus 24. This right here is minus 27. And then this right here, let's see, this is minus 10 right here. That is minus 10. Minus 4 times minus 10 is plus 40. And let's see if we can simplify this a little bit."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then this right here, let's see, this is minus 10 right here. That is minus 10. Minus 4 times minus 10 is plus 40. And let's see if we can simplify this a little bit. I don't want to make a careless mistake right at the end. So 18 minus 24. 24 minus 18 is 6."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's see if we can simplify this a little bit. I don't want to make a careless mistake right at the end. So 18 minus 24. 24 minus 18 is 6. So this is going to be equal to minus 6. 18 minus 24 is minus 6. And then we do it in green."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "24 minus 18 is 6. So this is going to be equal to minus 6. 18 minus 24 is minus 6. And then we do it in green. Now what's the difference if we have minus 27 plus 40? That's 13. That's positive 13."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we do it in green. Now what's the difference if we have minus 27 plus 40? That's 13. That's positive 13. So minus 6 plus positive 13 is equal to 7. And so we are done. After all of that computation, hopefully we haven't made a careless mistake, the determinant of this character right here is equal to 7."}, {"video_title": "n x n determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's positive 13. So minus 6 plus positive 13 is equal to 7. And so we are done. After all of that computation, hopefully we haven't made a careless mistake, the determinant of this character right here is equal to 7. The determinant is equal to 7. And so the one useful takeaway, we know that this is invertible because it has a non-zero determinant. Hopefully you found that useful."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I have an m by n matrix, the m is just the number of rows, and then the n is just the number of columns. So let me write out the m by n matrix. So I'll just specify, let's have the m by n matrix A. It's a capital bold A. And it's equal to, I'll be as general as possible, first entry is in, I'll just call that lowercase a. It's in row one, column one. The next entry is row one, column two."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's a capital bold A. And it's equal to, I'll be as general as possible, first entry is in, I'll just call that lowercase a. It's in row one, column one. The next entry is row one, column two. And you go all the way to row one, column n. You have n columns. And then when you go down, the next row will be row two, column one. And then you keep going all the way down to row m, column n, and then of course, this entry is going to be row two, let me write that a little smaller, row two, column two."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The next entry is row one, column two. And you go all the way to row one, column n. You have n columns. And then when you go down, the next row will be row two, column one. And then you keep going all the way down to row m, column n, and then of course, this entry is going to be row two, let me write that a little smaller, row two, column two. And you go all the way, and you're going to have row m, column n. And so if you think about it, you're going to have how many total entries here? You're going to have m entries this way, n that way. So you're going to have m times n total entries."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you keep going all the way down to row m, column n, and then of course, this entry is going to be row two, let me write that a little smaller, row two, column two. And you go all the way, and you're going to have row m, column n. And so if you think about it, you're going to have how many total entries here? You're going to have m entries this way, n that way. So you're going to have m times n total entries. And I think you're pretty familiar with this idea already of a matrix. You probably saw this in your Algebra 2 classes. So what we want to do now in this video is relate our notion of a matrix to everything we already know about vectors."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to have m times n total entries. And I think you're pretty familiar with this idea already of a matrix. You probably saw this in your Algebra 2 classes. So what we want to do now in this video is relate our notion of a matrix to everything we already know about vectors. Or maybe introduce some operations that allow matrix and vectors to interact with each other. And maybe the most natural one is multiplication, or taking the product. So what I'm going to do in this video is define what it means when we take the product of our matrix A, of any matrix A, I've written this as general as possible, with some vector x."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what we want to do now in this video is relate our notion of a matrix to everything we already know about vectors. Or maybe introduce some operations that allow matrix and vectors to interact with each other. And maybe the most natural one is multiplication, or taking the product. So what I'm going to do in this video is define what it means when we take the product of our matrix A, of any matrix A, I've written this as general as possible, with some vector x. And our definition will only work if only defined if x, if the vector we're multiplying A by, has the same number of components as A has columns. So this is only valid for an x that looks like this. x1, x2, all the way down to xn."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what I'm going to do in this video is define what it means when we take the product of our matrix A, of any matrix A, I've written this as general as possible, with some vector x. And our definition will only work if only defined if x, if the vector we're multiplying A by, has the same number of components as A has columns. So this is only valid for an x that looks like this. x1, x2, all the way down to xn. So let me very clear this. This vector right here, it could be a different, I guess you could view it at a different height than this vector. What matters is that the same number of A's you have in this direction, you have n A's here, then you have n components of this vector right here."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x1, x2, all the way down to xn. So let me very clear this. This vector right here, it could be a different, I guess you could view it at a different height than this vector. What matters is that the same number of A's you have in this direction, you have n A's here, then you have n components of this vector right here. And if you have that constraint, if the length of your vector, the number of components in vector, is equal to the number of columns in your matrix, then we define this product to be equal to, so that's my vector x. So this is a definition. There's nothing in nature that told us it had to be defined this way."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What matters is that the same number of A's you have in this direction, you have n A's here, then you have n components of this vector right here. And if you have that constraint, if the length of your vector, the number of components in vector, is equal to the number of columns in your matrix, then we define this product to be equal to, so that's my vector x. So this is a definition. There's nothing in nature that told us it had to be defined this way. It's just human beings or mathematicians decided that this is a useful convention to define the multiplication or the product of a matrix and a vector. So we're going to define A times our vector x. These are both bold, or this is a matrix, that's a vector."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's nothing in nature that told us it had to be defined this way. It's just human beings or mathematicians decided that this is a useful convention to define the multiplication or the product of a matrix and a vector. So we're going to define A times our vector x. These are both bold, or this is a matrix, that's a vector. And the convention, if I didn't draw the little vector symbol in your textbook, they'll just bold out the x. It'll be a lowercase x. Lowercase is vector, uppercase is matrix. Both of them are bolded."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These are both bold, or this is a matrix, that's a vector. And the convention, if I didn't draw the little vector symbol in your textbook, they'll just bold out the x. It'll be a lowercase x. Lowercase is vector, uppercase is matrix. Both of them are bolded. That tells you that you're not just dealing with regular numbers. So we're defining this to be equal to, let me write it out, fairly large. You're going to take each row, we're going to show you that there's multiple ways to kind of visualize this, but it's going to be a11 times x1."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Both of them are bolded. That tells you that you're not just dealing with regular numbers. So we're defining this to be equal to, let me write it out, fairly large. You're going to take each row, we're going to show you that there's multiple ways to kind of visualize this, but it's going to be a11 times x1. Let me write that down. So a11 times x1 plus a12 times x2 all the way to plus a1n times xn. So the product of this matrix, this m by n matrix, and this n component vector, will be a new vector, the first entry of which is essentially each of these entries times the corresponding entry here, and you add them all up."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're going to take each row, we're going to show you that there's multiple ways to kind of visualize this, but it's going to be a11 times x1. Let me write that down. So a11 times x1 plus a12 times x2 all the way to plus a1n times xn. So the product of this matrix, this m by n matrix, and this n component vector, will be a new vector, the first entry of which is essentially each of these entries times the corresponding entry here, and you add them all up. And as you can see, that's already looking fairly similar to a dot product. I'll discuss that in a second, but let me finish my definition before I start talking about what it means or what it might be related to. So that was that first row right there."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the product of this matrix, this m by n matrix, and this n component vector, will be a new vector, the first entry of which is essentially each of these entries times the corresponding entry here, and you add them all up. And as you can see, that's already looking fairly similar to a dot product. I'll discuss that in a second, but let me finish my definition before I start talking about what it means or what it might be related to. So that was that first row right there. It'll just look like that. We just multiply that times this thing to get that row there. Now the second row, I'll do it in this, I wanted to do it in a different color."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that was that first row right there. It'll just look like that. We just multiply that times this thing to get that row there. Now the second row, I'll do it in this, I wanted to do it in a different color. Remember, this is a definition. Human beings came up with this. Nothing about nature said we had to do it this way, but it's just nice and convenient."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now the second row, I'll do it in this, I wanted to do it in a different color. Remember, this is a definition. Human beings came up with this. Nothing about nature said we had to do it this way, but it's just nice and convenient. So our second row will have a21 times x1. We'll just do the whole thing over again, but this time we're multiplying this row times this column vector. So a21 times x1 plus a22 times x2 all the way until we get to a2n times xn."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Nothing about nature said we had to do it this way, but it's just nice and convenient. So our second row will have a21 times x1. We'll just do the whole thing over again, but this time we're multiplying this row times this column vector. So a21 times x1 plus a22 times x2 all the way until we get to a2n times xn. So we multiplied this entire row times that entire column, this term times that term plus this term plus this term, all the way down to plus this last term times that last term. And we keep doing this for every row until we get to the nth row, and then the nth row will be am1. Sorry, this is am1, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So a21 times x1 plus a22 times x2 all the way until we get to a2n times xn. So we multiplied this entire row times that entire column, this term times that term plus this term plus this term, all the way down to plus this last term times that last term. And we keep doing this for every row until we get to the nth row, and then the nth row will be am1. Sorry, this is am1, right? This is the nth row, first column. am1 times x1 plus am2 times x2 all the way until we get to amn times xn. So what is this vector going to look like?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Sorry, this is am1, right? This is the nth row, first column. am1 times x1 plus am2 times x2 all the way until we get to amn times xn. So what is this vector going to look like? It's essentially going to have, if we called this vector, let's say it's equal to vector b, what does vector b look like? How many entries is it going to have? Well, it has an entry for each row of this, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what is this vector going to look like? It's essentially going to have, if we called this vector, let's say it's equal to vector b, what does vector b look like? How many entries is it going to have? Well, it has an entry for each row of this, right? We're taking each row and we're multiplying it. We're essentially taking the dot product of this row vector with this column vector. And I'll be a little bit more formal with the notation in a second."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, it has an entry for each row of this, right? We're taking each row and we're multiplying it. We're essentially taking the dot product of this row vector with this column vector. And I'll be a little bit more formal with the notation in a second. But I think you understand that this is a dot product. We're taking each corresponding, the first component times the first component plus the second component times the second component plus the third component times the third component all the way to the nth component plus the nth component times the nth component. So this is essentially the dot product of this row vector."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I'll be a little bit more formal with the notation in a second. But I think you understand that this is a dot product. We're taking each corresponding, the first component times the first component plus the second component times the second component plus the third component times the third component all the way to the nth component plus the nth component times the nth component. So this is essentially the dot product of this row vector. Row vector is just a fan, it's just you're writing a row, we've been writing all of our vectors as columns. So we could call them column vectors. You're just writing them as rows."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is essentially the dot product of this row vector. Row vector is just a fan, it's just you're writing a row, we've been writing all of our vectors as columns. So we could call them column vectors. You're just writing them as rows. And we can be a little bit more specific with the notation in a second. But what's this going to look like? Well, we're doing this m times."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're just writing them as rows. And we can be a little bit more specific with the notation in a second. But what's this going to look like? Well, we're doing this m times. So we're going to have m entries. You're going to have b1, b2, all the way to bn. If you viewed these as all as matrices, you can kind of view it as you have a, and this will eventually work for the matrix math we're going to learn."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we're doing this m times. So we're going to have m entries. You're going to have b1, b2, all the way to bn. If you viewed these as all as matrices, you can kind of view it as you have a, and this will eventually work for the matrix math we're going to learn. This is an m by n matrix. And we're multiplying it by, how many rows does this guy have, he has n rows, right? He has n components."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you viewed these as all as matrices, you can kind of view it as you have a, and this will eventually work for the matrix math we're going to learn. This is an m by n matrix. And we're multiplying it by, how many rows does this guy have, he has n rows, right? He has n components. And he has one column. So m by n times an n by 1, you essentially, you can kind of ignore these middle two terms. And they will result with, how many rows does this guy have?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "He has n components. And he has one column. So m by n times an n by 1, you essentially, you can kind of ignore these middle two terms. And they will result with, how many rows does this guy have? He has m rows and one column. So if you take out these, these middle two terms have to be equal to each other, just for the multiplication to be defined. And then you're left with an m by 1 matrix."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And they will result with, how many rows does this guy have? He has m rows and one column. So if you take out these, these middle two terms have to be equal to each other, just for the multiplication to be defined. And then you're left with an m by 1 matrix. So this was all abstract. Let me actually apply it to some actual numbers. But it's important to actually set the definition."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you're left with an m by 1 matrix. So this was all abstract. Let me actually apply it to some actual numbers. But it's important to actually set the definition. Now that we have the definition, we can apply it to some actual matrices and vectors. So let's say we have the matrix. Let's say I want to multiply the matrix minus 3, 0, 3, 2."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But it's important to actually set the definition. Now that we have the definition, we can apply it to some actual matrices and vectors. So let's say we have the matrix. Let's say I want to multiply the matrix minus 3, 0, 3, 2. I'll do this one in yellow. 1, 7, minus 1, 9. And I want to multiply that by the vector."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I want to multiply the matrix minus 3, 0, 3, 2. I'll do this one in yellow. 1, 7, minus 1, 9. And I want to multiply that by the vector. Now how many components or rows does this vector have to have? Well, my matrix times vector product, or multiplication, is only defined if my vector has as many components as this matrix has columns. So we have 1, 2, 3, 4 columns."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want to multiply that by the vector. Now how many components or rows does this vector have to have? Well, my matrix times vector product, or multiplication, is only defined if my vector has as many components as this matrix has columns. So we have 1, 2, 3, 4 columns. So this guy's got to have 4 components for us to even be able to multiply them. Otherwise, it wouldn't be defined. So let me put 4 entries here."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we have 1, 2, 3, 4 columns. So this guy's got to have 4 components for us to even be able to multiply them. Otherwise, it wouldn't be defined. So let me put 4 entries here. Let's say it's 2, minus 3, 4, and then minus 1. So what is this going to be equal to? This is equal to, we essentially take the dot."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me put 4 entries here. Let's say it's 2, minus 3, 4, and then minus 1. So what is this going to be equal to? This is equal to, we essentially take the dot. The first term of this is going to be the dot product of this first row with this vector. And then the second entry here is going to be the dot product of this row vector with this column. So let's do it."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, we essentially take the dot. The first term of this is going to be the dot product of this first row with this vector. And then the second entry here is going to be the dot product of this row vector with this column. So let's do it. So it's going to be minus 3 times 2. I'm not going to color code it. Minus 3 times 2 plus 0 times minus 3 plus 3 times 4 plus 2 times minus 1."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's do it. So it's going to be minus 3 times 2. I'm not going to color code it. Minus 3 times 2 plus 0 times minus 3 plus 3 times 4 plus 2 times minus 1. Plus 2 times minus 1. And now my second row, or I guess my second component in this vector, is going to be 1 times 2. So 1 times 2 plus 7 times negative 3, 7 times minus 3, plus minus 1 times 4 plus 9 times minus 1."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 3 times 2 plus 0 times minus 3 plus 3 times 4 plus 2 times minus 1. Plus 2 times minus 1. And now my second row, or I guess my second component in this vector, is going to be 1 times 2. So 1 times 2 plus 7 times negative 3, 7 times minus 3, plus minus 1 times 4 plus 9 times minus 1. 9 times minus 1. And so what does this simplify to? This is equal to minus 3 times 2 is minus 6 plus 0 plus 12, this is 12, minus 2."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 1 times 2 plus 7 times negative 3, 7 times minus 3, plus minus 1 times 4 plus 9 times minus 1. 9 times minus 1. And so what does this simplify to? This is equal to minus 3 times 2 is minus 6 plus 0 plus 12, this is 12, minus 2. And then this is simplified to 2 minus 21 minus 4 minus 9. So this is equal to this top term. Let's see, I have a minus 6 plus 12 is 6, minus 2 is 4."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to minus 3 times 2 is minus 6 plus 0 plus 12, this is 12, minus 2. And then this is simplified to 2 minus 21 minus 4 minus 9. So this is equal to this top term. Let's see, I have a minus 6 plus 12 is 6, minus 2 is 4. And then I have 2 minus 21 is minus 19. I want to make sure I get the math right here. Let me see."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see, I have a minus 6 plus 12 is 6, minus 2 is 4. And then I have 2 minus 21 is minus 19. I want to make sure I get the math right here. Let me see. Minus 21 minus 9 is minus 30. And I have a minus 34. And then I have a plus 2."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me see. Minus 21 minus 9 is minus 30. And I have a minus 34. And then I have a plus 2. So minus 32. So that's my product right there. And like I said, you can view it."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I have a plus 2. So minus 32. So that's my product right there. And like I said, you can view it. And let me be very clear right here. Well, everything we've been used to right now, we've been writing our vectors as column vectors. But you can view each of these right here as a row vector."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And like I said, you can view it. And let me be very clear right here. Well, everything we've been used to right now, we've been writing our vectors as column vectors. But you can view each of these right here as a row vector. But let me be even better. Let's say that vector A1 is equal to minus 3, 0, 3, 2. And let me define vector A2 is equal to 1, 7, minus 1, 9."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But you can view each of these right here as a row vector. But let me be even better. Let's say that vector A1 is equal to minus 3, 0, 3, 2. And let me define vector A2 is equal to 1, 7, minus 1, 9. So all I did is I wrote these guys. But I wrote them in our standard vector form. I wrote them as column vectors."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And let me define vector A2 is equal to 1, 7, minus 1, 9. So all I did is I wrote these guys. But I wrote them in our standard vector form. I wrote them as column vectors. So what we can define to turn these guys into row vectors is the transpose function. And transpose, you just turn the rows into columns and the columns into rows. So if this is A1, then A1 transpose will just be the row version of this."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I wrote them as column vectors. So what we can define to turn these guys into row vectors is the transpose function. And transpose, you just turn the rows into columns and the columns into rows. So if this is A1, then A1 transpose will just be the row version of this. So it's minus 3, 0, 3, 2. And then A2 transpose would be equal to 1, 7, minus 1, and 9. And then this multiplication right here, we can rewrite it as, this matrix, we can rewrite it as, we have vector A1 transpose for the first row."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if this is A1, then A1 transpose will just be the row version of this. So it's minus 3, 0, 3, 2. And then A2 transpose would be equal to 1, 7, minus 1, and 9. And then this multiplication right here, we can rewrite it as, this matrix, we can rewrite it as, we have vector A1 transpose for the first row. That's a vector. These are vectors. Now, row vectors."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then this multiplication right here, we can rewrite it as, this matrix, we can rewrite it as, we have vector A1 transpose for the first row. That's a vector. These are vectors. Now, row vectors. And then this is A2 transpose. Transpose should be the superscript. This vector can be written exactly like this, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, row vectors. And then this is A2 transpose. Transpose should be the superscript. This vector can be written exactly like this, right? Because this is the first row, this is the second row. Let me just call this vector x. That right there is vector x."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This vector can be written exactly like this, right? Because this is the first row, this is the second row. Let me just call this vector x. That right there is vector x. So this is right here, vector x. We can now rewrite the definition as this would be equal to what? This first row right here that we wrote out, the first row that we wrote out right here, this was A1 dot x, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That right there is vector x. So this is right here, vector x. We can now rewrite the definition as this would be equal to what? This first row right here that we wrote out, the first row that we wrote out right here, this was A1 dot x, right? A1 dot x. And you know all about the dot products. This was, the first row was A1 dot x, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This first row right here that we wrote out, the first row that we wrote out right here, this was A1 dot x, right? A1 dot x. And you know all about the dot products. This was, the first row was A1 dot x, right? It's minus 3 times 2 plus 0 times minus 3 plus 3 times 4, right? It's A1 dot x. And this is useful because when I defined the dot product, I only defined it with column vectors like this."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This was, the first row was A1 dot x, right? It's minus 3 times 2 plus 0 times minus 3 plus 3 times 4, right? It's A1 dot x. And this is useful because when I defined the dot product, I only defined it with column vectors like this. So now I'm dotting two column vectors. I haven't formally defined a row vector times a column vector. So now I can say, look, if this is just a standard column vector like we've been working with, I can write my matrix as each row is the transpose of a column vector or it's a row vector, then I can write this product as just the dot products of each of these transpose or I guess you could say the inverse transpose with this vector right here."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is useful because when I defined the dot product, I only defined it with column vectors like this. So now I'm dotting two column vectors. I haven't formally defined a row vector times a column vector. So now I can say, look, if this is just a standard column vector like we've been working with, I can write my matrix as each row is the transpose of a column vector or it's a row vector, then I can write this product as just the dot products of each of these transpose or I guess you could say the inverse transpose with this vector right here. And then obviously the second row is going to be A2, vector A2 dot x, right? The second row is A2 dot x is 1 times 2 plus 7 times minus 3, right there, minus 1 times 4 plus 9 times minus 1. So just like that."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now I can say, look, if this is just a standard column vector like we've been working with, I can write my matrix as each row is the transpose of a column vector or it's a row vector, then I can write this product as just the dot products of each of these transpose or I guess you could say the inverse transpose with this vector right here. And then obviously the second row is going to be A2, vector A2 dot x, right? The second row is A2 dot x is 1 times 2 plus 7 times minus 3, right there, minus 1 times 4 plus 9 times minus 1. So just like that. So this is one way to view it. It's kind of matrix times a vector is just like the transpose of its rows dotted with the vector you're multiplying it by. This is one way to perceive matrix multiplication."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So just like that. So this is one way to view it. It's kind of matrix times a vector is just like the transpose of its rows dotted with the vector you're multiplying it by. This is one way to perceive matrix multiplication. Now the other way to perceive it, let me do it this way. Well, I'll do it with a different example. Those numbers are getting a little bit tiresome."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is one way to perceive matrix multiplication. Now the other way to perceive it, let me do it this way. Well, I'll do it with a different example. Those numbers are getting a little bit tiresome. Let's say I have the matrix A, nice and bold, is equal to oh, I don't know, let me say it's 3, 1, 0, 3, 2, 4, 7, 0, minus 1, 2, 3, and 4. And I need to multiply this. I have to multiply this times a four component vector."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Those numbers are getting a little bit tiresome. Let's say I have the matrix A, nice and bold, is equal to oh, I don't know, let me say it's 3, 1, 0, 3, 2, 4, 7, 0, minus 1, 2, 3, and 4. And I need to multiply this. I have to multiply this times a four component vector. So let me call vector x is equal to 5. Actually, let me keep it general. Let's say it's equal to x1, x2, x3, and x4."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have to multiply this times a four component vector. So let me call vector x is equal to 5. Actually, let me keep it general. Let's say it's equal to x1, x2, x3, and x4. Now, instead of viewing these as row vectors, we could view A as a set of column vectors. We could call this thing right here vector 1. We call this thing right here vector 2."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's equal to x1, x2, x3, and x4. Now, instead of viewing these as row vectors, we could view A as a set of column vectors. We could call this thing right here vector 1. We call this thing right here vector 2. We call this thing right here vector 3. And we call this thing right here vector 4. Then we could rewrite our matrix A as being equal to just a bunch of column vectors."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We call this thing right here vector 2. We call this thing right here vector 3. And we call this thing right here vector 4. Then we could rewrite our matrix A as being equal to just a bunch of column vectors. So we could rewrite it as vector 1, vector 2, vector 3, is vector 1, and vector 4. That's vector 4 right there. So how can the matrix multiplication be interpreted in this context?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then we could rewrite our matrix A as being equal to just a bunch of column vectors. So we could rewrite it as vector 1, vector 2, vector 3, is vector 1, and vector 4. That's vector 4 right there. So how can the matrix multiplication be interpreted in this context? Well, what did we do? We multiplied these guys. We always do all of the elements in here always get multiplied by x1, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So how can the matrix multiplication be interpreted in this context? Well, what did we do? We multiplied these guys. We always do all of the elements in here always get multiplied by x1, right? It's 3. Let me start some of the multiplication here, just from our definition. So if I multiply A times x, I'll start it off."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We always do all of the elements in here always get multiplied by x1, right? It's 3. Let me start some of the multiplication here, just from our definition. So if I multiply A times x, I'll start it off. Maybe I won't do the whole thing. I just want you to see the pattern. It's 3 times x1 plus 1 times x2 plus 0 times x3 plus 3 times x4."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I multiply A times x, I'll start it off. Maybe I won't do the whole thing. I just want you to see the pattern. It's 3 times x1 plus 1 times x2 plus 0 times x3 plus 3 times x4. That's the first entry. And then you have 2 times x1 plus 4 times x2, all the way. And then you finally have minus 1 times x1 plus 2 times x2."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's 3 times x1 plus 1 times x2 plus 0 times x3 plus 3 times x4. That's the first entry. And then you have 2 times x1 plus 4 times x2, all the way. And then you finally have minus 1 times x1 plus 2 times x2. You get the idea. But what's happening here? This guy, this first vector, is always being multiplied by the scalar x1."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you finally have minus 1 times x1 plus 2 times x2. You get the idea. But what's happening here? This guy, this first vector, is always being multiplied by the scalar x1. In fact, you can view this part of the entries right here, right? We're just multiplying this guy times the scalar of x1. In every case, you have 3, 2, minus 1, 3, 2, minus 1."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This guy, this first vector, is always being multiplied by the scalar x1. In fact, you can view this part of the entries right here, right? We're just multiplying this guy times the scalar of x1. In every case, you have 3, 2, minus 1, 3, 2, minus 1. We're multiplying by the scalar of x1. And then we're adding that to this guy times the scalar x2. And then we're adding that to this guy times the scalar x3."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In every case, you have 3, 2, minus 1, 3, 2, minus 1. We're multiplying by the scalar of x1. And then we're adding that to this guy times the scalar x2. And then we're adding that to this guy times the scalar x3. So we can rewrite a times x as being equal to the scalar x1 times the vector v1 plus the scalar x2. This is the scalar x1 times the vector v1 plus the scalar x2 times the vector v2. I want to do that in yellow."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we're adding that to this guy times the scalar x3. So we can rewrite a times x as being equal to the scalar x1 times the vector v1 plus the scalar x2. This is the scalar x1 times the vector v1 plus the scalar x2 times the vector v2. I want to do that in yellow. Times the vector v2 plus x3 times the vector v3 plus the scalar x4 times the vector v4. And obviously, if we had n terms here, we'd have to have n vectors here. And we could just make this more general to n. But what's interesting here is now the product Ax, it can be interpreted as a linear combination, right?"}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I want to do that in yellow. Times the vector v2 plus x3 times the vector v3 plus the scalar x4 times the vector v4. And obviously, if we had n terms here, we'd have to have n vectors here. And we could just make this more general to n. But what's interesting here is now the product Ax, it can be interpreted as a linear combination, right? These are just arbitrary numbers depending on what our vector x is. So depending on our vector x, we're taking a linear combination of the column vectors of A. So this is a linear combination of column vectors of A."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we could just make this more general to n. But what's interesting here is now the product Ax, it can be interpreted as a linear combination, right? These are just arbitrary numbers depending on what our vector x is. So depending on our vector x, we're taking a linear combination of the column vectors of A. So this is a linear combination of column vectors of A. So this is really interesting. I'm sure you've been exposed to matrix multiplication in the past. But I really want you to absorb these two ways of interpreting it."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is a linear combination of column vectors of A. So this is really interesting. I'm sure you've been exposed to matrix multiplication in the past. But I really want you to absorb these two ways of interpreting it. Because it'll be important when we talk about column spaces and things like that in the future, is that you can interpret matrix multiplication. And actually, there's other ways you can actually interpret it as a transformation of this vector x, but I won't cover that in this video just for brevity. But you can interpret it as a weighted combination or linear combination of the column vectors of A, where the matrix x dictates what the weights on each of the columns are."}, {"video_title": "Matrix vector products Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But I really want you to absorb these two ways of interpreting it. Because it'll be important when we talk about column spaces and things like that in the future, is that you can interpret matrix multiplication. And actually, there's other ways you can actually interpret it as a transformation of this vector x, but I won't cover that in this video just for brevity. But you can interpret it as a weighted combination or linear combination of the column vectors of A, where the matrix x dictates what the weights on each of the columns are. Or you can interpret it as essentially the dot product of the row vectors, or you could define the row vectors as a transpose of column vectors. The dot product of those column vectors, each of the corresponding column vectors, with your matrix x. So these are both completely valid interpretations."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, we defined a transformation that rotated any vector in R2 and just gave us another rotated version of that vector in R2. In this video, I'm essentially going to extend this, but I'm going to do it in R3. So I'm going to define a rotation transformation. Maybe I'll call it rotation... Well, I'll still call it theta, but it's going to be a mapping this time from R3 to R3. As you can imagine, the idea of a rotation and angle becomes a little bit more complicated when we're dealing in three dimensions. So in this case, we're going to rotate around the x-axis. So let me call it..."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I'll call it rotation... Well, I'll still call it theta, but it's going to be a mapping this time from R3 to R3. As you can imagine, the idea of a rotation and angle becomes a little bit more complicated when we're dealing in three dimensions. So in this case, we're going to rotate around the x-axis. So let me call it... So this is going to rotate around the x-axis. And what you see, what we do in this video, you can then just generalize that to other axes. And if you want to rotate around the x-axis and then the y-axis and then the z-axis by different angles, you can just apply the transformations one after another."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me call it... So this is going to rotate around the x-axis. And what you see, what we do in this video, you can then just generalize that to other axes. And if you want to rotate around the x-axis and then the y-axis and then the z-axis by different angles, you can just apply the transformations one after another. And we're going to cover that in a lot more detail in a future video. But this should kind of give you the tools to show you that this idea that we learned in the previous video is actually generalizable to multiple dimensions, and especially three dimensions. So let me just be clear what we're going to be doing here."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if you want to rotate around the x-axis and then the y-axis and then the z-axis by different angles, you can just apply the transformations one after another. And we're going to cover that in a lot more detail in a future video. But this should kind of give you the tools to show you that this idea that we learned in the previous video is actually generalizable to multiple dimensions, and especially three dimensions. So let me just be clear what we're going to be doing here. So let me draw some axes. So that's my x-axis. That is my y-axis."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me just be clear what we're going to be doing here. So let me draw some axes. So that's my x-axis. That is my y-axis. And this is my z-axis. And what I'm saying is any... Of course, this is R3. That any vector here in R3, I will be rotating it counterclockwise around the x-axis."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is my y-axis. And this is my z-axis. And what I'm saying is any... Of course, this is R3. That any vector here in R3, I will be rotating it counterclockwise around the x-axis. So it will be rotating like that. So if I had a vector, and I'm just drawing it in the zy-plane because it's a little bit easier to visualize. But if I have a vector sitting here in the zy-plane, it will still stay in the zy-plane, but it will be rotated counterclockwise by an angle of theta, just like that."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That any vector here in R3, I will be rotating it counterclockwise around the x-axis. So it will be rotating like that. So if I had a vector, and I'm just drawing it in the zy-plane because it's a little bit easier to visualize. But if I have a vector sitting here in the zy-plane, it will still stay in the zy-plane, but it will be rotated counterclockwise by an angle of theta, just like that. Now, a little harder to visualize is a vector that doesn't just sit in the zy-plane. Maybe we have some vector that has some x component, it comes out like that, and some y component, and some z component, it looks like that. Then when you rotate it, its z and its y components will change, but its x component will stay the same."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But if I have a vector sitting here in the zy-plane, it will still stay in the zy-plane, but it will be rotated counterclockwise by an angle of theta, just like that. Now, a little harder to visualize is a vector that doesn't just sit in the zy-plane. Maybe we have some vector that has some x component, it comes out like that, and some y component, and some z component, it looks like that. Then when you rotate it, its z and its y components will change, but its x component will stay the same. So then it might look something like this. Let me see if I can give it justice. So then the vector, when I rotate it around, might look something like that."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then when you rotate it, its z and its y components will change, but its x component will stay the same. So then it might look something like this. Let me see if I can give it justice. So then the vector, when I rotate it around, might look something like that. Anyway, I don't know if I'm giving it proper justice, but this was rotated around the x-axis, and I think you understand what that means. But just based on the last video, we want to build a transformation. We want to make this, let me call this rotation 3 theta, or let me call it 3 rotation theta, to know that we're dealing in R3."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So then the vector, when I rotate it around, might look something like that. Anyway, I don't know if I'm giving it proper justice, but this was rotated around the x-axis, and I think you understand what that means. But just based on the last video, we want to build a transformation. We want to make this, let me call this rotation 3 theta, or let me call it 3 rotation theta, to know that we're dealing in R3. And what we want to do is we want to find some matrix, so I can write my 3 rotation sub theta transformation of x as being some matrix A times the vector x. And since this is a transformation from R3 to R3, this is, of course, going to be a 3 by 3 matrix. Now in the last video, we learned that to figure this out, you just have to apply the transformation essentially to the identity matrix."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We want to make this, let me call this rotation 3 theta, or let me call it 3 rotation theta, to know that we're dealing in R3. And what we want to do is we want to find some matrix, so I can write my 3 rotation sub theta transformation of x as being some matrix A times the vector x. And since this is a transformation from R3 to R3, this is, of course, going to be a 3 by 3 matrix. Now in the last video, we learned that to figure this out, you just have to apply the transformation essentially to the identity matrix. So what we do is we start off with the identity matrix in R3, which is just going to be a 3 by 3. Let me draw it like, you're going to have 1, 1, 1, 0, 0, 0, 0, 0, 0. Each of these columns are the basis vectors for R3."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now in the last video, we learned that to figure this out, you just have to apply the transformation essentially to the identity matrix. So what we do is we start off with the identity matrix in R3, which is just going to be a 3 by 3. Let me draw it like, you're going to have 1, 1, 1, 0, 0, 0, 0, 0, 0. Each of these columns are the basis vectors for R3. That's E1, E2, E3. I'm writing it probably too small for you to see. But each of these are the basis vectors for R3, and what we need to do is just apply the transformation to each of these basis vectors in R3."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Each of these columns are the basis vectors for R3. That's E1, E2, E3. I'm writing it probably too small for you to see. But each of these are the basis vectors for R3, and what we need to do is just apply the transformation to each of these basis vectors in R3. So our matrix A will look like this. Our matrix A is going to be a 3 by 3 matrix where the first column is going to be our transformation, 3 rotations of theta applied to that column vector right there, 1, 0, 0. And then I'm going to apply it to this middle column vector right here."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But each of these are the basis vectors for R3, and what we need to do is just apply the transformation to each of these basis vectors in R3. So our matrix A will look like this. Our matrix A is going to be a 3 by 3 matrix where the first column is going to be our transformation, 3 rotations of theta applied to that column vector right there, 1, 0, 0. And then I'm going to apply it to this middle column vector right here. So I'm going to apply it. Well, you get the idea. I don't want to write that whole thing again."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then I'm going to apply it to this middle column vector right here. So I'm going to apply it. Well, you get the idea. I don't want to write that whole thing again. I'm going to apply 3 rotation sub theta to 0, 1, 0, and then I'm going to apply it. I'll do it here. 3 rotation sub theta."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to write that whole thing again. I'm going to apply 3 rotation sub theta to 0, 1, 0, and then I'm going to apply it. I'll do it here. 3 rotation sub theta. I'm going to apply it to this last column vector, so 0, 0, 1. We've seen this multiple times, so let's apply it. Let's rotate each of these basis vectors, the basis vectors for R3."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "3 rotation sub theta. I'm going to apply it to this last column vector, so 0, 0, 1. We've seen this multiple times, so let's apply it. Let's rotate each of these basis vectors, the basis vectors for R3. Let's rotate them around the x-axis. So the first guy, if I were to draw it in R3, what does he look like? He only has directionality in the x direction."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's rotate each of these basis vectors, the basis vectors for R3. Let's rotate them around the x-axis. So the first guy, if I were to draw it in R3, what does he look like? He only has directionality in the x direction. If we call this the x dimension, if the first entry corresponds to our x dimension, the second entry corresponds to our y dimension, and the third entry corresponds to our z dimension, then this guy only, this vector would just be a unit vector that just comes out like that. If I'm going to rotate this vector around the x-axis, what's going to happen to it? Nothing."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "He only has directionality in the x direction. If we call this the x dimension, if the first entry corresponds to our x dimension, the second entry corresponds to our y dimension, and the third entry corresponds to our z dimension, then this guy only, this vector would just be a unit vector that just comes out like that. If I'm going to rotate this vector around the x-axis, what's going to happen to it? Nothing. I'm just going to, I mean, it is the x-axis, so when you rotate it, it's not changing its direction or its magnitude or anything. This vector right here is just going to be the vector 1, 0, 0. Nothing happens when you rotate it."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Nothing. I'm just going to, I mean, it is the x-axis, so when you rotate it, it's not changing its direction or its magnitude or anything. This vector right here is just going to be the vector 1, 0, 0. Nothing happens when you rotate it. These are a little bit more interesting. To do these, let me just draw my z, y-axis. Let me just draw my z."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Nothing happens when you rotate it. These are a little bit more interesting. To do these, let me just draw my z, y-axis. Let me just draw my z. So that's my z-axis, and this is my y-axis right here. Now, this basis vector, it just goes in the y direction by 1. So this basis vector just looks like that, and it just goes, it has a length of 1."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me just draw my z. So that's my z-axis, and this is my y-axis right here. Now, this basis vector, it just goes in the y direction by 1. So this basis vector just looks like that, and it just goes, it has a length of 1. And then when you rotate it around the x-axis, when I draw it like this, you can imagine the x-axis is just popping out of your computer screen. So I could draw it, you know, like this is just like the tip of the arrow is just popping out. Instead of drawing it at an angle like this, I'm drawing it straight out of the computer screen."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this basis vector just looks like that, and it just goes, it has a length of 1. And then when you rotate it around the x-axis, when I draw it like this, you can imagine the x-axis is just popping out of your computer screen. So I could draw it, you know, like this is just like the tip of the arrow is just popping out. Instead of drawing it at an angle like this, I'm drawing it straight out of the computer screen. So if you were to rotate this vector right here, this blue vector right here, this vector right here, by an angle of theta, it'll look like this. It'll look like that. And we've done this in the previous video."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Instead of drawing it at an angle like this, I'm drawing it straight out of the computer screen. So if you were to rotate this vector right here, this blue vector right here, this vector right here, by an angle of theta, it'll look like this. It'll look like that. And we've done this in the previous video. What are its new coordinates? First of all, will its x-coordinate have changed at all? Its x-coordinate was 0 before because it doesn't break out into the x dimension."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we've done this in the previous video. What are its new coordinates? First of all, will its x-coordinate have changed at all? Its x-coordinate was 0 before because it doesn't break out into the x dimension. It just stays along the zy plane. It was 0 before. When you rotate it, it's still on the zy plane."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Its x-coordinate was 0 before because it doesn't break out into the x dimension. It just stays along the zy plane. It was 0 before. When you rotate it, it's still on the zy plane. So its x-direction or its x-component won't change at all. So the x-direction is still going to be 0. And then what's its new y-direction?"}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When you rotate it, it's still on the zy plane. So its x-direction or its x-component won't change at all. So the x-direction is still going to be 0. And then what's its new y-direction? Well, here we do exactly what we did in the last video. We figure out this is going to be its new, I don't want to draw a vector there necessarily, but this length right here is going to be its new y-component, and this length right here is going to be its new z-component. So what's its new y-component?"}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then what's its new y-direction? Well, here we do exactly what we did in the last video. We figure out this is going to be its new, I don't want to draw a vector there necessarily, but this length right here is going to be its new y-component, and this length right here is going to be its new z-component. So what's its new y-component? So this is going to be, and we did this in the last video, so I won't go into as much detail, but what is cosine of theta? The length of this vector is 1. These are the standard basis vectors, and one of the things that makes them a nice standard basis vector is that their lengths are 1."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what's its new y-component? So this is going to be, and we did this in the last video, so I won't go into as much detail, but what is cosine of theta? The length of this vector is 1. These are the standard basis vectors, and one of the things that makes them a nice standard basis vector is that their lengths are 1. So we know that the cosine of this angle is equal to the adjacent side over the hypotenuse. The adjacent side is this right here. And what is the hypotenuse?"}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are the standard basis vectors, and one of the things that makes them a nice standard basis vector is that their lengths are 1. So we know that the cosine of this angle is equal to the adjacent side over the hypotenuse. The adjacent side is this right here. And what is the hypotenuse? It's equal to 1. So this adjacent side, which we said is going to be our new second component or our second entry, is going to be equal to cosine of theta. That's a, because you can just ignore the 1s."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is the hypotenuse? It's equal to 1. So this adjacent side, which we said is going to be our new second component or our second entry, is going to be equal to cosine of theta. That's a, because you can just ignore the 1s. It's going to be equal to cosine of theta. And what's going to be its new z-component? Well, sine of theta is equal to the opposite side, this side, over 1."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's a, because you can just ignore the 1s. It's going to be equal to cosine of theta. And what's going to be its new z-component? Well, sine of theta is equal to the opposite side, this side, over 1. So it just equals this opposite side. And the length of that opposite side is this vector's, once it's rotated, is its new z-component. So that you get a sine theta right there."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, sine of theta is equal to the opposite side, this side, over 1. So it just equals this opposite side. And the length of that opposite side is this vector's, once it's rotated, is its new z-component. So that you get a sine theta right there. Now we just have to do everything in the z-direction. So this z-basis vector right there, what does it look like on this graph? Let me just actually redraw it, just to make things a little bit cleaner."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that you get a sine theta right there. Now we just have to do everything in the z-direction. So this z-basis vector right there, what does it look like on this graph? Let me just actually redraw it, just to make things a little bit cleaner. So that's my z-axis, and this is my y-axis. And my z-basis vector, E3, it starts off looking something like that. E3 looks like that."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me just actually redraw it, just to make things a little bit cleaner. So that's my z-axis, and this is my y-axis. And my z-basis vector, E3, it starts off looking something like that. E3 looks like that. It just goes only in the z-direction. So first of all, let's just rotate it by an angle of theta. So I'm going to rotate it like that."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "E3 looks like that. It just goes only in the z-direction. So first of all, let's just rotate it by an angle of theta. So I'm going to rotate it like that. That's an angle of theta. So its former x-entry was 0. It did not break out in the x-direction at all."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to rotate it like that. That's an angle of theta. So its former x-entry was 0. It did not break out in the x-direction at all. And of course, we're still just in the zy-plane, so it won't be moving out in the x-direction still. So it's still going to be a 0 up here. Now what about its new y-component?"}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It did not break out in the x-direction at all. And of course, we're still just in the zy-plane, so it won't be moving out in the x-direction still. So it's still going to be a 0 up here. Now what about its new y-component? Its new y-coordinate, I guess we can call it, is going to be this length, or it's going to be this coordinate right here. Can you figure that out? Well, that length is the same thing as that length."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what about its new y-component? Its new y-coordinate, I guess we can call it, is going to be this length, or it's going to be this coordinate right here. Can you figure that out? Well, that length is the same thing as that length. And if we call this the opposite side of the angle, we know that the sine of theta is equal to this opposite side over the length of this vector, which is just 1, so it's just equal to the opposite side. So the opposite side is equal to sine of theta, but our new coordinate is to the left of the z-axis. So this is going to be a negative version."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, that length is the same thing as that length. And if we call this the opposite side of the angle, we know that the sine of theta is equal to this opposite side over the length of this vector, which is just 1, so it's just equal to the opposite side. So the opposite side is equal to sine of theta, but our new coordinate is to the left of the z-axis. So this is going to be a negative version. We did this in the last video. So it's going to be a negative sine of theta. So this is going to be negative sine of theta, this point right here, that coordinate."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be a negative version. We did this in the last video. So it's going to be a negative sine of theta. So this is going to be negative sine of theta, this point right here, that coordinate. So it's minus sine of theta. And then finally, what's its new z-coordinate going to be? That's going to be this length right here."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be negative sine of theta, this point right here, that coordinate. So it's minus sine of theta. And then finally, what's its new z-coordinate going to be? That's going to be this length right here. And we know that this length, if we call that adjacent, we know that the cosine of our theta is equal to this divided by 1. So it's equal to that adjacent side. So we can just put a cosine of theta right there."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be this length right here. And we know that this length, if we call that adjacent, we know that the cosine of our theta is equal to this divided by 1. So it's equal to that adjacent side. So we can just put a cosine of theta right there. And we get our transformation matrix. We're done. Our transformation matrix A is this."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can just put a cosine of theta right there. And we get our transformation matrix. We're done. Our transformation matrix A is this. So we can now say our new transformation that this video is about, my 3, and I call it a 3 because it's a rotation in R3, but it's around, maybe I should call it 3 sub x because it's a rotation around the x-axis, but I think you get the idea. It is equal to this matrix right up here. Maybe I could rewrite it."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our transformation matrix A is this. So we can now say our new transformation that this video is about, my 3, and I call it a 3 because it's a rotation in R3, but it's around, maybe I should call it 3 sub x because it's a rotation around the x-axis, but I think you get the idea. It is equal to this matrix right up here. Maybe I could rewrite it. Well, actually, let me do it this way. Let me delete all of this so I don't have to rewrite. So my transformation that this video is about, the 3 rotation theta of x, that transformation is equal to this matrix times whatever vector x I have in R3."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I could rewrite it. Well, actually, let me do it this way. Let me delete all of this so I don't have to rewrite. So my transformation that this video is about, the 3 rotation theta of x, that transformation is equal to this matrix times whatever vector x I have in R3. And you might say, hey, Sal, that looks exactly like what you did in the second. If you remember the last video when we defined our rotation in R2, we had a transformation matrix that looked very similar to this. And that makes sense because we're essentially just rotating things counterclockwise in the zy-plane."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So my transformation that this video is about, the 3 rotation theta of x, that transformation is equal to this matrix times whatever vector x I have in R3. And you might say, hey, Sal, that looks exactly like what you did in the second. If you remember the last video when we defined our rotation in R2, we had a transformation matrix that looked very similar to this. And that makes sense because we're essentially just rotating things counterclockwise in the zy-plane. Now you might say, hey, Sal, why is this even useful? You extended it to the 3 dimensions or to R3. I saw what you did in R2."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that makes sense because we're essentially just rotating things counterclockwise in the zy-plane. Now you might say, hey, Sal, why is this even useful? You extended it to the 3 dimensions or to R3. I saw what you did in R2. Why is this useful? It's kind of a limited case where you're just rotating around the x-axis. And I did it for 2 reasons."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I saw what you did in R2. Why is this useful? It's kind of a limited case where you're just rotating around the x-axis. And I did it for 2 reasons. One, to show you that you can generalize to R3. But the other thing is, if you think about it, a lot of the rotations that you might want to do in R3 can be described by a rotation around the x-axis first, which we did in this video, then by rotation around the y-axis, and then maybe some rotation around the z-axis. So you can actually define, this is just the special case, this is just the special case where we're dealing with the x-axis, rotation around the x-axis, but you could do the exact same process to define transformation matrices for rotations around the y-axis or the z-axis."}, {"video_title": "Rotation in R3 around the x-axis Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I did it for 2 reasons. One, to show you that you can generalize to R3. But the other thing is, if you think about it, a lot of the rotations that you might want to do in R3 can be described by a rotation around the x-axis first, which we did in this video, then by rotation around the y-axis, and then maybe some rotation around the z-axis. So you can actually define, this is just the special case, this is just the special case where we're dealing with the x-axis, rotation around the x-axis, but you could do the exact same process to define transformation matrices for rotations around the y-axis or the z-axis. And then you can apply them one after another. And we'll talk a lot about that in the future when we start applying one transformation after the other. But anyway, hopefully you found this slightly useful."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You've already dealt with vectors, or I'm guessing that some of you all have already dealt with vectors in your calculus or your precalculus or your physics classes. But in this video, I hope to show you something that you're going to do in linear algebra that you've never done before, and that it would have been very hard to do had you not, I guess, been exposed to these videos. Well, I'm going to start with, once again, a different way of doing something that you already know how to do. So let me just define some vector here. Some vector, instead of making them bold, I'll just draw it with the arrow on top. I'm going to define my vector to be, I could do it with the arrow on top, or I could just make it super bold. I'm just going to define my vector."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me just define some vector here. Some vector, instead of making them bold, I'll just draw it with the arrow on top. I'm going to define my vector to be, I could do it with the arrow on top, or I could just make it super bold. I'm just going to define my vector. It's going to be a vector in R2. Let's just say that my vector is the vector 2, 1. If I were to draw it in standard position, it looks like this."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just going to define my vector. It's going to be a vector in R2. Let's just say that my vector is the vector 2, 1. If I were to draw it in standard position, it looks like this. You go 2 to the right, and up 1 like that. That's my, right there, that is my vector v. Now, if I were to ask you, what are all of the possible vectors I can create? So let me define a set."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to draw it in standard position, it looks like this. You go 2 to the right, and up 1 like that. That's my, right there, that is my vector v. Now, if I were to ask you, what are all of the possible vectors I can create? So let me define a set. Let me define a set S, and it's equal to all of the vectors I can create. If I were to multiply v times some constant, some scalar, times v, times my vector v. And just to be a little bit formal, let's say such that c is a member of the real numbers. Now, what would be a graphical representation of this set?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me define a set. Let me define a set S, and it's equal to all of the vectors I can create. If I were to multiply v times some constant, some scalar, times v, times my vector v. And just to be a little bit formal, let's say such that c is a member of the real numbers. Now, what would be a graphical representation of this set? Well, if we draw them on standard position, c could be any real number. So if I were to multiply, c could be 2. If c is 2, let me do it this way."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, what would be a graphical representation of this set? Well, if we draw them on standard position, c could be any real number. So if I were to multiply, c could be 2. If c is 2, let me do it this way. If I do 2 times our vector, I'm going to get the vector 4, 2. Let me draw that in standard position. 4, 2, it's right there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If c is 2, let me do it this way. If I do 2 times our vector, I'm going to get the vector 4, 2. Let me draw that in standard position. 4, 2, it's right there. It's this vector right there. That vector right there. It's collinear with this first vector."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4, 2, it's right there. It's this vector right there. That vector right there. It's collinear with this first vector. It's along the same line, but it just goes out further, too. Now, I could have done another. I could have done 1.5 times our vector v. Let me do that in a different color."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's collinear with this first vector. It's along the same line, but it just goes out further, too. Now, I could have done another. I could have done 1.5 times our vector v. Let me do that in a different color. And maybe that would be 1.5 times 2, which is 3, 1.5. Where would that vector get me? I'd go 1.5, and then I'd go 3, and then 1.5."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could have done 1.5 times our vector v. Let me do that in a different color. And maybe that would be 1.5 times 2, which is 3, 1.5. Where would that vector get me? I'd go 1.5, and then I'd go 3, and then 1.5. I'd get right there. And I could multiply it by anything. I could multiply 1.4999 times vector v and get right over here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'd go 1.5, and then I'd go 3, and then 1.5. I'd get right there. And I could multiply it by anything. I could multiply 1.4999 times vector v and get right over here. I could do minus 0.00001 times vector v. Let me write that down. I could do 0.001 times our vector v. And where would that put me? It would put me right, little super small vector right there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could multiply 1.4999 times vector v and get right over here. I could do minus 0.00001 times vector v. Let me write that down. I could do 0.001 times our vector v. And where would that put me? It would put me right, little super small vector right there. If I did minus 0.01, it would make a super small vector right there pointing in that direction. If I were to do minus 10, I would get a vector going in this direction that goes way like that. Or it goes way like that."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It would put me right, little super small vector right there. If I did minus 0.01, it would make a super small vector right there pointing in that direction. If I were to do minus 10, I would get a vector going in this direction that goes way like that. Or it goes way like that. But you can imagine that if I were to plot all of the vectors in standard position, all of them that could be represented by any c in real numbers, I'll essentially get, I'll end up drawing a bunch of vectors where their arrows are all lined up along this line right there. And all lined up in even the negative direction. Let me make sure I draw it properly."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or it goes way like that. But you can imagine that if I were to plot all of the vectors in standard position, all of them that could be represented by any c in real numbers, I'll essentially get, I'll end up drawing a bunch of vectors where their arrows are all lined up along this line right there. And all lined up in even the negative direction. Let me make sure I draw it properly. Along that line like that. I think you get the idea. So it's a set of collinear vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me make sure I draw it properly. Along that line like that. I think you get the idea. So it's a set of collinear vectors. So let me write that down. It's a set of collinear vectors. And if we view these vectors as position vectors, that this vector represents a point in space, in R2."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's a set of collinear vectors. So let me write that down. It's a set of collinear vectors. And if we view these vectors as position vectors, that this vector represents a point in space, in R2. This R2 is just our Cartesian coordinate plane right here in every direction. If we view this vector as a position vector, if we view it as kind of a coordinate in R2, then this set, if we visually represent it as a bunch of position vectors, it'll be represented by this whole line over here. And I want to make that point clear."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if we view these vectors as position vectors, that this vector represents a point in space, in R2. This R2 is just our Cartesian coordinate plane right here in every direction. If we view this vector as a position vector, if we view it as kind of a coordinate in R2, then this set, if we visually represent it as a bunch of position vectors, it'll be represented by this whole line over here. And I want to make that point clear. Because if I didn't, it's essentially a line of slope 2. Your rise, sorry, slope 1 half. Your rise is 1 for going over 2."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want to make that point clear. Because if I didn't, it's essentially a line of slope 2. Your rise, sorry, slope 1 half. Your rise is 1 for going over 2. But I don't want to go back to our algebra 1 notation too much. But I want to make this point that this line of slope 2 that goes through the origin, this is if we draw all of the vectors in this set as in their standard form. Or if we draw them all as position vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Your rise is 1 for going over 2. But I don't want to go back to our algebra 1 notation too much. But I want to make this point that this line of slope 2 that goes through the origin, this is if we draw all of the vectors in this set as in their standard form. Or if we draw them all as position vectors. If I didn't make that clarification or that qualification, I could have drawn these vectors anywhere. Because this 4, 2 vector, I could have drawn over here. And then to say that it's collinear probably wouldn't have made as much visual sense to you."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or if we draw them all as position vectors. If I didn't make that clarification or that qualification, I could have drawn these vectors anywhere. Because this 4, 2 vector, I could have drawn over here. And then to say that it's collinear probably wouldn't have made as much visual sense to you. But I think this collinearity of it makes more sense to you if you say, let's draw them all in standard form. All of them start at the origin and their tails are at the origin and their heads go essentially to the coordinate they represent. That's what I mean by their position vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then to say that it's collinear probably wouldn't have made as much visual sense to you. But I think this collinearity of it makes more sense to you if you say, let's draw them all in standard form. All of them start at the origin and their tails are at the origin and their heads go essentially to the coordinate they represent. That's what I mean by their position vectors. They don't necessarily have to be position vectors. But for I guess the visualization in this video, let's stick to that. Now, I was only able to represent something that goes through the origin with this slope."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's what I mean by their position vectors. They don't necessarily have to be position vectors. But for I guess the visualization in this video, let's stick to that. Now, I was only able to represent something that goes through the origin with this slope. So you can almost view that this vector kind of represented its slope. You almost want to view it as a slope vector if you wanted to tie it in to what you learned in algebra 1. What if we wanted to represent other lines that had that slope?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, I was only able to represent something that goes through the origin with this slope. So you can almost view that this vector kind of represented its slope. You almost want to view it as a slope vector if you wanted to tie it in to what you learned in algebra 1. What if we wanted to represent other lines that had that slope? What if we wanted to represent the same line, or I guess a parallel line, that goes through that point over there? This is the point 2, 4. Or if we're thinking in position vectors, we could say that point is represented by the vector."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What if we wanted to represent other lines that had that slope? What if we wanted to represent the same line, or I guess a parallel line, that goes through that point over there? This is the point 2, 4. Or if we're thinking in position vectors, we could say that point is represented by the vector. I'm going to call that x. It's represented by the vector x. And the vector x is equal to 2, 4."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or if we're thinking in position vectors, we could say that point is represented by the vector. I'm going to call that x. It's represented by the vector x. And the vector x is equal to 2, 4. That point right there. What if I want to represent the line that's parallel to this that goes through that point 2, 4? So I want to represent this line right here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the vector x is equal to 2, 4. That point right there. What if I want to represent the line that's parallel to this that goes through that point 2, 4? So I want to represent this line right here. I'll draw it as parallel to this as I can. I think you get the idea. And it just keeps going like that in every direction."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I want to represent this line right here. I'll draw it as parallel to this as I can. I think you get the idea. And it just keeps going like that in every direction. These two lines are parallel. How can I represent the set of all of these vectors, or drawn in standard form? Or all of the vectors, that if I were to draw them in standard form, would show this line?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And it just keeps going like that in every direction. These two lines are parallel. How can I represent the set of all of these vectors, or drawn in standard form? Or all of the vectors, that if I were to draw them in standard form, would show this line? Well, you could think about it this way. If every one of the vectors that represented this line, if I start with any vector that was on this line, and I add my x vector to it, I'll show up at a corresponding point on this line that I want to be at. If I start, let's say I do negative 2 times my vector v. That equaled what?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or all of the vectors, that if I were to draw them in standard form, would show this line? Well, you could think about it this way. If every one of the vectors that represented this line, if I start with any vector that was on this line, and I add my x vector to it, I'll show up at a corresponding point on this line that I want to be at. If I start, let's say I do negative 2 times my vector v. That equaled what? Minus 2, minus 4, minus 2. So that's that vector there. But if I were to add x to it, if I were to add my x vector, so if I were to do minus 2 times my vector v, but I were to add x to it."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I start, let's say I do negative 2 times my vector v. That equaled what? Minus 2, minus 4, minus 2. So that's that vector there. But if I were to add x to it, if I were to add my x vector, so if I were to do minus 2 times my vector v, but I were to add x to it. So plus x. I'm adding this vector 2, 4 to it. So from here, I'd go right to and up 4. So I'd go here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if I were to add x to it, if I were to add my x vector, so if I were to do minus 2 times my vector v, but I were to add x to it. So plus x. I'm adding this vector 2, 4 to it. So from here, I'd go right to and up 4. So I'd go here. Or visually, you could just say heads to tails. So I would go right there. So I would end up at a corresponding point over there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'd go here. Or visually, you could just say heads to tails. So I would go right there. So I would end up at a corresponding point over there. I would end up at a corresponding point over there. So the set of all points, so when I define my set S as the set of all points where I just multiply v times the scalar, I got this thing that went through the origin. But now let me define another set."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I would end up at a corresponding point over there. I would end up at a corresponding point over there. So the set of all points, so when I define my set S as the set of all points where I just multiply v times the scalar, I got this thing that went through the origin. But now let me define another set. Let me define a set L, maybe L for line, that's equal to the set of all vectors where it's x, the vector x, I could do it bold or I'll just do an arrow on it, plus some scalar. I could use c, but let me use t because I'm going to call this a parametrization of the line. So plus some scalar t times my vector v, where, or such that, t could be any member of the real numbers."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But now let me define another set. Let me define a set L, maybe L for line, that's equal to the set of all vectors where it's x, the vector x, I could do it bold or I'll just do an arrow on it, plus some scalar. I could use c, but let me use t because I'm going to call this a parametrization of the line. So plus some scalar t times my vector v, where, or such that, t could be any member of the real numbers. So what is this going to be? This is going to be this blue line. If I were to draw all of these vectors in standard position, I'm going to get my blue line."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So plus some scalar t times my vector v, where, or such that, t could be any member of the real numbers. So what is this going to be? This is going to be this blue line. If I were to draw all of these vectors in standard position, I'm going to get my blue line. For example, if I do minus 2, if this is minus 2 times my vector v I get here, then if I add x, I go there. So the vector, this vector right here, that has its end point right there, its end point sits on that line. I could do that with anything."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to draw all of these vectors in standard position, I'm going to get my blue line. For example, if I do minus 2, if this is minus 2 times my vector v I get here, then if I add x, I go there. So the vector, this vector right here, that has its end point right there, its end point sits on that line. I could do that with anything. If I take this vector, this is some scalar times my vector v, and I add x to it, I end up with this vector, whose end point, if I view it as a position vector, its end point dictates some coordinate in the xy-plane. So it'll look like that point. So I can get to any of these vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I could do that with anything. If I take this vector, this is some scalar times my vector v, and I add x to it, I end up with this vector, whose end point, if I view it as a position vector, its end point dictates some coordinate in the xy-plane. So it'll look like that point. So I can get to any of these vectors. This is a set of vectors right here, and all of these vectors are going to point, they're essentially going to point to something. When I draw them in standard form, if I draw them in standard form, they're going to point to a point on that blue line. Now you might say, hey Sal, this was a really obtuse way of defining a line."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I can get to any of these vectors. This is a set of vectors right here, and all of these vectors are going to point, they're essentially going to point to something. When I draw them in standard form, if I draw them in standard form, they're going to point to a point on that blue line. Now you might say, hey Sal, this was a really obtuse way of defining a line. I mean, we do it in algebra 1, where we just say, hey, y is equal to mx plus b, and we figure out the slope. By figuring out the difference of two points, and then we do a little substitution. And this is stuff you learned in seventh or eighth grade."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now you might say, hey Sal, this was a really obtuse way of defining a line. I mean, we do it in algebra 1, where we just say, hey, y is equal to mx plus b, and we figure out the slope. By figuring out the difference of two points, and then we do a little substitution. And this is stuff you learned in seventh or eighth grade. This was really straightforward. Why am I defining this obtuse set here, and making you think in terms of sets, and vectors, and adding vectors? And the reason is, is because this is very general."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is stuff you learned in seventh or eighth grade. This was really straightforward. Why am I defining this obtuse set here, and making you think in terms of sets, and vectors, and adding vectors? And the reason is, is because this is very general. This worked well in R2. So in R2, this was great. I mean, we just have to worry about x's and y's."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the reason is, is because this is very general. This worked well in R2. So in R2, this was great. I mean, we just have to worry about x's and y's. But what about the situation, I mean, notice in your algebra class, your teacher never really gave you, told you much, or at least the ones I took, about how do you represent lines in three dimensions? And I mean, maybe some classes go there, but they definitely didn't tell you how do you represent lines in four dimensions, or 100 dimensions. And that's what this is going to do for us."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I mean, we just have to worry about x's and y's. But what about the situation, I mean, notice in your algebra class, your teacher never really gave you, told you much, or at least the ones I took, about how do you represent lines in three dimensions? And I mean, maybe some classes go there, but they definitely didn't tell you how do you represent lines in four dimensions, or 100 dimensions. And that's what this is going to do for us. Right here, I defined x and v as vectors in R2. They're two-dimensional vectors. But we can extend it to an arbitrary number of dimensions."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's what this is going to do for us. Right here, I defined x and v as vectors in R2. They're two-dimensional vectors. But we can extend it to an arbitrary number of dimensions. So let me just, just to kind of hit the point home, let's do one more example in R2, where it's kind of the classic algebra problem, where you need to find the equation for the line. But here, we're going to call it the set definition for the line. Let's say we have two vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But we can extend it to an arbitrary number of dimensions. So let me just, just to kind of hit the point home, let's do one more example in R2, where it's kind of the classic algebra problem, where you need to find the equation for the line. But here, we're going to call it the set definition for the line. Let's say we have two vectors. Let's say we have the vector a, which I'll define as, let me just say it's 2, 1. So if I were to draw it in standard form, it's 2, 2, 1. That's my vector a right there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have two vectors. Let's say we have the vector a, which I'll define as, let me just say it's 2, 1. So if I were to draw it in standard form, it's 2, 2, 1. That's my vector a right there. And let's say I have vector b. Let me define vector b. I'm going to define it as, I don't know, let me define it as 0, 3. So my vector b is 0."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's my vector a right there. And let's say I have vector b. Let me define vector b. I'm going to define it as, I don't know, let me define it as 0, 3. So my vector b is 0. I don't move to the right at all, and I go up. So my vector b will look like that. Now, I'm going to say that these are position vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So my vector b is 0. I don't move to the right at all, and I go up. So my vector b will look like that. Now, I'm going to say that these are position vectors. We draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, I'm going to say that these are position vectors. We draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. I mean, all of these coordinate axes I draw are going to be R2. Now, what if I asked you, give me a parametrization of the line that goes through these two points?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you can almost view these as coordinate points in R2. This is R2. I mean, all of these coordinate axes I draw are going to be R2. Now, what if I asked you, give me a parametrization of the line that goes through these two points? So essentially, I want the equation, if you're thinking in algebra 1 terms, I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you'd have substituted back in. But instead, what we can do is we can say, hey, look."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, what if I asked you, give me a parametrization of the line that goes through these two points? So essentially, I want the equation, if you're thinking in algebra 1 terms, I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you'd have substituted back in. But instead, what we can do is we can say, hey, look. This line that goes through both of those points, you can almost say that both of those vectors lie on, I guess that's a better, both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar multiple, can represent any other vector on that line?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But instead, what we can do is we can say, hey, look. This line that goes through both of those points, you can almost say that both of those vectors lie on, I guess that's a better, both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar multiple, can represent any other vector on that line? And let me do it this way. What if I were to take, so this is vector b here, what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or even better, what vector, if I take any arbitrary scalar multiple, can represent any other vector on that line? And let me do it this way. What if I were to take, so this is vector b here, what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it, what do I have to add to a to get to b?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it, what do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a, we know how to do that. We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you just think about it, what do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a, we know how to do that. We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. We have to be careful. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a. What do we get then?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. We have to be careful. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a. What do we get then? So b minus a looks like that, but if we were to draw it in standard form, remember, in standard form, b minus a would look something like this. It would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What do we get then? So b minus a looks like that, but if we were to draw it in standard form, remember, in standard form, b minus a would look something like this. It would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or a parametrization, if you will, of this line, or this set. Let's call this set L. So we want to know what that set is equal to. So in order to get there, we have to start with this, which is this line here, and we have to shift it."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, that's not what we set out to do. We wanted to figure out an equation, or a parametrization, if you will, of this line, or this set. Let's call this set L. So we want to know what that set is equal to. So in order to get there, we have to start with this, which is this line here, and we have to shift it. And we could shift it either by shifting it straight up, we could add vector b to it. So we could take this line right here and add vector b to it. And so any point on here would have its corresponding point there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in order to get there, we have to start with this, which is this line here, and we have to shift it. And we could shift it either by shifting it straight up, we could add vector b to it. So we could take this line right here and add vector b to it. And so any point on here would have its corresponding point there. So when you add vector b, it essentially shifts it up. That would work. So we could say, we could add vector b to it."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so any point on here would have its corresponding point there. So when you add vector b, it essentially shifts it up. That would work. So we could say, we could add vector b to it. And now all of these points for any arbitrary t is a member of the real numbers will lie on this green line. Or the other option we could have done is we could have added vector a. Vector a would have taken any arbitrary point here and shifted it that way. You would be adding vector a to it."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we could say, we could add vector b to it. And now all of these points for any arbitrary t is a member of the real numbers will lie on this green line. Or the other option we could have done is we could have added vector a. Vector a would have taken any arbitrary point here and shifted it that way. You would be adding vector a to it. But either way, you're going to get to the green line that we cared about. So you could have also defined it as the set of vector a plus this line, essentially, t times vector b minus a, where t is a member of the reals. Now all of this, so my definition of my line could be either of these things."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You would be adding vector a to it. But either way, you're going to get to the green line that we cared about. So you could have also defined it as the set of vector a plus this line, essentially, t times vector b minus a, where t is a member of the reals. Now all of this, so my definition of my line could be either of these things. The definition of my line could be this set, or it could be this set. And all of this seems all very abstract. But when you actually deal with the numbers, it actually becomes very simple."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now all of this, so my definition of my line could be either of these things. The definition of my line could be this set, or it could be this set. And all of this seems all very abstract. But when you actually deal with the numbers, it actually becomes very simple. It becomes arguably simpler than what we did in algebra 1. So this L, for this particular case of a and b, let's figure it out. My line is equal to, let me just use the first example."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But when you actually deal with the numbers, it actually becomes very simple. It becomes arguably simpler than what we did in algebra 1. So this L, for this particular case of a and b, let's figure it out. My line is equal to, let me just use the first example. It's vector b. So it's the vector 0, 3, plus t times the vector b minus a. What's b minus a?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My line is equal to, let me just use the first example. It's vector b. So it's the vector 0, 3, plus t times the vector b minus a. What's b minus a? 0 minus 2 is minus 2. 3 minus 1 is 2. For t is a member of the reals."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What's b minus a? 0 minus 2 is minus 2. 3 minus 1 is 2. For t is a member of the reals. Now if this still seems kind of like a convoluted set definition for you, I could write it in terms that you might recognize better. If we want to plot points, if we call this the y-axis, and we call this the x-axis, and if we call this the x-coordinate, or maybe more properly that, the x-coordinate, and call this the y-coordinate, then we could set up an equation here. This actually is the x-slope."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For t is a member of the reals. Now if this still seems kind of like a convoluted set definition for you, I could write it in terms that you might recognize better. If we want to plot points, if we call this the y-axis, and we call this the x-axis, and if we call this the x-coordinate, or maybe more properly that, the x-coordinate, and call this the y-coordinate, then we could set up an equation here. This actually is the x-slope. Well, I don't want to. This is the x-coordinate, and that's the y-coordinate. Or actually even better, whatever, actually, let me be very careful there."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This actually is the x-slope. Well, I don't want to. This is the x-coordinate, and that's the y-coordinate. Or actually even better, whatever, actually, let me be very careful there. This is always going to end up becoming some vector, L1, L2. This is a set of vectors, and any member of the set is going to look something like this. So this could be Li."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or actually even better, whatever, actually, let me be very careful there. This is always going to end up becoming some vector, L1, L2. This is a set of vectors, and any member of the set is going to look something like this. So this could be Li. So this is the x-coordinate, and this is the y-coordinate. And just to get this coordinate, and just to get this in a form that you recognize, so we're saying that L is a set of this vector x plus t times this vector b minus a here. If we wanted to write it in kind of a parametric form, we can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2, plus 2t."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this could be Li. So this is the x-coordinate, and this is the y-coordinate. And just to get this coordinate, and just to get this in a form that you recognize, so we're saying that L is a set of this vector x plus t times this vector b minus a here. If we wanted to write it in kind of a parametric form, we can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2, plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we wanted to write it in kind of a parametric form, we can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2, plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time. This was convoluted. You had to define these sets and all of that. But now I'm going to show you something that you probably, well, unless you've done this before, I guess that's true of anything."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, you might have still viewed this as, Sal, this was a waste of time. This was convoluted. You had to define these sets and all of that. But now I'm going to show you something that you probably, well, unless you've done this before, I guess that's true of anything. But you probably haven't seen it in your traditional algebra class. Let's say I have two points. And now I'm going to deal in three dimensions."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But now I'm going to show you something that you probably, well, unless you've done this before, I guess that's true of anything. But you probably haven't seen it in your traditional algebra class. Let's say I have two points. And now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. So let's call it position 1."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. So let's call it position 1. And this is the point. This is in three dimensions. Let's make up some numbers."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's call it position 1. And this is the point. This is in three dimensions. Let's make up some numbers. Negative 1, 2, 7. Let's say I have point 2. And once again, this is in three dimensions."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's make up some numbers. Negative 1, 2, 7. Let's say I have point 2. And once again, this is in three dimensions. So you have to specify three coordinates. This could be the x, the y, and the z-coordinate. Point 2, I don't know, let's make it 0, 3, and 4."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And once again, this is in three dimensions. So you have to specify three coordinates. This could be the x, the y, and the z-coordinate. Point 2, I don't know, let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line, so I'll just call that, or the set of this line, let me just call this, you know, L, it's going to be equal to, we could just pick one of these guys."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Point 2, I don't know, let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line, so I'll just call that, or the set of this line, let me just call this, you know, L, it's going to be equal to, we could just pick one of these guys. It's going to be, it could be p1, the vector p1. These are all vectors. Be careful here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I just said that the equation of this line, so I'll just call that, or the set of this line, let me just call this, you know, L, it's going to be equal to, we could just pick one of these guys. It's going to be, it could be p1, the vector p1. These are all vectors. Be careful here. The vector p1 plus some random parameter t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors. Times p1, and it doesn't matter what order you take it. So that's a nice thing too."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Be careful here. The vector p1 plus some random parameter t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors. Times p1, and it doesn't matter what order you take it. So that's a nice thing too. p1 minus p2, it could be p2 minus p1, and then this, you know, because this can take on any positive or negative value, where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's a nice thing too. p1 minus p2, it could be p2 minus p1, and then this, you know, because this can take on any positive or negative value, where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is p1 minus p2? p1 minus p2 is equal to, let me get some space here, p1 minus p2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's apply it right here. What is p1 minus p2? p1 minus p2 is equal to, let me get some space here, p1 minus p2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1. 7 minus 4 is 3. So that thing is that vector. And so our line can be described as a set of vectors that if you were to plot it in standard position, it would be this set of position vectors."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 3 is minus 1. 7 minus 4 is 3. So that thing is that vector. And so our line can be described as a set of vectors that if you were to plot it in standard position, it would be this set of position vectors. It would be p1, it would be, let me do that in green, it would be minus 1, 2, 7. I could have put p2 there just as easily. Plus t times minus 1 minus 1, 3, where, or such that, t is a member of the real numbers."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so our line can be described as a set of vectors that if you were to plot it in standard position, it would be this set of position vectors. It would be p1, it would be, let me do that in green, it would be minus 1, 2, 7. I could have put p2 there just as easily. Plus t times minus 1 minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's?"}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Plus t times minus 1 minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is, let's say, let me go down here. Let's say that this is the z-axis, this is the x-axis, and let's say the y-axis, it kind of goes into our board like this. So the y-axis comes out like that."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is, let's say, let me go down here. Let's say that this is the z-axis, this is the x-axis, and let's say the y-axis, it kind of goes into our board like this. So the y-axis comes out like that. So what you can view this, and actually I probably won't grab, so the determinant for the x-coordinate, just our convention, is going to be this term right here. So we can write that x, let me write that down, so that term is going to determine our x-coordinate. So we can write that x is equal to minus 1, be careful with the colors, minus 1 plus minus 1 times t. Plus minus 1 times t. That's our x-coordinate."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the y-axis comes out like that. So what you can view this, and actually I probably won't grab, so the determinant for the x-coordinate, just our convention, is going to be this term right here. So we can write that x, let me write that down, so that term is going to determine our x-coordinate. So we can write that x is equal to minus 1, be careful with the colors, minus 1 plus minus 1 times t. Plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition, because these are the y-coordinates. So we could say that our y-coordinate is equal to, I'll just write it like this, 2 plus minus 1 times t. And then finally, our z-coordinate is determined by that there, and the t shows up, because it's t times 3, or I could just put this t into all of this. So the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we can write that x is equal to minus 1, be careful with the colors, minus 1 plus minus 1 times t. Plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition, because these are the y-coordinates. So we could say that our y-coordinate is equal to, I'll just write it like this, 2 plus minus 1 times t. And then finally, our z-coordinate is determined by that there, and the t shows up, because it's t times 3, or I could just put this t into all of this. So the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you could just have a regular y in terms of x. You don't have to have a parametric equation."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you could just have a regular y in terms of x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. We'll talk more about this in R3."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. We'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in three dimensions, it has to be a parametric equation. Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in three dimensions, it has to be a parametric equation. Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these, I guess we call it, these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you've never solved before."}, {"video_title": "Parametric representations of lines Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these, I guess we call it, these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you've never solved before. There's no reason why we had to just stop at three. Three coordinates right here. We could have done this with 50 dimensions."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And in that case, we dealt more particularly with projections onto lines that went through the origin. So if we had some line, let's say L, and let's say L is equal to the span of some vector v, or you could say alternately, you could say that L is equal to the set of all multiples of v, such that the scalar factors are just any real numbers. These are both representations of lines that go through the origin. We defined a projection of any vector onto that line. Let me just draw it real fast. So let me see. Let me draw some axes."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We defined a projection of any vector onto that line. Let me just draw it real fast. So let me see. Let me draw some axes. So if that is my vertical axis, and that is my horizontal axis, just like that. And let's say I have some line that goes through the origin. Let's say that that line right there goes through the origin."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw some axes. So if that is my vertical axis, and that is my horizontal axis, just like that. And let's say I have some line that goes through the origin. Let's say that that line right there goes through the origin. So that is L. We knew visually that a projection of some vector x onto L, so let's say that that is a vector x. Visually, it's essentially, if you were to draw, if you have some light coming straight down, it would be the shadow of x onto L. So this right here, that right there, was the projection onto the line L of the vector x. And we defined it more formally. We kind of took a perpendicular."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that that line right there goes through the origin. So that is L. We knew visually that a projection of some vector x onto L, so let's say that that is a vector x. Visually, it's essentially, if you were to draw, if you have some light coming straight down, it would be the shadow of x onto L. So this right here, that right there, was the projection onto the line L of the vector x. And we defined it more formally. We kind of took a perpendicular. We said that x minus the projection of x onto L is perpendicular to the line L, or perpendicular to everything, or orthogonal to everything on the line L. But this is how, at least, I visualize this, kind of the shadow as you go down onto the line L. And this was a special case in general of projections. You might notice that L is going to be a valid subspace. You could prove it to yourself."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We kind of took a perpendicular. We said that x minus the projection of x onto L is perpendicular to the line L, or perpendicular to everything, or orthogonal to everything on the line L. But this is how, at least, I visualize this, kind of the shadow as you go down onto the line L. And this was a special case in general of projections. You might notice that L is going to be a valid subspace. You could prove it to yourself. It contains the zero vector. It goes through the origin. It's closed under addition."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "You could prove it to yourself. It contains the zero vector. It goes through the origin. It's closed under addition. Any member of it plus any other member of it is going to be another member of it. And it's closed under scalar multiplication. You could take any member of it and scale it up or down."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "It's closed under addition. Any member of it plus any other member of it is going to be another member of it. And it's closed under scalar multiplication. You could take any member of it and scale it up or down. It's still going to be an L. So this was a subspace when we defined this. And just as a bit of a reminder of what it was, we were able to figure out what this projection is for some line L. If you have some spanning vector, the projection onto this line L that goes through the origin of the vector x, we figured out was, it was x dot your spanning vector for your line. So x dot v over v dot v, which is really just the length of v squared."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "You could take any member of it and scale it up or down. It's still going to be an L. So this was a subspace when we defined this. And just as a bit of a reminder of what it was, we were able to figure out what this projection is for some line L. If you have some spanning vector, the projection onto this line L that goes through the origin of the vector x, we figured out was, it was x dot your spanning vector for your line. So x dot v over v dot v, which is really just the length of v squared. So all of this was a number. And you want it to be in the same direction as your line. It's going to be another vector in your line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So x dot v over v dot v, which is really just the length of v squared. So all of this was a number. And you want it to be in the same direction as your line. It's going to be another vector in your line. So it's going to be times the vector v. So it's just going to be a scaled up or a scaled down version of your spanning vector. Maybe your spanning vector is like that. And really any vector in your line could be a spanning vector, any vector other than the 0 vector."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be another vector in your line. So it's going to be times the vector v. So it's just going to be a scaled up or a scaled down version of your spanning vector. Maybe your spanning vector is like that. And really any vector in your line could be a spanning vector, any vector other than the 0 vector. Now, that was a projection onto a line, which was a special kind of subspace. But now we're going to broaden our definition of a projection to any subspace. So we already know that if, let me draw a little dividing line to show that we're doing something slightly different."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And really any vector in your line could be a spanning vector, any vector other than the 0 vector. Now, that was a projection onto a line, which was a special kind of subspace. But now we're going to broaden our definition of a projection to any subspace. So we already know that if, let me draw a little dividing line to show that we're doing something slightly different. We already know that if v is a subspace of Rn, then v complement is also a subspace of Rn. So the orthogonal complement of v also a subspace. Now let's say we have some members, or let me write it this way."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So we already know that if, let me draw a little dividing line to show that we're doing something slightly different. We already know that if v is a subspace of Rn, then v complement is also a subspace of Rn. So the orthogonal complement of v also a subspace. Now let's say we have some members, or let me write it this way. Then if we have these two subspaces, you have a subspace and you have its orthogonal complement, we already learned that if you have any member of Rn, so let's say that x is a member of Rn, then x can be represented as the sum of a member of v and a member of the orthogonal complement of v. Let me write this, where the vector v is a member of the subspace v and the vector w is a member of the orthogonal complement of the subspace v. Just like that. We saw this several videos ago. We proved that this was true for any member of Rn."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's say we have some members, or let me write it this way. Then if we have these two subspaces, you have a subspace and you have its orthogonal complement, we already learned that if you have any member of Rn, so let's say that x is a member of Rn, then x can be represented as the sum of a member of v and a member of the orthogonal complement of v. Let me write this, where the vector v is a member of the subspace v and the vector w is a member of the orthogonal complement of the subspace v. Just like that. We saw this several videos ago. We proved that this was true for any member of Rn. Now, given that, we can define the projection of x onto the subspace v as being equal to just the part of x. You can kind of view it, these are two orthogonal parts of x. We define the projection onto v as a part of x that came from v. It's equal to just that vector v. Alternately, you could say that the projection of x onto the orthogonal complement of v. v is going to be equal to w. So this piece right here is a projection onto the subspace v. This piece right here is a projection onto the orthogonal complement of the subspace v. Now, what I want to do in this video is show you that these two definitions, that this definition right here, and it's in conjunction with this right here, this is equivalent to what we learned up here if the subspace v that we're dealing with is a line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We proved that this was true for any member of Rn. Now, given that, we can define the projection of x onto the subspace v as being equal to just the part of x. You can kind of view it, these are two orthogonal parts of x. We define the projection onto v as a part of x that came from v. It's equal to just that vector v. Alternately, you could say that the projection of x onto the orthogonal complement of v. v is going to be equal to w. So this piece right here is a projection onto the subspace v. This piece right here is a projection onto the orthogonal complement of the subspace v. Now, what I want to do in this video is show you that these two definitions, that this definition right here, and it's in conjunction with this right here, this is equivalent to what we learned up here if the subspace v that we're dealing with is a line. And because this was a valid subspace. But not all subspaces are going to be lines. And to see this, we can revisit an example that we saw several videos ago."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We define the projection onto v as a part of x that came from v. It's equal to just that vector v. Alternately, you could say that the projection of x onto the orthogonal complement of v. v is going to be equal to w. So this piece right here is a projection onto the subspace v. This piece right here is a projection onto the orthogonal complement of the subspace v. Now, what I want to do in this video is show you that these two definitions, that this definition right here, and it's in conjunction with this right here, this is equivalent to what we learned up here if the subspace v that we're dealing with is a line. And because this was a valid subspace. But not all subspaces are going to be lines. And to see this, we can revisit an example that we saw several videos ago. We had this matrix here, this 2 by 2 matrix, and then we had this other vector b that was a member of the column space of A. And we did this matrix, we did this problem to show you that the shortest solution to this right here was a unique member of the row space. Hopefully that gets your memory on track for when this problem, when we first did it."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And to see this, we can revisit an example that we saw several videos ago. We had this matrix here, this 2 by 2 matrix, and then we had this other vector b that was a member of the column space of A. And we did this matrix, we did this problem to show you that the shortest solution to this right here was a unique member of the row space. Hopefully that gets your memory on track for when this problem, when we first did it. But let me graph it and show you that the solution to that problem, we could have just as easily taken a projection onto a subspace. So let me graph everything in this problem. This might help you remember also about the problem."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Hopefully that gets your memory on track for when this problem, when we first did it. But let me graph it and show you that the solution to that problem, we could have just as easily taken a projection onto a subspace. So let me graph everything in this problem. This might help you remember also about the problem. So let me draw my axes just like that. So the first thing we learned, you could solve this, but I already did this in a video. I think it was two or three videos ago."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "This might help you remember also about the problem. So let me draw my axes just like that. So the first thing we learned, you could solve this, but I already did this in a video. I think it was two or three videos ago. The null space of A, or all of the x's that satisfy Ax is equal to 0, is a span of the vector 2, 3. So you go 2 to the right, 1, 2, and then you go 3 up, 1, 2, 3, and so it's a span of this vector. And so the span of that vector is just all of the points."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "I think it was two or three videos ago. The null space of A, or all of the x's that satisfy Ax is equal to 0, is a span of the vector 2, 3. So you go 2 to the right, 1, 2, and then you go 3 up, 1, 2, 3, and so it's a span of this vector. And so the span of that vector is just all of the points. Well, that vector specifies that point, but if you scale this vector up and down, you're going to specify all of the points on this line. Let me draw it like that. That's good enough."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And so the span of that vector is just all of the points. Well, that vector specifies that point, but if you scale this vector up and down, you're going to specify all of the points on this line. Let me draw it like that. That's good enough. It shouldn't curve down like that at the end, so let me draw it a little straighter. So this is the null space. That is our null space of that matrix right there."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "That's good enough. It shouldn't curve down like that at the end, so let me draw it a little straighter. So this is the null space. That is our null space of that matrix right there. And then the row space was a span of the vector 3 minus 2. You see that right here. 3 minus 2 is the first row."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "That is our null space of that matrix right there. And then the row space was a span of the vector 3 minus 2. You see that right here. 3 minus 2 is the first row. This guy is just a multiple of that one. That's why we don't have this guy right here in the span as well. And if we were to graph it, 3 minus 2, you go out 3, and then you go down 1, 2."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "3 minus 2 is the first row. This guy is just a multiple of that one. That's why we don't have this guy right here in the span as well. And if we were to graph it, 3 minus 2, you go out 3, and then you go down 1, 2. It would be the span of this vector right there. Let me draw it like that. And you take all of the scalar multiples of that vector, and you'd put those vectors in standard position."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And if we were to graph it, 3 minus 2, you go out 3, and then you go down 1, 2. It would be the span of this vector right there. Let me draw it like that. And you take all of the scalar multiples of that vector, and you'd put those vectors in standard position. They're going to specify, or their tips are going to be on points along this line right there. Along that line right there. I'm trying to make sure I draw them orthogonally."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And you take all of the scalar multiples of that vector, and you'd put those vectors in standard position. They're going to specify, or their tips are going to be on points along this line right there. Along that line right there. I'm trying to make sure I draw them orthogonally. So this right here is the row space. That right there is the row space of A, which is the same thing as the column space of A transpose. And we know that these guys are each other's orthogonal complements."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm trying to make sure I draw them orthogonally. So this right here is the row space. That right there is the row space of A, which is the same thing as the column space of A transpose. And we know that these guys are each other's orthogonal complements. We know, we've seen this in multiple videos, that the null space of A is the orthogonal complement of the row space. And we also know that the orthogonal complement of the null space is equal to the row space. Everything in this is orthogonal to everything in that, everything in that is orthogonal to everything in this."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And we know that these guys are each other's orthogonal complements. We know, we've seen this in multiple videos, that the null space of A is the orthogonal complement of the row space. And we also know that the orthogonal complement of the null space is equal to the row space. Everything in this is orthogonal to everything in that, everything in that is orthogonal to everything in this. And you can see it here in this graph, that these two spaces, which are represented by these lines that go through the origin, are orthogonal. And it makes sense that any, we said really at the beginning of the video, that anything in R2 in this situation, so this is R2, can be represented as some sum of a unique member of our row space and a unique member of its orthogonal complement. Let's say I have that point right there."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Everything in this is orthogonal to everything in that, everything in that is orthogonal to everything in this. And you can see it here in this graph, that these two spaces, which are represented by these lines that go through the origin, are orthogonal. And it makes sense that any, we said really at the beginning of the video, that anything in R2 in this situation, so this is R2, can be represented as some sum of a unique member of our row space and a unique member of its orthogonal complement. Let's say I have that point right there. How could I represent it as a sum of a member of this and a member of that? Well, if I go along this guy, I have this vector right here. I have that vector right there along that line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have that point right there. How could I represent it as a sum of a member of this and a member of that? Well, if I go along this guy, I have this vector right here. I have that vector right there along that line. And then I have this vector right here. If I were to shift it, this is just drawn in standard position, but I can draw a vector wherever I want. These lines are just all of the vectors drawn in standard position with their tails at the origin."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "I have that vector right there along that line. And then I have this vector right here. If I were to shift it, this is just drawn in standard position, but I can draw a vector wherever I want. These lines are just all of the vectors drawn in standard position with their tails at the origin. But we learned in really, I think, the first or second vector videos that I can draw them wherever I want. So if I add this vector and that vector, I can shift this vector over, and this vector will be right there. And there you have it."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "These lines are just all of the vectors drawn in standard position with their tails at the origin. But we learned in really, I think, the first or second vector videos that I can draw them wherever I want. So if I add this vector and that vector, I can shift this vector over, and this vector will be right there. And there you have it. I took an arbitrary point in R2, and I can represent it as a sum of a member of my row space and a member of the row space's orthogonal complement, or the null space. But just to review what we originally did in that problem, is we looked at the solution set of this. We said the solution set of this looks like this."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And there you have it. I took an arbitrary point in R2, and I can represent it as a sum of a member of my row space and a member of the row space's orthogonal complement, or the null space. But just to review what we originally did in that problem, is we looked at the solution set of this. We said the solution set of this looks like this. It has a particular solution plus members of your null space, plus homogeneous solutions. We've seen that multiple videos ago. So 3, 0, it looks like this, plus members of the null space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We said the solution set of this looks like this. It has a particular solution plus members of your null space, plus homogeneous solutions. We've seen that multiple videos ago. So 3, 0, it looks like this, plus members of the null space. So your solution set is going to be parallel to this, but shifted to the right by 3. So it looks, let me draw it a little neater than that. And no, let me draw it like that."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So 3, 0, it looks like this, plus members of the null space. So your solution set is going to be parallel to this, but shifted to the right by 3. So it looks, let me draw it a little neater than that. And no, let me draw it like that. And then it goes down like the second part. There you go. Oh, that's not good either."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And no, let me draw it like that. And then it goes down like the second part. There you go. Oh, that's not good either. Maybe I'm being too picky. OK, so this is your solution set. And if you remember in that video, we said, hey, there's some member of this solution set that is also a member of our row space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Oh, that's not good either. Maybe I'm being too picky. OK, so this is your solution set. And if you remember in that video, we said, hey, there's some member of this solution set that is also a member of our row space. And that member of the solution set that is a member of our row space is going to be the shortest solution. And we saw that. You could see it visually right here."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And if you remember in that video, we said, hey, there's some member of this solution set that is also a member of our row space. And that member of the solution set that is a member of our row space is going to be the shortest solution. And we saw that. You could see it visually right here. This vector right here. It is in our row space. It is a member of our row space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "You could see it visually right here. This vector right here. It is in our row space. It is a member of our row space. And it also specifies a point on our solution set. And you could see visually that it's going to be the shortest solution. And one way you could think about it is, this is the projection, any solution to, let me pick a good, different, new color, any solution on our solution set."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "It is a member of our row space. And it also specifies a point on our solution set. And you could see visually that it's going to be the shortest solution. And one way you could think about it is, this is the projection, any solution to, let me pick a good, different, new color, any solution on our solution set. Let me see, it's right there. Let's say that that is some arbitrary solution on our solution set. That's going to be a point in R2."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And one way you could think about it is, this is the projection, any solution to, let me pick a good, different, new color, any solution on our solution set. Let me see, it's right there. Let's say that that is some arbitrary solution on our solution set. That's going to be a point in R2. And any point in R2 can be represented as a sum of some vector in our row space and some vector in our null space. And so if I have this vector right here, how could I do that? Well, I could represent it as a sum of this guy right here and then this vector right here."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be a point in R2. And any point in R2 can be represented as a sum of some vector in our row space and some vector in our null space. And so if I have this vector right here, how could I do that? Well, I could represent it as a sum of this guy right here and then this vector right here. That vector right here. And this vector right here is clearly a member of my null space. I just shift it over."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I could represent it as a sum of this guy right here and then this vector right here. That vector right here. And this vector right here is clearly a member of my null space. I just shift it over. This line is only when I draw it in standard position. This vector right here, I'm just showing it heads to tails, if I add this member of my row space to this member of my null space, I get an arbitrary solution to my solution set. And if you think about it, the projection of my arbitrary solution onto my row space will be this guy right here."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "I just shift it over. This line is only when I draw it in standard position. This vector right here, I'm just showing it heads to tails, if I add this member of my row space to this member of my null space, I get an arbitrary solution to my solution set. And if you think about it, the projection of my arbitrary solution onto my row space will be this guy right here. And that just comes from our, well, there's two ways to think about it. We could say that, let's say that this is a solution right here. We could say our solution right here is equal to some member of my row space plus some member of my null space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And if you think about it, the projection of my arbitrary solution onto my row space will be this guy right here. And that just comes from our, well, there's two ways to think about it. We could say that, let's say that this is a solution right here. We could say our solution right here is equal to some member of my row space plus some member of my null space. This is the row space, that is the null space. And so by the definition of a projection onto a subspace I just gave you, we know that the projection of this solution onto my, let me write it a little bit, the projection onto my row space of my solution is just equal to this first thing. It's equal to the component of it that's in my row space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "We could say our solution right here is equal to some member of my row space plus some member of my null space. This is the row space, that is the null space. And so by the definition of a projection onto a subspace I just gave you, we know that the projection of this solution onto my, let me write it a little bit, the projection onto my row space of my solution is just equal to this first thing. It's equal to the component of it that's in my row space. Its other component, we could call it, is in the orthogonal complement of my row space, or it's in my null space. So this is just going to be equal to the r vector. Now I want to show you that that is essentially equal to the definition that we did before."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to the component of it that's in my row space. Its other component, we could call it, is in the orthogonal complement of my row space, or it's in my null space. So this is just going to be equal to the r vector. Now I want to show you that that is essentially equal to the definition that we did before. This is completely identical to the definition of a projection onto a line. Because in this case, the subspace is a line. So let's find a solution set."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to show you that that is essentially equal to the definition that we did before. This is completely identical to the definition of a projection onto a line. Because in this case, the subspace is a line. So let's find a solution set. And the easiest one, the easiest solution that we could find is if we set c is equal to 0 here, we know that x equals 3, 0 is one of the solutions. So x equals 3, 0 looks like that. So we know x equals 3, 0 is a solution."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's find a solution set. And the easiest one, the easiest solution that we could find is if we set c is equal to 0 here, we know that x equals 3, 0 is one of the solutions. So x equals 3, 0 looks like that. So we know x equals 3, 0 is a solution. And what we want to do is we want to find this shortest solution. Or we want to find the projection of x onto the row space, or we could also think of it as a projection of x onto this line. This line is equal to the row space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So we know x equals 3, 0 is a solution. And what we want to do is we want to find this shortest solution. Or we want to find the projection of x onto the row space, or we could also think of it as a projection of x onto this line. This line is equal to the row space. So let's do that. I'm doing this to show you that this definition of a projection onto a subspace that I've just introduced you to in this video, it is completely identical to the definition, or it's not identical. It's consistent with the definition of a projection onto a line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "This line is equal to the row space. So let's do that. I'm doing this to show you that this definition of a projection onto a subspace that I've just introduced you to in this video, it is completely identical to the definition, or it's not identical. It's consistent with the definition of a projection onto a line. Although this is more general, because a subspace doesn't have to be a line. But in this case, it is a line. So let's do that."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "It's consistent with the definition of a projection onto a line. Although this is more general, because a subspace doesn't have to be a line. But in this case, it is a line. So let's do that. So the projection of 3, 0 onto our row space, which is a line, so we can use that formula, it is equal to 3, 0 dot the spanning vector for our row space. So that's 3 minus 2. There's a bunch of spanning vectors for your row space."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So the projection of 3, 0 onto our row space, which is a line, so we can use that formula, it is equal to 3, 0 dot the spanning vector for our row space. So that's 3 minus 2. There's a bunch of spanning vectors for your row space. This is just the one we happened to pick. So dot 3 minus 2 all over the spanning vector dotted with itself, 3 minus 2 dot 3 minus 2. And then this is just going to be one big scalar."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "There's a bunch of spanning vectors for your row space. This is just the one we happened to pick. So dot 3 minus 2 all over the spanning vector dotted with itself, 3 minus 2 dot 3 minus 2. And then this is just going to be one big scalar. And then we want to multiply that, or essentially scale up our actual spanning vector by that. So this is a projection of this solution onto my row space, which should give me this vector right here. Because we're just taking a projection onto a line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And then this is just going to be one big scalar. And then we want to multiply that, or essentially scale up our actual spanning vector by that. So this is a projection of this solution onto my row space, which should give me this vector right here. Because we're just taking a projection onto a line. Because the row space and this subspace is a line. And so we use the linear projections that we first got introduced to, I think, when I first started doing linear transformations. So what is 3?"}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Because we're just taking a projection onto a line. Because the row space and this subspace is a line. And so we use the linear projections that we first got introduced to, I think, when I first started doing linear transformations. So what is 3? So let's see. This is 3 times 3 plus 0 times minus 2. This right here is equal to 9."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "So what is 3? So let's see. This is 3 times 3 plus 0 times minus 2. This right here is equal to 9. This is 3 times 3 plus minus 2 times minus 2. So that's 9 plus 4. That's 13."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "This right here is equal to 9. This is 3 times 3 plus minus 2 times minus 2. So that's 9 plus 4. That's 13. So it's 9 over 13 times this vector right here. So we're going to get it's going to be equal to 9 over 13 times the vector 3 minus 2, which is equal to the vector 27 over 13 and then minus 18 over 13, which is this vector right here. Which we got this exact answer when we first did it, although we just didn't use the projection onto a line."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "That's 13. So it's 9 over 13 times this vector right here. So we're going to get it's going to be equal to 9 over 13 times the vector 3 minus 2, which is equal to the vector 27 over 13 and then minus 18 over 13, which is this vector right here. Which we got this exact answer when we first did it, although we just didn't use the projection onto a line. But now we see that this is exactly consistent with what we did before. We just used the projection onto a line. And we see that this is consistent with our new, broader definition of a projection."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "Which we got this exact answer when we first did it, although we just didn't use the projection onto a line. But now we see that this is exactly consistent with what we did before. We just used the projection onto a line. And we see that this is consistent with our new, broader definition of a projection. Here we were able to do it because we did it onto a line. But here I'm calling a projection onto any subspace. We know how to do it if it's a line, but so far I've just kind of defined it onto an arbitrary subspace."}, {"video_title": "Projections onto subspaces Linear Algebra Khan Academy.mp3", "Sentence": "And we see that this is consistent with our new, broader definition of a projection. Here we were able to do it because we did it onto a line. But here I'm calling a projection onto any subspace. We know how to do it if it's a line, but so far I've just kind of defined it onto an arbitrary subspace. But I haven't given you a nice mathematical, I guess, or computational way to figure out what this is going to be if this isn't a line. In fact, I haven't even showed you when this is general, whether this is definitely a linear transformation. We know that when you take the projection onto a line it's a linear transformation, but I haven't shown you that when we take a projection onto an arbitrary subspace that it is a linear projection."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It only makes sense that we have something called a linear transformation because we're studying linear algebra. We already had linear combinations, so we might as well have a linear transformation. A linear transformation, by definition, is a transformation, which we know is just a function, we could say it's from the set Rm, let me say it from Rn to Rm, and it might be obvious in the next video why I'm being a little bit particular about that, although they are just arbitrary letters, where the following two things have to be true. If something is a linear transformation, if and only if the following thing is true. Let's say that we have two vectors, vector A and vector B, are both members of Rn. They're both in our domain. This is a linear transformation if and only if I take the transformation of the sum of our two vectors, if I add them up first, that's equivalent to taking the transformation of each of the vectors and then summing them."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If something is a linear transformation, if and only if the following thing is true. Let's say that we have two vectors, vector A and vector B, are both members of Rn. They're both in our domain. This is a linear transformation if and only if I take the transformation of the sum of our two vectors, if I add them up first, that's equivalent to taking the transformation of each of the vectors and then summing them. That's my first condition for this to be a linear transformation. The second one is, if I take the transformation of any scaled up version of a vector, so let me just multiply vector A times some scalar, some real number C, if this is a linear transformation, then this should be equal to C times the transformation of A. That seems pretty straightforward."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a linear transformation if and only if I take the transformation of the sum of our two vectors, if I add them up first, that's equivalent to taking the transformation of each of the vectors and then summing them. That's my first condition for this to be a linear transformation. The second one is, if I take the transformation of any scaled up version of a vector, so let me just multiply vector A times some scalar, some real number C, if this is a linear transformation, then this should be equal to C times the transformation of A. That seems pretty straightforward. Let's see if we can apply these rules to figure out if some actual transformations are linear or not. Let me define a transformation. Let's say that I have the transformation T. Part of my definition, I'm going to tell you it maps from R2 to R2."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That seems pretty straightforward. Let's see if we can apply these rules to figure out if some actual transformations are linear or not. Let me define a transformation. Let's say that I have the transformation T. Part of my definition, I'm going to tell you it maps from R2 to R2. If you give it a 2-tuple, its domain is 2-tuple, so you give it an x1 and an x2. Let's say it maps to, so this will be equal to, or it's associated with, let me just say x1 plus x2, and then let's just say it's 3 times x1 is the second 2-tuple. We could have written this more in vector form."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that I have the transformation T. Part of my definition, I'm going to tell you it maps from R2 to R2. If you give it a 2-tuple, its domain is 2-tuple, so you give it an x1 and an x2. Let's say it maps to, so this will be equal to, or it's associated with, let me just say x1 plus x2, and then let's just say it's 3 times x1 is the second 2-tuple. We could have written this more in vector form. This is kind of our tuple form. We could have written it, and it's good to see all the different notations that you might encounter. A transformation of some vector x where the vector looks like this, x1, x2, let me put a bracket there, it equals some new vector, x1 plus x2, and then the second component of the new vector would be 3x1."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could have written this more in vector form. This is kind of our tuple form. We could have written it, and it's good to see all the different notations that you might encounter. A transformation of some vector x where the vector looks like this, x1, x2, let me put a bracket there, it equals some new vector, x1 plus x2, and then the second component of the new vector would be 3x1. That's a completely legitimate way to express our transformation. The third way, which I never see, but to me it kind of captures the essence of what a transformation is. It's just a mapping or it's just a function."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A transformation of some vector x where the vector looks like this, x1, x2, let me put a bracket there, it equals some new vector, x1 plus x2, and then the second component of the new vector would be 3x1. That's a completely legitimate way to express our transformation. The third way, which I never see, but to me it kind of captures the essence of what a transformation is. It's just a mapping or it's just a function. We could say that the transformation is a mapping from any vector in R2 that looks like this, x1, x2, to, and I'll do this notation, to a vector that looks like this, x1 plus x2, and then 3x1. All of these statements are equivalent. But our whole point of writing this is to figure out whether T is linearly independent."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's just a mapping or it's just a function. We could say that the transformation is a mapping from any vector in R2 that looks like this, x1, x2, to, and I'll do this notation, to a vector that looks like this, x1 plus x2, and then 3x1. All of these statements are equivalent. But our whole point of writing this is to figure out whether T is linearly independent. Sorry, not linearly independent, whether it's a linear transformation. I was so obsessed with linear independence for so many videos, it's hard to get it out of my brain in this one. Whether it's a linear transformation."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But our whole point of writing this is to figure out whether T is linearly independent. Sorry, not linearly independent, whether it's a linear transformation. I was so obsessed with linear independence for so many videos, it's hard to get it out of my brain in this one. Whether it's a linear transformation. Let's test our two conditions. I have them up here. Let's take T of, let's say I have two vectors, A and B."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Whether it's a linear transformation. Let's test our two conditions. I have them up here. Let's take T of, let's say I have two vectors, A and B. They're members of R2, so let me write it. A is equal to a1, no, a1, a2. And B is equal to b1, b2."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's take T of, let's say I have two vectors, A and B. They're members of R2, so let me write it. A is equal to a1, no, a1, a2. And B is equal to b1, b2. Sorry, that's not a vector. I have to make sure that those are scalars. These are the components of a vector."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And B is equal to b1, b2. Sorry, that's not a vector. I have to make sure that those are scalars. These are the components of a vector. And b2. So, what is a1 plus b? Sorry, what is vector a plus vector b?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are the components of a vector. And b2. So, what is a1 plus b? Sorry, what is vector a plus vector b? Brain's malfunctioning. Alright, well you just add up their components. This is the definition of vector addition."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Sorry, what is vector a plus vector b? Brain's malfunctioning. Alright, well you just add up their components. This is the definition of vector addition. So it's a1 plus b1. Add up the first components. And the second components is just the sum of each of the vector's second components."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the definition of vector addition. So it's a1 plus b1. Add up the first components. And the second components is just the sum of each of the vector's second components. a2 plus b2. Nothing new here. But what is the transformation of this vector?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the second components is just the sum of each of the vector's second components. a2 plus b2. Nothing new here. But what is the transformation of this vector? So the transformation of vector a plus vector b, we could write it like this. That would be the same thing as the transformation of this vector, which is just a1 plus b1 and a2 plus b2. Which we know, it equals a vector."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But what is the transformation of this vector? So the transformation of vector a plus vector b, we could write it like this. That would be the same thing as the transformation of this vector, which is just a1 plus b1 and a2 plus b2. Which we know, it equals a vector. It equals this vector. Where what we do is, for the first component here, we add up the two components on this side. So the first component here is going to be these two guys added up."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which we know, it equals a vector. It equals this vector. Where what we do is, for the first component here, we add up the two components on this side. So the first component here is going to be these two guys added up. So it's a1 plus a2 plus b1 plus b2. And then the second component, by our transformation or our function definition, is just three times the first component in our domain, I guess you could say. So it's three times the first one."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the first component here is going to be these two guys added up. So it's a1 plus a2 plus b1 plus b2. And then the second component, by our transformation or our function definition, is just three times the first component in our domain, I guess you could say. So it's three times the first one. So it's going to be three times this first guy. So it's 3a1 plus 3b1. Fair enough."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's three times the first one. So it's going to be three times this first guy. So it's 3a1 plus 3b1. Fair enough. Now, what is the transformation individually of a and b? So the transformation of a is equal to the transformation of a1, a2 in brackets. That's another way of writing vector a."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now, what is the transformation individually of a and b? So the transformation of a is equal to the transformation of a1, a2 in brackets. That's another way of writing vector a. And what is that equal to? That's our definition of our transformation right up here. So this is going to be equal to the vector a1 plus a2 and then 3 times a1."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's another way of writing vector a. And what is that equal to? That's our definition of our transformation right up here. So this is going to be equal to the vector a1 plus a2 and then 3 times a1. Just comes straight out of this definition. I essentially just replaced an x with a's. By the same, I guess, argument, what is the transformation of our vector b?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to the vector a1 plus a2 and then 3 times a1. Just comes straight out of this definition. I essentially just replaced an x with a's. By the same, I guess, argument, what is the transformation of our vector b? Well, it's just going to be the same thing with the a's replaced by the b's. So the transformation of our vector b is going to be b is just b1, b2. So it's going to be b1 plus b2."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "By the same, I guess, argument, what is the transformation of our vector b? Well, it's just going to be the same thing with the a's replaced by the b's. So the transformation of our vector b is going to be b is just b1, b2. So it's going to be b1 plus b2. And then the second component in the transformation will be 3 times b1. Now, what is the transformation of vector a plus the transformation of vector b? What's this vector plus that vector?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be b1 plus b2. And then the second component in the transformation will be 3 times b1. Now, what is the transformation of vector a plus the transformation of vector b? What's this vector plus that vector? And what is that equal to? Well, this is just pure vector addition. So we just add up their components."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What's this vector plus that vector? And what is that equal to? Well, this is just pure vector addition. So we just add up their components. So it's a1 plus a2 plus b1 plus b2. That's just that component plus that component. The second component is 3a1."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we just add up their components. So it's a1 plus a2 plus b1 plus b2. That's just that component plus that component. The second component is 3a1. We're going to add it to that second component. So it's 3a1 plus 3b1. Now, we just showed you that if I take the transformation separately of each of the vectors and then add them up, I get the exact same thing as if I took the vectors and added them up first and then took the transformation."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The second component is 3a1. We're going to add it to that second component. So it's 3a1 plus 3b1. Now, we just showed you that if I take the transformation separately of each of the vectors and then add them up, I get the exact same thing as if I took the vectors and added them up first and then took the transformation. So we've met our first criteria, that the transformation of the sum of the vectors is the same thing as the sum of the transformations. Now, let's see if this works with a random scalar. So we know what the transformation of a looks like."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we just showed you that if I take the transformation separately of each of the vectors and then add them up, I get the exact same thing as if I took the vectors and added them up first and then took the transformation. So we've met our first criteria, that the transformation of the sum of the vectors is the same thing as the sum of the transformations. Now, let's see if this works with a random scalar. So we know what the transformation of a looks like. What does a transformation, well, what does ca look like, first of all? I guess that's a good place to start. c times our vector a is going to be equal to c times a1 and then c times a2."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we know what the transformation of a looks like. What does a transformation, well, what does ca look like, first of all? I guess that's a good place to start. c times our vector a is going to be equal to c times a1 and then c times a2. That's our definition of scalar multiplication times a vector. So what's our transformation? Let me go to a new color."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "c times our vector a is going to be equal to c times a1 and then c times a2. That's our definition of scalar multiplication times a vector. So what's our transformation? Let me go to a new color. What is our, let me do a color I haven't used in a long time, white. What is our transformation of ca going to be? Well, that's the same thing as our transformation of ca1, ca2, which is equal to a new vector where the first term, let's go to our definition, is you sum the first and second components and the second term is 3 times the first component."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me go to a new color. What is our, let me do a color I haven't used in a long time, white. What is our transformation of ca going to be? Well, that's the same thing as our transformation of ca1, ca2, which is equal to a new vector where the first term, let's go to our definition, is you sum the first and second components and the second term is 3 times the first component. So our first term, you sum them, so it's going to be ca1 plus ca2. And then our second term is 3 times our first term, so it's 3ca1. Now, what is this equal to?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's the same thing as our transformation of ca1, ca2, which is equal to a new vector where the first term, let's go to our definition, is you sum the first and second components and the second term is 3 times the first component. So our first term, you sum them, so it's going to be ca1 plus ca2. And then our second term is 3 times our first term, so it's 3ca1. Now, what is this equal to? This is the same thing, we can kind of, you can view it as factoring out the c. This is the same thing as c times the vector a1 plus a2 and then the second component is 3a1. But this thing right here, we already saw, this is the same thing as the transformation of a. So just like that, you see that the transformation of c times our vector a, for any vector a in R2, anything in R2 can be represented this way, it's the same thing as c times the transformation of a."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this equal to? This is the same thing, we can kind of, you can view it as factoring out the c. This is the same thing as c times the vector a1 plus a2 and then the second component is 3a1. But this thing right here, we already saw, this is the same thing as the transformation of a. So just like that, you see that the transformation of c times our vector a, for any vector a in R2, anything in R2 can be represented this way, it's the same thing as c times the transformation of a. So we've met our second condition. That it doesn't, that when you, well, I just stated it, so I don't have to restate it. So we meet both conditions, which tells us that this is a linear transformation."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So just like that, you see that the transformation of c times our vector a, for any vector a in R2, anything in R2 can be represented this way, it's the same thing as c times the transformation of a. So we've met our second condition. That it doesn't, that when you, well, I just stated it, so I don't have to restate it. So we meet both conditions, which tells us that this is a linear transformation. And you might be thinking, okay, Sal, fair enough, that was, you know, how do I know that all transformations aren't linear transformations? Show me something that won't work. And here I'll do a very simple example."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we meet both conditions, which tells us that this is a linear transformation. And you might be thinking, okay, Sal, fair enough, that was, you know, how do I know that all transformations aren't linear transformations? Show me something that won't work. And here I'll do a very simple example. Let me define my transformation. And just to make it, let me do one, let me make it, let me see, let me define a transformation. I'll do it from R2 to R2 just for, just to kind of compare the two."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And here I'll do a very simple example. Let me define my transformation. And just to make it, let me do one, let me make it, let me see, let me define a transformation. I'll do it from R2 to R2 just for, just to kind of compare the two. I could have done it from R to R if I wanted a simpler example. But I'm going to define my transformation, let's say my transformation of the vector x1, x2. Let's say it is equal to, let me just say x1 squared and then 0, just like that."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it from R2 to R2 just for, just to kind of compare the two. I could have done it from R to R if I wanted a simpler example. But I'm going to define my transformation, let's say my transformation of the vector x1, x2. Let's say it is equal to, let me just say x1 squared and then 0, just like that. Let me see if this is a linear transformation. So the first question is if, let's take, what's my transformation of a vector a? So my transformation of a vector a, where a is just the same a that I did before, it would look like this."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it is equal to, let me just say x1 squared and then 0, just like that. Let me see if this is a linear transformation. So the first question is if, let's take, what's my transformation of a vector a? So my transformation of a vector a, where a is just the same a that I did before, it would look like this. It would look like a1 squared and then a 0. Now, what would be my transformation if I took c times a? Well, this is the same thing as c times a1 and c times a2."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So my transformation of a vector a, where a is just the same a that I did before, it would look like this. It would look like a1 squared and then a 0. Now, what would be my transformation if I took c times a? Well, this is the same thing as c times a1 and c times a2. And by our transformation definition, if I take, sorry, the transformation of c times of this thing right here, because I'm taking the transformation on both sides, by our transformation definition, this will just be equal to a new vector that would be in our co-domain where the first term is just the first term of our input squared. So it's c a1 squared and the second term is 0. And what is this equal to?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is the same thing as c times a1 and c times a2. And by our transformation definition, if I take, sorry, the transformation of c times of this thing right here, because I'm taking the transformation on both sides, by our transformation definition, this will just be equal to a new vector that would be in our co-domain where the first term is just the first term of our input squared. So it's c a1 squared and the second term is 0. And what is this equal to? Let me switch colors. This is equal to c squared a1 squared and this is equal to 0. Now, let's see if we can factor out, well, if we can assume that c does not equal 0, this would be equal to what?"}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? Let me switch colors. This is equal to c squared a1 squared and this is equal to 0. Now, let's see if we can factor out, well, if we can assume that c does not equal 0, this would be equal to what? Actually, it doesn't even matter. We don't even have to make that assumption. So this is the same thing."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's see if we can factor out, well, if we can assume that c does not equal 0, this would be equal to what? Actually, it doesn't even matter. We don't even have to make that assumption. So this is the same thing. This is equal to c squared times the vector a1 squared 0, which is equal to what? This expression right here is a transformation of a. So this is equal to c squared times the transformation of a."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the same thing. This is equal to c squared times the vector a1 squared 0, which is equal to what? This expression right here is a transformation of a. So this is equal to c squared times the transformation of a. Let me do it in the same color, times the transformation of a. So what I've just showed you is if I take the transformation of a vector being multiplied by a scalar quantity first, that that's equal to, for this t, for this transformation that I've defined right here, that's equal to c squared times the transformation of a. So clearly, this statement right here for this choice of transformation conflicts with this requirement for a linear transformation."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is equal to c squared times the transformation of a. Let me do it in the same color, times the transformation of a. So what I've just showed you is if I take the transformation of a vector being multiplied by a scalar quantity first, that that's equal to, for this t, for this transformation that I've defined right here, that's equal to c squared times the transformation of a. So clearly, this statement right here for this choice of transformation conflicts with this requirement for a linear transformation. If I have a c here, I should see a c here. But in our case, I have a c here and I have a c squared here. So clearly, this negates that statement."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So clearly, this statement right here for this choice of transformation conflicts with this requirement for a linear transformation. If I have a c here, I should see a c here. But in our case, I have a c here and I have a c squared here. So clearly, this negates that statement. So this is not a linear transformation. And just to get a gut feel, if you're just looking at something, whether it's going to be a linear transformation or not, if the transformation just involves linear combinations of the different components of the inputs, you're probably dealing with a linear transformation. If you start seeing things where the components start getting multiplied by each other or you start seeing squares or exponents, you're probably not dealing with a linear transformation."}, {"video_title": "Linear transformations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So clearly, this negates that statement. So this is not a linear transformation. And just to get a gut feel, if you're just looking at something, whether it's going to be a linear transformation or not, if the transformation just involves linear combinations of the different components of the inputs, you're probably dealing with a linear transformation. If you start seeing things where the components start getting multiplied by each other or you start seeing squares or exponents, you're probably not dealing with a linear transformation. And then there are some functions that might be in a bit of a gray area. But it tends to be just linear combinations are going to lead to a linear transformation. But hopefully, that gives you a good sense of things."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have two linear transformations. Say I have the transformation S that's a mapping or a function from the set X to the set Y. And let's just say that X is a subset of Rn. Let's say that Y is a subset of Rm. Now we know if S is a linear transformation, it can be represented by a matrix vector product. So we can write S of X. Let me do it in the same color I was doing it before."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that Y is a subset of Rm. Now we know if S is a linear transformation, it can be represented by a matrix vector product. So we can write S of X. Let me do it in the same color I was doing it before. We can write that S of some vector X is equal to some matrix A times X. And the matrix A, it's going to be X, whatever X we input into the function, though we take the mapping of, it's going to be in this set right here, it's going to be a member of Rn. So this is going to be right here."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in the same color I was doing it before. We can write that S of some vector X is equal to some matrix A times X. And the matrix A, it's going to be X, whatever X we input into the function, though we take the mapping of, it's going to be in this set right here, it's going to be a member of Rn. So this is going to be right here. X is, let me do it like this, X is going to be a member of Rn, well it's actually going to be a member of X, which is a subset of Rn, but I'm just trying to figure out what the dimensions of matrix A are going to be. So this is going to have n components right here. Matrix A has to have n columns."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be right here. X is, let me do it like this, X is going to be a member of Rn, well it's actually going to be a member of X, which is a subset of Rn, but I'm just trying to figure out what the dimensions of matrix A are going to be. So this is going to have n components right here. Matrix A has to have n columns. So matrix A is going to be, let's just say it's an m by n matrix, m by n. Fair enough. Now let's say we have another linear transformation. So actually let me draw what I've done so far."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Matrix A has to have n columns. So matrix A is going to be, let's just say it's an m by n matrix, m by n. Fair enough. Now let's say we have another linear transformation. So actually let me draw what I've done so far. So we have some set X right here. That is set X. It is a subset of Rn."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So actually let me draw what I've done so far. So we have some set X right here. That is set X. It is a subset of Rn. I can draw it there, and we have this mapping, S, or this linear transformation from X to Y. So it goes to a new set Y right here. And Y is a member of Rm, so the mapping, X right here, you take some element here and you apply the transformation S, and I've told you it's a linear transformation, and you'll get to some value in set Y, which is in Rm."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It is a subset of Rn. I can draw it there, and we have this mapping, S, or this linear transformation from X to Y. So it goes to a new set Y right here. And Y is a member of Rm, so the mapping, X right here, you take some element here and you apply the transformation S, and I've told you it's a linear transformation, and you'll get to some value in set Y, which is in Rm. And I said that the matrix representation of our linear transformation is going to be an m by n matrix, right? Because you're going to start with something that has n entries, or a vector that's a member of Rn, and you want to end up with a vector that's in Rm. Fair enough."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And Y is a member of Rm, so the mapping, X right here, you take some element here and you apply the transformation S, and I've told you it's a linear transformation, and you'll get to some value in set Y, which is in Rm. And I said that the matrix representation of our linear transformation is going to be an m by n matrix, right? Because you're going to start with something that has n entries, or a vector that's a member of Rn, and you want to end up with a vector that's in Rm. Fair enough. Now let's say I have another linear transformation, T. And it's a mapping from the set Y to the set Z. So let me draw. So I have another set here called set Z, and I can map from elements of Y, so I can map from here, into elements of Z using the linear transformation, T. So similar to what I did before, we know that Y is a member of Rm."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now let's say I have another linear transformation, T. And it's a mapping from the set Y to the set Z. So let me draw. So I have another set here called set Z, and I can map from elements of Y, so I can map from here, into elements of Z using the linear transformation, T. So similar to what I did before, we know that Y is a member of Rm. We know that this is a subset, not a member, more of a subset of Rm. And these are just arbitrary letters. This could be 100 or 5, whatever."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I have another set here called set Z, and I can map from elements of Y, so I can map from here, into elements of Z using the linear transformation, T. So similar to what I did before, we know that Y is a member of Rm. We know that this is a subset, not a member, more of a subset of Rm. And these are just arbitrary letters. This could be 100 or 5, whatever. I'm just trying to stay abstract. Now let's say that Z is a member of Rl. Then what's the transformation T, what's its matrix representation going to be?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This could be 100 or 5, whatever. I'm just trying to stay abstract. Now let's say that Z is a member of Rl. Then what's the transformation T, what's its matrix representation going to be? And we know it's a linear transformation, I told you that, so we know it can be represented in this form. So we could say that T of X, where X is a member of Rm, is going to be equal to some matrix B times X. And what are the dimensions of matrix B going to be?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then what's the transformation T, what's its matrix representation going to be? And we know it's a linear transformation, I told you that, so we know it can be represented in this form. So we could say that T of X, where X is a member of Rm, is going to be equal to some matrix B times X. And what are the dimensions of matrix B going to be? X is going to be a member of Rm, so B is going to have to have m columns. And then it's a mapping into a set that's a member of Rl. So it's going to map from members of Rm to members of Rl."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what are the dimensions of matrix B going to be? X is going to be a member of Rm, so B is going to have to have m columns. And then it's a mapping into a set that's a member of Rl. So it's going to map from members of Rm to members of Rl. So it's going to be an L by m matrix right there. Now when you see this, a very natural question might arise in your head. Can we construct some mapping that goes all the way from set X all the way to set T?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to map from members of Rm to members of Rl. So it's going to be an L by m matrix right there. Now when you see this, a very natural question might arise in your head. Can we construct some mapping that goes all the way from set X all the way to set T? And maybe we'll call that the composition of, and maybe we can create that mapping using a combination of S and T. So let's just make up some word. Let's just call T with this little circle S, let's just call this a mapping from X all the way to Z. And we'll call this the composition of T with S. We're essentially just combining the two functions in order to try to create some mapping that takes us from set X all the way to set Z."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Can we construct some mapping that goes all the way from set X all the way to set T? And maybe we'll call that the composition of, and maybe we can create that mapping using a combination of S and T. So let's just make up some word. Let's just call T with this little circle S, let's just call this a mapping from X all the way to Z. And we'll call this the composition of T with S. We're essentially just combining the two functions in order to try to create some mapping that takes us from set X all the way to set Z. But we still haven't defined this. How can we actually construct this? Well, a natural thing might be to first apply transformation S, let's say that this is our X we're dealing with right here."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we'll call this the composition of T with S. We're essentially just combining the two functions in order to try to create some mapping that takes us from set X all the way to set Z. But we still haven't defined this. How can we actually construct this? Well, a natural thing might be to first apply transformation S, let's say that this is our X we're dealing with right here. Maybe the first thing we want to do is apply S, and that'll give us an S of X, that'll give us this value right here that's in set Y. And then what if we were to take that value and apply the transformation T to it? So we would take this value and apply the transformation T to it to maybe get to this value."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, a natural thing might be to first apply transformation S, let's say that this is our X we're dealing with right here. Maybe the first thing we want to do is apply S, and that'll give us an S of X, that'll give us this value right here that's in set Y. And then what if we were to take that value and apply the transformation T to it? So we would take this value and apply the transformation T to it to maybe get to this value. So this would be the linear transformation T applied to this value, this member of the set Y, which is in Rm. So we're just going to apply that transformation to this guy right here, which was the transformation S applied to X. This might look fancy, but all this is, remember, this is just a vector right here in the set Y, which is a subset of Rm."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we would take this value and apply the transformation T to it to maybe get to this value. So this would be the linear transformation T applied to this value, this member of the set Y, which is in Rm. So we're just going to apply that transformation to this guy right here, which was the transformation S applied to X. This might look fancy, but all this is, remember, this is just a vector right here in the set Y, which is a subset of Rm. This is a vector that's in X. So when you apply a mapping, you get another vector that's in Y. And then you apply the linear transformation T to that, and then you get another vector that's in set Z."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This might look fancy, but all this is, remember, this is just a vector right here in the set Y, which is a subset of Rm. This is a vector that's in X. So when you apply a mapping, you get another vector that's in Y. And then you apply the linear transformation T to that, and then you get another vector that's in set Z. So let's define the composition of T with S. So this is going to be a definition. Let's define the composition of T with S to be, first we apply S to some vector in X to get us here. And then we apply T to that vector to get us to set Z."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you apply the linear transformation T to that, and then you get another vector that's in set Z. So let's define the composition of T with S. So this is going to be a definition. Let's define the composition of T with S to be, first we apply S to some vector in X to get us here. And then we apply T to that vector to get us to set Z. So we apply T to this thing right there. Now, the first question might be, is this even a linear transformation? Is the composition of two linear transformations even a linear transformation?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we apply T to that vector to get us to set Z. So we apply T to this thing right there. Now, the first question might be, is this even a linear transformation? Is the composition of two linear transformations even a linear transformation? Is it a linear transformation? Well, there are two requirements to be a linear transformation. The sum of the linear transformation of the sum of two vectors should be the linear transformation of each of them summed together."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Is the composition of two linear transformations even a linear transformation? Is it a linear transformation? Well, there are two requirements to be a linear transformation. The sum of the linear transformation of the sum of two vectors should be the linear transformation of each of them summed together. So let's see. When I know when I just say that verbally, it probably doesn't make a lot of sense. So let's just try to take the composition of T with S of, let's say, the sum of two vectors in X."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The sum of the linear transformation of the sum of two vectors should be the linear transformation of each of them summed together. So let's see. When I know when I just say that verbally, it probably doesn't make a lot of sense. So let's just try to take the composition of T with S of, let's say, the sum of two vectors in X. So let's say I'm taking the vectors X and the vectors Y. Well, by definition, what is this equal to? This is equal to applying the linear transformation T to the linear transformation S applied to our two vectors, X plus Y."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's just try to take the composition of T with S of, let's say, the sum of two vectors in X. So let's say I'm taking the vectors X and the vectors Y. Well, by definition, what is this equal to? This is equal to applying the linear transformation T to the linear transformation S applied to our two vectors, X plus Y. And what is this equal to? I told you at the beginning of the video that S is a linear transformation. So by definition of a linear transformation, one of our requirements, we know that S of X plus Y is the same thing as S of X plus S of Y, because S is a linear transformation."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to applying the linear transformation T to the linear transformation S applied to our two vectors, X plus Y. And what is this equal to? I told you at the beginning of the video that S is a linear transformation. So by definition of a linear transformation, one of our requirements, we know that S of X plus Y is the same thing as S of X plus S of Y, because S is a linear transformation. We know that that is true. We know that we can replace this thing right there with that thing right there. Well, we also know that T is a linear transformation, which means that the transformation applied to the sum of two vectors is equal to the transformation of each of the vectors summed up."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So by definition of a linear transformation, one of our requirements, we know that S of X plus Y is the same thing as S of X plus S of Y, because S is a linear transformation. We know that that is true. We know that we can replace this thing right there with that thing right there. Well, we also know that T is a linear transformation, which means that the transformation applied to the sum of two vectors is equal to the transformation of each of the vectors summed up. So the transformation of S of X, or the transformation applied to the transformation of S applied to X. I know that terminology is getting confused. Plus T of S of Y. We can do this because we know that T is a linear transformation."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we also know that T is a linear transformation, which means that the transformation applied to the sum of two vectors is equal to the transformation of each of the vectors summed up. So the transformation of S of X, or the transformation applied to the transformation of S applied to X. I know that terminology is getting confused. Plus T of S of Y. We can do this because we know that T is a linear transformation. But what is this right here? This is equal to, this statement right here is equal to the composition of T with S applied to X plus the composition of T with S applied to Y. So given that both T and S are linear transformations, we got our first requirement."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can do this because we know that T is a linear transformation. But what is this right here? This is equal to, this statement right here is equal to the composition of T with S applied to X plus the composition of T with S applied to Y. So given that both T and S are linear transformations, we got our first requirement. That the composition applied to the sum of two vectors is equal to the composition applied to each of the vectors summed up. So that was our first requirement for a linear transformation. And then our second one is, we need to apply this to a scalar multiple of a vector in X."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So given that both T and S are linear transformations, we got our first requirement. That the composition applied to the sum of two vectors is equal to the composition applied to each of the vectors summed up. So that was our first requirement for a linear transformation. And then our second one is, we need to apply this to a scalar multiple of a vector in X. So T of S, or let me say it this way, the composition of T with S applied to some scalar multiple of some vector X that's in our set X. This is a vector X that's our set X. This should be a capital X."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then our second one is, we need to apply this to a scalar multiple of a vector in X. So T of S, or let me say it this way, the composition of T with S applied to some scalar multiple of some vector X that's in our set X. This is a vector X that's our set X. This should be a capital X. This is equal to what? Well, by our definition of our linear, of our composition, this is equal to the transformation T applied to the transformation S applied to C times our vector X. And what is this equal to?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This should be a capital X. This is equal to what? Well, by our definition of our linear, of our composition, this is equal to the transformation T applied to the transformation S applied to C times our vector X. And what is this equal to? We know that this is a linear transformation. So given that this is a linear transformation, that S is a linear transformation, we know that this can be rewritten as T times C times S applied to X. This little replacing that I did with S applied to C times X is the same thing as C times the linear transformation applied to X."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? We know that this is a linear transformation. So given that this is a linear transformation, that S is a linear transformation, we know that this can be rewritten as T times C times S applied to X. This little replacing that I did with S applied to C times X is the same thing as C times the linear transformation applied to X. This just comes out of the fact that S is a linear transformation. We've done that multiple times. Well, now we have T applied to some scalar multiple of some vector."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This little replacing that I did with S applied to C times X is the same thing as C times the linear transformation applied to X. This just comes out of the fact that S is a linear transformation. We've done that multiple times. Well, now we have T applied to some scalar multiple of some vector. And so we can do the same thing. We know that T is a linear transformation. So we know that this is equal to, I'll do it down here, this is equal to C times T applied to S applied to some vector X that's in there."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, now we have T applied to some scalar multiple of some vector. And so we can do the same thing. We know that T is a linear transformation. So we know that this is equal to, I'll do it down here, this is equal to C times T applied to S applied to some vector X that's in there. And what is this equal to? This is equal to the constant C times the composition T with S of our vector X right there. So we've met our second requirement for a linear transformation."}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we know that this is equal to, I'll do it down here, this is equal to C times T applied to S applied to some vector X that's in there. And what is this equal to? This is equal to the constant C times the composition T with S of our vector X right there. So we've met our second requirement for a linear transformation. So the composition as we've defined it is definitely a linear transformation. Now, that means that this thing right here can be written, this means that the composition of T with S can be written as some matrix, let me write it this way, the composition of T with S applied to, or the transformation of the, which is the composition of T with S applied to some vector X, can be written as some matrix times our vector X. And what will be the dimensions of our matrix?"}, {"video_title": "Compositions of linear transformations 1 Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we've met our second requirement for a linear transformation. So the composition as we've defined it is definitely a linear transformation. Now, that means that this thing right here can be written, this means that the composition of T with S can be written as some matrix, let me write it this way, the composition of T with S applied to, or the transformation of the, which is the composition of T with S applied to some vector X, can be written as some matrix times our vector X. And what will be the dimensions of our matrix? We're going from a n dimension space, so this is going to have n columns, to a l dimension space, so this is going to have l rows. So this is going to be an l by n matrix. Now, I'll leave you there in this video, because I realize I've been making too many 20-minute plus videos."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if that is R2, and this is R2, T just maps from any member of R2 to another member of R2, just like that. And I'm going to define T. It's a linear transformation. I'm going to define it when I take the transformation of some member of R2. It's equivalent to multiplying it by this matrix, by the matrix 1 minus 3 minus 1 3. So just to kind of understand this transformation a little bit more, let's think about all of the values that it can take on in our codomain. So let's set it equal to, so if we say that 1 minus 3 minus 1 3 times any vector in our domain, so x1, x2, it's going to be equal to some other vector in our codomain. Let me call that vector b. b is a member of R2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equivalent to multiplying it by this matrix, by the matrix 1 minus 3 minus 1 3. So just to kind of understand this transformation a little bit more, let's think about all of the values that it can take on in our codomain. So let's set it equal to, so if we say that 1 minus 3 minus 1 3 times any vector in our domain, so x1, x2, it's going to be equal to some other vector in our codomain. Let me call that vector b. b is a member of R2. So it's going to be equal to b1, b2. All I'm doing here is I'm essentially, if I call this right here, I call that matrix A. I'm just trying to find out what are all of the possible values for Ax is equal to b. I'm trying to find out all of the possible b's in this case. So if I were to solve this equation for any particular b, what I would do is I would put this in reduced row echelon form, well first I would make an augmented matrix of it."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me call that vector b. b is a member of R2. So it's going to be equal to b1, b2. All I'm doing here is I'm essentially, if I call this right here, I call that matrix A. I'm just trying to find out what are all of the possible values for Ax is equal to b. I'm trying to find out all of the possible b's in this case. So if I were to solve this equation for any particular b, what I would do is I would put this in reduced row echelon form, well first I would make an augmented matrix of it. So I would put a 1, a minus 3, a minus 1, and a 3. And I would augment it with the member of our codomain we're trying to be equal to. So b1, b2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I were to solve this equation for any particular b, what I would do is I would put this in reduced row echelon form, well first I would make an augmented matrix of it. So I would put a 1, a minus 3, a minus 1, and a 3. And I would augment it with the member of our codomain we're trying to be equal to. So b1, b2. And then I would put it in reduced row echelon form. So how can we do that? How can we put it in reduced row echelon form?"}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So b1, b2. And then I would put it in reduced row echelon form. So how can we do that? How can we put it in reduced row echelon form? I'll keep my first row the same. 1 minus 3, and then b1. And then my second row, I'll replace it with my second row plus my first row."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How can we put it in reduced row echelon form? I'll keep my first row the same. 1 minus 3, and then b1. And then my second row, I'll replace it with my second row plus my first row. So minus 1 plus 1 is 0. 3 plus minus 3 is 0. And then b2 plus b1 is, well I could just write that as b1 plus b2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then my second row, I'll replace it with my second row plus my first row. So minus 1 plus 1 is 0. 3 plus minus 3 is 0. And then b2 plus b1 is, well I could just write that as b1 plus b2. So this is only going to have solutions. I have an interesting situation here. We've seen it before."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then b2 plus b1 is, well I could just write that as b1 plus b2. So this is only going to have solutions. I have an interesting situation here. We've seen it before. I have a row of 0's. I have a row of 0's. And the only way that this is going to actually have solutions is if this thing right here is also going to be equal to 0."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've seen it before. I have a row of 0's. I have a row of 0's. And the only way that this is going to actually have solutions is if this thing right here is also going to be equal to 0. So the only members b that are members of Rm that have solutions are the ones where if you add up their two terms, so let's say b is equal to b1 and b2, are the ones where the two terms added up have to be equal to 0. b1 plus b2 have to be equal to 0. Or another way you could write it is that b2 has to be equal to minus b1. So if we would actually draw our codomain, let's do it."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the only way that this is going to actually have solutions is if this thing right here is also going to be equal to 0. So the only members b that are members of Rm that have solutions are the ones where if you add up their two terms, so let's say b is equal to b1 and b2, are the ones where the two terms added up have to be equal to 0. b1 plus b2 have to be equal to 0. Or another way you could write it is that b2 has to be equal to minus b1. So if we would actually draw our codomain, let's do it. We always stay in the abstract, but sometimes it's useful to draw an actual example. So let's say that our codomain is R2. Let me draw some axes right here."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we would actually draw our codomain, let's do it. We always stay in the abstract, but sometimes it's useful to draw an actual example. So let's say that our codomain is R2. Let me draw some axes right here. Let's say that this is my b1 axis and this is my b2 axis. I could have called that x and y, but I call that b1 and b2. What are all of the members of my codomain that have a solution, that have a mapping?"}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw some axes right here. Let's say that this is my b1 axis and this is my b2 axis. I could have called that x and y, but I call that b1 and b2. What are all of the members of my codomain that have a solution, that have a mapping? Well, b2 has to be equal to minus b1. So it's going to look like this. It's going to look like this."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What are all of the members of my codomain that have a solution, that have a mapping? Well, b2 has to be equal to minus b1. So it's going to look like this. It's going to look like this. It's just going to be a line with a slope of negative 1. This is all the b's that have solution. Because if you're not on this line, if you're a member of your codomain, this is the codomain right here, R2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to look like this. It's just going to be a line with a slope of negative 1. This is all the b's that have solution. Because if you're not on this line, if you're a member of your codomain, this is the codomain right here, R2. R2 is also our domain, but let me make it very clear that this is the codomain that I've drawn. This is what we're mapping into. It's very clear that if we're not on this line, if you pick somebody whose two terms don't add up to equaling 0 or aren't the negative of each other, if you pick someone over here in our codomain and then you try to solve the equation over here, you're going to have a 0 is going to equal some non-zero number here and you're not going to have a solution."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because if you're not on this line, if you're a member of your codomain, this is the codomain right here, R2. R2 is also our domain, but let me make it very clear that this is the codomain that I've drawn. This is what we're mapping into. It's very clear that if we're not on this line, if you pick somebody whose two terms don't add up to equaling 0 or aren't the negative of each other, if you pick someone over here in our codomain and then you try to solve the equation over here, you're going to have a 0 is going to equal some non-zero number here and you're not going to have a solution. And we touched on this in the last video. And so in this case, the image, we could say that this right here is the image of our transformation. Or even another way of thinking about it, obviously, if all of R2 is, this is our codomain."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's very clear that if we're not on this line, if you pick somebody whose two terms don't add up to equaling 0 or aren't the negative of each other, if you pick someone over here in our codomain and then you try to solve the equation over here, you're going to have a 0 is going to equal some non-zero number here and you're not going to have a solution. And we touched on this in the last video. And so in this case, the image, we could say that this right here is the image of our transformation. Or even another way of thinking about it, obviously, if all of R2 is, this is our codomain. Let me draw a domain. If our domain looks like this, if all of R2, so if you take any member of R2 and it's always mapping onto something onto that line, clearly each point on that line is going to be mapped to by more than one vector. So we're not dealing with an onto transformation."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or even another way of thinking about it, obviously, if all of R2 is, this is our codomain. Let me draw a domain. If our domain looks like this, if all of R2, so if you take any member of R2 and it's always mapping onto something onto that line, clearly each point on that line is going to be mapped to by more than one vector. So we're not dealing with an onto transformation. And we saw that in the last video. In order for something to be onto, when you put it in reduced row echelon form, you cannot have all 0's in one of the rows, or another way to say it, in reduced row echelon form, every row has to have a pivot entry. But let's focus on the b's that actually do have a solution."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we're not dealing with an onto transformation. And we saw that in the last video. In order for something to be onto, when you put it in reduced row echelon form, you cannot have all 0's in one of the rows, or another way to say it, in reduced row echelon form, every row has to have a pivot entry. But let's focus on the b's that actually do have a solution. So let's focus on these b's so that when you take a b1 plus b2, it actually is going to be equal to 0. So we could have the b, I don't know, that could be the b5, 5, minus 5. Or you could have the b, well, obviously the b00 is going to work."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But let's focus on the b's that actually do have a solution. So let's focus on these b's so that when you take a b1 plus b2, it actually is going to be equal to 0. So we could have the b, I don't know, that could be the b5, 5, minus 5. Or you could have the b, well, obviously the b00 is going to work. You could have 1, negative 1, maybe that's right there. Let's focus on those for a second and solve and see how many guys, how many members of our domain map into them. So if we take this right here, and then we apply this equation up here, we only have one constraint."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or you could have the b, well, obviously the b00 is going to work. You could have 1, negative 1, maybe that's right there. Let's focus on those for a second and solve and see how many guys, how many members of our domain map into them. So if we take this right here, and then we apply this equation up here, we only have one constraint. We're assuming that this is going to be equal to 0. So let's assume that we're dealing with the b's in our image. So let's assume that we're dealing with something where we can get a solution, that b1 plus b2 is equal to 0."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we take this right here, and then we apply this equation up here, we only have one constraint. We're assuming that this is going to be equal to 0. So let's assume that we're dealing with the b's in our image. So let's assume that we're dealing with something where we can get a solution, that b1 plus b2 is equal to 0. Then what is our constraints? What will map to our vector b that we're dealing with? So if we just take this top equation right here, we have 1 times x1, 1 times x1, let me switch colors, we have 1 times x1, 1 times x1, minus 3 times x2, minus 3 times x2, is equal to b1."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's assume that we're dealing with something where we can get a solution, that b1 plus b2 is equal to 0. Then what is our constraints? What will map to our vector b that we're dealing with? So if we just take this top equation right here, we have 1 times x1, 1 times x1, let me switch colors, we have 1 times x1, 1 times x1, minus 3 times x2, minus 3 times x2, is equal to b1. And then this row will give us no constraints, because this is just going to be a bunch of 0's. So this is the only constraint for a member of our domain that will map to some particular b now that we are picking, some particular b that satisfies this constraint. So we could write this solution set, let me rewrite it as x1 is equal to b1 plus 3 x2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we just take this top equation right here, we have 1 times x1, 1 times x1, let me switch colors, we have 1 times x1, 1 times x1, minus 3 times x2, minus 3 times x2, is equal to b1. And then this row will give us no constraints, because this is just going to be a bunch of 0's. So this is the only constraint for a member of our domain that will map to some particular b now that we are picking, some particular b that satisfies this constraint. So we could write this solution set, let me rewrite it as x1 is equal to b1 plus 3 x2. Or if we wanted to write the entire solution set, it would look like this. x1, x2 is equal to the vector b1, 0, plus, let's see, x2, x1 is b1 plus 3 times x2. So plus x2 times 3."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could write this solution set, let me rewrite it as x1 is equal to b1 plus 3 x2. Or if we wanted to write the entire solution set, it would look like this. x1, x2 is equal to the vector b1, 0, plus, let's see, x2, x1 is b1 plus 3 times x2. So plus x2 times 3. And x2 is just going to be equal to x2. That's a free variable. So x2 is equal to 0 plus x2 times 1."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So plus x2 times 3. And x2 is just going to be equal to x2. That's a free variable. So x2 is equal to 0 plus x2 times 1. So what we're saying here is, look, this transformation just maps to this line here for all of the vectors in our codomain where their two entries add up to each other. Now let's assume that we actually have one of those vectors. And so first of all, this is definitely not an onto transformation."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So x2 is equal to 0 plus x2 times 1. So what we're saying here is, look, this transformation just maps to this line here for all of the vectors in our codomain where their two entries add up to each other. Now let's assume that we actually have one of those vectors. And so first of all, this is definitely not an onto transformation. But let's assume that we are dealing with one of those guys. So if we are picking a particular one of those guys, for a particular b, let me write this down, for a particular b that has a solution to ax is equal to b, the solution set will be equal to this thing right here. It'll be equal to x1, x2 is equal to the first entry of your b, b1, 0 plus x2 times the vector 3, 1."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so first of all, this is definitely not an onto transformation. But let's assume that we are dealing with one of those guys. So if we are picking a particular one of those guys, for a particular b, let me write this down, for a particular b that has a solution to ax is equal to b, the solution set will be equal to this thing right here. It'll be equal to x1, x2 is equal to the first entry of your b, b1, 0 plus x2 times the vector 3, 1. And if you think about this, if you pick a particular b, so let's say we pick our, let me draw this out because I think it's nice to visualize it all. Actually, maybe I'll draw it like this. I don't want to draw our blobs anymore."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It'll be equal to x1, x2 is equal to the first entry of your b, b1, 0 plus x2 times the vector 3, 1. And if you think about this, if you pick a particular b, so let's say we pick our, let me draw this out because I think it's nice to visualize it all. Actually, maybe I'll draw it like this. I don't want to draw our blobs anymore. So let me draw my axes. So my axes look like this. We know that the image of our transformation is aligned with a negative 1 slope, because the two entries have to be equal to the negative of each other."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to draw our blobs anymore. So let me draw my axes. So my axes look like this. We know that the image of our transformation is aligned with a negative 1 slope, because the two entries have to be equal to the negative of each other. Let's pick a particular b. Let's pick a particular b that has a solution. So let's say we pick that b right there."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know that the image of our transformation is aligned with a negative 1 slope, because the two entries have to be equal to the negative of each other. Let's pick a particular b. Let's pick a particular b that has a solution. So let's say we pick that b right there. In order for it to have a solution, its entries have to be the negative of each other. So let's say that's the entry 5 minus 5. That is our b."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we pick that b right there. In order for it to have a solution, its entries have to be the negative of each other. So let's say that's the entry 5 minus 5. That is our b. So what we just showed is that the solution set, if we want to say, hey, what in our domain maps to this guy? So let's think about what in our domain maps to this right here. Let's think about what in our domain maps to this point right here, to this particular b."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is our b. So what we just showed is that the solution set, if we want to say, hey, what in our domain maps to this guy? So let's think about what in our domain maps to this right here. Let's think about what in our domain maps to this point right here, to this particular b. So that's going to be all of the x's that satisfy ax is equal to 5 minus 5. And what this is telling us is this is going to be equal to. So the solution set is going to be equal to, so x1, x2 is going to be equal to b1 is going to be equal to 5, 0, plus x2, plus any scalar multiple of the vector 3, 1."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's think about what in our domain maps to this point right here, to this particular b. So that's going to be all of the x's that satisfy ax is equal to 5 minus 5. And what this is telling us is this is going to be equal to. So the solution set is going to be equal to, so x1, x2 is going to be equal to b1 is going to be equal to 5, 0, plus x2, plus any scalar multiple of the vector 3, 1. So our solution set is going to be, you take the vector 5, 0. So maybe the vector 5, 0 specifies this position right here. And then you're going to add to it multiples of the vector 3, 1."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the solution set is going to be equal to, so x1, x2 is going to be equal to b1 is going to be equal to 5, 0, plus x2, plus any scalar multiple of the vector 3, 1. So our solution set is going to be, you take the vector 5, 0. So maybe the vector 5, 0 specifies this position right here. And then you're going to add to it multiples of the vector 3, 1. The vector 3, 1 looks like this. 1, 2, 3, and you go up 1. The vector 3, 1 is going to look like this."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you're going to add to it multiples of the vector 3, 1. The vector 3, 1 looks like this. 1, 2, 3, and you go up 1. The vector 3, 1 is going to look like this. So if you add multiples of this, multiples of that could stretch out like that or it could go negative like that to this vector 5, 0. You're essentially going to, let me see if I can draw this neatly. You're going to end up with a solution set that looks like that right there."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The vector 3, 1 is going to look like this. So if you add multiples of this, multiples of that could stretch out like that or it could go negative like that to this vector 5, 0. You're essentially going to, let me see if I can draw this neatly. You're going to end up with a solution set that looks like that right there. So if you pick a particular b right there that has a solution, we just said that everything on this line will map to that point in our solution set. Everything on that line would map to our solution set. And in fact, if you picked another point, let's say you picked the point minus 5, 5."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're going to end up with a solution set that looks like that right there. So if you pick a particular b right there that has a solution, we just said that everything on this line will map to that point in our solution set. Everything on that line would map to our solution set. And in fact, if you picked another point, let's say you picked the point minus 5, 5. Let's say you picked, let me write it this way, minus 5, 5. Then the solution set that maps to that would actually be, you would take minus, this first term would be the minus 5 would be here, and all of these guys would map to that. Well, this is all interesting."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And in fact, if you picked another point, let's say you picked the point minus 5, 5. Let's say you picked, let me write it this way, minus 5, 5. Then the solution set that maps to that would actually be, you would take minus, this first term would be the minus 5 would be here, and all of these guys would map to that. Well, this is all interesting. I mean, we've been doing a lot of abstract things and I think it might be satisfying that you're actually seeing something kind of more concrete in this example. But I'm doing all of this for a reason. Because I want to understand what the solution set is to a general non-homogeneous equation like this."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is all interesting. I mean, we've been doing a lot of abstract things and I think it might be satisfying that you're actually seeing something kind of more concrete in this example. But I'm doing all of this for a reason. Because I want to understand what the solution set is to a general non-homogeneous equation like this. And to understand a little bit better, let's imagine what is the solution set if I were to pick this guy, if I were to pick the 0 vector. If I pick the vector, this is the 0 vector right there. Then what's going to be the solution set?"}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because I want to understand what the solution set is to a general non-homogeneous equation like this. And to understand a little bit better, let's imagine what is the solution set if I were to pick this guy, if I were to pick the 0 vector. If I pick the vector, this is the 0 vector right there. Then what's going to be the solution set? It's going to be the vector. So if we say that Ax is equal to 0, then our solution set is going to be the vector 0, 0 plus, right, 0, 0 plus x2 times 3, 1. So it's going to be what?"}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then what's going to be the solution set? It's going to be the vector. So if we say that Ax is equal to 0, then our solution set is going to be the vector 0, 0 plus, right, 0, 0 plus x2 times 3, 1. So it's going to be what? This is just the 0 vector, so it's going to be here. And it's going to just be multiples of 3, 1. So it's going to look like something like that."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be what? This is just the 0 vector, so it's going to be here. And it's going to just be multiples of 3, 1. So it's going to look like something like that. But what's this? What is the solution set to the equation of Ax equal to 0? This is the null space."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to look like something like that. But what's this? What is the solution set to the equation of Ax equal to 0? This is the null space. This, by definition, is the null space of A. So notice, and this is the big kind of takeaway from this video, is that for any solution, we're picking b's that actually do have solutions, because we're picking them on this line. We're picking them in the image of our codomain."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the null space. This, by definition, is the null space of A. So notice, and this is the big kind of takeaway from this video, is that for any solution, we're picking b's that actually do have solutions, because we're picking them on this line. We're picking them in the image of our codomain. But the solution set to any Ax is equal to some b, where b does have a solution, it's essentially equal to a shifted version of the null set. This right here, or the null space, this right here is the null space. That right there is a null space for any real number x2."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're picking them in the image of our codomain. But the solution set to any Ax is equal to some b, where b does have a solution, it's essentially equal to a shifted version of the null set. This right here, or the null space, this right here is the null space. That right there is a null space for any real number x2. Any scalar multiple of 3, 1 is the null space. I just showed it right there. It's going to be that."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That right there is a null space for any real number x2. Any scalar multiple of 3, 1 is the null space. I just showed it right there. It's going to be that. And so all of these other solution sets are just some particular vector, some x particular, plus the null space, plus the null space. Obviously, this vector by itself would also be a solution to Ax is equal to b, because you could just set x2 to be equal to 0. So in general, and I haven't proven this to you rigorously, but hopefully you kind of get the intuition behind it, the solution, and I'll do this in the next video, just because I realize I'm running long on time."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be that. And so all of these other solution sets are just some particular vector, some x particular, plus the null space, plus the null space. Obviously, this vector by itself would also be a solution to Ax is equal to b, because you could just set x2 to be equal to 0. So in general, and I haven't proven this to you rigorously, but hopefully you kind of get the intuition behind it, the solution, and I'll do this in the next video, just because I realize I'm running long on time. Assuming Axb has a solution, in the example we just did, we can assume it has if we pick one of these points here. So assuming it has a solution, if we pick a point off of our image, we're not going to have a solution. But assuming Axb has a solution, the solution set is going to be equal to some particular vector."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So in general, and I haven't proven this to you rigorously, but hopefully you kind of get the intuition behind it, the solution, and I'll do this in the next video, just because I realize I'm running long on time. Assuming Axb has a solution, in the example we just did, we can assume it has if we pick one of these points here. So assuming it has a solution, if we pick a point off of our image, we're not going to have a solution. But assuming Axb has a solution, the solution set is going to be equal to some particular vector. So you could just think of it as a set with just one vector right there, with, or combined with, or the union of that set with your null space of this matrix right here. I haven't proven this to you yet, but hopefully you get the intuition why this is true. We just solved it for particular cases that do have solutions, we say, hey, it's going to take this form, and I just showed you that this is the form of the null space."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But assuming Axb has a solution, the solution set is going to be equal to some particular vector. So you could just think of it as a set with just one vector right there, with, or combined with, or the union of that set with your null space of this matrix right here. I haven't proven this to you yet, but hopefully you get the intuition why this is true. We just solved it for particular cases that do have solutions, we say, hey, it's going to take this form, and I just showed you that this is the form of the null space. This thing right here is the null space. And the reason why we're doing that is because we've been talking about invertibility, and in order to be invertible you have to be onto and one-to-one. And for something to be one-to-one, you have to have at most one solution that maps to a particular vector."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just solved it for particular cases that do have solutions, we say, hey, it's going to take this form, and I just showed you that this is the form of the null space. This thing right here is the null space. And the reason why we're doing that is because we've been talking about invertibility, and in order to be invertible you have to be onto and one-to-one. And for something to be one-to-one, you have to have at most one solution that maps to a particular vector. You might have none, but you might have at most one. So in order to have at most one solution, and the solution set is always going to be equal to this, so you're always going to have this solution. So in order to have at most one solution, your null space can't have anything in it, or it can just have the zero vector."}, {"video_title": "Exploring the solution set of Ax = b Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And for something to be one-to-one, you have to have at most one solution that maps to a particular vector. You might have none, but you might have at most one. So in order to have at most one solution, and the solution set is always going to be equal to this, so you're always going to have this solution. So in order to have at most one solution, your null space can't have anything in it, or it can just have the zero vector. So it's just going to have to be. So that means that your null space, the null space of A, has to be trivial, or just to have to be empty, or just to just have the zero vector. I'll do this a little bit more rigorously in the next video, but I think when you do it sometimes with the rigor, you don't necessarily get the intuition."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "If I have some linear transformation that's a mapping from Rn to Rn, and if we're dealing with standard coordinates, that transformation applied to some vector x in standard coordinates will be equal to the matrix A times x. So let me write this down. If we are dealing with the standard coordinates, so I have x in standard coordinates. If I apply the transformation that is equivalent to multiplying x by A, then if I multiply x by A, then I'm going to get the transformation of x in standard coordinates. This is a world that we're very, very familiar with. Now let's say that we have an alternate basis to Rn. So let's say that B is equal to v1, v2, all the way to vn."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "If I apply the transformation that is equivalent to multiplying x by A, then if I multiply x by A, then I'm going to get the transformation of x in standard coordinates. This is a world that we're very, very familiar with. Now let's say that we have an alternate basis to Rn. So let's say that B is equal to v1, v2, all the way to vn. So it has n linearly independent vectors. Let's say B is a basis for Rn. So it's a basis for Rn, but it's the non-standard basis."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that B is equal to v1, v2, all the way to vn. So it has n linearly independent vectors. Let's say B is a basis for Rn. So it's a basis for Rn, but it's the non-standard basis. These aren't just our standard basis vectors. So B is a basis for Rn, and let's say that C, which just has these guys as its column vectors, C is v1, v2, all the way to vn, is the change of basis matrix for this basis, for the basis B. Now we've learned, we've seen this several times already, that if I have some vector x in Rn represented in B coordinates, or in coordinates with respect to B, I can multiply it by the change of basis matrix, and then I'll get just the standard coordinates for x."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So it's a basis for Rn, but it's the non-standard basis. These aren't just our standard basis vectors. So B is a basis for Rn, and let's say that C, which just has these guys as its column vectors, C is v1, v2, all the way to vn, is the change of basis matrix for this basis, for the basis B. Now we've learned, we've seen this several times already, that if I have some vector x in Rn represented in B coordinates, or in coordinates with respect to B, I can multiply it by the change of basis matrix, and then I'll get just the standard coordinates for x. Or if you multiply both sides of this equation by C inverse, you can get that if I start with the standard coordinates for x, I can multiply it by C inverse, and then I could get the B coordinates for x, or the alternate non-standard coordinates for x. So we've seen both of these before. So let's apply that to this little diagram here."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Now we've learned, we've seen this several times already, that if I have some vector x in Rn represented in B coordinates, or in coordinates with respect to B, I can multiply it by the change of basis matrix, and then I'll get just the standard coordinates for x. Or if you multiply both sides of this equation by C inverse, you can get that if I start with the standard coordinates for x, I can multiply it by C inverse, and then I could get the B coordinates for x, or the alternate non-standard coordinates for x. So we've seen both of these before. So let's apply that to this little diagram here. So if I want to get x, and if I wanted to write it in non-standard coordinates, what do I do? Well, if I have x, and if I want to write it, so if I have x, let me go right here, so if I have x, what do I multiply it by if I want to go to non-standard coordinates? Well, I multiply it by C inverse."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let's apply that to this little diagram here. So if I want to get x, and if I wanted to write it in non-standard coordinates, what do I do? Well, if I have x, and if I want to write it, so if I have x, let me go right here, so if I have x, what do I multiply it by if I want to go to non-standard coordinates? Well, I multiply it by C inverse. If I multiply it by C inverse, whatever I write next to this line, you could say, what matrix do you have to multiply by to get to the end point on your line? So I multiply x by C inverse, then I get the B coordinates for x. So these are, let's see, coordinates with respect to B. I could do the same thing here with the transformation of x."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Well, I multiply it by C inverse. If I multiply it by C inverse, whatever I write next to this line, you could say, what matrix do you have to multiply by to get to the end point on your line? So I multiply x by C inverse, then I get the B coordinates for x. So these are, let's see, coordinates with respect to B. I could do the same thing here with the transformation of x. This is just the standard representation of the transformation of x. So I could multiply it by C inverse if we want to go in that direction, and then we're going to get the transformation of x represented in B coordinates. Now, in the last video, what we saw is, hey, why do these separately?"}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So these are, let's see, coordinates with respect to B. I could do the same thing here with the transformation of x. This is just the standard representation of the transformation of x. So I could multiply it by C inverse if we want to go in that direction, and then we're going to get the transformation of x represented in B coordinates. Now, in the last video, what we saw is, hey, why do these separately? Maybe there is some matrix, and we found out what it is. Maybe there's some matrix D that if we multiply this guy times it, I can go straight from the B coordinates of x to the B coordinates of the transformation of x. And we said that is matrix D. And in that last video, we showed that D can be represented by A."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Now, in the last video, what we saw is, hey, why do these separately? Maybe there is some matrix, and we found out what it is. Maybe there's some matrix D that if we multiply this guy times it, I can go straight from the B coordinates of x to the B coordinates of the transformation of x. And we said that is matrix D. And in that last video, we showed that D can be represented by A. Actually, you could go around the circle and re-derive it if you like. But we found out that, let me write it in another color, that D is equal to C inverse times A times C. Now, this is all a review of everything that we learned in the last video. Hopefully it clarified things up a little bit."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And we said that is matrix D. And in that last video, we showed that D can be represented by A. Actually, you could go around the circle and re-derive it if you like. But we found out that, let me write it in another color, that D is equal to C inverse times A times C. Now, this is all a review of everything that we learned in the last video. Hopefully it clarified things up a little bit. It's nice to just realize that these are just alternate ways of doing the same thing. Both of these are the transformation. When you multiply by A, you're applying the same transformation when you multiply by D. You're just doing it in a different coordinate system."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Hopefully it clarified things up a little bit. It's nice to just realize that these are just alternate ways of doing the same thing. Both of these are the transformation. When you multiply by A, you're applying the same transformation when you multiply by D. You're just doing it in a different coordinate system. Different coordinate systems are just different ways of representing the same vector. This and this are different labels for the same vector. This and this are different labels for the same vector."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "When you multiply by A, you're applying the same transformation when you multiply by D. You're just doing it in a different coordinate system. Different coordinate systems are just different ways of representing the same vector. This and this are different labels for the same vector. This and this are different labels for the same vector. So these are both performing the transformation T. Now, this was a relation we got in the last video, that if we have our change of basis matrix, we have its inverse, and we have just our standard basis linear transformation matrix, we're able to get this. Let's see if we can go the other way. If we have D, can we solve for A?"}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "This and this are different labels for the same vector. So these are both performing the transformation T. Now, this was a relation we got in the last video, that if we have our change of basis matrix, we have its inverse, and we have just our standard basis linear transformation matrix, we're able to get this. Let's see if we can go the other way. If we have D, can we solve for A? Well, if you multiply both sides of this equation on the right by C inverse, you get D C inverse is equal to C inverse A C C inverse. I just put a C inverse on the right-hand side of both sides of this equation. This is going to be the identity matrix, so we can ignore it."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "If we have D, can we solve for A? Well, if you multiply both sides of this equation on the right by C inverse, you get D C inverse is equal to C inverse A C C inverse. I just put a C inverse on the right-hand side of both sides of this equation. This is going to be the identity matrix, so we can ignore it. And then let's multiply both sides on the left by C. So then you get C D C inverse is equal to C C inverse A. And this is going to be the identity matrix. And then you're left with A is equal to C times D C inverse, which is another interesting result."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be the identity matrix, so we can ignore it. And then let's multiply both sides on the left by C. So then you get C D C inverse is equal to C C inverse A. And this is going to be the identity matrix. And then you're left with A is equal to C times D C inverse, which is another interesting result. It's another thing to put in our toolkit. Now, everything I've been doing has been fairly abstract. Let's actually apply some of these principles with a real concrete example."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then you're left with A is equal to C times D C inverse, which is another interesting result. It's another thing to put in our toolkit. Now, everything I've been doing has been fairly abstract. Let's actually apply some of these principles with a real concrete example. So let's say that I have a transformation T. I'll keep these guys around just because they might be useful. That is a mapping from R2 to R2. And let's say that the transformation matrix for T, so let's say that T of x in standard coordinates is equal to the matrix 3, 2, minus 2, minus 2, minus 2 times x."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's actually apply some of these principles with a real concrete example. So let's say that I have a transformation T. I'll keep these guys around just because they might be useful. That is a mapping from R2 to R2. And let's say that the transformation matrix for T, so let's say that T of x in standard coordinates is equal to the matrix 3, 2, minus 2, minus 2, minus 2 times x. So this is, in the example we just said, this would be our transformation matrix with respect to the standard basis. And we could call that A right there. Now let's say we have some alternate basis."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that the transformation matrix for T, so let's say that T of x in standard coordinates is equal to the matrix 3, 2, minus 2, minus 2, minus 2 times x. So this is, in the example we just said, this would be our transformation matrix with respect to the standard basis. And we could call that A right there. Now let's say we have some alternate basis. Let's say we have some alternate basis. So alternate R2 basis. Let's call that B because we've been calling it B so far."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Now let's say we have some alternate basis. Let's say we have some alternate basis. So alternate R2 basis. Let's call that B because we've been calling it B so far. And let's say this alternate R2 basis, the vectors 1, 2 and 2, 1. So let's see, given this alternate basis, whether we can come up for a transformation matrix in that coordinate world. So what we're looking for, we're looking for some matrix D, we're looking for a matrix D such that if I apply my transformation to x in B coordinates, so if I apply it to x in B coordinates, or in coordinates with respect to this alternate basis, it should be equal to this matrix, this matrix, it should be equal to D times x. x in the B coordinates."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that B because we've been calling it B so far. And let's say this alternate R2 basis, the vectors 1, 2 and 2, 1. So let's see, given this alternate basis, whether we can come up for a transformation matrix in that coordinate world. So what we're looking for, we're looking for some matrix D, we're looking for a matrix D such that if I apply my transformation to x in B coordinates, so if I apply it to x in B coordinates, or in coordinates with respect to this alternate basis, it should be equal to this matrix, this matrix, it should be equal to D times x. x in the B coordinates. So this is what I'm looking for. I'm looking for that. Or if we go back to our diagram, I'm looking for that."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So what we're looking for, we're looking for some matrix D, we're looking for a matrix D such that if I apply my transformation to x in B coordinates, so if I apply it to x in B coordinates, or in coordinates with respect to this alternate basis, it should be equal to this matrix, this matrix, it should be equal to D times x. x in the B coordinates. So this is what I'm looking for. I'm looking for that. Or if we go back to our diagram, I'm looking for that. You give me x in B coordinates and you multiply it by D and I'm going to give you the transformation of x in B coordinates. Now just applying it to this concrete example here, we have this formula right here. This is a formula for D, which we proved in the last video."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Or if we go back to our diagram, I'm looking for that. You give me x in B coordinates and you multiply it by D and I'm going to give you the transformation of x in B coordinates. Now just applying it to this concrete example here, we have this formula right here. This is a formula for D, which we proved in the last video. So we have to figure out C inverse. So what is the change of basis matrix for B? I want to leave this up here."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "This is a formula for D, which we proved in the last video. So we have to figure out C inverse. So what is the change of basis matrix for B? I want to leave this up here. So let me, so change of basis matrix for B is just going to be, let's just call it C, and it's going to be the basis vectors for B within the column. So 1, 2, and 2, 1. And then we're going to want to figure out its inverse."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I want to leave this up here. So let me, so change of basis matrix for B is just going to be, let's just call it C, and it's going to be the basis vectors for B within the column. So 1, 2, and 2, 1. And then we're going to want to figure out its inverse. So let's figure out its determinant first. So the determinant of C is equal to 1 times 1 minus 2 times 2. So 1 minus 4 is minus 3."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to want to figure out its inverse. So let's figure out its determinant first. So the determinant of C is equal to 1 times 1 minus 2 times 2. So 1 minus 4 is minus 3. And so C inverse is going to be equal to 1 over the determinant, 1 over minus 3 or minus 1 third, times, we switch these two guys, so we switch the 1 and the 1, and then we make these two guys negative. Minus 2, minus 2. That is C inverse."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So 1 minus 4 is minus 3. And so C inverse is going to be equal to 1 over the determinant, 1 over minus 3 or minus 1 third, times, we switch these two guys, so we switch the 1 and the 1, and then we make these two guys negative. Minus 2, minus 2. That is C inverse. So D, this D vector right here, is going to be equal to C inverse times A, times the transformation matrix with respect to the standard basis, times C. So let me write it down here, just because, let me write it. So D, the D that we're looking for, D is going to be equal to C inverse times A times C. Which is equal to, C inverse is minus 1 third times 1 minus 2 minus 2, 1, times A. Let me do this in a different color."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That is C inverse. So D, this D vector right here, is going to be equal to C inverse times A, times the transformation matrix with respect to the standard basis, times C. So let me write it down here, just because, let me write it. So D, the D that we're looking for, D is going to be equal to C inverse times A times C. Which is equal to, C inverse is minus 1 third times 1 minus 2 minus 2, 1, times A. Let me do this in a different color. I like to switch colors. So C inverse times A, A is right there, so times 3 minus 2, 2 minus 2, times C. C is right there. I'll do it in yellow."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let me do this in a different color. I like to switch colors. So C inverse times A, A is right there, so times 3 minus 2, 2 minus 2, times C. C is right there. I'll do it in yellow. Times C, which is 1, 2, and then 2, 1. So let's do this piece by piece. Let's work through this."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it in yellow. Times C, which is 1, 2, and then 2, 1. So let's do this piece by piece. Let's work through this. So what is this piece going to be equal to? We have a 2 by 2 times a 2 by 2. That's going to give us another 2 by 2 matrix."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's work through this. So what is this piece going to be equal to? We have a 2 by 2 times a 2 by 2. That's going to give us another 2 by 2 matrix. So this first term right here is going to be 3 times 1 plus minus 2 times 2. So 3, 3 minus 4. So it's going to be minus 1."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That's going to give us another 2 by 2 matrix. So this first term right here is going to be 3 times 1 plus minus 2 times 2. So 3, 3 minus 4. So it's going to be minus 1. 3 times 1 plus minus 2 times 2, right, is minus 1. Then you have 3 times 2, which is 6, minus 2. Minus 2 times 1."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be minus 1. 3 times 1 plus minus 2 times 2, right, is minus 1. Then you have 3 times 2, which is 6, minus 2. Minus 2 times 1. So that is 4. 3 times 2 minus 2 is 4. And then when you go down here, 2 times 1 minus 2 times 2, that's 2 minus 4, that's minus 2."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 times 1. So that is 4. 3 times 2 minus 2 is 4. And then when you go down here, 2 times 1 minus 2 times 2, that's 2 minus 4, that's minus 2. And then 2 times 2 is 4, minus 2 times 1. So 4 minus 2 is just 2. So our matrix D is going to be equal to minus 1 third times this guy, 1 minus 2 minus 2, 1, times this guy, which was just the product of those two matrices."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then when you go down here, 2 times 1 minus 2 times 2, that's 2 minus 4, that's minus 2. And then 2 times 2 is 4, minus 2 times 1. So 4 minus 2 is just 2. So our matrix D is going to be equal to minus 1 third times this guy, 1 minus 2 minus 2, 1, times this guy, which was just the product of those two matrices. Now let's figure out what this is. If I take the product of these two guys, it's going to be another 2 by 2 matrix. So I have 1 times minus 1, which is minus 1, plus minus 2 times minus 2."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So our matrix D is going to be equal to minus 1 third times this guy, 1 minus 2 minus 2, 1, times this guy, which was just the product of those two matrices. Now let's figure out what this is. If I take the product of these two guys, it's going to be another 2 by 2 matrix. So I have 1 times minus 1, which is minus 1, plus minus 2 times minus 2. So let me make sure. So 1 times minus 1 minus 2 times minus 2 is 4. And then 1 times minus 1 is minus 1."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So I have 1 times minus 1, which is minus 1, plus minus 2 times minus 2. So let me make sure. So 1 times minus 1 minus 2 times minus 2 is 4. And then 1 times minus 1 is minus 1. So it's going to be 3. And then we go to the next term. We have 1 times 4 plus minus 2 times 2."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then 1 times minus 1 is minus 1. So it's going to be 3. And then we go to the next term. We have 1 times 4 plus minus 2 times 2. So that's 4 minus 4, which is 0. And then we have minus 2 times minus 1, which is 2, plus 1 times minus 2. So that is 0."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "We have 1 times 4 plus minus 2 times 2. So that's 4 minus 4, which is 0. And then we have minus 2 times minus 1, which is 2, plus 1 times minus 2. So that is 0. And then finally, we have minus 2 times 4, which is minus 8. Minus 2 times 4 is minus 8. Plus 1 times 2."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So that is 0. And then finally, we have minus 2 times 4, which is minus 8. Minus 2 times 4 is minus 8. Plus 1 times 2. So minus 8, minus 2 times 4 is minus 8, plus 2 is minus 6. And all of that times minus 1 third. So this is going to be equal to 3 times minus 1 third is minus 1, 0, and then 0 minus 6 times minus 1 third is 2."}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Plus 1 times 2. So minus 8, minus 2 times 4 is minus 8, plus 2 is minus 6. And all of that times minus 1 third. So this is going to be equal to 3 times minus 1 third is minus 1, 0, and then 0 minus 6 times minus 1 third is 2. So D is now our transformation matrix with respect to the basis B. So we were able to figure it out just applying this formula here. Now, what happens?"}, {"video_title": "Alternate basis transformation matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to 3 times minus 1 third is minus 1, 0, and then 0 minus 6 times minus 1 third is 2. So D is now our transformation matrix with respect to the basis B. So we were able to figure it out just applying this formula here. Now, what happens? Let's actually do it with some, actually I'll save that for the next video, where we actually show that it works. That we can actually take some vectors x, apply the transformation, or apply the change of coordinates to get to this, and then apply D. And then maybe we could go up that way, multiply by C to get the transformation. It's going to be equivalent to A. I'll do that in the next video."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You just perform a bunch of row operations. But what I want to show you in this video is that those row operations are equivalent to linear transformations on the column vectors of A. So let me show you by example. So if we just want to put A into reduced row echelon form, the first step that we might want to do if we want to zero out these entries right here is we'll keep our first entry the same. So for each of these column vectors, we're going to keep the first entry the same. So there's going to be 1, minus 1, minus 1. And actually, let me simultaneously construct my transformation."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we just want to put A into reduced row echelon form, the first step that we might want to do if we want to zero out these entries right here is we'll keep our first entry the same. So for each of these column vectors, we're going to keep the first entry the same. So there's going to be 1, minus 1, minus 1. And actually, let me simultaneously construct my transformation. So I'm saying that my row operation I'm going to perform is equivalent to a linear transformation on the column vector. So it's going to be a transformation that's going to take some column vector, A1, A2, and A3. It's going to take each of these and then do something to them in a linear way."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And actually, let me simultaneously construct my transformation. So I'm saying that my row operation I'm going to perform is equivalent to a linear transformation on the column vector. So it's going to be a transformation that's going to take some column vector, A1, A2, and A3. It's going to take each of these and then do something to them in a linear way. There'll be linear transformations. So we're keeping the first entry of our column vector the same. So this is just going to be A1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to take each of these and then do something to them in a linear way. There'll be linear transformations. So we're keeping the first entry of our column vector the same. So this is just going to be A1. This is a line right here. That's going to be A1. Now what could we do if we want to get to reduced row echelon form?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is just going to be A1. This is a line right here. That's going to be A1. Now what could we do if we want to get to reduced row echelon form? We'd want to make this equal to 0. So we want to replace our second row with the second row plus the first row. Because then these guys would turn out to be 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what could we do if we want to get to reduced row echelon form? We'd want to make this equal to 0. So we want to replace our second row with the second row plus the first row. Because then these guys would turn out to be 0. So let me write that in my transformation. I'm going to replace the second row with the second row plus the first row. And let me write it out here."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because then these guys would turn out to be 0. So let me write that in my transformation. I'm going to replace the second row with the second row plus the first row. And let me write it out here. Minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let me write it out here. Minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2. Now, we also want to get a 0 here. So let me replace my third row with my third row minus my first row. So I'm going to replace my third row with my third row minus my first row."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "3 plus minus 1 is 2. Now, we also want to get a 0 here. So let me replace my third row with my third row minus my first row. So I'm going to replace my third row with my third row minus my first row. So 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to replace my third row with my third row minus my first row. So 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5. Just like that. So you see, this was just a linear transformation. And any linear transformation you could actually represent as a matrix vector product."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "4 minus minus 1 is 5. Just like that. So you see, this was just a linear transformation. And any linear transformation you could actually represent as a matrix vector product. So for example, this transformation, I could represent it to figure out its transformation matrix. If we say that T of x is equal to, I don't know, let's call it some matrix S times x. We already used the matrix A, so I have to pick another letter."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And any linear transformation you could actually represent as a matrix vector product. So for example, this transformation, I could represent it to figure out its transformation matrix. If we say that T of x is equal to, I don't know, let's call it some matrix S times x. We already used the matrix A, so I have to pick another letter. So how do we find S? Well, we just apply the transformation to all of the column vectors, or the standard basis vectors, of the identity matrix. So let's do that."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We already used the matrix A, so I have to pick another letter. So how do we find S? Well, we just apply the transformation to all of the column vectors, or the standard basis vectors, of the identity matrix. So let's do that. So the identity matrix, I'll draw it really small like this. The identity matrix looks like this. 1, 0, 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So the identity matrix, I'll draw it really small like this. The identity matrix looks like this. 1, 0, 0. 0, 1, 0. 0, 0, 1. That's what the identity matrix looks like."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1, 0, 0. 0, 1, 0. 0, 0, 1. That's what the identity matrix looks like. To find the transformation matrix, we just apply this guy to each of the column vectors of this. So what do we get? Do it a little bit bigger."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what the identity matrix looks like. To find the transformation matrix, we just apply this guy to each of the column vectors of this. So what do we get? Do it a little bit bigger. So the first, we apply it to each of these column vectors, but we see the first row always stays the same. So the first row is always going to be the same thing. So 1, 0, 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Do it a little bit bigger. So the first, we apply it to each of these column vectors, but we see the first row always stays the same. So the first row is always going to be the same thing. So 1, 0, 0. I'm essentially applying it simultaneously to each of these column vectors, saying look, when you transform each of these column vectors, their first entry stays the same. The second entry becomes the second entry plus the first entry. So 0 plus 1 is 1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 1, 0, 0. I'm essentially applying it simultaneously to each of these column vectors, saying look, when you transform each of these column vectors, their first entry stays the same. The second entry becomes the second entry plus the first entry. So 0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. And then the third entry gets replaced with the third entry minus the first entry."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. And then the third entry gets replaced with the third entry minus the first entry. So 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the third entry gets replaced with the third entry minus the first entry. So 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1. Now notice, when I apply this transformation to the column vectors of our identity matrix, I essentially just perform those same row operations that I did up there. I perform those exact same row operations on this identity matrix. But we know that this is actually the transformation matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then 1 minus 0 is 1. Now notice, when I apply this transformation to the column vectors of our identity matrix, I essentially just perform those same row operations that I did up there. I perform those exact same row operations on this identity matrix. But we know that this is actually the transformation matrix. That if we multiply it by each of these column vectors, we will get these column vectors. So you could view it this way. Let me call this, this is right here."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But we know that this is actually the transformation matrix. That if we multiply it by each of these column vectors, we will get these column vectors. So you could view it this way. Let me call this, this is right here. This is equal to S. This is our transformation matrix. So we could say that S, if we create a new matrix whose columns are S times this column vector, S times 1 minus 1, 1, and then the next column is S times this guy, minus 1, 2, 1, and then the third column is going to be S times this third column vector, minus 1, 3, 4. This product, we now know we're applying this transformation, this is S, times each of these column vectors."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me call this, this is right here. This is equal to S. This is our transformation matrix. So we could say that S, if we create a new matrix whose columns are S times this column vector, S times 1 minus 1, 1, and then the next column is S times this guy, minus 1, 2, 1, and then the third column is going to be S times this third column vector, minus 1, 3, 4. This product, we now know we're applying this transformation, this is S, times each of these column vectors. That is the matrix representation of this transformation. This guy right here will be transformed to this right here, that'll become, let me do it down here. This guy, I wanted to show that stuff that I had above here as well."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This product, we now know we're applying this transformation, this is S, times each of these column vectors. That is the matrix representation of this transformation. This guy right here will be transformed to this right here, that'll become, let me do it down here. This guy, I wanted to show that stuff that I had above here as well. Well, I'll just draw an arrow. That's probably the simplest thing. This matrix right here will become that matrix right there."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This guy, I wanted to show that stuff that I had above here as well. Well, I'll just draw an arrow. That's probably the simplest thing. This matrix right here will become that matrix right there. So another way you could write it, this is equivalent to what? What is this equivalent to? When you take a matrix and you multiply it times each of the column vectors, when you transform each of the column vectors by this matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This matrix right here will become that matrix right there. So another way you could write it, this is equivalent to what? What is this equivalent to? When you take a matrix and you multiply it times each of the column vectors, when you transform each of the column vectors by this matrix. This is the definition of a matrix-matrix product. This is equal to our matrix S, I'll do it in pink. This is equal to our matrix S, which is 1, 0, 0, 1, 1, 0, minus 1, 0, 1, times our matrix A."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When you take a matrix and you multiply it times each of the column vectors, when you transform each of the column vectors by this matrix. This is the definition of a matrix-matrix product. This is equal to our matrix S, I'll do it in pink. This is equal to our matrix S, which is 1, 0, 0, 1, 1, 0, minus 1, 0, 1, times our matrix A. Times 1, minus 1, 1, minus 1, 2, 1, minus 1, 3, 4. So let me make this very clear. This is our matrix, this is our transformation matrix S. This is our matrix A."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to our matrix S, which is 1, 0, 0, 1, 1, 0, minus 1, 0, 1, times our matrix A. Times 1, minus 1, 1, minus 1, 2, 1, minus 1, 3, 4. So let me make this very clear. This is our matrix, this is our transformation matrix S. This is our matrix A. And when you perform this product, you're going to get this guy right over here. Let me just copy and paste it. Edit, let me copy, let me paste it."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is our matrix, this is our transformation matrix S. This is our matrix A. And when you perform this product, you're going to get this guy right over here. Let me just copy and paste it. Edit, let me copy, let me paste it. You're going to get that guy, just like that. Now the whole reason why I'm doing that is just to remind you that when we perform each of these row operations, we're just multiplying, we're performing a linear transformation on each of these columns. And it is completely equivalent to just multiplying this guy by some matrix S. In this case, we took the trouble of figuring out what that matrix S is."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Edit, let me copy, let me paste it. You're going to get that guy, just like that. Now the whole reason why I'm doing that is just to remind you that when we perform each of these row operations, we're just multiplying, we're performing a linear transformation on each of these columns. And it is completely equivalent to just multiplying this guy by some matrix S. In this case, we took the trouble of figuring out what that matrix S is. But any of these row operations that we've been doing, you can always represent them by a matrix multiplication. You can always represent them by a matrix multiplication. So this leads to a very interesting idea."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it is completely equivalent to just multiplying this guy by some matrix S. In this case, we took the trouble of figuring out what that matrix S is. But any of these row operations that we've been doing, you can always represent them by a matrix multiplication. You can always represent them by a matrix multiplication. So this leads to a very interesting idea. When you put something in reduced row echelon form, let me do it up here. So let me, actually let's just finish what we started with this guy. Let's put this guy in reduced row echelon form."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this leads to a very interesting idea. When you put something in reduced row echelon form, let me do it up here. So let me, actually let's just finish what we started with this guy. Let's put this guy in reduced row echelon form. So this, we already said, this is equal to, let me call this first S, let's call that S1. So this guy right here is equal to that first S1 times a. We already showed that that's true."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's put this guy in reduced row echelon form. So this, we already said, this is equal to, let me call this first S, let's call that S1. So this guy right here is equal to that first S1 times a. We already showed that that's true. Now let's perform another transformation. Or let's just do another set of row operations to get us to reduced row echelon form. So let's keep our middle row the same, 0, 1, 2."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We already showed that that's true. Now let's perform another transformation. Or let's just do another set of row operations to get us to reduced row echelon form. So let's keep our middle row the same, 0, 1, 2. And let's replace the first row with the first row plus the second row, because I want to make this a 0. So 1 plus 0 is 1. Let me do it in another color."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's keep our middle row the same, 0, 1, 2. And let's replace the first row with the first row plus the second row, because I want to make this a 0. So 1 plus 0 is 1. Let me do it in another color. 1 plus 0 is 1. Minus 1 plus 1 is 0. Minus 1 plus 2 is 1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in another color. 1 plus 0 is 1. Minus 1 plus 1 is 0. Minus 1 plus 2 is 1. Now I want to replace the third row with, let's say, the third row minus 2 times the first row. So that's 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 plus 2 is 1. Now I want to replace the third row with, let's say, the third row minus 2 times the first row. So that's 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 5 minus 2 times 2 is 1. 5 minus 4 is 1. And we're almost there."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 2 times 1 is 0. 5 minus 2 times 2 is 1. 5 minus 4 is 1. And we're almost there. We just have to zero out these guys right there. See if we can get this into reduced row echelon form. So what is this?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we're almost there. We just have to zero out these guys right there. See if we can get this into reduced row echelon form. So what is this? I just performed another linear transformation. Actually, let me write this. Let's say if this was our first linear transformation, what I just did is I performed another linear transformation."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is this? I just performed another linear transformation. Actually, let me write this. Let's say if this was our first linear transformation, what I just did is I performed another linear transformation. T2, I'll write it in a different notation, where you give me some vector, some column vector, x1, x2, x3. What did I just do? What was the transformation that I just performed?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say if this was our first linear transformation, what I just did is I performed another linear transformation. T2, I'll write it in a different notation, where you give me some vector, some column vector, x1, x2, x3. What did I just do? What was the transformation that I just performed? My new vector, I made the top row equal to the top row plus the second row. So that's x1 plus x2. I kept the second row the same."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What was the transformation that I just performed? My new vector, I made the top row equal to the top row plus the second row. So that's x1 plus x2. I kept the second row the same. And then the third row, I replaced it with the third row minus 2 times the second row. That was a linear transformation we just did. And we could represent this linear transformation as being, we could say T2 applied to some vector x is equal to some transformation vector s2 times our vector x."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I kept the second row the same. And then the third row, I replaced it with the third row minus 2 times the second row. That was a linear transformation we just did. And we could represent this linear transformation as being, we could say T2 applied to some vector x is equal to some transformation vector s2 times our vector x. Now, we could say that this is equal to, we could say, because if we applied this transformation matrix to each of these columns, it's equivalent to multiplying this guy by this transformation matrix. So you could say that this guy right here, we haven't figured out what this is, but I think you get the idea. This matrix right here is going to be equal to this guy."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we could represent this linear transformation as being, we could say T2 applied to some vector x is equal to some transformation vector s2 times our vector x. Now, we could say that this is equal to, we could say, because if we applied this transformation matrix to each of these columns, it's equivalent to multiplying this guy by this transformation matrix. So you could say that this guy right here, we haven't figured out what this is, but I think you get the idea. This matrix right here is going to be equal to this guy. It's going to be equal to s2 times this guy. And what is this guy right here? Well, this guy is equal to s1 times a."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This matrix right here is going to be equal to this guy. It's going to be equal to s2 times this guy. And what is this guy right here? Well, this guy is equal to s1 times a. It's going to be s2 times s1 times a. Fair enough. So this is just s2 times s1 times a."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, this guy is equal to s1 times a. It's going to be s2 times s1 times a. Fair enough. So this is just s2 times s1 times a. And you could have gotten straight here if you created a new, if you just multiplied s2 times s1, this could be some other matrix, and you just multiplied it by a, you'd go straight from there to there. Fair enough. Now, we still haven't gotten this guy in reduced row echelon form."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is just s2 times s1 times a. And you could have gotten straight here if you created a new, if you just multiplied s2 times s1, this could be some other matrix, and you just multiplied it by a, you'd go straight from there to there. Fair enough. Now, we still haven't gotten this guy in reduced row echelon form. So let's try to get there. And I've run out of space below him, so I'm going to have to go up. So let's go upwards."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we still haven't gotten this guy in reduced row echelon form. So let's try to get there. And I've run out of space below him, so I'm going to have to go up. So let's go upwards. Let's go upwards like this. And what I want to do is I'm going to keep the third row the same. 0, 0, 1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's go upwards. Let's go upwards like this. And what I want to do is I'm going to keep the third row the same. 0, 0, 1. And let me replace the second row with the second row minus 2 times the third row. So we get a 0, we get a 1 minus 2 times 0, and we get a 2 minus 2 times 1. So that's a 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0, 0, 1. And let me replace the second row with the second row minus 2 times the third row. So we get a 0, we get a 1 minus 2 times 0, and we get a 2 minus 2 times 1. So that's a 0. And let's replace the first row with the first row minus the third row. So 1 minus 0 is 1. 0 minus 0 is 0."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's a 0. And let's replace the first row with the first row minus the third row. So 1 minus 0 is 1. 0 minus 0 is 0. And 1 minus 1 is 1. 0 and 1 minus 1 is 0. Just like that."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 0 is 0. And 1 minus 1 is 1. 0 and 1 minus 1 is 0. Just like that. And since we did them for the ellipse, let's just actually write what our transformation was. Let's call it T3. I'll do it in purple."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. And since we did them for the ellipse, let's just actually write what our transformation was. Let's call it T3. I'll do it in purple. T3 is the transformation on some vector x. Let me write it like this. On some vector x1, x2, x3."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it in purple. T3 is the transformation on some vector x. Let me write it like this. On some vector x1, x2, x3. It was equal to, what did we do? We replaced the first row with the first row minus the third row. x1 minus x3."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "On some vector x1, x2, x3. It was equal to, what did we do? We replaced the first row with the first row minus the third row. x1 minus x3. We replaced the second row with the second row minus 2 times the third row. So it's x2 minus 2 times x3. And then the third row just stayed the same."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "x1 minus x3. We replaced the second row with the second row minus 2 times the third row. So it's x2 minus 2 times x3. And then the third row just stayed the same. So obviously, this could also be represented. T3 of x could be equal to some other transformation matrix, S3 times x. So this transformation, when you multiply it to each of these columns, is equivalent to multiplying this guy times this transformation matrix, which we haven't found yet."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the third row just stayed the same. So obviously, this could also be represented. T3 of x could be equal to some other transformation matrix, S3 times x. So this transformation, when you multiply it to each of these columns, is equivalent to multiplying this guy times this transformation matrix, which we haven't found yet. Well, we can write it. So this is going to be equal to S3 times this matrix right here, which is S2, S1, A. And what do we have here?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this transformation, when you multiply it to each of these columns, is equivalent to multiplying this guy times this transformation matrix, which we haven't found yet. Well, we can write it. So this is going to be equal to S3 times this matrix right here, which is S2, S1, A. And what do we have here? We got the identity matrix. We put it in reduced row echelon form. We got the identity matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what do we have here? We got the identity matrix. We put it in reduced row echelon form. We got the identity matrix. We already know from previous videos, the reduced row echelon form of something is the identity matrix. Then we are dealing with an invertible transformation or an invertible matrix. Because this obviously could be the transformation for some transformation."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We got the identity matrix. We already know from previous videos, the reduced row echelon form of something is the identity matrix. Then we are dealing with an invertible transformation or an invertible matrix. Because this obviously could be the transformation for some transformation. Let's just call this transformation, I don't know, let's just call it T. Did I already use T? Well, let's just call it T naught for our transformation. Apply it to some vector x."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because this obviously could be the transformation for some transformation. Let's just call this transformation, I don't know, let's just call it T. Did I already use T? Well, let's just call it T naught for our transformation. Apply it to some vector x. It might be equal to Ax. So we know that this is invertible. Because we put it in reduced row echelon form."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Apply it to some vector x. It might be equal to Ax. So we know that this is invertible. Because we put it in reduced row echelon form. We put its transformation matrix in reduced row echelon form, and we got the identity matrix. So that tells us that it's an invertible. But something even more interesting happened."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because we put it in reduced row echelon form. We put its transformation matrix in reduced row echelon form, and we got the identity matrix. So that tells us that it's an invertible. But something even more interesting happened. We got here by performing some row operations. And we said those row operations were equivalent, were completely equivalent, to multiplying this guy right here, by multiplying our original transformation matrix by a series of transformation matrices that represent our row operations. When we multiplied all this, this was equal to the identity matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But something even more interesting happened. We got here by performing some row operations. And we said those row operations were equivalent, were completely equivalent, to multiplying this guy right here, by multiplying our original transformation matrix by a series of transformation matrices that represent our row operations. When we multiplied all this, this was equal to the identity matrix. Now, in the last video, we said that the inverse matrix, so if this is T naught, T naught inverse could be represented. It's also linear transformation. It can be represented by some inverse matrix that we just called A inverse times x."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When we multiplied all this, this was equal to the identity matrix. Now, in the last video, we said that the inverse matrix, so if this is T naught, T naught inverse could be represented. It's also linear transformation. It can be represented by some inverse matrix that we just called A inverse times x. And we saw that A inverse times, or the inverse transformation matrix times our transformation matrix is equal to the identity matrix. We saw this last time. We proved this to you."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It can be represented by some inverse matrix that we just called A inverse times x. And we saw that A inverse times, or the inverse transformation matrix times our transformation matrix is equal to the identity matrix. We saw this last time. We proved this to you. Now, something very interesting here. We have a series of matrix products times this guy, times this guy, that also got me the identity matrix. So this guy right here, the series of matrix products, this must be the same thing as my inverse matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We proved this to you. Now, something very interesting here. We have a series of matrix products times this guy, times this guy, that also got me the identity matrix. So this guy right here, the series of matrix products, this must be the same thing as my inverse matrix. As my inverse transformation matrix. And so we could actually calculate it if we wanted to. We could actually, just like we did, we actually figured out what S1 was."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this guy right here, the series of matrix products, this must be the same thing as my inverse matrix. As my inverse transformation matrix. And so we could actually calculate it if we wanted to. We could actually, just like we did, we actually figured out what S1 was. We did it down here. We could do a similar operation to figure out what S2 was, S3 was, and then multiply them all out, and we would have actually constructed A inverse. But something, I guess, something more interesting we could do instead of doing that."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could actually, just like we did, we actually figured out what S1 was. We did it down here. We could do a similar operation to figure out what S2 was, S3 was, and then multiply them all out, and we would have actually constructed A inverse. But something, I guess, something more interesting we could do instead of doing that. What if we started off, what if we applied the same matrix products to the identity matrix? So the whole time we did here, when we did our first row operation, so we have here, we have the matrix A. And let's say we have an identity matrix on the right."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But something, I guess, something more interesting we could do instead of doing that. What if we started off, what if we applied the same matrix products to the identity matrix? So the whole time we did here, when we did our first row operation, so we have here, we have the matrix A. And let's say we have an identity matrix on the right. Let's call that I, right there. Now, our first linear transformation we did, we saw that right here, that was equivalent to multiplying S1 times A. The first set of row operations was this."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say we have an identity matrix on the right. Let's call that I, right there. Now, our first linear transformation we did, we saw that right here, that was equivalent to multiplying S1 times A. The first set of row operations was this. It got us here. Now, if we perform that same set of row operations on the identity matrix, what are we going to get? We're going to get the matrix S1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The first set of row operations was this. It got us here. Now, if we perform that same set of row operations on the identity matrix, what are we going to get? We're going to get the matrix S1. S1 times the identity matrix is just S1. All of the columns of anything times the identity, times the standard basis columns, it'll just be equal to itself, and you'll just be left with that S1. Or you could call this as S1 times I, but that's just S1."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're going to get the matrix S1. S1 times the identity matrix is just S1. All of the columns of anything times the identity, times the standard basis columns, it'll just be equal to itself, and you'll just be left with that S1. Or you could call this as S1 times I, but that's just S1. Fair enough. Now you performed your next row operation, and you ended up with S2 times S1 times A. Now, if you performed that same row operation on this guy right there, what would you have?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or you could call this as S1 times I, but that's just S1. Fair enough. Now you performed your next row operation, and you ended up with S2 times S1 times A. Now, if you performed that same row operation on this guy right there, what would you have? You would have S2 times S1 times the identity matrix. Now, our last row operation we represented with the matrix product S3, or multiplying it by the transformation matrix S3. So if you did that, if you have S3, S2, S1, A, but if you performed the same exact row operations on this guy right here, you have S3, S2, S1 times the identity matrix."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, if you performed that same row operation on this guy right there, what would you have? You would have S2 times S1 times the identity matrix. Now, our last row operation we represented with the matrix product S3, or multiplying it by the transformation matrix S3. So if you did that, if you have S3, S2, S1, A, but if you performed the same exact row operations on this guy right here, you have S3, S2, S1 times the identity matrix. Now, when you did this, when you performed these row operations here, this got you to the identity matrix. But what are these going to get you to? When you just performed the same exact row operations you perform on A to get to the identity matrix, if you perform those same exact row operations on the identity matrix, what do you get?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you did that, if you have S3, S2, S1, A, but if you performed the same exact row operations on this guy right here, you have S3, S2, S1 times the identity matrix. Now, when you did this, when you performed these row operations here, this got you to the identity matrix. But what are these going to get you to? When you just performed the same exact row operations you perform on A to get to the identity matrix, if you perform those same exact row operations on the identity matrix, what do you get? You get this guy right here. Anything times the identity matrix is going to be equal to itself. So what is that right there?"}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When you just performed the same exact row operations you perform on A to get to the identity matrix, if you perform those same exact row operations on the identity matrix, what do you get? You get this guy right here. Anything times the identity matrix is going to be equal to itself. So what is that right there? That is A inverse. So we have a generalized way of figuring out the inverse for a transformation matrix. What I can do is, let's say I have some transformation matrix A. I can set up an augmented matrix where I put the identity matrix right there, just like that, and I perform a bunch of row operations."}, {"video_title": "Deriving a method for determining inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is that right there? That is A inverse. So we have a generalized way of figuring out the inverse for a transformation matrix. What I can do is, let's say I have some transformation matrix A. I can set up an augmented matrix where I put the identity matrix right there, just like that, and I perform a bunch of row operations. I perform a bunch of row operations, and you could represent them as matrix products, but you perform a bunch of row operations on all of them. You perform the same row operations you perform on A as you would do on the identity matrix, and by the time you have A as an identity matrix, you have A in reduced row echelon form, at the time A is like that, your identity matrix is going to, having performed the same exact operations on it, it is going to be transformed into A's inverse. And this is a very useful tool for solving actual inverses."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, we evaluated this 4 by 4 determinant, and we found out that it was equal to 7. And the way we did it is we went down this first row. We used the definition I gave you in the last video, where you use this first row. I could even write it here. We said that this is equal to 1 times the determinant of 0, 2, 0, 1, 2, 3, 3, 0, 0, minus 2 times the determinant. You cross these guys, cross that row and that column out 1, 0, 2, 1, 0, 2, 2, 2, 0, 0, 3, 0. And then we went to the plus the 3 times its submatrix."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I could even write it here. We said that this is equal to 1 times the determinant of 0, 2, 0, 1, 2, 3, 3, 0, 0, minus 2 times the determinant. You cross these guys, cross that row and that column out 1, 0, 2, 1, 0, 2, 2, 2, 0, 0, 3, 0. And then we went to the plus the 3 times its submatrix. I don't have to figure that out. Just cross out that row, that column. And then minus 4, we just keep switching the signs, times the determinant of its submatrix."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we went to the plus the 3 times its submatrix. I don't have to figure that out. Just cross out that row, that column. And then minus 4, we just keep switching the signs, times the determinant of its submatrix. So this one had a bunch of terms, and this one's going to have a bunch of terms. Cross that row and that column out, you get 1, 0, 2, 0, 1, 2, 2, 3, 0. I'll just write it here."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then minus 4, we just keep switching the signs, times the determinant of its submatrix. So this one had a bunch of terms, and this one's going to have a bunch of terms. Cross that row and that column out, you get 1, 0, 2, 0, 1, 2, 2, 3, 0. I'll just write it here. 1, 0, 2, 0, 1, 2, 2, 3, 0. And that is a completely legitimate way to figure out a determinant, and that was our definition for how to find a determinant. But I want to show you in this video is that there's more than one way to solve for a determinant."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll just write it here. 1, 0, 2, 0, 1, 2, 2, 3, 0. And that is a completely legitimate way to figure out a determinant, and that was our definition for how to find a determinant. But I want to show you in this video is that there's more than one way to solve for a determinant. And what I'm going to show you this way is the same thing that we did down this first row, we can actually do down any row or any column of this determinant, of this matrix. And the reason why that's useful is because we can pick rows or columns that have a unusually large number of zeros, because that tends to simplify our computation. So the first thing that you have to do before you embark on picking an arbitrary row or a column, let's say, for example, we want to pick this."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But I want to show you in this video is that there's more than one way to solve for a determinant. And what I'm going to show you this way is the same thing that we did down this first row, we can actually do down any row or any column of this determinant, of this matrix. And the reason why that's useful is because we can pick rows or columns that have a unusually large number of zeros, because that tends to simplify our computation. So the first thing that you have to do before you embark on picking an arbitrary row or a column, let's say, for example, we want to pick this. Let's do one row and one column in this example. So let's say we want to go down that row instead, because we like the fact that it has a lot of zeros there. The first thing you have to do is remember the pattern."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing that you have to do before you embark on picking an arbitrary row or a column, let's say, for example, we want to pick this. Let's do one row and one column in this example. So let's say we want to go down that row instead, because we like the fact that it has a lot of zeros there. The first thing you have to do is remember the pattern. Remember, you switch signs on the coefficients. And you don't just switch signs as you go down a row. You also switch signs as you go down a column."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The first thing you have to do is remember the pattern. Remember, you switch signs on the coefficients. And you don't just switch signs as you go down a row. You also switch signs as you go down a column. So the general pattern for a 4 by 4 will look like this. It'll be plus, minus, plus, minus, minus, plus, minus, plus, and then you get a plus, minus, minus, plus, plus, minus, minus, plus. It's really this checkerboard pattern."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You also switch signs as you go down a column. So the general pattern for a 4 by 4 will look like this. It'll be plus, minus, plus, minus, minus, plus, minus, plus, and then you get a plus, minus, minus, plus, plus, minus, minus, plus. It's really this checkerboard pattern. If you wanted to figure out the sign for any i, j, so let's say you wanted to figure out the sign for, this is 2, 2, so if you want to find the sign, let me write it this way, let's say we have a function. Let me define a function. I mean, I think the checkerboard pattern's pretty clear to you, but I'll just write it down."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's really this checkerboard pattern. If you wanted to figure out the sign for any i, j, so let's say you wanted to figure out the sign for, this is 2, 2, so if you want to find the sign, let me write it this way, let's say we have a function. Let me define a function. I mean, I think the checkerboard pattern's pretty clear to you, but I'll just write it down. So let's say I wanted to figure out the sign, not the trig identity, not the trig ratio. I want to figure out the sign of any entry where you give me an i and a j. What you could do is you just take negative 1 to the i plus j power."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I mean, I think the checkerboard pattern's pretty clear to you, but I'll just write it down. So let's say I wanted to figure out the sign, not the trig identity, not the trig ratio. I want to figure out the sign of any entry where you give me an i and a j. What you could do is you just take negative 1 to the i plus j power. So if you wanted to figure out the sign for the, let's say you are in row 4, column 2. So what will it be? You do 4 plus 2."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What you could do is you just take negative 1 to the i plus j power. So if you wanted to figure out the sign for the, let's say you are in row 4, column 2. So what will it be? You do 4 plus 2. Negative 1 to the 6th power is equal to 1. So that's going to be a positive 1. Now, if you take this guy, let's say you want to take this guy."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You do 4 plus 2. Negative 1 to the 6th power is equal to 1. So that's going to be a positive 1. Now, if you take this guy, let's say you want to take this guy. So this is i is equal to 2, j is equal to 3. You're in the second row, third column. 2 plus 3 is 5."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, if you take this guy, let's say you want to take this guy. So this is i is equal to 2, j is equal to 3. You're in the second row, third column. 2 plus 3 is 5. Minus 1 to the 5th power is minus 1. And you have a minus there. So that's another way to think about it."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 plus 3 is 5. Minus 1 to the 5th power is minus 1. And you have a minus there. So that's another way to think about it. But the checkerboard pattern's pretty straightforward. So now that you have the checkerboard pattern in your mind, let's go down this row. So we start with a 2, but notice that we have to multiply it by a minus, because you go plus, minus, plus, minus."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's another way to think about it. But the checkerboard pattern's pretty straightforward. So now that you have the checkerboard pattern in your mind, let's go down this row. So we start with a 2, but notice that we have to multiply it by a minus, because you go plus, minus, plus, minus. So you have a minus 2 times the determinant of its submatrix. So you cross out this row and that column, and you're left with this matrix up here. So it's 2, 3, 4, 0, 2, 0, 1, 2, 3."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we start with a 2, but notice that we have to multiply it by a minus, because you go plus, minus, plus, minus. So you have a minus 2 times the determinant of its submatrix. So you cross out this row and that column, and you're left with this matrix up here. So it's 2, 3, 4, 0, 2, 0, 1, 2, 3. And then you go, this was a minus, so then you have a plus. So then you have a plus 3 times, get rid of this guy's column and row, and you have 1, 1, 0. That's that right there."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's 2, 3, 4, 0, 2, 0, 1, 2, 3. And then you go, this was a minus, so then you have a plus. So then you have a plus 3 times, get rid of this guy's column and row, and you have 1, 1, 0. That's that right there. 3, 2, 2. And you have 4, 0, 3. And then you would have a minus of 0 times its submatrix plus a 0."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's that right there. 3, 2, 2. And you have 4, 0, 3. And then you would have a minus of 0 times its submatrix plus a 0. But we can ignore those, because 0 times anything is a 0. So already we've simplified our determinant a good bit. So let's see if we can evaluate this and get the same number, because only then will it be reasonably satisfying."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you would have a minus of 0 times its submatrix plus a 0. But we can ignore those, because 0 times anything is a 0. So already we've simplified our determinant a good bit. So let's see if we can evaluate this and get the same number, because only then will it be reasonably satisfying. So what's the determinant of this guy? Well, we can do it the exact same principle. We can go down any row or column that seems to be especially simple."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can evaluate this and get the same number, because only then will it be reasonably satisfying. So what's the determinant of this guy? Well, we can do it the exact same principle. We can go down any row or column that seems to be especially simple. So let's go down that row, because that row seems especially simple. So this is going to be minus 2. That's this minus 2 right here, times the determinant of this guy."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can go down any row or column that seems to be especially simple. So let's go down that row, because that row seems especially simple. So this is going to be minus 2. That's this minus 2 right here, times the determinant of this guy. So the determinant of this guy, we just have to go and say, OK, we have a plus, and we have a minus, and then we have a plus. So it's going to be minus 0 times its subdeterminant. I guess we could call it."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's this minus 2 right here, times the determinant of this guy. So the determinant of this guy, we just have to go and say, OK, we have a plus, and we have a minus, and then we have a plus. So it's going to be minus 0 times its subdeterminant. I guess we could call it. We get rid of that row, or that column and that row. So it would be minus 0 times anything. That's just going to be 0."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I guess we could call it. We get rid of that row, or that column and that row. So it would be minus 0 times anything. That's just going to be 0. Plus 2. So plus 2 times the determinant, get rid of its row and its column. So it's 2, 4, 1, 3."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's just going to be 0. Plus 2. So plus 2 times the determinant, get rid of its row and its column. So it's 2, 4, 1, 3. And then you would have a minus 0 times this thing. But who cares what that is, because you have a 0 times it. So that just simplified to that, which is nice."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's 2, 4, 1, 3. And then you would have a minus 0 times this thing. But who cares what that is, because you have a 0 times it. So that just simplified to that, which is nice. Let me write it like that. And then you have plus 3 times this thing right here. Let's pick a nice, we don't want to do the first row."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that just simplified to that, which is nice. Let me write it like that. And then you have plus 3 times this thing right here. Let's pick a nice, we don't want to do the first row. We have no non-zero terms here. Let's at least do, let me do this row right here for a little bit of variety. None of the columns actually seem that interesting."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's pick a nice, we don't want to do the first row. We have no non-zero terms here. Let's at least do, let me do this row right here for a little bit of variety. None of the columns actually seem that interesting. They all have at most one 0. So if we do that one right here, this is a plus, minus, plus, minus, plus. So we'll have a plus 0 times 3, 4, 2, 0."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "None of the columns actually seem that interesting. They all have at most one 0. So if we do that one right here, this is a plus, minus, plus, minus, plus. So we'll have a plus 0 times 3, 4, 2, 0. We can ignore that. Minus 2 times the determinant, get rid of this column, that row. So 1, I have to be very careful."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we'll have a plus 0 times 3, 4, 2, 0. We can ignore that. Minus 2 times the determinant, get rid of this column, that row. So 1, I have to be very careful. I put this minus there, but there wasn't a minus 1 there. So let me write it, let me make sure, I want to make sure I don't make any careless errors right here. This is a plus 1."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 1, I have to be very careful. I put this minus there, but there wasn't a minus 1 there. So let me write it, let me make sure, I want to make sure I don't make any careless errors right here. This is a plus 1. I just do a minus 1 there to show you how things switch signs. So this is going to be a 3 times, so we're going to go down, this is this 3 right here. I lost my bearings with that minus there."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a plus 1. I just do a minus 1 there to show you how things switch signs. So this is going to be a 3 times, so we're going to go down, this is this 3 right here. I lost my bearings with that minus there. But we're trying to find the determinant of this. So it's a 0 times this matrix. We could ignore that."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I lost my bearings with that minus there. But we're trying to find the determinant of this. So it's a 0 times this matrix. We could ignore that. Minus 2 times its submatrix, which is that and that. So it's 1, 1, 4, 0. And then you have a plus 3 times its submatrix."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could ignore that. Minus 2 times its submatrix, which is that and that. So it's 1, 1, 4, 0. And then you have a plus 3 times its submatrix. So it's 1, 3, 1, 2. Just like that. Let's see if we can simplify this."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have a plus 3 times its submatrix. So it's 1, 3, 1, 2. Just like that. Let's see if we can simplify this. 2 times 3 is 6 minus 1 times 4, so this becomes 6 minus 4. So that's 2. So this whole thing simplifies to 2 times 2, which is 4."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if we can simplify this. 2 times 3 is 6 minus 1 times 4, so this becomes 6 minus 4. So that's 2. So this whole thing simplifies to 2 times 2, which is 4. 4 times minus 2, the whole thing simplifies to minus 8. Now, this guy right here, we have 1 times 0, which is 0, minus 1 times 4, so this is minus 4. Times minus 2, this whole thing becomes a positive 8."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this whole thing simplifies to 2 times 2, which is 4. 4 times minus 2, the whole thing simplifies to minus 8. Now, this guy right here, we have 1 times 0, which is 0, minus 1 times 4, so this is minus 4. Times minus 2, this whole thing becomes a positive 8. And you have 1 times 2, which is 2, minus 1 times 3, which is minus 3. So you get a minus 1. So this becomes a minus, you get 2 minus 3."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times minus 2, this whole thing becomes a positive 8. And you have 1 times 2, which is 2, minus 1 times 3, which is minus 3. So you get a minus 1. So this becomes a minus, you get 2 minus 3. Minus 1 times 3, so this becomes a minus 3. So you have an 8 minus a 3. So this becomes a minus 5."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this becomes a minus, you get 2 minus 3. Minus 1 times 3, so this becomes a minus 3. So you have an 8 minus a 3. So this becomes a minus 5. And then you have a 3 times a minus 5 right there. So 3 times minus 5 is going to be equal to minus 15. Let me make sure I got, oh, I made a silly mistake."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this becomes a minus 5. And then you have a 3 times a minus 5 right there. So 3 times minus 5 is going to be equal to minus 15. Let me make sure I got, oh, I made a silly mistake. If you have an 8 minus a 3, this is 8, this is minus 3, this is going to be 5. Very easy, your brain starts to get fried if you do this long enough. And then you have a 3 times a 5, you get a 15."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me make sure I got, oh, I made a silly mistake. If you have an 8 minus a 3, this is 8, this is minus 3, this is going to be 5. Very easy, your brain starts to get fried if you do this long enough. And then you have a 3 times a 5, you get a 15. You get a 15. And then you have a 15, this term plus this term is a 15 minus an 8, which is equal to 7. Which lucky for us, I barely evaded making a careless mistake there, we got the right answer."}, {"video_title": "Determinants along other rows cols Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have a 3 times a 5, you get a 15. You get a 15. And then you have a 15, this term plus this term is a 15 minus an 8, which is equal to 7. Which lucky for us, I barely evaded making a careless mistake there, we got the right answer. But this is a much simpler computation than we did in the last video. And it was much simpler because we picked the row that happened to have a lot of 0's on it. So we only had 2 of these terms instead of 4 terms like we had in the last video."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Let's take a little bit of a hiatus from our more rigorous math where we're building the mathematics of vector algebra and just think a little bit about something that you'll probably encounter if you have to write a three-dimensional computer program or have to do any mathematics dealing with three dimensions. And that's the idea of just the equation of a plane in R3. And you know what a plane is. I mean, we live in a three-dimensional world and we see planes all around us. The surface of your computer monitor is a plane regardless of what angle you hold it at. So I can draw it here in three dimensions. Let me do a little bit better job than that."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "I mean, we live in a three-dimensional world and we see planes all around us. The surface of your computer monitor is a plane regardless of what angle you hold it at. So I can draw it here in three dimensions. Let me do a little bit better job than that. So let's say that that is the x-axis. This is my y-axis. And then that's my z-axis."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Let me do a little bit better job than that. So let's say that that is the x-axis. This is my y-axis. And then that's my z-axis. And we know what a plane looks like. It looks something like that. I'm just drawing it at an arbitrary angle and it goes off in every direction."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And then that's my z-axis. And we know what a plane looks like. It looks something like that. I'm just drawing it at an arbitrary angle and it goes off in every direction. Now, the equation of a plane, and you've probably seen this before, it's a linear function of x, y, and z. So it's ax plus by plus cz is equal to d. If this is the graph of that plane, then that means that every point on this plane, every x, y, and z on this plane, satisfies this equation. Now, another way, just as valid way, to specify a plane is to give you an actual point on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "I'm just drawing it at an arbitrary angle and it goes off in every direction. Now, the equation of a plane, and you've probably seen this before, it's a linear function of x, y, and z. So it's ax plus by plus cz is equal to d. If this is the graph of that plane, then that means that every point on this plane, every x, y, and z on this plane, satisfies this equation. Now, another way, just as valid way, to specify a plane is to give you an actual point on the plane. So say, look, that's a point on the plane. So let's just say that this is the point x naught, y naught, and z naught. Different, it could be one of the instances of this point right here, but I'm just saying this is another point."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Now, another way, just as valid way, to specify a plane is to give you an actual point on the plane. So say, look, that's a point on the plane. So let's just say that this is the point x naught, y naught, and z naught. Different, it could be one of the instances of this point right here, but I'm just saying this is another point. It's on the plane. That obviously by itself isn't going to specify the plane. You could pivot the plane around that point in an infinite number of ways."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Different, it could be one of the instances of this point right here, but I'm just saying this is another point. It's on the plane. That obviously by itself isn't going to specify the plane. You could pivot the plane around that point in an infinite number of ways. But if you specify that point and you specify a vector that's perpendicular to the plane, and I could draw it starting from here, but I can shift a vector wherever, but let me just draw it right there. So if I also specify a normal vector to the plane, normal, and I just used a word that I haven't defined for you yet, but when I say a normal vector, so n is a normal vector, n is normal, or, which just means that it's perpendicular to the plane. It's perpendicular to everything on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "You could pivot the plane around that point in an infinite number of ways. But if you specify that point and you specify a vector that's perpendicular to the plane, and I could draw it starting from here, but I can shift a vector wherever, but let me just draw it right there. So if I also specify a normal vector to the plane, normal, and I just used a word that I haven't defined for you yet, but when I say a normal vector, so n is a normal vector, n is normal, or, which just means that it's perpendicular to the plane. It's perpendicular to everything on the plane. It's perpendicular to every vector on the plane. Perpendicular to, I guess the best way to say it, I'll just say it in very imprecise terms, everything on the plane. So if you have some vectors lying on the plane, if I have some vector here, let's say that's lying on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "It's perpendicular to everything on the plane. It's perpendicular to every vector on the plane. Perpendicular to, I guess the best way to say it, I'll just say it in very imprecise terms, everything on the plane. So if you have some vectors lying on the plane, if I have some vector here, let's say that's lying on the plane. If you imagine the plane as a piece of cardboard, that yellow arrow I just drew, I would have actually drawn that on the cardboard. It's sitting on the plane. If this yellow vector, let me call it vector A, then if this is just some arbitrary vector sitting on the plane, and this is the normal vector through the plane, we know from our definition of vector angles that this is perpendicular to this if and only if n dot A, only if the dot product of these two things, are equal to 0."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So if you have some vectors lying on the plane, if I have some vector here, let's say that's lying on the plane. If you imagine the plane as a piece of cardboard, that yellow arrow I just drew, I would have actually drawn that on the cardboard. It's sitting on the plane. If this yellow vector, let me call it vector A, then if this is just some arbitrary vector sitting on the plane, and this is the normal vector through the plane, we know from our definition of vector angles that this is perpendicular to this if and only if n dot A, only if the dot product of these two things, are equal to 0. And that's true for any vector that we pick that actually lies on the plane. So let's see if we can use this definition of a plane, if we can use the, I'll call it the normal, or I'll call it the n plus some x naught, y naught, z naught definition. And if I can go from that to just the standard linear equation definition, Ax plus By plus Cz is equal to 0."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "If this yellow vector, let me call it vector A, then if this is just some arbitrary vector sitting on the plane, and this is the normal vector through the plane, we know from our definition of vector angles that this is perpendicular to this if and only if n dot A, only if the dot product of these two things, are equal to 0. And that's true for any vector that we pick that actually lies on the plane. So let's see if we can use this definition of a plane, if we can use the, I'll call it the normal, or I'll call it the n plus some x naught, y naught, z naught definition. And if I can go from that to just the standard linear equation definition, Ax plus By plus Cz is equal to 0. Let's see if there's some way, based on what we already know and using this information, that we can do that. So the way to think about it, this point, this little blue point that lies on the plane, I can specify it by a position vector. So let me set some position vector, x naught, to be equal to, so I'm going to define x naught to be equal to the scalar x naught, y naught, z naught."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And if I can go from that to just the standard linear equation definition, Ax plus By plus Cz is equal to 0. Let's see if there's some way, based on what we already know and using this information, that we can do that. So the way to think about it, this point, this little blue point that lies on the plane, I can specify it by a position vector. So let me set some position vector, x naught, to be equal to, so I'm going to define x naught to be equal to the scalar x naught, y naught, z naught. Now I want to be very clear. This specifies the coordinate that lies on the plane. This vector does not lie entirely on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So let me set some position vector, x naught, to be equal to, so I'm going to define x naught to be equal to the scalar x naught, y naught, z naught. Now I want to be very clear. This specifies the coordinate that lies on the plane. This vector does not lie entirely on the plane. It starts, the way I drew it right here, it's starting at the origin, it's a position vector. And the way I drew it, it's behind the plane, and the tip of its arrow sits on the plane. But this vector itself, it's not necessarily drawn on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "This vector does not lie entirely on the plane. It starts, the way I drew it right here, it's starting at the origin, it's a position vector. And the way I drew it, it's behind the plane, and the tip of its arrow sits on the plane. But this vector itself, it's not necessarily drawn on the plane. This plane might not even go through the origin, while this vector does touch on the origin. It just specifies some point on the plane. Similarly, let me define another vector."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "But this vector itself, it's not necessarily drawn on the plane. This plane might not even go through the origin, while this vector does touch on the origin. It just specifies some point on the plane. Similarly, let me define another vector. I said this was some other arbitrary point on the plane, x, y, z. And so this is true for any point on the plane. Let me define another vector, x, and I'm going to define that as x, y, and z."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Similarly, let me define another vector. I said this was some other arbitrary point on the plane, x, y, z. And so this is true for any point on the plane. Let me define another vector, x, and I'm going to define that as x, y, and z. So once again, like x naught, the vector x, let me draw it right there, this vector x does not lie on the plane. It goes from the origin. It's a position vector that specifies a point on the plane, so it goes from the origin and it goes out."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Let me define another vector, x, and I'm going to define that as x, y, and z. So once again, like x naught, the vector x, let me draw it right there, this vector x does not lie on the plane. It goes from the origin. It's a position vector that specifies a point on the plane, so it goes from the origin and it goes out. And you can almost view them as if the plane was like a coffee table, this would kind of be these vectors. Let me see if I can draw it. If this was the flat surface of the plane, that the vector x naught is going from the origin to specify some point on the plane, and then the vector x is also going from the origin to specify, let me do it in a different color, the vector x is also going from the origin to specify some other point on the plane right there."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "It's a position vector that specifies a point on the plane, so it goes from the origin and it goes out. And you can almost view them as if the plane was like a coffee table, this would kind of be these vectors. Let me see if I can draw it. If this was the flat surface of the plane, that the vector x naught is going from the origin to specify some point on the plane, and then the vector x is also going from the origin to specify, let me do it in a different color, the vector x is also going from the origin to specify some other point on the plane right there. I just took the plane and made it flat, so you see it right along its side. If you can imagine sitting right at the surface of the plane, and then you can see that these guys clearly do not lie in the plane. But using these guys, I can construct a vector that does lie on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "If this was the flat surface of the plane, that the vector x naught is going from the origin to specify some point on the plane, and then the vector x is also going from the origin to specify, let me do it in a different color, the vector x is also going from the origin to specify some other point on the plane right there. I just took the plane and made it flat, so you see it right along its side. If you can imagine sitting right at the surface of the plane, and then you can see that these guys clearly do not lie in the plane. But using these guys, I can construct a vector that does lie on the plane. What does the vector x minus x naught look like? Well, I just drew a little triangle here. x minus x naught, I'll do it in this green color, it'll look exactly like this, x minus x naught will be this green line right here."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "But using these guys, I can construct a vector that does lie on the plane. What does the vector x minus x naught look like? Well, I just drew a little triangle here. x minus x naught, I'll do it in this green color, it'll look exactly like this, x minus x naught will be this green line right here. This is x minus x naught. We could do x naught plus this vector, plus x minus x naught is going to be equal to x. If I were to do it on this graph, it's going to look like this."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "x minus x naught, I'll do it in this green color, it'll look exactly like this, x minus x naught will be this green line right here. This is x minus x naught. We could do x naught plus this vector, plus x minus x naught is going to be equal to x. If I were to do it on this graph, it's going to look like this. It's going to be like this. Let me draw it better than that. It's going to go from that point, from x naught, the point specified by x naught, to the point specified by x."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "If I were to do it on this graph, it's going to look like this. It's going to be like this. Let me draw it better than that. It's going to go from that point, from x naught, the point specified by x naught, to the point specified by x. And it's going to lie along the plane. So this right here is x minus x naught. I know this drawing is getting very dirty, but you can see that this is definitely lying on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "It's going to go from that point, from x naught, the point specified by x naught, to the point specified by x. And it's going to lie along the plane. So this right here is x minus x naught. I know this drawing is getting very dirty, but you can see that this is definitely lying on the plane. So this vector right here must be perpendicular to n. Perpendicular to our normal vector. Now, if my normal vector, let's say my normal vector, and so this vector is perpendicular to this guy right here, it's perpendicular to the vector n1, n2, n3. Now using this information, how can we get to this type of an expression, just this linear equation of x, y, and z?"}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "I know this drawing is getting very dirty, but you can see that this is definitely lying on the plane. So this vector right here must be perpendicular to n. Perpendicular to our normal vector. Now, if my normal vector, let's say my normal vector, and so this vector is perpendicular to this guy right here, it's perpendicular to the vector n1, n2, n3. Now using this information, how can we get to this type of an expression, just this linear equation of x, y, and z? Well, we know that n, let me switch to a neutral color. We know that n, actually I didn't want to do this caret here, n is not a unit vector. But let's say n, so it's going perpendicular to this, so the dot product of it, we saw that right there, we saw it in the previous video, the dot product of n with this vector right here."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Now using this information, how can we get to this type of an expression, just this linear equation of x, y, and z? Well, we know that n, let me switch to a neutral color. We know that n, actually I didn't want to do this caret here, n is not a unit vector. But let's say n, so it's going perpendicular to this, so the dot product of it, we saw that right there, we saw it in the previous video, the dot product of n with this vector right here. Actually, let me draw it, I already drew the plane sideways so I can actually draw my n vector. My n vector is going to look something like this. It's going to be popping straight out of the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "But let's say n, so it's going perpendicular to this, so the dot product of it, we saw that right there, we saw it in the previous video, the dot product of n with this vector right here. Actually, let me draw it, I already drew the plane sideways so I can actually draw my n vector. My n vector is going to look something like this. It's going to be popping straight out of the plane. And I could shift it over, but it's always going to be in that same direction. But it's going to be perpendicular to this vector right here. So n is perpendicular to x minus x naught, which means that their dot product is equal to 0."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be popping straight out of the plane. And I could shift it over, but it's always going to be in that same direction. But it's going to be perpendicular to this vector right here. So n is perpendicular to x minus x naught, which means that their dot product is equal to 0. Well, what does x minus x naught look like? So this is going to look like this expression, if I write out the vectors themselves, it's the vector n1, n2, n3 being dotted with, well, if I take x minus x naught, that's just the scalar x minus the scalar x naught, the first term subtracted. And then the scalar y minus the scalar y naught."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So n is perpendicular to x minus x naught, which means that their dot product is equal to 0. Well, what does x minus x naught look like? So this is going to look like this expression, if I write out the vectors themselves, it's the vector n1, n2, n3 being dotted with, well, if I take x minus x naught, that's just the scalar x minus the scalar x naught, the first term subtracted. And then the scalar y minus the scalar y naught. And then the scalar z minus the scalar z naught. And we know that this whole thing has to be equal to 0 because they're perpendicular. And then if we take the dot product here, we get n1 times x minus x naught plus n2 times y minus y naught plus n3 times z minus z naught is equal to 0."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And then the scalar y minus the scalar y naught. And then the scalar z minus the scalar z naught. And we know that this whole thing has to be equal to 0 because they're perpendicular. And then if we take the dot product here, we get n1 times x minus x naught plus n2 times y minus y naught plus n3 times z minus z naught is equal to 0. And you might not completely recognize it, but this is, you'll have to do a little algebra to clean it up, but this is the form Ax plus By plus Cz is equal to D. And actually, I think I made a mistake here. This should be not 0. This is equal to D. This is the general form for a plane in R3, for a plane is just a linear surface in R3."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And then if we take the dot product here, we get n1 times x minus x naught plus n2 times y minus y naught plus n3 times z minus z naught is equal to 0. And you might not completely recognize it, but this is, you'll have to do a little algebra to clean it up, but this is the form Ax plus By plus Cz is equal to D. And actually, I think I made a mistake here. This should be not 0. This is equal to D. This is the general form for a plane in R3, for a plane is just a linear surface in R3. I shouldn't have written a 0 there. So this does take this form. And if you don't believe me, we can do it with an actual example."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to D. This is the general form for a plane in R3, for a plane is just a linear surface in R3. I shouldn't have written a 0 there. So this does take this form. And if you don't believe me, we can do it with an actual example. So let's say we have, I gave you my normal, I specify a plane, but I give you a normal vector. And I tell you that normal vector is the 0.13 minus 2. And I say that it intersects the point, or a point that lies on the plane, the normal vector and the point don't necessarily have to intersect."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And if you don't believe me, we can do it with an actual example. So let's say we have, I gave you my normal, I specify a plane, but I give you a normal vector. And I tell you that normal vector is the 0.13 minus 2. And I say that it intersects the point, or a point that lies on the plane, the normal vector and the point don't necessarily have to intersect. But let's say for a point that lies on the plane, I have the 0.12 and 3. And I say give me the equation for this plane. Well, I would say, well, if I take any other point on that plane, x, y, z, and it's specified by this vector, the vector that's defined by the difference between these two is going to lie on the plane, right?"}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "And I say that it intersects the point, or a point that lies on the plane, the normal vector and the point don't necessarily have to intersect. But let's say for a point that lies on the plane, I have the 0.12 and 3. And I say give me the equation for this plane. Well, I would say, well, if I take any other point on that plane, x, y, z, and it's specified by this vector, the vector that's defined by the difference between these two is going to lie on the plane, right? This point and this point lie on the plane. So the difference between these two vectors, the whole vector will lie on the plane. So let me take the difference."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "Well, I would say, well, if I take any other point on that plane, x, y, z, and it's specified by this vector, the vector that's defined by the difference between these two is going to lie on the plane, right? This point and this point lie on the plane. So the difference between these two vectors, the whole vector will lie on the plane. So let me take the difference. So x minus x naught is equal to x minus 1, y minus 2, and then z minus 3. I'm saying this will lie on the plane. So this is on the plane."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So let me take the difference. So x minus x naught is equal to x minus 1, y minus 2, and then z minus 3. I'm saying this will lie on the plane. So this is on the plane. And it's going to be perpendicular to our normal vector. So if I dot my normal vector, 1, 3, minus 2, with this thing right here, with x minus 1, y minus 2, z minus 3, I should get 0, because this has to be perpendicular to anything that lies on the plane. So what do we get?"}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So this is on the plane. And it's going to be perpendicular to our normal vector. So if I dot my normal vector, 1, 3, minus 2, with this thing right here, with x minus 1, y minus 2, z minus 3, I should get 0, because this has to be perpendicular to anything that lies on the plane. So what do we get? We get 1 times x minus 1 is x minus 1, plus 3 times y minus 2, just taking the dot product, minus 2 times z minus 3 is equal to 0. Let's see if we can do a little bit of algebra here to clean this up a little bit. I get x minus 1 plus 3y minus 6 minus 2z plus 6 is equal to 0, and let's see, minus 6 and a plus 6 cancel out, and then I can take this minus 1, I can add 1 to both sides, and I get x plus 3y minus 2z, add that 1 to both sides, is equal to 1."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "So what do we get? We get 1 times x minus 1 is x minus 1, plus 3 times y minus 2, just taking the dot product, minus 2 times z minus 3 is equal to 0. Let's see if we can do a little bit of algebra here to clean this up a little bit. I get x minus 1 plus 3y minus 6 minus 2z plus 6 is equal to 0, and let's see, minus 6 and a plus 6 cancel out, and then I can take this minus 1, I can add 1 to both sides, and I get x plus 3y minus 2z, add that 1 to both sides, is equal to 1. And there you have it. Just by using the simple fact that this is a point on the plane and this is a normal vector, I was able to use the idea that this has to be normal or it's a dot product with any vector that lies on the plane, I was able to get this right here. I didn't have to go through this whole business right here, you could have just used this formula right here."}, {"video_title": "Defining a plane in R3 with a point and normal vector Linear Algebra Khan Academy.mp3", "Sentence": "I get x minus 1 plus 3y minus 6 minus 2z plus 6 is equal to 0, and let's see, minus 6 and a plus 6 cancel out, and then I can take this minus 1, I can add 1 to both sides, and I get x plus 3y minus 2z, add that 1 to both sides, is equal to 1. And there you have it. Just by using the simple fact that this is a point on the plane and this is a normal vector, I was able to use the idea that this has to be normal or it's a dot product with any vector that lies on the plane, I was able to get this right here. I didn't have to go through this whole business right here, you could have just used this formula right here. You could have just said n1 is 1 times x minus x1, or x0 I could call it, so x minus this 1, plus n2, 3 times y minus 2, plus minus 2 times z minus 3 is equal to 0. And then if you just did a little bit of math, a little bit of algebra, you would have gotten there. So hopefully you find this reasonably useful."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This first one is minus 1, 0. I tried to draw them ahead of time. So minus 1, 0 is this point right there. Do it in these new colors. The next point is 0, 1, which is going to be 0, 1, which is that point right there. Then the next point is 1, 2, which is that point right up there. And then the last point is 2, 1, which is this point, that point there."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Do it in these new colors. The next point is 0, 1, which is going to be 0, 1, which is that point right there. Then the next point is 1, 2, which is that point right up there. And then the last point is 2, 1, which is this point, that point there. Now my goal in this video is to find some line, y equals mx plus v, that goes through these points. Now the first thing you might say, hey Sal, there's not going to be any line that goes through these points. And you can see that immediately."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then the last point is 2, 1, which is this point, that point there. Now my goal in this video is to find some line, y equals mx plus v, that goes through these points. Now the first thing you might say, hey Sal, there's not going to be any line that goes through these points. And you can see that immediately. You could find a line that maybe goes through these points, but it's not going to go through this point over here. If you try to make a line that go through these two points, it's not going to go through those points there. So you're not going to be able to find a solution that goes through those points."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And you can see that immediately. You could find a line that maybe goes through these points, but it's not going to go through this point over here. If you try to make a line that go through these two points, it's not going to go through those points there. So you're not going to be able to find a solution that goes through those points. But let's at least set up the equation that we know we can't find the solution to. And maybe we can use our least squares approximation to find a line that almost goes through all of these points. Or is at least the best approximation for a line that goes through those points."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you're not going to be able to find a solution that goes through those points. But let's at least set up the equation that we know we can't find the solution to. And maybe we can use our least squares approximation to find a line that almost goes through all of these points. Or is at least the best approximation for a line that goes through those points. So this first one, I can express my line. Instead of y equals mx plus v, let me just express it as f of x is equal to mx plus v, or y is equal to f of x. We could write it like that way."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or is at least the best approximation for a line that goes through those points. So this first one, I can express my line. Instead of y equals mx plus v, let me just express it as f of x is equal to mx plus v, or y is equal to f of x. We could write it like that way. So our first point right there, let me do it in that color, that orange. That tells us that f of minus 1, which is equal to m times let me just write this way, minus 1 times m, so it's minus m plus v, that that is going to be equal to 0. That's what that first equation tells us."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We could write it like that way. So our first point right there, let me do it in that color, that orange. That tells us that f of minus 1, which is equal to m times let me just write this way, minus 1 times m, so it's minus m plus v, that that is going to be equal to 0. That's what that first equation tells us. The second equation tells us that f of 0, which is equal to 0 times m, which is just 0, plus v is equal to 1. f of 0 is 1. This is the x, this is f of x. The next one, let me do it in this yellow color, tells us that f of 1, which is equal to 1 times m, or just m, plus v, is going to be equal to 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's what that first equation tells us. The second equation tells us that f of 0, which is equal to 0 times m, which is just 0, plus v is equal to 1. f of 0 is 1. This is the x, this is f of x. The next one, let me do it in this yellow color, tells us that f of 1, which is equal to 1 times m, or just m, plus v, is going to be equal to 2. And then this last one down here tells us that f of 2, which is, of course, 2 times m plus v, that that is going to be equal to 1. These are the constraints. If we assume that our line can go through all of these points, then all of these things must be true."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The next one, let me do it in this yellow color, tells us that f of 1, which is equal to 1 times m, or just m, plus v, is going to be equal to 2. And then this last one down here tells us that f of 2, which is, of course, 2 times m plus v, that that is going to be equal to 1. These are the constraints. If we assume that our line can go through all of these points, then all of these things must be true. Now you can immediately, if you wish, try to solve this equation, but you'll find that you won't find a solution. We want to find some m's and b's that satisfy all of these equations. Or another way of writing this, we want to write it as a matrix vector or matrix equation."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "If we assume that our line can go through all of these points, then all of these things must be true. Now you can immediately, if you wish, try to solve this equation, but you'll find that you won't find a solution. We want to find some m's and b's that satisfy all of these equations. Or another way of writing this, we want to write it as a matrix vector or matrix equation. We could write it like this, minus 1, minus 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 1, times the vector mv has got to be equal to the vector 0, 1, 2, 1. 0, 1, 2, 1. These two systems, this system and this system right here, are equivalent statements, right?"}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or another way of writing this, we want to write it as a matrix vector or matrix equation. We could write it like this, minus 1, minus 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 1, times the vector mv has got to be equal to the vector 0, 1, 2, 1. 0, 1, 2, 1. These two systems, this system and this system right here, are equivalent statements, right? Minus 1 times m plus 1 times v has got to be equal to that 0. 0 times m plus 1 times v has got to be equal to that 1. That's equivalent to this statement right here."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "These two systems, this system and this system right here, are equivalent statements, right? Minus 1 times m plus 1 times v has got to be equal to that 0. 0 times m plus 1 times v has got to be equal to that 1. That's equivalent to this statement right here. And this isn't going to have a solution. The solution would have to go through all of those points. So let's at least try to find a least square solution."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's equivalent to this statement right here. And this isn't going to have a solution. The solution would have to go through all of those points. So let's at least try to find a least square solution. So if we call this a, if we call that x, unless we call this b, there is no solution to ax is equal to b. Now, maybe we can find a least square, we can definitely find a least square solution. So let's find our least square solution such that a transpose a times our least squares solution is equal to a transpose times b."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's at least try to find a least square solution. So if we call this a, if we call that x, unless we call this b, there is no solution to ax is equal to b. Now, maybe we can find a least square, we can definitely find a least square solution. So let's find our least square solution such that a transpose a times our least squares solution is equal to a transpose times b. Our least squares solution is the one that satisfies this equation. We proved it two videos ago. So let's figure out what a transpose a is and what a transpose b is and then we can solve."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's find our least square solution such that a transpose a times our least squares solution is equal to a transpose times b. Our least squares solution is the one that satisfies this equation. We proved it two videos ago. So let's figure out what a transpose a is and what a transpose b is and then we can solve. So a transpose will look like this. It would be minus 1, 1, 0, 1, 1, 1, and then 2, 1. This first column becomes this first row."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's figure out what a transpose a is and what a transpose b is and then we can solve. So a transpose will look like this. It would be minus 1, 1, 0, 1, 1, 1, and then 2, 1. This first column becomes this first row. The second column becomes the second row. So we're going to take the product of a transpose and then a. a is that thing right there. Minus 1, 0, 1, 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This first column becomes this first row. The second column becomes the second row. So we're going to take the product of a transpose and then a. a is that thing right there. Minus 1, 0, 1, 2. And then we just get a bunch of 1's. So what is this equal to? We have a 2 by 4 times a 4 by 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1, 0, 1, 2. And then we just get a bunch of 1's. So what is this equal to? We have a 2 by 4 times a 4 by 2. So we're going to have a 2 by 2 matrix. So this is going to be, let's do it this way. Well, we're going to have minus 1 times minus 1, which is 1, plus 0 times 0, which is 0."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We have a 2 by 4 times a 4 by 2. So we're going to have a 2 by 2 matrix. So this is going to be, let's do it this way. Well, we're going to have minus 1 times minus 1, which is 1, plus 0 times 0, which is 0. So we're at 1 right now. Plus 1 times 1. So that's 1 plus the other 1 up there."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, we're going to have minus 1 times minus 1, which is 1, plus 0 times 0, which is 0. So we're at 1 right now. Plus 1 times 1. So that's 1 plus the other 1 up there. So that's 2. Plus 2 times 2. 2 times 2 is 4."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that's 1 plus the other 1 up there. So that's 2. Plus 2 times 2. 2 times 2 is 4. So we get 6. So that's equal to 6. That's that row dotted with that column was equal to 6."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "2 times 2 is 4. So we get 6. So that's equal to 6. That's that row dotted with that column was equal to 6. Now let's take this row dotted with this column. So it's going to be negative 1 times 1 plus 0 times 1. So all of these guys times 1 plus each other."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's that row dotted with that column was equal to 6. Now let's take this row dotted with this column. So it's going to be negative 1 times 1 plus 0 times 1. So all of these guys times 1 plus each other. So minus 1 plus 0 plus 1. That's all 0. So it's going to plus 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So all of these guys times 1 plus each other. So minus 1 plus 0 plus 1. That's all 0. So it's going to plus 2. So it's going to get a 2. I just dotted that guy with that guy. Now I need to take the dot of this guy with this column."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to plus 2. So it's going to get a 2. I just dotted that guy with that guy. Now I need to take the dot of this guy with this column. So it's just going to be 1 times minus 1 plus 1 times 0 plus 1 times 1 plus 1 times 2. Well, these are all 1 times everything. So it's minus 1 plus 0 plus 1, which is 0 plus 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now I need to take the dot of this guy with this column. So it's just going to be 1 times minus 1 plus 1 times 0 plus 1 times 1 plus 1 times 2. Well, these are all 1 times everything. So it's minus 1 plus 0 plus 1, which is 0 plus 2. So it's going to be 2. And then finally, I think you see some symmetry here. We're going to have to take the dot of this guy and this guy over here."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's minus 1 plus 0 plus 1, which is 0 plus 2. So it's going to be 2. And then finally, I think you see some symmetry here. We're going to have to take the dot of this guy and this guy over here. So what is that? That's 1 times 1, which is 1, plus 1 times 1, which is 2, plus 1 times 1. So we're going to have 1 plus itself 4 times."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We're going to have to take the dot of this guy and this guy over here. So what is that? That's 1 times 1, which is 1, plus 1 times 1, which is 2, plus 1 times 1. So we're going to have 1 plus itself 4 times. So we're going to get it's equal to 4. So this is A transpose A. And let's figure out what A transpose B looks like."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to have 1 plus itself 4 times. So we're going to get it's equal to 4. So this is A transpose A. And let's figure out what A transpose B looks like. Let's scroll down a little bit. So A transpose is this matrix again. Let me switch colors."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And let's figure out what A transpose B looks like. Let's scroll down a little bit. So A transpose is this matrix again. Let me switch colors. Minus 1, 0, 1, 2. We get all of our 1's just like that. And then the matrix B is 0, 1, 2, 1."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me switch colors. Minus 1, 0, 1, 2. We get all of our 1's just like that. And then the matrix B is 0, 1, 2, 1. We have a 2 by 4 times a 4 by 1. So we're just going to get a 2 by 1 matrix. So this is going to be equal to a 2 by 1 matrix."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then the matrix B is 0, 1, 2, 1. We have a 2 by 4 times a 4 by 1. So we're just going to get a 2 by 1 matrix. So this is going to be equal to a 2 by 1 matrix. We have here, see, minus 1 times 0 is 0, plus 0 times 1 is still 0. Plus 1 times 2, which is 2, plus 2 times 1, which is 4. This is 2 plus 2, so it's going to be 4 right there."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to a 2 by 1 matrix. We have here, see, minus 1 times 0 is 0, plus 0 times 1 is still 0. Plus 1 times 2, which is 2, plus 2 times 1, which is 4. This is 2 plus 2, so it's going to be 4 right there. And then we have 1 times 0, plus 1 times 2, plus 1. So 1 times all of these guys added up. So 0 plus 1 is 1."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is 2 plus 2, so it's going to be 4 right there. And then we have 1 times 0, plus 1 times 2, plus 1. So 1 times all of these guys added up. So 0 plus 1 is 1. 1 plus 2 is 3. 3 plus 1 is 4. So this right here is A transpose B."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 0 plus 1 is 1. 1 plus 2 is 3. 3 plus 1 is 4. So this right here is A transpose B. So just like that, we know that the least square solution will be the solution to this system. 6, 2, 2, 4 times our least squares solution is going to be equal to 4, 4. Or we could write it this way."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is A transpose B. So just like that, we know that the least square solution will be the solution to this system. 6, 2, 2, 4 times our least squares solution is going to be equal to 4, 4. Or we could write it this way. We could write it 6, 2, 2, 4 times our least squares solution, which I'll write. Remember, the first entry was m. I'll write it as m star. That's our least square m. And this is our least square B is equal to 4, 4."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or we could write it this way. We could write it 6, 2, 2, 4 times our least squares solution, which I'll write. Remember, the first entry was m. I'll write it as m star. That's our least square m. And this is our least square B is equal to 4, 4. And I could do this as an augmented matrix. Or I could just write this as a system of 2 unknowns, which is actually probably easier. So let's do it that way."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's our least square m. And this is our least square B is equal to 4, 4. And I could do this as an augmented matrix. Or I could just write this as a system of 2 unknowns, which is actually probably easier. So let's do it that way. So this, if I were to write it as a system of equations, is 6 times m star plus 2 times B star is equal to 4. And then I get 2 times m star plus 4 times B star is equal to this 4, is equal to that 4. So let me solve for my m stars and my B stars."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's do it that way. So this, if I were to write it as a system of equations, is 6 times m star plus 2 times B star is equal to 4. And then I get 2 times m star plus 4 times B star is equal to this 4, is equal to that 4. So let me solve for my m stars and my B stars. So let's multiply this second equation. Actually, let's multiply that top equation by 2. This is a straight algebra 1."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let me solve for my m stars and my B stars. So let's multiply this second equation. Actually, let's multiply that top equation by 2. This is a straight algebra 1. So times 2, what do we get? We get 12 m star plus 4 B star is equal to 8. Just multiply that top guy by 2."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is a straight algebra 1. So times 2, what do we get? We get 12 m star plus 4 B star is equal to 8. Just multiply that top guy by 2. Now, let's multiply this magenta one by negative 1. So this becomes a minus. That becomes a minus."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Just multiply that top guy by 2. Now, let's multiply this magenta one by negative 1. So this becomes a minus. That becomes a minus. And now we can add these two equations. So we get minus 2 plus 12 m star, that's 10 m star. And then the minus 4B and the 4B cancel out is equal to 4."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That becomes a minus. And now we can add these two equations. So we get minus 2 plus 12 m star, that's 10 m star. And then the minus 4B and the 4B cancel out is equal to 4. Or m star is equal to 4 over 10, which is equal to 2 fifths. Now we can just go and back substitute into this. We could say 6 times m star."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then the minus 4B and the 4B cancel out is equal to 4. Or m star is equal to 4 over 10, which is equal to 2 fifths. Now we can just go and back substitute into this. We could say 6 times m star. This is just straight algebra 1. So 6 times our m star. So 6 times 2 over 5 plus 2 times our B star is equal to 4."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We could say 6 times m star. This is just straight algebra 1. So 6 times our m star. So 6 times 2 over 5 plus 2 times our B star is equal to 4. That white, let me use yellow. So we get 12 over 5 plus 2B star is equal to 4. Or we could say 2B star, let me scroll down a little bit, 2B star is equal to 4, which is the same thing as 20 over 5 minus 12 over 5, which is equal to, I'm just subtracting the 12 over 5 from both sides, which is equal to 8 over 5."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 6 times 2 over 5 plus 2 times our B star is equal to 4. That white, let me use yellow. So we get 12 over 5 plus 2B star is equal to 4. Or we could say 2B star, let me scroll down a little bit, 2B star is equal to 4, which is the same thing as 20 over 5 minus 12 over 5, which is equal to, I'm just subtracting the 12 over 5 from both sides, which is equal to 8 over 5. And you divide both sides of this equation by 2, you get B star is equal to 4 fifths. And just like that, we got our m star and our B star. Our least squares solution is equal to 2 fifths and 4 fifths."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or we could say 2B star, let me scroll down a little bit, 2B star is equal to 4, which is the same thing as 20 over 5 minus 12 over 5, which is equal to, I'm just subtracting the 12 over 5 from both sides, which is equal to 8 over 5. And you divide both sides of this equation by 2, you get B star is equal to 4 fifths. And just like that, we got our m star and our B star. Our least squares solution is equal to 2 fifths and 4 fifths. So m is equal to 2 fifths and B is equal to 4 fifths. And remember, the whole point of this was to find an equation of the line. y is equal to mx plus B."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Our least squares solution is equal to 2 fifths and 4 fifths. So m is equal to 2 fifths and B is equal to 4 fifths. And remember, the whole point of this was to find an equation of the line. y is equal to mx plus B. Now, we can't find a line that went through all of those points up there, that went through all of these points. But this is going to be our least squares solution. This is the one that minimizes the distance between A times our vector and B."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "y is equal to mx plus B. Now, we can't find a line that went through all of those points up there, that went through all of these points. But this is going to be our least squares solution. This is the one that minimizes the distance between A times our vector and B. No vector. When you multiply it times that matrix A, that's not A, that's A star A transpose A. No other solution is going to give us a closer solution to B than when we put our newly found x star into this equation."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is the one that minimizes the distance between A times our vector and B. No vector. When you multiply it times that matrix A, that's not A, that's A star A transpose A. No other solution is going to give us a closer solution to B than when we put our newly found x star into this equation. This is going to give us our best solution. It's going to minimize the distance to B. So let's write it out."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "No other solution is going to give us a closer solution to B than when we put our newly found x star into this equation. This is going to give us our best solution. It's going to minimize the distance to B. So let's write it out. A is equal to mx plus B. So y is equal to 2 fifths x plus 2 fifths, let's graph that out. y is equal to 2 fifths x plus 2 fifths."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's write it out. A is equal to mx plus B. So y is equal to 2 fifths x plus 2 fifths, let's graph that out. y is equal to 2 fifths x plus 2 fifths. y is equal to 2 fifths x plus 2 fifths. So its y-intercept is 2 fifths, which is about there. This is at 1."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "y is equal to 2 fifths x plus 2 fifths. y is equal to 2 fifths x plus 2 fifths. So its y-intercept is 2 fifths, which is about there. This is at 1. 2 fifths is right about there. And then its slope is 2 fifths. So for every, let's think of it this way, for every 2 and 1 half, you go to the right, you're going to go up 1."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is at 1. 2 fifths is right about there. And then its slope is 2 fifths. So for every, let's think of it this way, for every 2 and 1 half, you go to the right, you're going to go up 1. So if we go 1, 2, and 1 half, we're going to go up 1. We're going to go up 1 like that. So our line, and obviously this isn't precise, but our line is going to look something like this."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So for every, let's think of it this way, for every 2 and 1 half, you go to the right, you're going to go up 1. So if we go 1, 2, and 1 half, we're going to go up 1. We're going to go up 1 like that. So our line, and obviously this isn't precise, but our line is going to look something like this. Our line is going to look something like this. I want to do my best shot at drawing it, because this is the fun part. It's going to look something like that."}, {"video_title": "Another least squares example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So our line, and obviously this isn't precise, but our line is going to look something like this. Our line is going to look something like this. I want to do my best shot at drawing it, because this is the fun part. It's going to look something like that. And that right there is my least squares estimate for a line that goes through all of those points. And you're not going to find a line that minimizes the error in a better way. At least when you measure the error, it's the distance between this vector and the vector a times our least squares estimate."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "So the vectors that have the form, the transformation of my vector is just equal to some scaled up version of the vector. If this doesn't look familiar, I can jog your memory a little bit. When we were looking for basis vectors for the transformation, let me draw it, this was from R2 to R2. From R2 to R2, so let me draw R2 right here. Now let's say I had the vector, let's say v1 was equal to the vector 1, 2, and we had the line spanned by that vector, we did this problem several videos ago. And I had the transformation that flipped across this line. So if you call that line L, T was the transformation from R2 to R2 that flipped vectors across this line."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "From R2 to R2, so let me draw R2 right here. Now let's say I had the vector, let's say v1 was equal to the vector 1, 2, and we had the line spanned by that vector, we did this problem several videos ago. And I had the transformation that flipped across this line. So if you call that line L, T was the transformation from R2 to R2 that flipped vectors across this line. So it flipped vectors across L. So if you remember that transformation, if I had some random vector that looked like that, let's say that's x, that's vector x, then the transformation of x looks something like this. It was just flipped across that line. That was the transformation of x."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "So if you call that line L, T was the transformation from R2 to R2 that flipped vectors across this line. So it flipped vectors across L. So if you remember that transformation, if I had some random vector that looked like that, let's say that's x, that's vector x, then the transformation of x looks something like this. It was just flipped across that line. That was the transformation of x. And if you remember that video, we were looking for a change of basis that would allow us to at least figure out the matrix for the transformation, at least an alternate basis, and then we could figure out the matrix for the transformation in the standard basis. And the basis we picked were basis vectors that didn't get changed much by the transformation, or ones that only got scaled by the transformation. For example, when I took the transformation of v1, it just equaled v1, or we could say that the transformation of v1 just equaled 1 times v1."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "That was the transformation of x. And if you remember that video, we were looking for a change of basis that would allow us to at least figure out the matrix for the transformation, at least an alternate basis, and then we could figure out the matrix for the transformation in the standard basis. And the basis we picked were basis vectors that didn't get changed much by the transformation, or ones that only got scaled by the transformation. For example, when I took the transformation of v1, it just equaled v1, or we could say that the transformation of v1 just equaled 1 times v1. So if you just follow this little format that I set up here, lambda in this case would be 1, and of course the vector in this case is v1. The transformation just scaled up v1 by 1. Now if that same problem, we had the other vector that we also looked at."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "For example, when I took the transformation of v1, it just equaled v1, or we could say that the transformation of v1 just equaled 1 times v1. So if you just follow this little format that I set up here, lambda in this case would be 1, and of course the vector in this case is v1. The transformation just scaled up v1 by 1. Now if that same problem, we had the other vector that we also looked at. Let's say it was the vector minus, let's say it's the vector v2, which is, let's say it's 2 minus 1. And then if you take the transformation of it, since it was orthogonal to the line, it just got flipped over like that. And that was a pretty interesting vector for us as well, because the transformation of v2 in this situation is equal to what?"}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "Now if that same problem, we had the other vector that we also looked at. Let's say it was the vector minus, let's say it's the vector v2, which is, let's say it's 2 minus 1. And then if you take the transformation of it, since it was orthogonal to the line, it just got flipped over like that. And that was a pretty interesting vector for us as well, because the transformation of v2 in this situation is equal to what? Just minus v2. It's equal to minus v2, or you could say that the transformation of v2 is equal to minus 1 times v2. And these were interesting vectors for us, because when we defined a new basis with these guys as the basis vector, it was very easy to figure out our transformation matrix."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "And that was a pretty interesting vector for us as well, because the transformation of v2 in this situation is equal to what? Just minus v2. It's equal to minus v2, or you could say that the transformation of v2 is equal to minus 1 times v2. And these were interesting vectors for us, because when we defined a new basis with these guys as the basis vector, it was very easy to figure out our transformation matrix. And actually that basis was very easy to compute with, and we'll explore that a little bit more in the future. But hopefully you realize that these are interesting vectors. There's also the cases where we had the planes spanned by some vectors, and then we had another vector that was popping out of the plane like that, and we were transforming things by taking the mirror image across this."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "And these were interesting vectors for us, because when we defined a new basis with these guys as the basis vector, it was very easy to figure out our transformation matrix. And actually that basis was very easy to compute with, and we'll explore that a little bit more in the future. But hopefully you realize that these are interesting vectors. There's also the cases where we had the planes spanned by some vectors, and then we had another vector that was popping out of the plane like that, and we were transforming things by taking the mirror image across this. And we're like, well, in that transformation, these red vectors don't change at all, and this guy gets flipped over. So maybe those would make for good basis vectors, and they did. So in general, we're always interested with the vectors that just get scaled up by a transformation."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "There's also the cases where we had the planes spanned by some vectors, and then we had another vector that was popping out of the plane like that, and we were transforming things by taking the mirror image across this. And we're like, well, in that transformation, these red vectors don't change at all, and this guy gets flipped over. So maybe those would make for good basis vectors, and they did. So in general, we're always interested with the vectors that just get scaled up by a transformation. And it's not going to be all vectors, right? This vector that I drew here, this vector x, it doesn't just get scaled up. It actually gets changed."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "So in general, we're always interested with the vectors that just get scaled up by a transformation. And it's not going to be all vectors, right? This vector that I drew here, this vector x, it doesn't just get scaled up. It actually gets changed. Its direction gets changed. The vectors that get scaled up might go from this direction to that direction, or maybe they go from that. Maybe that's x, and then the transformation of x might be a scaled up version of x."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "It actually gets changed. Its direction gets changed. The vectors that get scaled up might go from this direction to that direction, or maybe they go from that. Maybe that's x, and then the transformation of x might be a scaled up version of x. Maybe it's that. But the actual line that they span will not change. And so that's what we're going to concern ourselves with, because these have a special name."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "Maybe that's x, and then the transformation of x might be a scaled up version of x. Maybe it's that. But the actual line that they span will not change. And so that's what we're going to concern ourselves with, because these have a special name. And they have the special name. I want to make this very clear, because they're useful. It's not just some mathematical game we're playing, although sometimes we do fall into that trap."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "And so that's what we're going to concern ourselves with, because these have a special name. And they have the special name. I want to make this very clear, because they're useful. It's not just some mathematical game we're playing, although sometimes we do fall into that trap. But they're actually useful. They're useful for defining bases, because in those bases it's easier to find transformation matrices. They're more natural coordinate systems."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "It's not just some mathematical game we're playing, although sometimes we do fall into that trap. But they're actually useful. They're useful for defining bases, because in those bases it's easier to find transformation matrices. They're more natural coordinate systems. And oftentimes the transformation matrices in those bases are easier to compute with. And so these have special names. Any vector that satisfies this right here is called an eigenvector for the transformation T. And the lambda, the multiple that it becomes, this is the eigenvalue associated with that eigenvector."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "They're more natural coordinate systems. And oftentimes the transformation matrices in those bases are easier to compute with. And so these have special names. Any vector that satisfies this right here is called an eigenvector for the transformation T. And the lambda, the multiple that it becomes, this is the eigenvalue associated with that eigenvector. So in the example I just gave where the transformation is flipping around this line, v1, the vector 1, 2, is an eigenvector of our transformation. So 1, 2 is an eigenvector. And its corresponding eigenvalue is 1."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "Any vector that satisfies this right here is called an eigenvector for the transformation T. And the lambda, the multiple that it becomes, this is the eigenvalue associated with that eigenvector. So in the example I just gave where the transformation is flipping around this line, v1, the vector 1, 2, is an eigenvector of our transformation. So 1, 2 is an eigenvector. And its corresponding eigenvalue is 1. So eigenvalue is 1. This guy is also an eigenvector, the vector 2 minus 1, he's also an eigenvector. A very fancy word, but all it means is a vector that's just scaled up by a transformation."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "And its corresponding eigenvalue is 1. So eigenvalue is 1. This guy is also an eigenvector, the vector 2 minus 1, he's also an eigenvector. A very fancy word, but all it means is a vector that's just scaled up by a transformation. It doesn't get changed in any more meaningful way than just a scaling factor. And its corresponding eigenvalue is minus 1. If this transformation, I don't know what its transformation matrix is, I forgot what it was, we actually figured it out a while ago."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "A very fancy word, but all it means is a vector that's just scaled up by a transformation. It doesn't get changed in any more meaningful way than just a scaling factor. And its corresponding eigenvalue is minus 1. If this transformation, I don't know what its transformation matrix is, I forgot what it was, we actually figured it out a while ago. If this transformation matrix, this transformation can be represented as a matrix vector product, and it should be, it's a linear transformation. Then any v that satisfies the transformation of, I'll say transformation of v, is equal to lambda v, which is also would be, the transformation of v would just be a times v. These are also called eigenvectors of A, because A is just really the matrix representation of the transformation. So in this case, this would be an eigenvector of A, and this would be the eigenvalue associated with the eigenvector."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "If this transformation, I don't know what its transformation matrix is, I forgot what it was, we actually figured it out a while ago. If this transformation matrix, this transformation can be represented as a matrix vector product, and it should be, it's a linear transformation. Then any v that satisfies the transformation of, I'll say transformation of v, is equal to lambda v, which is also would be, the transformation of v would just be a times v. These are also called eigenvectors of A, because A is just really the matrix representation of the transformation. So in this case, this would be an eigenvector of A, and this would be the eigenvalue associated with the eigenvector. Eigenvalue associated with the eigenvector. So if you give me a matrix, it represents some linear transformation, you can also figure these things out. Now in the next video, we're actually going to figure out a way to figure these things out, but what I want you to appreciate in this video is that it's easy to say, oh, the vectors don't get changed much, but I want you to understand what that means."}, {"video_title": "Introduction to eigenvalues and eigenvectors Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, this would be an eigenvector of A, and this would be the eigenvalue associated with the eigenvector. Eigenvalue associated with the eigenvector. So if you give me a matrix, it represents some linear transformation, you can also figure these things out. Now in the next video, we're actually going to figure out a way to figure these things out, but what I want you to appreciate in this video is that it's easy to say, oh, the vectors don't get changed much, but I want you to understand what that means. They literally just get scaled up, or maybe they get reversed, but their direction or the lines they span fundamentally don't change. And the reason why they're interesting for us is that they make for interesting basis vectors. Basis vectors whose transformation matrices are maybe computationally more simpler, or ones that make for better coordinate systems."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So let's say v1, v2, all the way through vk. These are the basis vectors, basis orthonormal, maybe I'll write like this, orthonormal basis vectors for v. We saw this in the last video, and that was another reason why we like orthonormal bases. Let's do this with an actual concrete example. So let's say v is equal to the span of the vector, let's say, 1 third, 2 third, and 2 thirds, and the vector 2 thirds, 1 third, and minus 2 thirds. Now, we've already seen that these two guys are linearly independent, and they both have length 1, and they're both orthogonal to each other. So if we say B, so we could say the set, let me write it this way, we'll call this vector 1 for shorthand, if that's v1, and that this is v2, we know that the set of v1 and v2 is an orthonormal basis for v. Now, we want to use this result to find the projection. We want to find the transformation matrix for the projection of any vector x in R, well, in this case, it's going to be in R3, onto our subspace, onto v. And the subspace is going to be a plane in R3."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So let's say v is equal to the span of the vector, let's say, 1 third, 2 third, and 2 thirds, and the vector 2 thirds, 1 third, and minus 2 thirds. Now, we've already seen that these two guys are linearly independent, and they both have length 1, and they're both orthogonal to each other. So if we say B, so we could say the set, let me write it this way, we'll call this vector 1 for shorthand, if that's v1, and that this is v2, we know that the set of v1 and v2 is an orthonormal basis for v. Now, we want to use this result to find the projection. We want to find the transformation matrix for the projection of any vector x in R, well, in this case, it's going to be in R3, onto our subspace, onto v. And the subspace is going to be a plane in R3. So what's it going to be? Well, we found the result that we just have to construct a matrix A, which has these guys as the column vectors. So 1 third, 2 thirds, 2 thirds, and 2 thirds, 1 third, and minus 2 thirds."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "We want to find the transformation matrix for the projection of any vector x in R, well, in this case, it's going to be in R3, onto our subspace, onto v. And the subspace is going to be a plane in R3. So what's it going to be? Well, we found the result that we just have to construct a matrix A, which has these guys as the column vectors. So 1 third, 2 thirds, 2 thirds, and 2 thirds, 1 third, and minus 2 thirds. And if we construct A in that way, then the projection of x onto v, this linear transformation, can be represented as A times A transpose times x. So to find our transformation matrix, we just have to multiply this guy times his transpose. So let's do that."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So 1 third, 2 thirds, 2 thirds, and 2 thirds, 1 third, and minus 2 thirds. And if we construct A in that way, then the projection of x onto v, this linear transformation, can be represented as A times A transpose times x. So to find our transformation matrix, we just have to multiply this guy times his transpose. So let's do that. Let me just copy and paste this. OK, copy. Let me do it right here."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. Let me just copy and paste this. OK, copy. Let me do it right here. Paste it. So that's A. I need to multiply that times A transpose. A transpose is just going to be 1 third, 2 thirds, 2 thirds, 1 third, and then 2 thirds minus 2 thirds."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it right here. Paste it. So that's A. I need to multiply that times A transpose. A transpose is just going to be 1 third, 2 thirds, 2 thirds, 1 third, and then 2 thirds minus 2 thirds. That's A transpose. So what is this going to be equal to? So we have a 3 by 2 times a 2 by 3 matrix."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "A transpose is just going to be 1 third, 2 thirds, 2 thirds, 1 third, and then 2 thirds minus 2 thirds. That's A transpose. So what is this going to be equal to? So we have a 3 by 2 times a 2 by 3 matrix. So it's going to result in a 3 by 3 matrix, which makes sense because this thing right here should be a mapping from R3 to R3. You give me some member of R3, and I'm going to give you another member of R3 that is in my subspace V and is kind of the projection of x onto V. And we've also seen it's the closest member of the 2x. So what is this going to be?"}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So we have a 3 by 2 times a 2 by 3 matrix. So it's going to result in a 3 by 3 matrix, which makes sense because this thing right here should be a mapping from R3 to R3. You give me some member of R3, and I'm going to give you another member of R3 that is in my subspace V and is kind of the projection of x onto V. And we've also seen it's the closest member of the 2x. So what is this going to be? It's going to be a 3 by 3 matrix. And so this first term right here, we're going to dot this guy with this guy. So it's going to be 1 third times 1 third, which is 1 ninth, plus 2 thirds times 2 thirds."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So what is this going to be? It's going to be a 3 by 3 matrix. And so this first term right here, we're going to dot this guy with this guy. So it's going to be 1 third times 1 third, which is 1 ninth, plus 2 thirds times 2 thirds. So it's going to be 1 ninth plus 4 ninths. So I think we're going to be dealing with a lot of ninths here. So let me just divide everything by ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be 1 third times 1 third, which is 1 ninth, plus 2 thirds times 2 thirds. So it's going to be 1 ninth plus 4 ninths. So I think we're going to be dealing with a lot of ninths here. So let me just divide everything by ninths. It's going to be 1 ninth plus 4 ninths, which is 5 ninths. But I'll just write a 5 here and just know that we're going to divide everything by 9 in the end. So that's that guy dotted with that guy."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So let me just divide everything by ninths. It's going to be 1 ninth plus 4 ninths, which is 5 ninths. But I'll just write a 5 here and just know that we're going to divide everything by 9 in the end. So that's that guy dotted with that guy. Now let's take the dot of this guy with this guy. I'm going to get 2 ninths plus 2 ninths. That's 4 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So that's that guy dotted with that guy. Now let's take the dot of this guy with this guy. I'm going to get 2 ninths plus 2 ninths. That's 4 ninths. Now I'm going to dot this guy with this guy. 2 ninths minus 4 ninths. That is minus 2 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "That's 4 ninths. Now I'm going to dot this guy with this guy. 2 ninths minus 4 ninths. That is minus 2 ninths. Now let's take the dot of this row. We're in the second row. And then we're going to have to do the dot of 2 thirds and 1 ninth, that is 2 ninths plus 2 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "That is minus 2 ninths. Now let's take the dot of this row. We're in the second row. And then we're going to have to do the dot of 2 thirds and 1 ninth, that is 2 ninths plus 2 ninths. That is 4 ninths. So we put a 4 there. We have 4 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to have to do the dot of 2 thirds and 1 ninth, that is 2 ninths plus 2 ninths. That is 4 ninths. So we put a 4 there. We have 4 ninths. Then we have 2 ninths plus 1 ninth is 3 ninths. Let me make sure I got that right. Oh no, sorry."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "We have 4 ninths. Then we have 2 ninths plus 1 ninth is 3 ninths. Let me make sure I got that right. Oh no, sorry. 2 thirds times 2 thirds is 4 thirds. So it's 4, sorry. 2 thirds times 2 thirds is 4 ninths plus 1 ninth is 5 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "Oh no, sorry. 2 thirds times 2 thirds is 4 thirds. So it's 4, sorry. 2 thirds times 2 thirds is 4 ninths plus 1 ninth is 5 ninths. And then we have 4 ninths minus 2 ninths is 2 ninths. And then let's do this last one. We're almost there."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "2 thirds times 2 thirds is 4 ninths plus 1 ninth is 5 ninths. And then we have 4 ninths minus 2 ninths is 2 ninths. And then let's do this last one. We're almost there. I hope you already appreciate this is a lot less painful than we had to take A transpose A and then inverse it in between. We're just taking A times A transpose. So 2 thirds times 1 third."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "We're almost there. I hope you already appreciate this is a lot less painful than we had to take A transpose A and then inverse it in between. We're just taking A times A transpose. So 2 thirds times 1 third. So that's 2 ninths minus 4 ninths. So that's minus 2 ninths. And then we have 4 ninths minus 2 ninths."}, {"video_title": "Finding projection onto subspace with orthonormal basis example Linear Algebra Khan Academy.mp3", "Sentence": "So 2 thirds times 1 third. So that's 2 ninths minus 4 ninths. So that's minus 2 ninths. And then we have 4 ninths minus 2 ninths. That's 2 ninths. And then we have 4 ninths plus 4 ninths. So that is 8 ninths."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And it's equal to the span of some set of vectors. And I showed in that video that the span of any set of vectors is a valid subspace. So this is going to be the span of V1, V2, all the way, so it's going to be n vectors. So each of these are vectors. Now let me also say that all of these vectors are linearly independent. So V1, V2, all the way to Vn, this set of vectors are linearly independent. Now before I kind of give you the punchline, let's review what exactly span meant."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So each of these are vectors. Now let me also say that all of these vectors are linearly independent. So V1, V2, all the way to Vn, this set of vectors are linearly independent. Now before I kind of give you the punchline, let's review what exactly span meant. Span meant that this set, this subspace, represents all of the possible linear combinations of all of these vectors. So I could have all of the combinations for all of the different C's. So C1 times V1 plus C2 times V2, all the way to Cn times Vn."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now before I kind of give you the punchline, let's review what exactly span meant. Span meant that this set, this subspace, represents all of the possible linear combinations of all of these vectors. So I could have all of the combinations for all of the different C's. So C1 times V1 plus C2 times V2, all the way to Cn times Vn. For all of the possible C's in the real numbers, if you take all of the possibilities of these and you put all of those vectors into a set, that is the span and that's what we're defining the subspace V as. Now the definition of linear independence meant that the only solution to C1V1 plus C2V2 plus all the way to CnVn, that the only solution to this equaling the zero vector, maybe I should put a little vector sign up there, is when all of these terms are equal to zero. C1 is equal to C2 is equal to all of these Cn, all of them are equal to zero."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So C1 times V1 plus C2 times V2, all the way to Cn times Vn. For all of the possible C's in the real numbers, if you take all of the possibilities of these and you put all of those vectors into a set, that is the span and that's what we're defining the subspace V as. Now the definition of linear independence meant that the only solution to C1V1 plus C2V2 plus all the way to CnVn, that the only solution to this equaling the zero vector, maybe I should put a little vector sign up there, is when all of these terms are equal to zero. C1 is equal to C2 is equal to all of these Cn, all of them are equal to zero. Or kind of a more common sense way to think of it is that you can't represent any one of these vectors as a combination of the other vectors. Now if both of these conditions are true, that this set of vectors, the span of this set of vectors is equal to this subspace, or creates this subspace, or spans this subspace, and that all of these vectors are linearly independent, then we can say that the set of vectors, we can call this set of vectors S, let me say S is equal to V1, V2, all the way to Vn, it's equal to that set of vectors. We can then say, and this is the punchline, we can then say that S, the set S, is a basis for V. And this is the definition I wanted to make."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "C1 is equal to C2 is equal to all of these Cn, all of them are equal to zero. Or kind of a more common sense way to think of it is that you can't represent any one of these vectors as a combination of the other vectors. Now if both of these conditions are true, that this set of vectors, the span of this set of vectors is equal to this subspace, or creates this subspace, or spans this subspace, and that all of these vectors are linearly independent, then we can say that the set of vectors, we can call this set of vectors S, let me say S is equal to V1, V2, all the way to Vn, it's equal to that set of vectors. We can then say, and this is the punchline, we can then say that S, the set S, is a basis for V. And this is the definition I wanted to make. If something is a basis for a set, that means that those vectors, if you take the span of those vectors, you can construct, you can get to any of the vectors in that subspace, and that those vectors are linearly independent. So there's a couple of ways to think about it. One is there's a lot of things that might span for something."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can then say, and this is the punchline, we can then say that S, the set S, is a basis for V. And this is the definition I wanted to make. If something is a basis for a set, that means that those vectors, if you take the span of those vectors, you can construct, you can get to any of the vectors in that subspace, and that those vectors are linearly independent. So there's a couple of ways to think about it. One is there's a lot of things that might span for something. For example, if this spans for V, then so would, let me add another vector, let me define another set. Let me define set T to be all of set S, V1, V2, all the way to Vn, but it also contains this other vector, I'm going to call it the V special vector, so it's going to be essentially the set S plus one more vector, where this vector, I'm just saying, is equal to V1 plus V2. So clearly this is not a linearly independent set, but if I asked you what the span of T is, the span of T is still going to be this subspace, it's still going to be this subspace V. But I have this extra vector in here that made it nonlinearly independent."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "One is there's a lot of things that might span for something. For example, if this spans for V, then so would, let me add another vector, let me define another set. Let me define set T to be all of set S, V1, V2, all the way to Vn, but it also contains this other vector, I'm going to call it the V special vector, so it's going to be essentially the set S plus one more vector, where this vector, I'm just saying, is equal to V1 plus V2. So clearly this is not a linearly independent set, but if I asked you what the span of T is, the span of T is still going to be this subspace, it's still going to be this subspace V. But I have this extra vector in here that made it nonlinearly independent. This set is not linearly independent, so this is linearly, so T is linearly dependent. So in this case, T is not a basis for V. And I showed you this example because the way my head thinks about basis is, the basis is really the minimum set of vectors that I need, the minimum set, and I'll write this down, this isn't a formal definition, but I view a basis, let me switch colors, as really the, let me get a good color here, as a basis is the minimum, I'll put it in quotes because I haven't defined that, the minimum set of vectors that spans the space that it's a basis of, spans the subspace. So in this case, this is the minimum set of vectors, and I'm not going to prove it just yet, but you can see that, look, this set of vectors right here, it does span the subspace, but it's clearly not the minimum set of vectors, because the span of this thing, I could still remove this last vector here, I could still remove that guy, and still, and then the span of what's left over is still going to get me to my subspace, is still going to span my subspace V. So this guy right here was redundant."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So clearly this is not a linearly independent set, but if I asked you what the span of T is, the span of T is still going to be this subspace, it's still going to be this subspace V. But I have this extra vector in here that made it nonlinearly independent. This set is not linearly independent, so this is linearly, so T is linearly dependent. So in this case, T is not a basis for V. And I showed you this example because the way my head thinks about basis is, the basis is really the minimum set of vectors that I need, the minimum set, and I'll write this down, this isn't a formal definition, but I view a basis, let me switch colors, as really the, let me get a good color here, as a basis is the minimum, I'll put it in quotes because I haven't defined that, the minimum set of vectors that spans the space that it's a basis of, spans the subspace. So in this case, this is the minimum set of vectors, and I'm not going to prove it just yet, but you can see that, look, this set of vectors right here, it does span the subspace, but it's clearly not the minimum set of vectors, because the span of this thing, I could still remove this last vector here, I could still remove that guy, and still, and then the span of what's left over is still going to get me to my subspace, is still going to span my subspace V. So this guy right here was redundant. In a basis, you have no redundancy. Each one of these guys is needed to be able to construct any of the vectors in the subspace V. Let me do some examples. So let's just take some vectors here."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, this is the minimum set of vectors, and I'm not going to prove it just yet, but you can see that, look, this set of vectors right here, it does span the subspace, but it's clearly not the minimum set of vectors, because the span of this thing, I could still remove this last vector here, I could still remove that guy, and still, and then the span of what's left over is still going to get me to my subspace, is still going to span my subspace V. So this guy right here was redundant. In a basis, you have no redundancy. Each one of these guys is needed to be able to construct any of the vectors in the subspace V. Let me do some examples. So let's just take some vectors here. Let's say I have, let's say I define my set of vectors, and I'll deal in R2. So let's say I have the vector 2 and 2, 3, and I have the other vector 7, 0. So first of all, let's just think about the span of this set of vectors."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's just take some vectors here. Let's say I have, let's say I define my set of vectors, and I'll deal in R2. So let's say I have the vector 2 and 2, 3, and I have the other vector 7, 0. So first of all, let's just think about the span of this set of vectors. So what's the span of S? What's all of the linear combinations of this? Well, let's see if it's all of R2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So first of all, let's just think about the span of this set of vectors. So what's the span of S? What's all of the linear combinations of this? Well, let's see if it's all of R2. So if it's all of R2, that means the linear combination of this could be, we could always construct anything in R2 with a linear combination of this. So if we have C1 times 2, 3 plus C2 times 7, 0. If it is true that this spans all of R2, then we should be able to construct, we should always be able to find a C1 and a C2 to construct any point in R2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, let's see if it's all of R2. So if it's all of R2, that means the linear combination of this could be, we could always construct anything in R2 with a linear combination of this. So if we have C1 times 2, 3 plus C2 times 7, 0. If it is true that this spans all of R2, then we should be able to construct, we should always be able to find a C1 and a C2 to construct any point in R2. And let's see if we can show that. So we get 2C1 plus 7C2 is equal to X1. And then we get 3C1 plus 0C2, plus 0 is equal to X2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If it is true that this spans all of R2, then we should be able to construct, we should always be able to find a C1 and a C2 to construct any point in R2. And let's see if we can show that. So we get 2C1 plus 7C2 is equal to X1. And then we get 3C1 plus 0C2, plus 0 is equal to X2. And if we take this second equation and divide both sides by 3, we get C1 is equal to X2 over 3. And if we substitute that back into this first equation, we get 2 thirds, 2 thirds, I'm just substituting C1 in there, so 2 thirds X2, right? 2 times X2 over 3 is 2 thirds X2, plus 7C2 is equal to X1."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we get 3C1 plus 0C2, plus 0 is equal to X2. And if we take this second equation and divide both sides by 3, we get C1 is equal to X2 over 3. And if we substitute that back into this first equation, we get 2 thirds, 2 thirds, I'm just substituting C1 in there, so 2 thirds X2, right? 2 times X2 over 3 is 2 thirds X2, plus 7C2 is equal to X1. And then what can we do? We can subtract the 2 thirds X2 from both sides. Let me do it right here."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times X2 over 3 is 2 thirds X2, plus 7C2 is equal to X1. And then what can we do? We can subtract the 2 thirds X2 from both sides. Let me do it right here. So we get 7C2 is equal to X1 minus 2 thirds X2. Divide both sides of this by 7 and you get C2, let me do it in yellow, you get C2 is equal to X1 over 7 minus 2 over 21 X2. So if you give me any X1 and any X2, where either of, you know, X1 or X2 are a member of the real numbers, we're talking about, everything we're going to be dealing with right now is real numbers, you give me any two real numbers, I just put them into, I take my X2 divided by 3 and I'll give you your C1, and I take the X1 divided by 7 and subtract 2 over 21 times your X2, and I'll get you your C2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it right here. So we get 7C2 is equal to X1 minus 2 thirds X2. Divide both sides of this by 7 and you get C2, let me do it in yellow, you get C2 is equal to X1 over 7 minus 2 over 21 X2. So if you give me any X1 and any X2, where either of, you know, X1 or X2 are a member of the real numbers, we're talking about, everything we're going to be dealing with right now is real numbers, you give me any two real numbers, I just put them into, I take my X2 divided by 3 and I'll give you your C1, and I take the X1 divided by 7 and subtract 2 over 21 times your X2, and I'll get you your C2. This will never break, there's no division by any of these, so you don't have to worry about these equaling a zero. These two formulas will always work. So you give me any X1 and any X2, I can always find you a C1 or a C2, which is essentially finding a linear combination that will equal your vector."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you give me any X1 and any X2, where either of, you know, X1 or X2 are a member of the real numbers, we're talking about, everything we're going to be dealing with right now is real numbers, you give me any two real numbers, I just put them into, I take my X2 divided by 3 and I'll give you your C1, and I take the X1 divided by 7 and subtract 2 over 21 times your X2, and I'll get you your C2. This will never break, there's no division by any of these, so you don't have to worry about these equaling a zero. These two formulas will always work. So you give me any X1 and any X2, I can always find you a C1 or a C2, which is essentially finding a linear combination that will equal your vector. So the span of S is R2. Now, the second question is, are these two vectors linearly independent? Linear independence means that the only solution to the equation C, let me switch colors, the only solution to the equation C1 times the first vector plus C2 times the second vector, equaling the zero vector, that the only solution to this is when both of these equal zero."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you give me any X1 and any X2, I can always find you a C1 or a C2, which is essentially finding a linear combination that will equal your vector. So the span of S is R2. Now, the second question is, are these two vectors linearly independent? Linear independence means that the only solution to the equation C, let me switch colors, the only solution to the equation C1 times the first vector plus C2 times the second vector, equaling the zero vector, that the only solution to this is when both of these equal zero. So let's see if that's true. We've already solved for it. So if X1, in this case X1 is equal to zero and X2 is equal to zero, this is just a special case where I'm making them equal to zero vector."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Linear independence means that the only solution to the equation C, let me switch colors, the only solution to the equation C1 times the first vector plus C2 times the second vector, equaling the zero vector, that the only solution to this is when both of these equal zero. So let's see if that's true. We've already solved for it. So if X1, in this case X1 is equal to zero and X2 is equal to zero, this is just a special case where I'm making them equal to zero vector. If I want to get the zero vector, C1 is equal to zero over 3, so C1 must be equal to zero, and C2 is equal to zero over 7 minus 221 times zero, so C2 must also be equal to zero. So the only solution to this was setting both of these guys equal to zero. So these two vectors, so S is also a linearly independent set."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if X1, in this case X1 is equal to zero and X2 is equal to zero, this is just a special case where I'm making them equal to zero vector. If I want to get the zero vector, C1 is equal to zero over 3, so C1 must be equal to zero, and C2 is equal to zero over 7 minus 221 times zero, so C2 must also be equal to zero. So the only solution to this was setting both of these guys equal to zero. So these two vectors, so S is also a linearly independent set. So it spans R2, it's linearly independent, so we can say definitively that the set S, the set of vectors S, is a basis for R2. Now, is this the only basis for R2? Well, I could draw a trivially simple set of vectors."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So these two vectors, so S is also a linearly independent set. So it spans R2, it's linearly independent, so we can say definitively that the set S, the set of vectors S, is a basis for R2. Now, is this the only basis for R2? Well, I could draw a trivially simple set of vectors. I could do this one. Let me call it T. If I define T to be the set 1, 0 and 0, 1, does this span R2? Let's say I want to generate the, I want to get to X1 and X2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I could draw a trivially simple set of vectors. I could do this one. Let me call it T. If I define T to be the set 1, 0 and 0, 1, does this span R2? Let's say I want to generate the, I want to get to X1 and X2. How can I construct that out of these two vectors? Well, if I always just do X1 times 1, 0 plus X2 times 0, 1, that will always give me X1, X2. So this definitely does span R2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I want to generate the, I want to get to X1 and X2. How can I construct that out of these two vectors? Well, if I always just do X1 times 1, 0 plus X2 times 0, 1, that will always give me X1, X2. So this definitely does span R2. Is it linearly independent? I could show it to you. If you wanted to make this equal to the zero vector, if this is a zero and this is a zero, then this has to be a zero and this has to be a zero."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this definitely does span R2. Is it linearly independent? I could show it to you. If you wanted to make this equal to the zero vector, if this is a zero and this is a zero, then this has to be a zero and this has to be a zero. And that's kind of obvious. You could get one of the other vectors by some multiple of the other one. There's no way you could get a 1 here by multiplying this by anything, and vice versa."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you wanted to make this equal to the zero vector, if this is a zero and this is a zero, then this has to be a zero and this has to be a zero. And that's kind of obvious. You could get one of the other vectors by some multiple of the other one. There's no way you could get a 1 here by multiplying this by anything, and vice versa. So it's also linearly independent. And the whole reason why I showed you this is because I wanted to show you that this set T spans R2. It's also linearly independent."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's no way you could get a 1 here by multiplying this by anything, and vice versa. So it's also linearly independent. And the whole reason why I showed you this is because I wanted to show you that this set T spans R2. It's also linearly independent. So T is also a basis for R2. And I wanted to show you this to show that if I look at a vector subspace, and R2 is a valid subspace of itself, you can verify that. But if I have a subspace, it doesn't have just one basis."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's also linearly independent. So T is also a basis for R2. And I wanted to show you this to show that if I look at a vector subspace, and R2 is a valid subspace of itself, you can verify that. But if I have a subspace, it doesn't have just one basis. It could have multiple bases. In fact, it normally has infinite bases. So in this case, S is a valid basis and T is also a valid basis for R2."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if I have a subspace, it doesn't have just one basis. It could have multiple bases. In fact, it normally has infinite bases. So in this case, S is a valid basis and T is also a valid basis for R2. And actually, just so you know what T is, the situation here, this is called a standard basis. And this is the standard basis. And this is what you're used to dealing with in just regular calculus or physics class."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, S is a valid basis and T is also a valid basis for R2. And actually, just so you know what T is, the situation here, this is called a standard basis. And this is the standard basis. And this is what you're used to dealing with in just regular calculus or physics class. If you remember from physics class, this is the unit vector I, and then this is the unit vector J. And it's the standard basis for two-dimensional Cartesian coordinates. And what's useful about a basis is that you can always, and it's not just true of the standard basis, is that you can represent any vector in your subspace by some unique combination of the vectors in your basis."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is what you're used to dealing with in just regular calculus or physics class. If you remember from physics class, this is the unit vector I, and then this is the unit vector J. And it's the standard basis for two-dimensional Cartesian coordinates. And what's useful about a basis is that you can always, and it's not just true of the standard basis, is that you can represent any vector in your subspace by some unique combination of the vectors in your basis. So let me show you that. So let's say that the set V1, V2, all the way to Vn, let's say that this is a basis for some subspace U. So that means that these guys are linearly independent, and it also means that the span of these guys, or all of the linear combinations of these vectors, will get you all of the vectors, all of the possible components, all of the different members of U."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what's useful about a basis is that you can always, and it's not just true of the standard basis, is that you can represent any vector in your subspace by some unique combination of the vectors in your basis. So let me show you that. So let's say that the set V1, V2, all the way to Vn, let's say that this is a basis for some subspace U. So that means that these guys are linearly independent, and it also means that the span of these guys, or all of the linear combinations of these vectors, will get you all of the vectors, all of the possible components, all of the different members of U. Now what I want to show you is each member of U can only be uniquely defined by a unique combination of these guys. Let me be clear about that. Let's say my vector A is a member of our subspace U."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that means that these guys are linearly independent, and it also means that the span of these guys, or all of the linear combinations of these vectors, will get you all of the vectors, all of the possible components, all of the different members of U. Now what I want to show you is each member of U can only be uniquely defined by a unique combination of these guys. Let me be clear about that. Let's say my vector A is a member of our subspace U. That means that A can be represented by some linear combination of these guys. These guys span U. So that means that we can represent our vector A as being C1 times V1 plus C2 times V2, these are vectors, all the way to Cn times Vn."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say my vector A is a member of our subspace U. That means that A can be represented by some linear combination of these guys. These guys span U. So that means that we can represent our vector A as being C1 times V1 plus C2 times V2, these are vectors, all the way to Cn times Vn. Now I want to show you that this is a unique combination. To show that, I'm going to prove contradiction. Let's say that there's another combination."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that means that we can represent our vector A as being C1 times V1 plus C2 times V2, these are vectors, all the way to Cn times Vn. Now I want to show you that this is a unique combination. To show that, I'm going to prove contradiction. Let's say that there's another combination. Let's say I could also represent A by some other combination, D1 times V1 plus D2 times V2 plus all the way to Dn times Vn. Now, what happens if I subtract A from A? I'm going to get the zero vector."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that there's another combination. Let's say I could also represent A by some other combination, D1 times V1 plus D2 times V2 plus all the way to Dn times Vn. Now, what happens if I subtract A from A? I'm going to get the zero vector. Let me subtract these two things. If I subtract A from A, A minus A is clearly the zero vector. If I subtract this side from that side, what do we get?"}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to get the zero vector. Let me subtract these two things. If I subtract A from A, A minus A is clearly the zero vector. If I subtract this side from that side, what do we get? I'll do it in a different color. We get C1 minus D1 times V1 plus C2 minus D2 times V2, all the way to Cn minus Vn. The zero vector is equal to C1 minus D1 times V1 plus all the way up to Cn minus Dn times Vn."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I subtract this side from that side, what do we get? I'll do it in a different color. We get C1 minus D1 times V1 plus C2 minus D2 times V2, all the way to Cn minus Vn. The zero vector is equal to C1 minus D1 times V1 plus all the way up to Cn minus Dn times Vn. I just subtracted the vector from by itself. Now, I told you that these are a basis. There's two things when you say a basis."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The zero vector is equal to C1 minus D1 times V1 plus all the way up to Cn minus Dn times Vn. I just subtracted the vector from by itself. Now, I told you that these are a basis. There's two things when you say a basis. It says that the span of these guys makes this subspace, or the span of these guys is this subspace, and also tells you that these guys are linearly independent. If they're linearly independent, the only solution to this equation, this is just a constant times V1 plus another constant times V2 all the way to a constant times Vn. The only solution to this equation is if each of these constants equals zero."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's two things when you say a basis. It says that the span of these guys makes this subspace, or the span of these guys is this subspace, and also tells you that these guys are linearly independent. If they're linearly independent, the only solution to this equation, this is just a constant times V1 plus another constant times V2 all the way to a constant times Vn. The only solution to this equation is if each of these constants equals zero. All of those constants have to be equal to zero. Over here, before it messed up, this has to equal zero, this has to equal zero. That was the definition of linear independence."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The only solution to this equation is if each of these constants equals zero. All of those constants have to be equal to zero. Over here, before it messed up, this has to equal zero, this has to equal zero. That was the definition of linear independence. We know that this is a linearly independent set. If all of those constants are equal to zero, then we know that if this is equal to zero, then C1 is equal to D1, C2 is equal to D2, all the way to Cn is equal to Dn. By the fact that it's linearly independent, each of these constants have to be equal to each other."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That was the definition of linear independence. We know that this is a linearly independent set. If all of those constants are equal to zero, then we know that if this is equal to zero, then C1 is equal to D1, C2 is equal to D2, all the way to Cn is equal to Dn. By the fact that it's linearly independent, each of these constants have to be equal to each other. That's our contradiction. I assume they're different, but the linear independence forced them to be the same. If you have a basis for some subspace, any member of that subspace can be uniquely determined by a unique combination of those vectors."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "By the fact that it's linearly independent, each of these constants have to be equal to each other. That's our contradiction. I assume they're different, but the linear independence forced them to be the same. If you have a basis for some subspace, any member of that subspace can be uniquely determined by a unique combination of those vectors. Just to hit the point home, I told you that this was a basis for R2. My next question is, and I just want to backtrack a bit, if I just added another vector here, if I just added the vector 1, 0, is S now a basis for R2? No."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you have a basis for some subspace, any member of that subspace can be uniquely determined by a unique combination of those vectors. Just to hit the point home, I told you that this was a basis for R2. My next question is, and I just want to backtrack a bit, if I just added another vector here, if I just added the vector 1, 0, is S now a basis for R2? No. It clearly will continue to span R2, but this guy is redundant. This guy is in R2, and I already told you that these two guys alone span R2, that anything in R2 can be represented by a linear combination of these two guys. This guy is clearly in R2, so he can be represented by a linear combination of these two guys."}, {"video_title": "Basis of a subspace Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "No. It clearly will continue to span R2, but this guy is redundant. This guy is in R2, and I already told you that these two guys alone span R2, that anything in R2 can be represented by a linear combination of these two guys. This guy is clearly in R2, so he can be represented by a linear combination of these two guys. Therefore, this is not a linearly independent set. This is linearly dependent. Because it's linearly dependent, I have redundant information here, and then this would no longer be a basis."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "It's an n by k matrix. And I have the equation Ax is equal to b. So in this case, x would have to be a member of Rk, because we have k columns here. And b is a member of Rn. Now, let's say that it just so happens that there is no solution. So there's no solution to Ax is equal to b. What does that mean?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And b is a member of Rn. Now, let's say that it just so happens that there is no solution. So there's no solution to Ax is equal to b. What does that mean? Let's just expand out A just to, I think you already know what that means. If I write A like this, A1, A2, if I just write it as its column vectors right there, all the way through Ak, and then I multiply it times x1, x2, all the way through xk, this is the same thing as that equation there. I just kind of wrote out the two matrices."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "What does that mean? Let's just expand out A just to, I think you already know what that means. If I write A like this, A1, A2, if I just write it as its column vectors right there, all the way through Ak, and then I multiply it times x1, x2, all the way through xk, this is the same thing as that equation there. I just kind of wrote out the two matrices. Now, this is the same thing as x1 times A1 plus x2 times A2, all the way to plus xk times Ak is equal to the vector b. Now, if this has no solution, then that means that there's no set of weights here on the column vectors of A where we can get to b. Or another way to say it is no linear combinations of the column vectors of A will be equal to b."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "I just kind of wrote out the two matrices. Now, this is the same thing as x1 times A1 plus x2 times A2, all the way to plus xk times Ak is equal to the vector b. Now, if this has no solution, then that means that there's no set of weights here on the column vectors of A where we can get to b. Or another way to say it is no linear combinations of the column vectors of A will be equal to b. Or an even further way of saying it is that b is not in the column space of A. No linear combination of these guys can equal to that. So let's see if we can visualize it a bit."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is no linear combinations of the column vectors of A will be equal to b. Or an even further way of saying it is that b is not in the column space of A. No linear combination of these guys can equal to that. So let's see if we can visualize it a bit. So let me draw the column space of A. So maybe the column space of A looks something like this right here. I'll just assume it's a plane in Rn."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can visualize it a bit. So let me draw the column space of A. So maybe the column space of A looks something like this right here. I'll just assume it's a plane in Rn. It doesn't have to be a plane. Things can be very general. But let's say that this is the column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "I'll just assume it's a plane in Rn. It doesn't have to be a plane. Things can be very general. But let's say that this is the column space. This is the column space of A. Now, if that's the column space and b is not in the column space, maybe we can draw b like this. Maybe b, let's say this is the origin right there, and b just pops out right there."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "But let's say that this is the column space. This is the column space of A. Now, if that's the column space and b is not in the column space, maybe we can draw b like this. Maybe b, let's say this is the origin right there, and b just pops out right there. So this is the 0 vector, this is my vector b. Clearly not in my column space. It's clearly not in this plane."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Maybe b, let's say this is the origin right there, and b just pops out right there. So this is the 0 vector, this is my vector b. Clearly not in my column space. It's clearly not in this plane. Now, up until now, we would get an equation like that. We would make an augmented matrix, put it in reduced row echelon form, and get a line that said 0 equals 1. And we'd say no solution, nothing we can do here."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "It's clearly not in this plane. Now, up until now, we would get an equation like that. We would make an augmented matrix, put it in reduced row echelon form, and get a line that said 0 equals 1. And we'd say no solution, nothing we can do here. But what if we can do better? We clearly can't find a solution to this. But what if we can find a solution that gets us close to this?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And we'd say no solution, nothing we can do here. But what if we can do better? We clearly can't find a solution to this. But what if we can find a solution that gets us close to this? So what if I want to find some x, I'll call it x star for now, where A times x star is, and this is a vector, is as close as possible to b as possible. Or another way to view it, when I say close, I'm talking about length. So I want to minimize the length of b minus A times x star."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "But what if we can find a solution that gets us close to this? So what if I want to find some x, I'll call it x star for now, where A times x star is, and this is a vector, is as close as possible to b as possible. Or another way to view it, when I say close, I'm talking about length. So I want to minimize the length of b minus A times x star. Now, some of you all might already know where this is going. But when you take the difference between 2 and then take its length, what does that look like? Let me just call Ax."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So I want to minimize the length of b minus A times x star. Now, some of you all might already know where this is going. But when you take the difference between 2 and then take its length, what does that look like? Let me just call Ax. Ax is going to be a member of my column space. Let me just call that v. Ax is equal to v. Since we're taking some linear, you multiply any vector in Rk times your matrix A, you're going to get a member of your column space. So any Ax is going to be in your column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Let me just call Ax. Ax is going to be a member of my column space. Let me just call that v. Ax is equal to v. Since we're taking some linear, you multiply any vector in Rk times your matrix A, you're going to get a member of your column space. So any Ax is going to be in your column space. So maybe that is the vector v is equal to A times x star. We want this vector to get as close as possible to this as long as it stays, I mean it has to be in my column space. But we want the distance between this vector and this vector to be minimized."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So any Ax is going to be in your column space. So maybe that is the vector v is equal to A times x star. We want this vector to get as close as possible to this as long as it stays, I mean it has to be in my column space. But we want the distance between this vector and this vector to be minimized. I just want to show you where the terminology for this will come from. I haven't given it its proper title yet. If you were to take this vector, I'm saying that this is, let's just call this vector v for simplicity."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "But we want the distance between this vector and this vector to be minimized. I just want to show you where the terminology for this will come from. I haven't given it its proper title yet. If you were to take this vector, I'm saying that this is, let's just call this vector v for simplicity. This is equivalent to the length of the vector. You take the difference between each of the elements. So b1 minus v1, b2 minus v2, all the way to bn minus vn."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "If you were to take this vector, I'm saying that this is, let's just call this vector v for simplicity. This is equivalent to the length of the vector. You take the difference between each of the elements. So b1 minus v1, b2 minus v2, all the way to bn minus vn. And if you take the length of this vector, this is the same thing as this. This is going to be equal to the square root. Let me take the length squared, actually."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So b1 minus v1, b2 minus v2, all the way to bn minus vn. And if you take the length of this vector, this is the same thing as this. This is going to be equal to the square root. Let me take the length squared, actually. The length squared of this is just going to be b1 minus v1 squared plus b2 minus v2 squared plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the least value that it can be possible."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Let me take the length squared, actually. The length squared of this is just going to be b1 minus v1 squared plus b2 minus v2 squared plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the least value that it can be possible. I want to get the least squares estimate here. And that's why I'll have this last minute or two when I was just explaining this. That was just to give you the motivation for why this right here is called the least squares estimate."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So I want to make this value the least value that it can be possible. I want to get the least squares estimate here. And that's why I'll have this last minute or two when I was just explaining this. That was just to give you the motivation for why this right here is called the least squares estimate. Least squares estimate. Or the least squares solution. Or the least squares approximation for the equation Ax equals b."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "That was just to give you the motivation for why this right here is called the least squares estimate. Least squares estimate. Or the least squares solution. Or the least squares approximation for the equation Ax equals b. There is no solution to this. But maybe we can find some x star where if I multiply a times x star, this is clearly going to be in my column space and I want to get this vector to be as close to b as possible. Now, we've already seen in several videos what is the closest vector in any subspace to a vector that's not in my subspace."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Or the least squares approximation for the equation Ax equals b. There is no solution to this. But maybe we can find some x star where if I multiply a times x star, this is clearly going to be in my column space and I want to get this vector to be as close to b as possible. Now, we've already seen in several videos what is the closest vector in any subspace to a vector that's not in my subspace. Well, the closest vector to it is the projection. The closest vector to b that's in my subspace is going to be the projection of b onto my column space. That is the closest vector there."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Now, we've already seen in several videos what is the closest vector in any subspace to a vector that's not in my subspace. Well, the closest vector to it is the projection. The closest vector to b that's in my subspace is going to be the projection of b onto my column space. That is the closest vector there. So if I want to minimize this, I'm essentially, I want to figure out my x star where Ax star is equal to the projection of my vector b onto my subspace or onto the column space of a. Remember what we're doing here. We said Axb has no solution."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "That is the closest vector there. So if I want to minimize this, I'm essentially, I want to figure out my x star where Ax star is equal to the projection of my vector b onto my subspace or onto the column space of a. Remember what we're doing here. We said Axb has no solution. But maybe we can find some x that gets us as close as possible. So I'm calling that my least squares solution or my least squares approximation. And this guy right here is clearly going to be in my column space, right?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "We said Axb has no solution. But maybe we can find some x that gets us as close as possible. So I'm calling that my least squares solution or my least squares approximation. And this guy right here is clearly going to be in my column space, right? Because you take some vector x times a, that's going to be a linear combination of these column vectors. So it's going to be in the column space. And I want this guy to be as close as possible to this guy."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And this guy right here is clearly going to be in my column space, right? Because you take some vector x times a, that's going to be a linear combination of these column vectors. So it's going to be in the column space. And I want this guy to be as close as possible to this guy. It would be as close as possible to that guy. Well, the closest vector in my column space to that guy is the projection. So Ax needs to be equal to the projection of b on my column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And I want this guy to be as close as possible to this guy. It would be as close as possible to that guy. Well, the closest vector in my column space to that guy is the projection. So Ax needs to be equal to the projection of b on my column space. It needs to be equal to that. But this is still pretty hard to find. To find that you saw how you take a times the inverse of a transpose a times a transpose."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So Ax needs to be equal to the projection of b on my column space. It needs to be equal to that. But this is still pretty hard to find. To find that you saw how you take a times the inverse of a transpose a times a transpose. It's hard to find that transformation matrix. So let's see if we can find an easier way to figure out the least squares solution or kind of our best solution. It's not the solution."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "To find that you saw how you take a times the inverse of a transpose a times a transpose. It's hard to find that transformation matrix. So let's see if we can find an easier way to figure out the least squares solution or kind of our best solution. It's not the solution. It's our best solution to this right here. That's what we call it the least squares solution or approximation. So let's just subtract b from both sides of this and we might get something interesting."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "It's not the solution. It's our best solution to this right here. That's what we call it the least squares solution or approximation. So let's just subtract b from both sides of this and we might get something interesting. So what happens if we take Ax minus the vector b on both sides of this equation? I'll do it up here on the right. So if we, on the left-hand side, we get a times x star."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So let's just subtract b from both sides of this and we might get something interesting. So what happens if we take Ax minus the vector b on both sides of this equation? I'll do it up here on the right. So if we, on the left-hand side, we get a times x star. It's hard to write the x and then the star because they're very similar. And then we subtract b from it. We subtract our vector b."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So if we, on the left-hand side, we get a times x star. It's hard to write the x and then the star because they're very similar. And then we subtract b from it. We subtract our vector b. That's going to be equal to the projection of b onto our column space minus b. All I did is I subtracted b from both sides of this equation. Now, what is the projection of b minus our vector b?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "We subtract our vector b. That's going to be equal to the projection of b onto our column space minus b. All I did is I subtracted b from both sides of this equation. Now, what is the projection of b minus our vector b? If we draw it right here, it's going to be this vector, right? Let me do it in this orange color. It's going to be this right here."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is the projection of b minus our vector b? If we draw it right here, it's going to be this vector, right? Let me do it in this orange color. It's going to be this right here. It's going to be that vector right there. If I take the projection of b, which is that, minus b, I'm going to get this vector. You could say b plus this vector is equal to my projection of b onto my subspace."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be this right here. It's going to be that vector right there. If I take the projection of b, which is that, minus b, I'm going to get this vector. You could say b plus this vector is equal to my projection of b onto my subspace. So this vector right here is orthogonal. It's actually part of the definition of a projection. That this guy is going to be orthogonal to my subspace or to my column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "You could say b plus this vector is equal to my projection of b onto my subspace. So this vector right here is orthogonal. It's actually part of the definition of a projection. That this guy is going to be orthogonal to my subspace or to my column space. And so this guy is orthogonal to my column space. So I could write ax star minus b. It's orthogonal to my column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "That this guy is going to be orthogonal to my subspace or to my column space. And so this guy is orthogonal to my column space. So I could write ax star minus b. It's orthogonal to my column space. Or we could say it's a member of the orthogonal complement of my column space. The orthogonal complement is just the set of everything, all of the vectors that are orthogonal to everything in your subspace, in your column space right here. So this vector right here, this kind of pointing straight down onto my plane, is clearly a member of the orthogonal complement of my column space."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "It's orthogonal to my column space. Or we could say it's a member of the orthogonal complement of my column space. The orthogonal complement is just the set of everything, all of the vectors that are orthogonal to everything in your subspace, in your column space right here. So this vector right here, this kind of pointing straight down onto my plane, is clearly a member of the orthogonal complement of my column space. Now, this might look familiar to you already. What is the orthogonal complement of my column space? The orthogonal complement of my column space is equal to the null space of A transpose."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So this vector right here, this kind of pointing straight down onto my plane, is clearly a member of the orthogonal complement of my column space. Now, this might look familiar to you already. What is the orthogonal complement of my column space? The orthogonal complement of my column space is equal to the null space of A transpose. Or the left null space of A. We've done this in many, many videos. So we can say that A times my least squares estimate of the equation ax is equal to b. I wrote that."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "The orthogonal complement of my column space is equal to the null space of A transpose. Or the left null space of A. We've done this in many, many videos. So we can say that A times my least squares estimate of the equation ax is equal to b. I wrote that. So x star is my least squares solution to ax is equal to b. So A times that minus b is a member of the null space of A transpose. Now, what does that mean?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So we can say that A times my least squares estimate of the equation ax is equal to b. I wrote that. So x star is my least squares solution to ax is equal to b. So A times that minus b is a member of the null space of A transpose. Now, what does that mean? Well, that means that if I multiply A transpose times this guy right here, times ax star. And I don't want to lose the vector signs there on the x. This is a vector."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Now, what does that mean? Well, that means that if I multiply A transpose times this guy right here, times ax star. And I don't want to lose the vector signs there on the x. This is a vector. Don't want to forget that. Ax star minus b. So if I multiply A transpose times this right there, that is the same thing as that."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "This is a vector. Don't want to forget that. Ax star minus b. So if I multiply A transpose times this right there, that is the same thing as that. What am I going to get? Well, this is a member of the null space of A transpose. So this times A transpose has got to be equal to 0."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So if I multiply A transpose times this right there, that is the same thing as that. What am I going to get? Well, this is a member of the null space of A transpose. So this times A transpose has got to be equal to 0. It is a solution to A transpose times something is equal to the 0 vector. Now, let's see if we can simplify this a little bit. We get A transpose A times x star minus A transpose b is equal to 0."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So this times A transpose has got to be equal to 0. It is a solution to A transpose times something is equal to the 0 vector. Now, let's see if we can simplify this a little bit. We get A transpose A times x star minus A transpose b is equal to 0. And then if we add this term to both sides of the equation, we are left with A transpose A times the least squares solution to Ax equal to b is equal to A transpose b. That's what we get. Now, why did we do all of this work?"}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "We get A transpose A times x star minus A transpose b is equal to 0. And then if we add this term to both sides of the equation, we are left with A transpose A times the least squares solution to Ax equal to b is equal to A transpose b. That's what we get. Now, why did we do all of this work? Remember what we started with. We said we're trying to find a solution to Ax is equal to b, but there was no solution. So we said, well, let's find at least an x star."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Now, why did we do all of this work? Remember what we started with. We said we're trying to find a solution to Ax is equal to b, but there was no solution. So we said, well, let's find at least an x star. Let's at least find an x star that minimizes the distance between b and Ax star. And we call this the least squares solution because when you actually take the length, or when you're minimizing the length, you're minimizing the squares of the differences right there. So it's the least squares solution."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So we said, well, let's find at least an x star. Let's at least find an x star that minimizes the distance between b and Ax star. And we call this the least squares solution because when you actually take the length, or when you're minimizing the length, you're minimizing the squares of the differences right there. So it's the least squares solution. Now, to find this, we know that this has to be the closest vector in our subspace to b. And we know that the closest vector in our subspace to b is the projection of b onto our subspace, onto our column space of A. And so we know that A times our least squares solution should be equal to the projection of b onto the column space of A."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "So it's the least squares solution. Now, to find this, we know that this has to be the closest vector in our subspace to b. And we know that the closest vector in our subspace to b is the projection of b onto our subspace, onto our column space of A. And so we know that A times our least squares solution should be equal to the projection of b onto the column space of A. If we can find some x in Rk that satisfies this, that is our least squares solution. But we've seen before that the projection b is easier said than done. There's a lot of work to it."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And so we know that A times our least squares solution should be equal to the projection of b onto the column space of A. If we can find some x in Rk that satisfies this, that is our least squares solution. But we've seen before that the projection b is easier said than done. There's a lot of work to it. So maybe we can do it a simpler way. And this is our simpler way. If we're looking for this, alternately we could just find a solution to this equation."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "There's a lot of work to it. So maybe we can do it a simpler way. And this is our simpler way. If we're looking for this, alternately we could just find a solution to this equation. So you give me an Ax equal to b. There is no solution. Well, what I'm going to do is I'm just going to multiply both sides of this equation times A transpose."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "If we're looking for this, alternately we could just find a solution to this equation. So you give me an Ax equal to b. There is no solution. Well, what I'm going to do is I'm just going to multiply both sides of this equation times A transpose. If I multiply both sides of this equation by A transpose, I get A transpose times Ax. Ax is equal to A transpose. Now I want to do that the same blue."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Well, what I'm going to do is I'm just going to multiply both sides of this equation times A transpose. If I multiply both sides of this equation by A transpose, I get A transpose times Ax. Ax is equal to A transpose. Now I want to do that the same blue. No, that's not the same blue. A transpose b. All I did is I multiplied both sides of this."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to do that the same blue. No, that's not the same blue. A transpose b. All I did is I multiplied both sides of this. Now this solution is not going to be, the solution to this equation will not be the same as the solution to this equation. This right here will always have a solution. And this right here is our least squares solution."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I multiplied both sides of this. Now this solution is not going to be, the solution to this equation will not be the same as the solution to this equation. This right here will always have a solution. And this right here is our least squares solution. So this right here is our least squares solution. And notice, this is some matrix. And then this right here is some vector."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And this right here is our least squares solution. So this right here is our least squares solution. And notice, this is some matrix. And then this right here is some vector. So as long as we can find a solution here, we've given our best shot at finding a solution to Ax equal to b. We've minimized the error. We're going to get Ax star."}, {"video_title": "Least squares approximation Linear Algebra Khan Academy.mp3", "Sentence": "And then this right here is some vector. So as long as we can find a solution here, we've given our best shot at finding a solution to Ax equal to b. We've minimized the error. We're going to get Ax star. The difference between Ax star and b is going to be minimized. It's going to be our least squares solution. It's all a little bit abstract right now in this video, but hopefully in the next video we'll realize that it's actually a very, very useful concept."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'll just always call it a subspace of Rn. Everything we're doing is linear. Subspace of Rn. I'm going to make a definition here. I'm going to say that a set of vectors, V, so V is some subset of vectors. Some subset of Rn. We already said Rn, when we think about it, it's really just an infinitely large set of vectors where each of those vectors have n components."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to make a definition here. I'm going to say that a set of vectors, V, so V is some subset of vectors. Some subset of Rn. We already said Rn, when we think about it, it's really just an infinitely large set of vectors where each of those vectors have n components. I'm going to not formally define it, but this is just a set of vectors. Sometimes we visualize it as multidimensional space and all of that, but if we wanted to be just as abstract about it as possible, it's just all the set of all of the, we could call it x1, x2, all the way to xn, where each of the xi's are a member of the real numbers for all of the i's. That was our definition of Rn."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We already said Rn, when we think about it, it's really just an infinitely large set of vectors where each of those vectors have n components. I'm going to not formally define it, but this is just a set of vectors. Sometimes we visualize it as multidimensional space and all of that, but if we wanted to be just as abstract about it as possible, it's just all the set of all of the, we could call it x1, x2, all the way to xn, where each of the xi's are a member of the real numbers for all of the i's. That was our definition of Rn. It's just a huge set of vectors, an infinitely large set of vectors. V, I'm going to call that a subset of Rn, which means it's just some, it could be all of these vectors, and I'll talk about it in a second, or it could be some subset of these vectors. Maybe it's all of them but one particular vector."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That was our definition of Rn. It's just a huge set of vectors, an infinitely large set of vectors. V, I'm going to call that a subset of Rn, which means it's just some, it could be all of these vectors, and I'll talk about it in a second, or it could be some subset of these vectors. Maybe it's all of them but one particular vector. In order for this V to be a subspace, I'm already saying it's a subset of Rn. Maybe this will help you. If I draw all of Rn here, this big blob, these are all of the vectors that are in Rn."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe it's all of them but one particular vector. In order for this V to be a subspace, I'm already saying it's a subset of Rn. Maybe this will help you. If I draw all of Rn here, this big blob, these are all of the vectors that are in Rn. V is some subset of it. It could be all of Rn, and I'll show that in a second. Let's just say this is V. V is a subset of vectors."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I draw all of Rn here, this big blob, these are all of the vectors that are in Rn. V is some subset of it. It could be all of Rn, and I'll show that in a second. Let's just say this is V. V is a subset of vectors. In order for V to be a subspace, and this is a definition, if V is a subspace or linear subspace of Rn, this means, this is my definition, this means three things. This means that V contains the zero vector. V contains, I'll do it really, that's the zero vector, or you could, this is equal to zero all the way, and you have n zeros."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's just say this is V. V is a subset of vectors. In order for V to be a subspace, and this is a definition, if V is a subspace or linear subspace of Rn, this means, this is my definition, this means three things. This means that V contains the zero vector. V contains, I'll do it really, that's the zero vector, or you could, this is equal to zero all the way, and you have n zeros. V contains the zero vector, and this is a big V right there. If we have some vector x in V, so let me write this. If my vector x is in V, if x is one of these vectors that's included in my V, then when I multiply x times any member of the reals, so if x is in V, then if V is a subspace of Rn, then x times any scalar is also in V. This has to be the case."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "V contains, I'll do it really, that's the zero vector, or you could, this is equal to zero all the way, and you have n zeros. V contains the zero vector, and this is a big V right there. If we have some vector x in V, so let me write this. If my vector x is in V, if x is one of these vectors that's included in my V, then when I multiply x times any member of the reals, so if x is in V, then if V is a subspace of Rn, then x times any scalar is also in V. This has to be the case. For those of you who are familiar with the term, this term is called closure. If I have any element of a set, and this is closure under multiplication. Let me write that down."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If my vector x is in V, if x is one of these vectors that's included in my V, then when I multiply x times any member of the reals, so if x is in V, then if V is a subspace of Rn, then x times any scalar is also in V. This has to be the case. For those of you who are familiar with the term, this term is called closure. If I have any element of a set, and this is closure under multiplication. Let me write that down. I'll do a new color. This is closure under scalar multiplication. That's just a fancy way of saying, look, if I take some member of my set, and I multiply it by some scalar, I'm still going to be in my set."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. I'll do a new color. This is closure under scalar multiplication. That's just a fancy way of saying, look, if I take some member of my set, and I multiply it by some scalar, I'm still going to be in my set. If I multiply it by some scalar, and I end up outside of my set, if I ended up with some other vector that's not included in my subset, then this wouldn't be a subspace. In order for it to be a subspace, if I multiply any vector in my subset by a real scalar, I'm defining this subspace over real numbers, if I multiply it by any real number, I should also get another member of this subset. This is one of the requirements."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's just a fancy way of saying, look, if I take some member of my set, and I multiply it by some scalar, I'm still going to be in my set. If I multiply it by some scalar, and I end up outside of my set, if I ended up with some other vector that's not included in my subset, then this wouldn't be a subspace. In order for it to be a subspace, if I multiply any vector in my subset by a real scalar, I'm defining this subspace over real numbers, if I multiply it by any real number, I should also get another member of this subset. This is one of the requirements. The other requirement is if I take two vectors, let's say I have vector A, it's in here, and I have vector B in here. This is my other requirement for V being a subspace. If vector A is in my set V, and vector B is in my set V, then if V is a subspace of Rn, that tells me that A and B must be in V as well."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is one of the requirements. The other requirement is if I take two vectors, let's say I have vector A, it's in here, and I have vector B in here. This is my other requirement for V being a subspace. If vector A is in my set V, and vector B is in my set V, then if V is a subspace of Rn, that tells me that A and B must be in V as well. This is closure under addition. Let me write that down. Closure under addition."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If vector A is in my set V, and vector B is in my set V, then if V is a subspace of Rn, that tells me that A and B must be in V as well. This is closure under addition. Let me write that down. Closure under addition. Once again, just a very fancy way of saying, look, if you give me two elements that's in my subset, and if I add them to each other, and these could be any two arbitrary elements in my subset, and I add them to each other, I'm going to get another element in my subset. That's what closure under addition means, that when you add two vectors in your set, you still end up with another vector in your set. You don't somehow end up with a vector that's outside of your set."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Closure under addition. Once again, just a very fancy way of saying, look, if you give me two elements that's in my subset, and if I add them to each other, and these could be any two arbitrary elements in my subset, and I add them to each other, I'm going to get another element in my subset. That's what closure under addition means, that when you add two vectors in your set, you still end up with another vector in your set. You don't somehow end up with a vector that's outside of your set. If I have a subset of Rn, some subset of vectors of Rn, that contains the zero vector, and it's closed under multiplication and addition, then I have a subspace. A subspace implies all of these things, and all of these things imply a subspace. This is the definition of a subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You don't somehow end up with a vector that's outside of your set. If I have a subset of Rn, some subset of vectors of Rn, that contains the zero vector, and it's closed under multiplication and addition, then I have a subspace. A subspace implies all of these things, and all of these things imply a subspace. This is the definition of a subspace. This might seem all abstract to you right now, so let's do a couple of examples. I don't know if these examples will make it any more concrete, but I think if we do it enough, you'll kind of get the intuitive sense of what a space implies. Let me just do some examples, because I want to stay relatively mathematically formal."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is the definition of a subspace. This might seem all abstract to you right now, so let's do a couple of examples. I don't know if these examples will make it any more concrete, but I think if we do it enough, you'll kind of get the intuitive sense of what a space implies. Let me just do some examples, because I want to stay relatively mathematically formal. Let's just say I have the almost trivially basic set. Let's say my set of vectors, I only have one vector in it, and I have the zero vector. I'll just do a really bold zero there, or I could write it like this."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me just do some examples, because I want to stay relatively mathematically formal. Let's just say I have the almost trivially basic set. Let's say my set of vectors, I only have one vector in it, and I have the zero vector. I'll just do a really bold zero there, or I could write it like this. The only vector in my set is the zero vector. Let's say we're talking about R3. Let's say my zero vector in R3 looks like that."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'll just do a really bold zero there, or I could write it like this. The only vector in my set is the zero vector. Let's say we're talking about R3. Let's say my zero vector in R3 looks like that. What I want to know is, is my set V a subspace of R3? In order for it to be a subspace, three conditions. It has to contain the zero vector, well, the only thing it does contain is the zero vector, so it definitely contains the zero vector."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say my zero vector in R3 looks like that. What I want to know is, is my set V a subspace of R3? In order for it to be a subspace, three conditions. It has to contain the zero vector, well, the only thing it does contain is the zero vector, so it definitely contains the zero vector. Zero vector, check. Now, is it closed under multiplication? That means if I take any member of the set, there's only one of them, and I multiply it by any scalar, I should get another member of the set, or I should get maybe itself."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It has to contain the zero vector, well, the only thing it does contain is the zero vector, so it definitely contains the zero vector. Zero vector, check. Now, is it closed under multiplication? That means if I take any member of the set, there's only one of them, and I multiply it by any scalar, I should get another member of the set, or I should get maybe itself. Let's see, there's only one member of the set. The one member of the set is the zero vector. If I multiply it times any scalar, what am I going to get?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That means if I take any member of the set, there's only one of them, and I multiply it by any scalar, I should get another member of the set, or I should get maybe itself. Let's see, there's only one member of the set. The one member of the set is the zero vector. If I multiply it times any scalar, what am I going to get? I'm going to get C times zero, which is zero, C times zero, which is zero, and C times zero. I'm going to get its only member, but it is closed. It is closed under multiplication."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply it times any scalar, what am I going to get? I'm going to get C times zero, which is zero, C times zero, which is zero, and C times zero. I'm going to get its only member, but it is closed. It is closed under multiplication. You can multiply this one vector times any scalar, and you're just going to get this vector again. You're going to end up being in your zero vector set. That's a check."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It is closed under multiplication. You can multiply this one vector times any scalar, and you're just going to get this vector again. You're going to end up being in your zero vector set. That's a check. Is it closed under addition? Clearly, if I add any member of this set to itself, or there's only one member, or to another member of the set, there's only one option here. If I just add that to that, what do I get?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's a check. Is it closed under addition? Clearly, if I add any member of this set to itself, or there's only one member, or to another member of the set, there's only one option here. If I just add that to that, what do I get? I just get that. I just get it again. It definitely is closed under addition."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I just add that to that, what do I get? I just get that. I just get it again. It definitely is closed under addition. Check. It does turn out that this trivially basic subset of R3 that just contains the zero vector, it is a subspace. Maybe a trivially simple subspace, but it satisfies our constraints of a subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It definitely is closed under addition. Check. It does turn out that this trivially basic subset of R3 that just contains the zero vector, it is a subspace. Maybe a trivially simple subspace, but it satisfies our constraints of a subspace. You can't do anything with the vectors in it that will somehow get you out of that subspace, or at least if you're dealing with scalar multiplication or addition. Let me do one that maybe the idea will be a little clearer. If I show you an example of something that is not a subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe a trivially simple subspace, but it satisfies our constraints of a subspace. You can't do anything with the vectors in it that will somehow get you out of that subspace, or at least if you're dealing with scalar multiplication or addition. Let me do one that maybe the idea will be a little clearer. If I show you an example of something that is not a subspace. Let me get my coordinate axes over here. Let's say I were to find some subspace, some subset. I don't know whether it's a subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I show you an example of something that is not a subspace. Let me get my coordinate axes over here. Let's say I were to find some subspace, some subset. I don't know whether it's a subspace. Let me call my set S. It equals all the vectors, let me say x1, x2, that are a member of R2, such that I'm going to make a little constraint here. Such that x1 is greater than or equal to zero. It contains all of the vectors in R2 that are at least zero or greater for the first term."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I don't know whether it's a subspace. Let me call my set S. It equals all the vectors, let me say x1, x2, that are a member of R2, such that I'm going to make a little constraint here. Such that x1 is greater than or equal to zero. It contains all of the vectors in R2 that are at least zero or greater for the first term. If we were to graph that here, what do you get? We can move up or down in any direction. We can go up and down in any direction, but we're constraining ourselves."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It contains all of the vectors in R2 that are at least zero or greater for the first term. If we were to graph that here, what do you get? We can move up or down in any direction. We can go up and down in any direction, but we're constraining ourselves. These are all going to be zero or greater. All of these first coordinates are going to be zero or greater. This one, we can go up and down arbitrarily."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can go up and down in any direction, but we're constraining ourselves. These are all going to be zero or greater. All of these first coordinates are going to be zero or greater. This one, we can go up and down arbitrarily. Essentially, this subset of R2, R2 is my entire Cartesian plane, but this subset of R2, it'll include the vertical axis, often referred to as the y-axis, it'll include the vertical axis, and essentially the first and fourth quadrants. If you remember your quadrant labeling, that's the first quadrant and that's the fourth quadrant. My question to you is, is S a subspace of R2?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This one, we can go up and down arbitrarily. Essentially, this subset of R2, R2 is my entire Cartesian plane, but this subset of R2, it'll include the vertical axis, often referred to as the y-axis, it'll include the vertical axis, and essentially the first and fourth quadrants. If you remember your quadrant labeling, that's the first quadrant and that's the fourth quadrant. My question to you is, is S a subspace of R2? The first question, does it contain the zero vector? In the case of R2, does it contain zero, zero? Sure, it includes zero, zero right there."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My question to you is, is S a subspace of R2? The first question, does it contain the zero vector? In the case of R2, does it contain zero, zero? Sure, it includes zero, zero right there. We included, we said x is greater than or equal to zero, so this could be zero. Obviously, there's no constraint on this, so definitely the zero, zero vector is definitely contained in our set S. That is a check. Now, let's try another one."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Sure, it includes zero, zero right there. We included, we said x is greater than or equal to zero, so this could be zero. Obviously, there's no constraint on this, so definitely the zero, zero vector is definitely contained in our set S. That is a check. Now, let's try another one. If I add any two vectors in the set, is that also going to show up in our set? Let me just do a couple of examples. Maybe this isn't a proof."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's try another one. If I add any two vectors in the set, is that also going to show up in our set? Let me just do a couple of examples. Maybe this isn't a proof. Let's see, if I add that vector to that vector, what am I going to get? If I put this up here, I'm going to get that vector. If I add that vector to that vector, what am I going to get?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe this isn't a proof. Let's see, if I add that vector to that vector, what am I going to get? If I put this up here, I'm going to get that vector. If I add that vector to that vector, what am I going to get? I could put this one as tail, I would get a vector that looks like that. If I did it formally, if I add, let's say that I have two vectors that are a member of our set, let's say the first one is AB, and I add it to CD, what do I get? I get A plus C over, this was a D, over B plus D. This thing is going to be greater than zero."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I add that vector to that vector, what am I going to get? I could put this one as tail, I would get a vector that looks like that. If I did it formally, if I add, let's say that I have two vectors that are a member of our set, let's say the first one is AB, and I add it to CD, what do I get? I get A plus C over, this was a D, over B plus D. This thing is going to be greater than zero. This thing is also going to be greater than zero. That was my requirement for being in the set. If both of these are greater than zero, and we add them to each other, this thing is also going to be greater than zero."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I get A plus C over, this was a D, over B plus D. This thing is going to be greater than zero. This thing is also going to be greater than zero. That was my requirement for being in the set. If both of these are greater than zero, and we add them to each other, this thing is also going to be greater than zero. We don't care what these, these can be anything. I didn't put any constraints on the second component of my vector. It does seem like it is closed under addition."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If both of these are greater than zero, and we add them to each other, this thing is also going to be greater than zero. We don't care what these, these can be anything. I didn't put any constraints on the second component of my vector. It does seem like it is closed under addition. What about scalar multiplication? Let's take a particular case here. Let's take my AB again."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It does seem like it is closed under addition. What about scalar multiplication? Let's take a particular case here. Let's take my AB again. I have my vector AB. I can pick any real scalar. Remember, any real scalar, what if I just multiply it by minus one?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's take my AB again. I have my vector AB. I can pick any real scalar. Remember, any real scalar, what if I just multiply it by minus one? Minus one. If I multiply it by minus one, I get minus A minus B. If I were to draw it visually, if this is, let's say AB was the vector 2,4, so it's like this."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, any real scalar, what if I just multiply it by minus one? Minus one. If I multiply it by minus one, I get minus A minus B. If I were to draw it visually, if this is, let's say AB was the vector 2,4, so it's like this. When I multiply it by minus one, what do I get? I get minus A minus B. I get this vector, which you can visually clearly see, falls out of, if we view these as kind of position vectors, it falls out of our subspace. If you just view it, not even visually, if you just do it mathematically, clearly if this is positive, then this is going to, and let's say we assume this is positive and definitely not zero, so it's definitely a positive number, so this is definitely going to be a negative number."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to draw it visually, if this is, let's say AB was the vector 2,4, so it's like this. When I multiply it by minus one, what do I get? I get minus A minus B. I get this vector, which you can visually clearly see, falls out of, if we view these as kind of position vectors, it falls out of our subspace. If you just view it, not even visually, if you just do it mathematically, clearly if this is positive, then this is going to, and let's say we assume this is positive and definitely not zero, so it's definitely a positive number, so this is definitely going to be a negative number. When we multiply it by negative one, for really any element of this that doesn't have a zero there, you're going to end up with something that falls out of it. This is not a member of the set because to be a member of the set, your first component had to be greater than zero. This first component is less than zero."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you just view it, not even visually, if you just do it mathematically, clearly if this is positive, then this is going to, and let's say we assume this is positive and definitely not zero, so it's definitely a positive number, so this is definitely going to be a negative number. When we multiply it by negative one, for really any element of this that doesn't have a zero there, you're going to end up with something that falls out of it. This is not a member of the set because to be a member of the set, your first component had to be greater than zero. This first component is less than zero. This subset that I drew out here, the subset of R2, is not a subspace because it's not closed under multiplication or scalar multiplication, not closed under scalar multiplication. Not closed. This is not a subspace of R2."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This first component is less than zero. This subset that I drew out here, the subset of R2, is not a subspace because it's not closed under multiplication or scalar multiplication, not closed under scalar multiplication. Not closed. This is not a subspace of R2. I'll ask you one interesting question. What if I ask you just the span of some set of vectors? Is that a valid, let's say I have the span of some, let's say I want to know the span of, I don't know, let's say I have vector V1, V2, and V3."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is not a subspace of R2. I'll ask you one interesting question. What if I ask you just the span of some set of vectors? Is that a valid, let's say I have the span of some, let's say I want to know the span of, I don't know, let's say I have vector V1, V2, and V3. I'm not even going to tell you how many elements each of these vectors have. Is this a valid subspace of Rn where n is the number of elements that each of these have of Rn? Let's pick one of the elements of, let me define, let me just call U to be the set, or the set of all linear combinations of this is the span."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Is that a valid, let's say I have the span of some, let's say I want to know the span of, I don't know, let's say I have vector V1, V2, and V3. I'm not even going to tell you how many elements each of these vectors have. Is this a valid subspace of Rn where n is the number of elements that each of these have of Rn? Let's pick one of the elements of, let me define, let me just call U to be the set, or the set of all linear combinations of this is the span. Let me just define U to be the span. I want to know is U a valid subspace? Let's think about it this way."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's pick one of the elements of, let me define, let me just call U to be the set, or the set of all linear combinations of this is the span. Let me just define U to be the span. I want to know is U a valid subspace? Let's think about it this way. Let me just pick out a random element of U. Let me pick out a random element of U. Actually, let me, does this contain the zero vector?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's think about it this way. Let me just pick out a random element of U. Let me pick out a random element of U. Actually, let me, does this contain the zero vector? Well, sure. If we just multiply all of these times zero, so if we just say zero times V1 plus zero times V2, these are all vectors, I didn't write them bold, plus zero times V3, we get the zero vector. Everything just is zeroed out."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me, does this contain the zero vector? Well, sure. If we just multiply all of these times zero, so if we just say zero times V1 plus zero times V2, these are all vectors, I didn't write them bold, plus zero times V3, we get the zero vector. Everything just is zeroed out. It definitely contains the zeroed vector. This is a linear combination of those three vectors, so it's included in the span. Now, what if I, let me just pick some arbitrary member of this span."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Everything just is zeroed out. It definitely contains the zeroed vector. This is a linear combination of those three vectors, so it's included in the span. Now, what if I, let me just pick some arbitrary member of this span. In order to be a member of this set, it just means that you can be represented, let me just call the vector X, it means that you can be represented as a linear combination of these vectors. Some combination, C1 times V1 plus C2 times V2 plus C3 times V3. I'm just representing this vector X, it's a member of this, so it can be represented as a linear combination of those three vectors."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, what if I, let me just pick some arbitrary member of this span. In order to be a member of this set, it just means that you can be represented, let me just call the vector X, it means that you can be represented as a linear combination of these vectors. Some combination, C1 times V1 plus C2 times V2 plus C3 times V3. I'm just representing this vector X, it's a member of this, so it can be represented as a linear combination of those three vectors. Now, is this set closed under multiplication? Well, let's just multiply this times some arbitrary constant. What is C times X?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just representing this vector X, it's a member of this, so it can be represented as a linear combination of those three vectors. Now, is this set closed under multiplication? Well, let's just multiply this times some arbitrary constant. What is C times X? Let me scroll down a little bit. What is C times X equal? Let me do a different constant, actually."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is C times X? Let me scroll down a little bit. What is C times X equal? Let me do a different constant, actually. Let me multiply it times some arbitrary constant, A. What is A times X? Well, it's A times C1 times V1."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do a different constant, actually. Let me multiply it times some arbitrary constant, A. What is A times X? Well, it's A times C1 times V1. I'm just multiplying both sides of this equation times A. A times C2 times V2 plus A times C3 V3. Right?"}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's A times C1 times V1. I'm just multiplying both sides of this equation times A. A times C2 times V2 plus A times C3 V3. Right? Well, you could just, I mean, if this was an arbitrary constant, you could just write this as another arbitrary constant. This is another arbitrary constant, and this is another arbitrary constant. I want to be clear."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Right? Well, you could just, I mean, if this was an arbitrary constant, you could just write this as another arbitrary constant. This is another arbitrary constant, and this is another arbitrary constant. I want to be clear. All I did is I just multiplied both sides of this equation times a scalar. But clearly, this expression right here, I mean, I could write this, I could rewrite this as C4 times V1 plus C5 times V2, where this is C5, this is C4, plus C6 times V3. This is clearly another linear combination of these three vectors."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I want to be clear. All I did is I just multiplied both sides of this equation times a scalar. But clearly, this expression right here, I mean, I could write this, I could rewrite this as C4 times V1 plus C5 times V2, where this is C5, this is C4, plus C6 times V3. This is clearly another linear combination of these three vectors. So the span is the set of all of the linear combinations of these three vectors. So clearly, this is one of the linear combinations, so it's also included in the span. So this is also in U."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is clearly another linear combination of these three vectors. So the span is the set of all of the linear combinations of these three vectors. So clearly, this is one of the linear combinations, so it's also included in the span. So this is also in U. It's also in the span of those three vectors. So it is closed under multiplication. Now we just have to show that it's closed under addition, and then we know that the span of, and we added three here, but you can extend it to an arbitrary n number of vectors, that the span of any set of vectors is a valid subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is also in U. It's also in the span of those three vectors. So it is closed under multiplication. Now we just have to show that it's closed under addition, and then we know that the span of, and we added three here, but you can extend it to an arbitrary n number of vectors, that the span of any set of vectors is a valid subspace. So let me prove that. So we already defined one x here. Let me define another vector that's in U or that's in the span of these vectors, and it equals, I don't know, let's say it equals D1 times V1 plus D2 times V2 plus D3 times V3."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now we just have to show that it's closed under addition, and then we know that the span of, and we added three here, but you can extend it to an arbitrary n number of vectors, that the span of any set of vectors is a valid subspace. So let me prove that. So we already defined one x here. Let me define another vector that's in U or that's in the span of these vectors, and it equals, I don't know, let's say it equals D1 times V1 plus D2 times V2 plus D3 times V3. Now what is x plus y? If I add these two vectors, what are they equal to? Well, I could just add, x plus y means all of this stuff plus all of this stuff."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me define another vector that's in U or that's in the span of these vectors, and it equals, I don't know, let's say it equals D1 times V1 plus D2 times V2 plus D3 times V3. Now what is x plus y? If I add these two vectors, what are they equal to? Well, I could just add, x plus y means all of this stuff plus all of this stuff. And so what does that equal? It means if you just add these together, you get C1 plus D1 times V1 plus C2 plus D2 times V2 plus C3 plus D3 times V3. You had a V3 here, you had a V3 there, you just add up their coefficients."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I could just add, x plus y means all of this stuff plus all of this stuff. And so what does that equal? It means if you just add these together, you get C1 plus D1 times V1 plus C2 plus D2 times V2 plus C3 plus D3 times V3. You had a V3 here, you had a V3 there, you just add up their coefficients. Clearly, this is just another linear combination. These are just constants again. That's an arbitrary constant, that's an arbitrary constant, that's an arbitrary constant."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You had a V3 here, you had a V3 there, you just add up their coefficients. Clearly, this is just another linear combination. These are just constants again. That's an arbitrary constant, that's an arbitrary constant, that's an arbitrary constant. So this thing is just a linear combination of V1, V2, and V3. So it must be, by definition, in the span of V1, V2, and V3. So we are definitely closed under addition."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's an arbitrary constant, that's an arbitrary constant, that's an arbitrary constant. So this thing is just a linear combination of V1, V2, and V3. So it must be, by definition, in the span of V1, V2, and V3. So we are definitely closed under addition. Now you might say, hey, Sal, the span of any vector is a valid subspace, but let me show you an example that, clearly, if I just took the span of one vector, let me just define u to be equal to the span of just the vector, let me just do a really simple one. Let's say it's just the vector 1, 1. Clearly, this can't be a valid subspace."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we are definitely closed under addition. Now you might say, hey, Sal, the span of any vector is a valid subspace, but let me show you an example that, clearly, if I just took the span of one vector, let me just define u to be equal to the span of just the vector, let me just do a really simple one. Let's say it's just the vector 1, 1. Clearly, this can't be a valid subspace. Let's think about this visually. So what does a vector 1, 1 look like? A vector 1, 1 looks like this."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Clearly, this can't be a valid subspace. Let's think about this visually. So what does a vector 1, 1 look like? A vector 1, 1 looks like this. And the span of vector 1, 1 is all of, and this is in its standard position, the span of vector 1, 1 is all of the linear combinations of this vector. Well, there's nothing else to add it to, so it's really just going to be all of the scaled up and scaled down versions of this. So if you scale it up, you get things that look more like that."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A vector 1, 1 looks like this. And the span of vector 1, 1 is all of, and this is in its standard position, the span of vector 1, 1 is all of the linear combinations of this vector. Well, there's nothing else to add it to, so it's really just going to be all of the scaled up and scaled down versions of this. So if you scale it up, you get things that look more like that. If you scale it down, you get things that look more like that if you go into the negative domain. So just by multiplying this vector times different values, and if you were to put them all into standard position, you would essentially get a line that looks like that. And you say, gee, you know, that doesn't look like a whole subspace, but a couple of things."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you scale it up, you get things that look more like that. If you scale it down, you get things that look more like that if you go into the negative domain. So just by multiplying this vector times different values, and if you were to put them all into standard position, you would essentially get a line that looks like that. And you say, gee, you know, that doesn't look like a whole subspace, but a couple of things. Clearly, it contains the vector 0, it contains the 0 vector. We can just multiply 0 times 1, we can just scale it by 0. The span is just all of the different scales of this, and if there were other vectors, you would add it to those as well, but this is clearly going to be the 0 vector."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you say, gee, you know, that doesn't look like a whole subspace, but a couple of things. Clearly, it contains the vector 0, it contains the 0 vector. We can just multiply 0 times 1, we can just scale it by 0. The span is just all of the different scales of this, and if there were other vectors, you would add it to those as well, but this is clearly going to be the 0 vector. So it contains the 0 vector. Is it closed under multiplication? Well, the span is the set of all of the vectors, where if you take all of the real numbers for c, and you multiply it times 1, 1, that is the span."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The span is just all of the different scales of this, and if there were other vectors, you would add it to those as well, but this is clearly going to be the 0 vector. So it contains the 0 vector. Is it closed under multiplication? Well, the span is the set of all of the vectors, where if you take all of the real numbers for c, and you multiply it times 1, 1, that is the span. So clearly, you multiply this times anything, it's going to equal another thing that's definitely in your span. Now the last thing, is it closed under addition? So any two vectors in the span could, let's say that I have one vector a, that's in my span, I can represent it as c1, some scalar times my vector there, and then I have another vector, b, and I can represent it as c2 times my one vector in my set right there."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, the span is the set of all of the vectors, where if you take all of the real numbers for c, and you multiply it times 1, 1, that is the span. So clearly, you multiply this times anything, it's going to equal another thing that's definitely in your span. Now the last thing, is it closed under addition? So any two vectors in the span could, let's say that I have one vector a, that's in my span, I can represent it as c1, some scalar times my vector there, and then I have another vector, b, and I can represent it as c2 times my one vector in my set right there. And so what is this going to be equal to? This is going to be equal to c1 plus c2 times my vector. This is almost trivially obvious."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So any two vectors in the span could, let's say that I have one vector a, that's in my span, I can represent it as c1, some scalar times my vector there, and then I have another vector, b, and I can represent it as c2 times my one vector in my set right there. And so what is this going to be equal to? This is going to be equal to c1 plus c2 times my vector. This is almost trivially obvious. But clearly, this is in the span, it's just a scaled up version of this. This is in the span, it's in a scaled up version of this. And this is also going to be in the span of this vector, because this is just another scalar, we could call that c3."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is almost trivially obvious. But clearly, this is in the span, it's just a scaled up version of this. This is in the span, it's in a scaled up version of this. And this is also going to be in the span of this vector, because this is just another scalar, we could call that c3. And if you just do it visually, if I take this vector right there, let's say I take that vector, and I were to add it to this vector, if you put them head to tails, you would end up with this vector, right there in green. I don't know if you can see it, maybe I'll do it in red right there. You will end up with that vector."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is also going to be in the span of this vector, because this is just another scalar, we could call that c3. And if you just do it visually, if I take this vector right there, let's say I take that vector, and I were to add it to this vector, if you put them head to tails, you would end up with this vector, right there in green. I don't know if you can see it, maybe I'll do it in red right there. You will end up with that vector. And you could do that, any vector plus any other vector on this line is going to equal another vector on this line. Any vector on this line multiplied by some scalar is just going to be another vector on this line. So you're closed under multiplication, you're closed under addition, and you include the zero vector."}, {"video_title": "Linear subspaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You will end up with that vector. And you could do that, any vector plus any other vector on this line is going to equal another vector on this line. Any vector on this line multiplied by some scalar is just going to be another vector on this line. So you're closed under multiplication, you're closed under addition, and you include the zero vector. So even this trivially simple span is a valid subspace. And that just backs up the idea that we showed here, that in general, I could have just made this a set of n vectors. I picked three vectors right here, but it could have been n vectors, and I could have used the same argument, that the span of n vectors is a valid subspace of Rn."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Let me just see if I can do it in general terms. In the first row and first column in that entry, it has a 1, and then everything else, the rest of the n minus 1 rows in that first column, are all going to be 0s. It's going to be 0s all the way down to the nth term. And in the second column, we have a 0 in the first component, but then a 1 in the second component, and then it goes 0s all the way down. And you keep doing this. In the third row, in the third or let me say third column, although it would have applied to the third row as well, the 1 shows up in the third component, and then it's 0s all the way down. And this, essentially, you have the 1s filling up the diagonal of this matrix right here."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And in the second column, we have a 0 in the first component, but then a 1 in the second component, and then it goes 0s all the way down. And you keep doing this. In the third row, in the third or let me say third column, although it would have applied to the third row as well, the 1 shows up in the third component, and then it's 0s all the way down. And this, essentially, you have the 1s filling up the diagonal of this matrix right here. So if you go all the way to the nth column or the nth column vector, you have a bunch of 0s until you have n minus 1 0s, and then the very last component, the nth component, there will be a 1. So you have essentially a matrix with 1s down the diagonal. Now this matrix has a bunch of neat properties, and we'll explore it more in the future."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And this, essentially, you have the 1s filling up the diagonal of this matrix right here. So if you go all the way to the nth column or the nth column vector, you have a bunch of 0s until you have n minus 1 0s, and then the very last component, the nth component, there will be a 1. So you have essentially a matrix with 1s down the diagonal. Now this matrix has a bunch of neat properties, and we'll explore it more in the future. But I'm just exposing you to this because it has one very neat property relative to linear transformations. But I'm going to call this the identity matrix, and I'll call this I sub n. And I call that sub n because it's an n by n identity matrix. If I, you know, I sub 2 would be equal to a 2 by 2 identity matrix, it would look like that."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Now this matrix has a bunch of neat properties, and we'll explore it more in the future. But I'm just exposing you to this because it has one very neat property relative to linear transformations. But I'm going to call this the identity matrix, and I'll call this I sub n. And I call that sub n because it's an n by n identity matrix. If I, you know, I sub 2 would be equal to a 2 by 2 identity matrix, it would look like that. And I sub 3 would look like this. 1 0 0, 0 1 0, 0 0 1. I think you get the point."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "If I, you know, I sub 2 would be equal to a 2 by 2 identity matrix, it would look like that. And I sub 3 would look like this. 1 0 0, 0 1 0, 0 0 1. I think you get the point. Now the neat thing about this identity matrix becomes evident when you multiply it times any vector. So we can multiply this guy times an n component vector or a member of Rn. So let's do that."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "I think you get the point. Now the neat thing about this identity matrix becomes evident when you multiply it times any vector. So we can multiply this guy times an n component vector or a member of Rn. So let's do that. So if we multiply this matrix times, let's call this vector x, this is x1, x2, all the way down to xn. What is this going to be equal to? So this is vector x right here."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So if we multiply this matrix times, let's call this vector x, this is x1, x2, all the way down to xn. What is this going to be equal to? So this is vector x right here. So if I multiply matrix I, my identity matrix, I sub n, and I multiply it times my vector x, where x is a member of Rn, it has n components, what am I going to get? Well, I'm going to get 1 times x1 plus 0 times x2 plus 0 times x3 plus 0 times x4, all of that. So essentially, you can kind of view it as this row dotted with the vector."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So this is vector x right here. So if I multiply matrix I, my identity matrix, I sub n, and I multiply it times my vector x, where x is a member of Rn, it has n components, what am I going to get? Well, I'm going to get 1 times x1 plus 0 times x2 plus 0 times x3 plus 0 times x4, all of that. So essentially, you can kind of view it as this row dotted with the vector. So the only non-zero term is going to be the 1 times the x1. So it's going to be x1. Sorry, let me do it like this."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So essentially, you can kind of view it as this row dotted with the vector. So the only non-zero term is going to be the 1 times the x1. So it's going to be x1. Sorry, let me do it like this. So you're going to get another vector in Rn like that. And so the first term is that row essentially being dotted with that column. And so you just get x1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Sorry, let me do it like this. So you're going to get another vector in Rn like that. And so the first term is that row essentially being dotted with that column. And so you just get x1. And then the next entry is going to be this row, or you could view it as the transpose of this row dotted with that column, so 0 times x1 plus 1 times x2 plus 0 times everything else. So the only non-zero term is the 1 times x2. So you get an x2 there."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And so you just get x1. And then the next entry is going to be this row, or you could view it as the transpose of this row dotted with that column, so 0 times x1 plus 1 times x2 plus 0 times everything else. So the only non-zero term is the 1 times x2. So you get an x2 there. And then you keep doing that. And what are you going to get? You're going to get an x3, because the only non-zero term here is the third one."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So you get an x2 there. And then you keep doing that. And what are you going to get? You're going to get an x3, because the only non-zero term here is the third one. And you're going to go all the way down until you get an xn. But what is this thing equal to? This is just equal to x."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "You're going to get an x3, because the only non-zero term here is the third one. And you're going to go all the way down until you get an xn. But what is this thing equal to? This is just equal to x. So the neat thing about this identity matrix that we've created is that when you multiply it times any vector, you get the vector again. The identity matrix times any vector in Rn, it's only defined for vectors in Rn, is equal to that vector again. And actually, the columns of the identity matrix have a special, I guess, the set of columns has a special name."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "This is just equal to x. So the neat thing about this identity matrix that we've created is that when you multiply it times any vector, you get the vector again. The identity matrix times any vector in Rn, it's only defined for vectors in Rn, is equal to that vector again. And actually, the columns of the identity matrix have a special, I guess, the set of columns has a special name. They are called, so if we call this first column e1 and this second column e2 and the third column e3 and we go all the way to en, these vectors, these column vectors here, the set of these, so let's say e1, e2, all the way to en, this is called the standard basis for Rn. So why is it called that? Well, the word basis is there, so two things must be true."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And actually, the columns of the identity matrix have a special, I guess, the set of columns has a special name. They are called, so if we call this first column e1 and this second column e2 and the third column e3 and we go all the way to en, these vectors, these column vectors here, the set of these, so let's say e1, e2, all the way to en, this is called the standard basis for Rn. So why is it called that? Well, the word basis is there, so two things must be true. These things must span Rn and they must be linearly independent. It's pretty obvious from inspection they're linearly independent. If this guy has a 1 here and no one else has a 1 there, there's no way you can construct that 1 with some combination of the rest of the guys."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Well, the word basis is there, so two things must be true. These things must span Rn and they must be linearly independent. It's pretty obvious from inspection they're linearly independent. If this guy has a 1 here and no one else has a 1 there, there's no way you can construct that 1 with some combination of the rest of the guys. And you can make that same argument for each of the 1's in each of the components. So it's clearly linearly independent. And then to see that you can span, that you can construct any vector with a linear combination of these guys, you really just have to, whatever vector you want to construct, if you want to construct x1, let me put it this way, if you want to construct this vector, let me write it this way, let me pick a different one."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "If this guy has a 1 here and no one else has a 1 there, there's no way you can construct that 1 with some combination of the rest of the guys. And you can make that same argument for each of the 1's in each of the components. So it's clearly linearly independent. And then to see that you can span, that you can construct any vector with a linear combination of these guys, you really just have to, whatever vector you want to construct, if you want to construct x1, let me put it this way, if you want to construct this vector, let me write it this way, let me pick a different one. Let's say you want to construct the vector a1, a2, a3, all the way down to an. So this is some member of Rn. You want to construct this vector."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And then to see that you can span, that you can construct any vector with a linear combination of these guys, you really just have to, whatever vector you want to construct, if you want to construct x1, let me put it this way, if you want to construct this vector, let me write it this way, let me pick a different one. Let's say you want to construct the vector a1, a2, a3, all the way down to an. So this is some member of Rn. You want to construct this vector. Well, the linear combination that would get you this is literally a1 times e1 plus a2 times e2 plus all the way to an times en. This scalar times this first column vector will essentially just get you, what will this look like? This will look like a1 and then you'd have a bunch of 0's."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "You want to construct this vector. Well, the linear combination that would get you this is literally a1 times e1 plus a2 times e2 plus all the way to an times en. This scalar times this first column vector will essentially just get you, what will this look like? This will look like a1 and then you'd have a bunch of 0's. You'd have n minus 1 0's plus 0 and you'd have an a2 and then you'd have a bunch of 0's. And then you keep doing that and then you would have 0's, a bunch of 0's and you would have an an. And obviously, by our definition of vector addition, you add all of these things up, you get this guy right here, and it's kind of obvious because this right here is the same thing as our identity matrix times a1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "This will look like a1 and then you'd have a bunch of 0's. You'd have n minus 1 0's plus 0 and you'd have an a2 and then you'd have a bunch of 0's. And then you keep doing that and then you would have 0's, a bunch of 0's and you would have an an. And obviously, by our definition of vector addition, you add all of these things up, you get this guy right here, and it's kind of obvious because this right here is the same thing as our identity matrix times a1. I just wanted to expose you to that idea. Now, let's apply what we already know about linear transformations to what we've just learned about this identity matrix. I just told you that I can represent any vector like this, let me rewrite it in maybe terms of x. I can write any vector x as a linear combination of the standard basis, which are really just the columns of the identity matrix."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And obviously, by our definition of vector addition, you add all of these things up, you get this guy right here, and it's kind of obvious because this right here is the same thing as our identity matrix times a1. I just wanted to expose you to that idea. Now, let's apply what we already know about linear transformations to what we've just learned about this identity matrix. I just told you that I can represent any vector like this, let me rewrite it in maybe terms of x. I can write any vector x as a linear combination of the standard basis, which are really just the columns of the identity matrix. So I can write that as x1 times e1 plus x2 times e2 all the way to xn times en. And remember, each of these column vectors right here, like for e1, is just 1 in the first entry and then all the rest are 0's. e2 is a 1 in the second entry and everything else is 0. en is, or e5 is a 1 in the fifth entry and everything else is 0."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "I just told you that I can represent any vector like this, let me rewrite it in maybe terms of x. I can write any vector x as a linear combination of the standard basis, which are really just the columns of the identity matrix. So I can write that as x1 times e1 plus x2 times e2 all the way to xn times en. And remember, each of these column vectors right here, like for e1, is just 1 in the first entry and then all the rest are 0's. e2 is a 1 in the second entry and everything else is 0. en is, or e5 is a 1 in the fifth entry and everything else is 0. And this I just showed you and this is a bit obvious from this right here. Now, we know that by definition a linear transformation of x is the same thing as taking the linear transformation of this whole thing, let me do it in another color, is equal to the linear transformation of, let me say, actually instead of using the t, let me use t. Instead of using l, let me use t. I used l by accident because I was thinking linear. But if I were to take the linear transformation of x, because that's the notation we're used to, that's the same thing as taking a linear transformation of this thing."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "e2 is a 1 in the second entry and everything else is 0. en is, or e5 is a 1 in the fifth entry and everything else is 0. And this I just showed you and this is a bit obvious from this right here. Now, we know that by definition a linear transformation of x is the same thing as taking the linear transformation of this whole thing, let me do it in another color, is equal to the linear transformation of, let me say, actually instead of using the t, let me use t. Instead of using l, let me use t. I used l by accident because I was thinking linear. But if I were to take the linear transformation of x, because that's the notation we're used to, that's the same thing as taking a linear transformation of this thing. They're equivalent. So x1 times e1 plus x2 times e2 all the way to plus xn times en, equivalent statements. Now, from the definition of linear transformations, we know that this is the same thing, that the transformation of the sum is equal to the sum of the transformation."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "But if I were to take the linear transformation of x, because that's the notation we're used to, that's the same thing as taking a linear transformation of this thing. They're equivalent. So x1 times e1 plus x2 times e2 all the way to plus xn times en, equivalent statements. Now, from the definition of linear transformations, we know that this is the same thing, that the transformation of the sum is equal to the sum of the transformation. So this is equal to the transformation of x1 e1 plus the transformation of x2 e2. Well, this is just any linear transformation. Let me make that very clear."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Now, from the definition of linear transformations, we know that this is the same thing, that the transformation of the sum is equal to the sum of the transformation. So this is equal to the transformation of x1 e1 plus the transformation of x2 e2. Well, this is just any linear transformation. Let me make that very clear. This is any linear transformation. Any linear transformation. By definition, linear transformations have to satisfy these properties."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Let me make that very clear. This is any linear transformation. Any linear transformation. By definition, linear transformations have to satisfy these properties. So the transformation times x2 e2 all the way to this transformation times this last entry. The scalar xn times my standard basis vector en. And we know from the other property of linear transformations that the transformation of a vector multiplied by the scalar is the same thing as the scalar multiplied by the transformation of the vector."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "By definition, linear transformations have to satisfy these properties. So the transformation times x2 e2 all the way to this transformation times this last entry. The scalar xn times my standard basis vector en. And we know from the other property of linear transformations that the transformation of a vector multiplied by the scalar is the same thing as the scalar multiplied by the transformation of the vector. That's just from our definition of linear transformations. Plus x2 times the transformation of e2. Plus all the way to xn times the transformation of en."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And we know from the other property of linear transformations that the transformation of a vector multiplied by the scalar is the same thing as the scalar multiplied by the transformation of the vector. That's just from our definition of linear transformations. Plus x2 times the transformation of e2. Plus all the way to xn times the transformation of en. Now, what is this? I could rewrite this, so everything I've done so far, so the transformation of x is equal to that, which I'm just using our properties of linear transformations. All linear transformations, this has to be true for them."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Plus all the way to xn times the transformation of en. Now, what is this? I could rewrite this, so everything I've done so far, so the transformation of x is equal to that, which I'm just using our properties of linear transformations. All linear transformations, this has to be true for them. I get to this, and this is equivalent. If we view each of these as a column vector, this is equal to what? This is equal to the matrix where this is the first column, TE1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "All linear transformations, this has to be true for them. I get to this, and this is equivalent. If we view each of these as a column vector, this is equal to what? This is equal to the matrix where this is the first column, TE1. And then the second column is TE2. And then we go all the way to TEN times our x1, x2, all the way to xn. We've seen this multiple, multiple times."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to the matrix where this is the first column, TE1. And then the second column is TE2. And then we go all the way to TEN times our x1, x2, all the way to xn. We've seen this multiple, multiple times. Now, what's really, really, really neat about this is I just started with an arbitrary transformation. And I just showed that an arbitrary linear transformation of x can be rewritten as a product of a matrix where I'm taking that same linear transformation of each of our standard basis vectors, and I can construct that matrix, and multiplying that matrix times my x vector is the same thing as this transformation. So this is essentially showing you that all transformations, all linear transformations can be represented by, can be a matrix vector product."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "We've seen this multiple, multiple times. Now, what's really, really, really neat about this is I just started with an arbitrary transformation. And I just showed that an arbitrary linear transformation of x can be rewritten as a product of a matrix where I'm taking that same linear transformation of each of our standard basis vectors, and I can construct that matrix, and multiplying that matrix times my x vector is the same thing as this transformation. So this is essentially showing you that all transformations, all linear transformations can be represented by, can be a matrix vector product. Not only did I show you that you can do it, but it's actually a fairly straightforward thing to do. This is actually a pretty simple operation to do. Let me show you an example."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So this is essentially showing you that all transformations, all linear transformations can be represented by, can be a matrix vector product. Not only did I show you that you can do it, but it's actually a fairly straightforward thing to do. This is actually a pretty simple operation to do. Let me show you an example. And this is what, I mean, I don't know, I think this is super neat. Let's say that I just, I'm just going to make up some transformation, let's say I have a transformation, and it's a mapping between, let's make it extra interesting, between R2 and R3. And let's say my transformation, say, let's say that T of x1, x2 is equal to, let's say the first entry is x1 plus 3x2, the second entry is 5x2 minus x1, and let's say the third entry is 4x1 plus x2."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you an example. And this is what, I mean, I don't know, I think this is super neat. Let's say that I just, I'm just going to make up some transformation, let's say I have a transformation, and it's a mapping between, let's make it extra interesting, between R2 and R3. And let's say my transformation, say, let's say that T of x1, x2 is equal to, let's say the first entry is x1 plus 3x2, the second entry is 5x2 minus x1, and let's say the third entry is 4x1 plus x2. This is a mapping. I could have written it like this. I could write T of any vector in R2, x1, x2, is equal to, and maybe this is just redundant, but I think you get the idea, I like this notation better, x1 plus 3x2, 5x2 minus x1, and then 4x1 plus x2."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And let's say my transformation, say, let's say that T of x1, x2 is equal to, let's say the first entry is x1 plus 3x2, the second entry is 5x2 minus x1, and let's say the third entry is 4x1 plus x2. This is a mapping. I could have written it like this. I could write T of any vector in R2, x1, x2, is equal to, and maybe this is just redundant, but I think you get the idea, I like this notation better, x1 plus 3x2, 5x2 minus x1, and then 4x1 plus x2. This statement and this statement I just wrote are equivalent, and I like to visualize this a little bit more. Now I just told you that I can represent this transformation as a matrix vector product. How do I do it?"}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "I could write T of any vector in R2, x1, x2, is equal to, and maybe this is just redundant, but I think you get the idea, I like this notation better, x1 plus 3x2, 5x2 minus x1, and then 4x1 plus x2. This statement and this statement I just wrote are equivalent, and I like to visualize this a little bit more. Now I just told you that I can represent this transformation as a matrix vector product. How do I do it? Well, what I do is I take the transformation of this guy. So what am I, my domain right here is R2, and I produce a vector that's going to be an Rn. So what I do is, let's say, so I'm concerned with multiplying things times vectors in R2."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "How do I do it? Well, what I do is I take the transformation of this guy. So what am I, my domain right here is R2, and I produce a vector that's going to be an Rn. So what I do is, let's say, so I'm concerned with multiplying things times vectors in R2. So what we're going to do is we're going to start with the identity matrix, identity 2, because that's my domain, and it just looks like this, 1, 0, 0, 1. I'm just going to start with that. And all I do is I apply my transformation to each of the columns, each of my standard bases."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So what I do is, let's say, so I'm concerned with multiplying things times vectors in R2. So what we're going to do is we're going to start with the identity matrix, identity 2, because that's my domain, and it just looks like this, 1, 0, 0, 1. I'm just going to start with that. And all I do is I apply my transformation to each of the columns, each of my standard bases. These are the standard bases for R2. And you might be wondering, I showed you that they're bases, how do I know that they're standard? Why are they called the standard bases?"}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And all I do is I apply my transformation to each of the columns, each of my standard bases. These are the standard bases for R2. And you might be wondering, I showed you that they're bases, how do I know that they're standard? Why are they called the standard bases? And I haven't covered this in a lot of detail right yet, but you could take the dot product of any of these guys with any of the other guys, and you'll see that they're all orthogonal to each other. The dot product of any one of these columns with the other is always 0. So that's a nice clue."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Why are they called the standard bases? And I haven't covered this in a lot of detail right yet, but you could take the dot product of any of these guys with any of the other guys, and you'll see that they're all orthogonal to each other. The dot product of any one of these columns with the other is always 0. So that's a nice clue. And they all have length of 1, so that's a nice reason why they're called the standard bases. But anyway, back to our attempt to represent this transformation as a matrix vector product. So we say, look, our domain is in R2."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So that's a nice clue. And they all have length of 1, so that's a nice reason why they're called the standard bases. But anyway, back to our attempt to represent this transformation as a matrix vector product. So we say, look, our domain is in R2. So let's start with I2, or we could call it our 2 by 2 identity matrix. And let's apply the transformation to each of its column vectors, where each of its column vectors are a vector in the standard bases for R2. So I'm going to write it like this."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So we say, look, our domain is in R2. So let's start with I2, or we could call it our 2 by 2 identity matrix. And let's apply the transformation to each of its column vectors, where each of its column vectors are a vector in the standard bases for R2. So I'm going to write it like this. T of, the first column is T of this column, and then the second column is going to be T of 0, 1. And I know I'm getting messier with my handwriting, but what is T of the vector 1, 0? Well, we just go here, and we construct another vector."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to write it like this. T of, the first column is T of this column, and then the second column is going to be T of 0, 1. And I know I'm getting messier with my handwriting, but what is T of the vector 1, 0? Well, we just go here, and we construct another vector. So we get 1 plus 3 times 0 is 1. Then we get 5 times 0 minus 1, so that's minus 1, right? x2 is 0 in this case."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Well, we just go here, and we construct another vector. So we get 1 plus 3 times 0 is 1. Then we get 5 times 0 minus 1, so that's minus 1, right? x2 is 0 in this case. And then we get 4 times 1 plus 0, so that's just 4. So that's T of 1, 0. And then what is T of 0, 1?"}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "x2 is 0 in this case. And then we get 4 times 1 plus 0, so that's just 4. So that's T of 1, 0. And then what is T of 0, 1? T of 0, 1 is equal to, so we have 0 plus 3 times 1 is 3. Then we have 0 minus 1 is minus 1. Let me make sure I did this one right."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And then what is T of 0, 1? T of 0, 1 is equal to, so we have 0 plus 3 times 1 is 3. Then we have 0 minus 1 is minus 1. Let me make sure I did this one right. What was this? This was 5 times 0 minus 1. And then 5 times 0 minus x1, which is 1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "Let me make sure I did this one right. What was this? This was 5 times 0 minus 1. And then 5 times 0 minus x1, which is 1. Now this case, it's 5 times, oh, I have to be careful. This is 5 times x2. x2 is 1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And then 5 times 0 minus x1, which is 1. Now this case, it's 5 times, oh, I have to be careful. This is 5 times x2. x2 is 1. So 5 times 1 minus 0. So it's 5. And then I have 4 times 0 plus x2, plus 1."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "x2 is 1. So 5 times 1 minus 0. So it's 5. And then I have 4 times 0 plus x2, plus 1. And I just showed you, if I replace each of these standard basis vectors with the transformation of them, what do I get? I get this vector right here. So I already figured out what they are."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And then I have 4 times 0 plus x2, plus 1. And I just showed you, if I replace each of these standard basis vectors with the transformation of them, what do I get? I get this vector right here. So I already figured out what they are. If I take this guy and evaluate it, it's the vector 1 minus 1, 4. And then this guy is a vector 3, 5, and 1. So what we just did, and this is, I don't know, for some reason I find this to be pretty amazing."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So I already figured out what they are. If I take this guy and evaluate it, it's the vector 1 minus 1, 4. And then this guy is a vector 3, 5, and 1. So what we just did, and this is, I don't know, for some reason I find this to be pretty amazing. We can now rewrite this transformation here as the product of any vector. So if we define this to be equal to a, we could write it this way. We can now write our transformation."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "So what we just did, and this is, I don't know, for some reason I find this to be pretty amazing. We can now rewrite this transformation here as the product of any vector. So if we define this to be equal to a, we could write it this way. We can now write our transformation. Our transformation of x1, x2 can now be rewritten as the product of this vector. I'll write it in green. The vector 1, 3, minus 1, 5, 4, 1 times our input vector, x1, x2, which is super cool."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "We can now write our transformation. Our transformation of x1, x2 can now be rewritten as the product of this vector. I'll write it in green. The vector 1, 3, minus 1, 5, 4, 1 times our input vector, x1, x2, which is super cool. Because now we just have to do a matrix multiplication instead of this. And if we have some processor that does this super fast, we can then use that. And what's really, I don't know, I think it's especially elegant because what happens here is we applied the transformations to each of the columns of a 2 by 2 matrix."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "The vector 1, 3, minus 1, 5, 4, 1 times our input vector, x1, x2, which is super cool. Because now we just have to do a matrix multiplication instead of this. And if we have some processor that does this super fast, we can then use that. And what's really, I don't know, I think it's especially elegant because what happens here is we applied the transformations to each of the columns of a 2 by 2 matrix. And we got a 3 by 2 matrix. And we know what happens when you multiply a 3 by 2 matrix times a vector that's in R2, or you can almost view this as a 2 by 1 matrix. You're going to get a vector that is in R3."}, {"video_title": "Linear transformations as matrix vector products Linear Algebra Khan Academy.mp3", "Sentence": "And what's really, I don't know, I think it's especially elegant because what happens here is we applied the transformations to each of the columns of a 2 by 2 matrix. And we got a 3 by 2 matrix. And we know what happens when you multiply a 3 by 2 matrix times a vector that's in R2, or you can almost view this as a 2 by 1 matrix. You're going to get a vector that is in R3. Because you're going to have these guys times that guy is going to be the first term, these guys are going to be the second term, these guys times those guys are going to be the third term. So by kind of creating this 3 by 2 matrix, we have actually created a mapping from R2 to R3. Anyway, for some reason I find this to be especially neat."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I made a bit of a hand-wavy argument. It was at the end of the video, and I was tired. It was actually at the end of the day. And I thought it'd be worthwhile to maybe flush this out a little bit, because it's an important takeaway, and it'll help us understand everything we've learned a little bit better. So let's just understand what the, actually I'm going to start with the rank of A transpose. The rank of A transpose is equal to the dimension of the column space of A transpose. That's the definition of the rank."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I thought it'd be worthwhile to maybe flush this out a little bit, because it's an important takeaway, and it'll help us understand everything we've learned a little bit better. So let's just understand what the, actually I'm going to start with the rank of A transpose. The rank of A transpose is equal to the dimension of the column space of A transpose. That's the definition of the rank. And what is it? The dimension of the column space of A transpose is the number of basis vectors for the column space of A transpose. That's what dimension is."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the definition of the rank. And what is it? The dimension of the column space of A transpose is the number of basis vectors for the column space of A transpose. That's what dimension is. For any subspace, you figure out how many basis vectors you need in that subspace, and you count them, and that's your dimension. So it's the number of basis vectors for the column space of A transpose, which is, of course, the same thing. This thing we've seen multiple times is the same thing as the row space of A."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's what dimension is. For any subspace, you figure out how many basis vectors you need in that subspace, and you count them, and that's your dimension. So it's the number of basis vectors for the column space of A transpose, which is, of course, the same thing. This thing we've seen multiple times is the same thing as the row space of A. The columns of A transpose are the same thing as the rows of A, just because you switch the rows and the columns. Now, how can we figure out the number of basis vectors we need for the column space of A transpose, or the row space of A? So let's just think about what the column space of A transpose is telling us."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This thing we've seen multiple times is the same thing as the row space of A. The columns of A transpose are the same thing as the rows of A, just because you switch the rows and the columns. Now, how can we figure out the number of basis vectors we need for the column space of A transpose, or the row space of A? So let's just think about what the column space of A transpose is telling us. So it's equivalent to, so let's say, let me draw A like this. Let me draw A. I have some matrix A. Let's say it's an m by n matrix."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's just think about what the column space of A transpose is telling us. So it's equivalent to, so let's say, let me draw A like this. Let me draw A. I have some matrix A. Let's say it's an m by n matrix. And let me just write it as a bunch of row vectors. I could also write it as a bunch of column vectors, but right now let's just stick to the row vectors. So we'd have row 1, the transpose of column vectors."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's an m by n matrix. And let me just write it as a bunch of row vectors. I could also write it as a bunch of column vectors, but right now let's just stick to the row vectors. So we'd have row 1, the transpose of column vectors. We could just write that's row 1, and we're going to have row 2. And we're going to go all the way down to row m. It's an m by n matrix. Each of these vectors are members of Rn, because they're going to have n entries in them, because we have n columns."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we'd have row 1, the transpose of column vectors. We could just write that's row 1, and we're going to have row 2. And we're going to go all the way down to row m. It's an m by n matrix. Each of these vectors are members of Rn, because they're going to have n entries in them, because we have n columns. So that's what A is going to look like. A is going to look like that. And then A transpose, all of these rows are going to become columns."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Each of these vectors are members of Rn, because they're going to have n entries in them, because we have n columns. So that's what A is going to look like. A is going to look like that. And then A transpose, all of these rows are going to become columns. A transpose is going to look like this. R1, R2, all the way to Rm. And this is, of course, going to be an n by m matrix."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then A transpose, all of these rows are going to become columns. A transpose is going to look like this. R1, R2, all the way to Rm. And this is, of course, going to be an n by m matrix. You swap these out. So all these rows are going to be columns. And obviously, the column space, or maybe not so obviously, the column space of A transpose is equal to the span of R1, R2, all the way to Rm."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is, of course, going to be an n by m matrix. You swap these out. So all these rows are going to be columns. And obviously, the column space, or maybe not so obviously, the column space of A transpose is equal to the span of R1, R2, all the way to Rm. It's equal to the span of these things, or you could equivalently call it, it's equal to the span of the rows of A, and that's why it's also called the row space. So this is equal to the span of the rows of A. These two things are equivalent."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And obviously, the column space, or maybe not so obviously, the column space of A transpose is equal to the span of R1, R2, all the way to Rm. It's equal to the span of these things, or you could equivalently call it, it's equal to the span of the rows of A, and that's why it's also called the row space. So this is equal to the span of the rows of A. These two things are equivalent. Now, these are the span. That means this is some subspace that's all of the linear combinations of these columns, or all the linear combinations of these rows. If we want the basis for it, we want to find a minimum set of linearly independent vectors that we could use to construct any of these columns, or that we could use to construct any of these rows right here."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These two things are equivalent. Now, these are the span. That means this is some subspace that's all of the linear combinations of these columns, or all the linear combinations of these rows. If we want the basis for it, we want to find a minimum set of linearly independent vectors that we could use to construct any of these columns, or that we could use to construct any of these rows right here. Now, what happens when we put A into reduced row echelon form? So let's just, we do a bunch of row operations to put it into reduced row echelon form. You do a bunch of row operations, and you eventually you'll get something like this."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we want the basis for it, we want to find a minimum set of linearly independent vectors that we could use to construct any of these columns, or that we could use to construct any of these rows right here. Now, what happens when we put A into reduced row echelon form? So let's just, we do a bunch of row operations to put it into reduced row echelon form. You do a bunch of row operations, and you eventually you'll get something like this. You'll get the reduced row echelon form of A. The reduced row echelon form of A is going to look something like this. You're going to have some pivot rows, some rows that have pivot entries."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You do a bunch of row operations, and you eventually you'll get something like this. You'll get the reduced row echelon form of A. The reduced row echelon form of A is going to look something like this. You're going to have some pivot rows, some rows that have pivot entries. Let's say that's one of them. And let's say that's one of them. This will all have 0's all the way down."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're going to have some pivot rows, some rows that have pivot entries. Let's say that's one of them. And let's say that's one of them. This will all have 0's all the way down. This one will have 0's. Your pivot entry has to be the only non-zero entry in its column. And everything to the left of it also has to be 0."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This will all have 0's all the way down. This one will have 0's. Your pivot entry has to be the only non-zero entry in its column. And everything to the left of it also has to be 0. Let's say that this one isn't. These are some non-zero values. These are 0."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And everything to the left of it also has to be 0. Let's say that this one isn't. These are some non-zero values. These are 0. So we have another pivot entry over here. Everything else is 0. And let's say everything else are non-pivot entries."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are 0. So we have another pivot entry over here. Everything else is 0. And let's say everything else are non-pivot entries. So you come here, and you add a certain number of pivot rows, or a certain number of pivot entries. And you got there by performing linear row operations on these guys. So those linear row operations, I take 3 times row 2, and I add it to row 1, and that's going to become my new row 2."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say everything else are non-pivot entries. So you come here, and you add a certain number of pivot rows, or a certain number of pivot entries. And you got there by performing linear row operations on these guys. So those linear row operations, I take 3 times row 2, and I add it to row 1, and that's going to become my new row 2. And you keep doing that, and you get these things here. So these things here are linear combinations of those guys. Or another way to do it, you could reverse those row operations."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So those linear row operations, I take 3 times row 2, and I add it to row 1, and that's going to become my new row 2. And you keep doing that, and you get these things here. So these things here are linear combinations of those guys. Or another way to do it, you could reverse those row operations. I could start with these guys right here, and I could just as easily perform the reverse row operations. Any linear operation, you could perform the reverse of it. We've seen that multiple times."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to do it, you could reverse those row operations. I could start with these guys right here, and I could just as easily perform the reverse row operations. Any linear operation, you could perform the reverse of it. We've seen that multiple times. You could perform row operations with these guys to get all of these guys. Or another way to view it is, these vectors here, these row vectors right here, they span all of these. Or all of these row vectors can be represented as linear combinations of your pivot rows right here."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've seen that multiple times. You could perform row operations with these guys to get all of these guys. Or another way to view it is, these vectors here, these row vectors right here, they span all of these. Or all of these row vectors can be represented as linear combinations of your pivot rows right here. Obviously, you're going to have your non-pivot rows are going to be all 0's. And those are useless. But your pivot rows, if you take linear combinations of them, you can clearly do reverse row echelon form and get back to your matrix."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or all of these row vectors can be represented as linear combinations of your pivot rows right here. Obviously, you're going to have your non-pivot rows are going to be all 0's. And those are useless. But your pivot rows, if you take linear combinations of them, you can clearly do reverse row echelon form and get back to your matrix. So all of these guys can be represented as linear combinations of them. And all of these pivot entries are by definition, well, almost by definition, they're linearly independent. Because I've got a 1 here, no one else has a 1 there, so this guy can definitely not be represented as a linear combination of the other guy."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But your pivot rows, if you take linear combinations of them, you can clearly do reverse row echelon form and get back to your matrix. So all of these guys can be represented as linear combinations of them. And all of these pivot entries are by definition, well, almost by definition, they're linearly independent. Because I've got a 1 here, no one else has a 1 there, so this guy can definitely not be represented as a linear combination of the other guy. So why am I going through this whole exercise? Well, we started off saying we wanted a basis for the row space, we wanted some minimum set of linearly independent vectors that spans everything that these guys can span. Well, if all of these guys can be represented as linear combinations of these row vectors in reduced row echelon form, or these pivot rows in reduced row echelon form, and these guys are all linearly independent, then they are a reasonable basis."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because I've got a 1 here, no one else has a 1 there, so this guy can definitely not be represented as a linear combination of the other guy. So why am I going through this whole exercise? Well, we started off saying we wanted a basis for the row space, we wanted some minimum set of linearly independent vectors that spans everything that these guys can span. Well, if all of these guys can be represented as linear combinations of these row vectors in reduced row echelon form, or these pivot rows in reduced row echelon form, and these guys are all linearly independent, then they are a reasonable basis. So these pivot rows right here, that's one of them. This is the second one. This is the third one."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, if all of these guys can be represented as linear combinations of these row vectors in reduced row echelon form, or these pivot rows in reduced row echelon form, and these guys are all linearly independent, then they are a reasonable basis. So these pivot rows right here, that's one of them. This is the second one. This is the third one. Maybe they're the only three. This is just my particular example. That would be a suitable basis for the row space."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the third one. Maybe they're the only three. This is just my particular example. That would be a suitable basis for the row space. Let me write this down. The pivot rows in reduced row echelon form of A are a basis for the row space of A. And the row space of A is the same thing, or the column space of A transpose."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That would be a suitable basis for the row space. Let me write this down. The pivot rows in reduced row echelon form of A are a basis for the row space of A. And the row space of A is the same thing, or the column space of A transpose. Row space of A is the same thing as the column space of A transpose. We've seen that multiple times. Now, so if we want to know the dimension of your column space, we just count the number of pivot rows you have."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the row space of A is the same thing, or the column space of A transpose. Row space of A is the same thing as the column space of A transpose. We've seen that multiple times. Now, so if we want to know the dimension of your column space, we just count the number of pivot rows you have. So you just count the number of pivot rows. So the dimension of your row space, which is the same thing as a column space of A transpose, is going to be the number of pivot rows you have in reduced row echelon form. Or, even simpler, the number of pivot entries you have, because every pivot entry has a pivot row."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, so if we want to know the dimension of your column space, we just count the number of pivot rows you have. So you just count the number of pivot rows. So the dimension of your row space, which is the same thing as a column space of A transpose, is going to be the number of pivot rows you have in reduced row echelon form. Or, even simpler, the number of pivot entries you have, because every pivot entry has a pivot row. So we can say, we can write that the rank of A transpose is equal to the number of pivot entries in reduced row echelon form of A. Right? Because every pivot entry corresponds to a pivot row."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or, even simpler, the number of pivot entries you have, because every pivot entry has a pivot row. So we can say, we can write that the rank of A transpose is equal to the number of pivot entries in reduced row echelon form of A. Right? Because every pivot entry corresponds to a pivot row. Those pivot rows are a suitable basis for the entire row space, because every row can be made with a linear combination of these guys. And since all of these can be, then anything that these guys can construct, these guys can construct. Fair enough."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because every pivot entry corresponds to a pivot row. Those pivot rows are a suitable basis for the entire row space, because every row can be made with a linear combination of these guys. And since all of these can be, then anything that these guys can construct, these guys can construct. Fair enough. Now what is the rank of A? This is the rank of A transpose that we've been dealing with so far. The rank of A is equal to the dimension of the column space of A."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now what is the rank of A? This is the rank of A transpose that we've been dealing with so far. The rank of A is equal to the dimension of the column space of A. Or, you could say, it's the number of vectors in the basis for the column space of A. So if we take that same matrix A that we used above, and instead we write it as a bunch of column vectors. So C1, C2, all the way to Cn."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The rank of A is equal to the dimension of the column space of A. Or, you could say, it's the number of vectors in the basis for the column space of A. So if we take that same matrix A that we used above, and instead we write it as a bunch of column vectors. So C1, C2, all the way to Cn. We have n columns right there. The column space is essentially the subspace that's spanned by all of these characters right here. Right?"}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So C1, C2, all the way to Cn. We have n columns right there. The column space is essentially the subspace that's spanned by all of these characters right here. Right? Spanned by each of these column vectors. So the column space of A is equal to the span of C1, C2, all the way to Cn. That's the definition of it."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Right? Spanned by each of these column vectors. So the column space of A is equal to the span of C1, C2, all the way to Cn. That's the definition of it. But we want to know the number of basis vectors. And we've seen before, we've done this multiple times, the basis vectors, a suitable basis vectors could be, if you put this into reduced row echelon form, and you have some pivot entries, and their corresponding pivot columns. So some pivot entries with their corresponding pivot columns, just like that."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the definition of it. But we want to know the number of basis vectors. And we've seen before, we've done this multiple times, the basis vectors, a suitable basis vectors could be, if you put this into reduced row echelon form, and you have some pivot entries, and their corresponding pivot columns. So some pivot entries with their corresponding pivot columns, just like that. Maybe that's like that, and then maybe this one isn't one, and then this one is. So you have a certain number of pivot columns. Let me do them in another color right here."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So some pivot entries with their corresponding pivot columns, just like that. Maybe that's like that, and then maybe this one isn't one, and then this one is. So you have a certain number of pivot columns. Let me do them in another color right here. When you put A into reduced row echelon form, we learned that the basis vectors, or the basis columns that form a basis for your column space, are the columns that correspond to the pivot columns. So the first column here is a pivot column, so this guy could be a basis vector. The second column is, so this guy could be a pivot vector."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do them in another color right here. When you put A into reduced row echelon form, we learned that the basis vectors, or the basis columns that form a basis for your column space, are the columns that correspond to the pivot columns. So the first column here is a pivot column, so this guy could be a basis vector. The second column is, so this guy could be a pivot vector. And maybe the fourth one right here, so this guy could be a pivot vector. So in general, you just say, hey, if you want to count the number of basis vectors, because we don't even have to know what they are to figure out the rank, we just have to know the number they are. Well, you say, well, for every pivot column here, we have a basis vector over there, so we could just count the number of pivot columns."}, {"video_title": "rank(a) = rank(transpose of a) Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The second column is, so this guy could be a pivot vector. And maybe the fourth one right here, so this guy could be a pivot vector. So in general, you just say, hey, if you want to count the number of basis vectors, because we don't even have to know what they are to figure out the rank, we just have to know the number they are. Well, you say, well, for every pivot column here, we have a basis vector over there, so we could just count the number of pivot columns. But the number of pivot columns is equivalent to just the number of pivot entries we have, because every pivot entry gets its own column. So we could say that the rank of A is equal to the number of pivot entries in the reduced row echelon form of A. And as you can see very clearly, that's the exact same thing that we deduced was equivalent to the rank of A transpose, or the dimension of the column space of A transpose, or the dimension of the row space of A."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "One term you're going to hear a lot of in these videos and in linear algebra in general is the idea of a linear combination. Linear combination. And all a linear combination of vectors are, they're just a linear combination. Let me show you what that means. So let's say I have a couple of vectors, v1, v2, and it goes all the way to vn. And they're all in, you know, it can be in R2 or Rn. Let's say that they're all in Rn."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you what that means. So let's say I have a couple of vectors, v1, v2, and it goes all the way to vn. And they're all in, you know, it can be in R2 or Rn. Let's say that they're all in Rn. You know, they're in some dimension of real space, I guess we could call it. But the idea is fairly simple. A linear combination of these vectors means you just add up the vectors."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that they're all in Rn. You know, they're in some dimension of real space, I guess we could call it. But the idea is fairly simple. A linear combination of these vectors means you just add up the vectors. It's some combination of a sum of the vectors. So v1 plus v2 plus all the way to vn. But you scale them by arbitrary constants."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A linear combination of these vectors means you just add up the vectors. It's some combination of a sum of the vectors. So v1 plus v2 plus all the way to vn. But you scale them by arbitrary constants. So you scale them by c1, c2, all the way to cn. Where everything from c1 to cn are all a member of the real numbers. That's all a linear combination is."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But you scale them by arbitrary constants. So you scale them by c1, c2, all the way to cn. Where everything from c1 to cn are all a member of the real numbers. That's all a linear combination is. Let me show you a concrete example of linear combinations. Let me make the vector, let me define the vector a to be equal to, and these are all bolded. These purple v's are all bolded just because those are vectors."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's all a linear combination is. Let me show you a concrete example of linear combinations. Let me make the vector, let me define the vector a to be equal to, and these are all bolded. These purple v's are all bolded just because those are vectors. But sometimes it's kind of onerous to keep bolding things. So let's just say I define the vector a to be equal to 1, 2. And I define the vector b to be equal to 0, 3."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These purple v's are all bolded just because those are vectors. But sometimes it's kind of onerous to keep bolding things. So let's just say I define the vector a to be equal to 1, 2. And I define the vector b to be equal to 0, 3. What is a linear combination of a and b? Well, I could just, it could be any constant times a plus any constant times b. So it could be 0 times a plus, well it could be 0 times a plus 0 times b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I define the vector b to be equal to 0, 3. What is a linear combination of a and b? Well, I could just, it could be any constant times a plus any constant times b. So it could be 0 times a plus, well it could be 0 times a plus 0 times b. Which of course would be what? That would be 0 times a would be 0, 0. And 0, 0, that would be the 0 vector."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it could be 0 times a plus, well it could be 0 times a plus 0 times b. Which of course would be what? That would be 0 times a would be 0, 0. And 0, 0, that would be the 0 vector. But this is a completely valid linear combination. And we can denote the 0 vector by just a big bold 0 like that. We could also, this is almost, I could do 3 times a. I'm just picking these numbers at random."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And 0, 0, that would be the 0 vector. But this is a completely valid linear combination. And we can denote the 0 vector by just a big bold 0 like that. We could also, this is almost, I could do 3 times a. I'm just picking these numbers at random. 3 times a plus, let me do a negative number just for fun. So let's do plus minus 2 times b. What is that equal to?"}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We could also, this is almost, I could do 3 times a. I'm just picking these numbers at random. 3 times a plus, let me do a negative number just for fun. So let's do plus minus 2 times b. What is that equal to? Well, let's figure it out. Let me write it out. It's minus 3 minus 2 times 0."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is that equal to? Well, let's figure it out. Let me write it out. It's minus 3 minus 2 times 0. So minus 0. And this 3 times 2 is 6. 6 minus 2 times 3."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's minus 3 minus 2 times 0. So minus 0. And this 3 times 2 is 6. 6 minus 2 times 3. So minus 6. So it's the vector 3, 0. This is a linear combination of a and b. I could keep putting in a bunch of random real numbers here and here."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "6 minus 2 times 3. So minus 6. So it's the vector 3, 0. This is a linear combination of a and b. I could keep putting in a bunch of random real numbers here and here. And I'll just get a bunch of different linear combinations of my vectors a and b. If I had a third vector here, if I had vector c, and maybe that was just 7, 2. Then I could add that to the mix."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a linear combination of a and b. I could keep putting in a bunch of random real numbers here and here. And I'll just get a bunch of different linear combinations of my vectors a and b. If I had a third vector here, if I had vector c, and maybe that was just 7, 2. Then I could add that to the mix. And I could throw in plus 8 times vector c. These are all just linear combinations. Now why do we just call them combinations? Why do we have to add that little linear prefix there?"}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then I could add that to the mix. And I could throw in plus 8 times vector c. These are all just linear combinations. Now why do we just call them combinations? Why do we have to add that little linear prefix there? Because we're just scaling them up. We're not multiplying the vectors times each other. We haven't even defined what it means to multiply a vector."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Why do we have to add that little linear prefix there? Because we're just scaling them up. We're not multiplying the vectors times each other. We haven't even defined what it means to multiply a vector. And there's actually several ways to do it. But we can't square a vector. And we haven't even defined what this means yet."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We haven't even defined what it means to multiply a vector. And there's actually several ways to do it. But we can't square a vector. And we haven't even defined what this means yet. But this would all of a sudden make it nonlinear in some form. So all we're doing is we're adding the vectors and we're just scaling them up by some scaling factor. So that's why it's called a linear combination."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we haven't even defined what this means yet. But this would all of a sudden make it nonlinear in some form. So all we're doing is we're adding the vectors and we're just scaling them up by some scaling factor. So that's why it's called a linear combination. Now you might say, hey Sal, why are you even introducing this idea of a linear combination? Because I want to introduce the idea, and this is an idea that confounds most students when it's first taught. I think it's just the very nature that it's taught."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's why it's called a linear combination. Now you might say, hey Sal, why are you even introducing this idea of a linear combination? Because I want to introduce the idea, and this is an idea that confounds most students when it's first taught. I think it's just the very nature that it's taught. Over here I just kept putting different numbers for the weights. I guess we could call them for c1 and c2 in this combination of a and b. Let's ignore c for a little bit."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I think it's just the very nature that it's taught. Over here I just kept putting different numbers for the weights. I guess we could call them for c1 and c2 in this combination of a and b. Let's ignore c for a little bit. I just put in a bunch of different numbers there. But it begs the question, what is the set of all of the vectors I could have created? This is just one member of that set."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's ignore c for a little bit. I just put in a bunch of different numbers there. But it begs the question, what is the set of all of the vectors I could have created? This is just one member of that set. But what is the set of all of the vectors I could have created by taking linear combinations of a and b? So let me draw a and b here. Maybe we could think about it visually."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is just one member of that set. But what is the set of all of the vectors I could have created by taking linear combinations of a and b? So let me draw a and b here. Maybe we could think about it visually. And then maybe we could think about it mathematically. So let's say a and b. So a is 1, 2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe we could think about it visually. And then maybe we could think about it mathematically. So let's say a and b. So a is 1, 2. So 1, 2 looks like that. That's vector a. Let me do vector b in a different color."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So a is 1, 2. So 1, 2 looks like that. That's vector a. Let me do vector b in a different color. Let me do it in yellow. Vector b is 0, 3. So vector b looks like that."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do vector b in a different color. Let me do it in yellow. Vector b is 0, 3. So vector b looks like that. 0, 3. So what's the set of all of the vectors that I can represent by adding and subtracting these vectors? And we saw if we multiply them both by 0 and add them to each other, we end up there."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So vector b looks like that. 0, 3. So what's the set of all of the vectors that I can represent by adding and subtracting these vectors? And we saw if we multiply them both by 0 and add them to each other, we end up there. If we take 3 times a, that's the equivalent of scaling up a by 3. So you go 1a, 2a, 3a. So that's 3a."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we saw if we multiply them both by 0 and add them to each other, we end up there. If we take 3 times a, that's the equivalent of scaling up a by 3. So you go 1a, 2a, 3a. So that's 3a. 3 times a will look like that. So this vector is 3a. And then we add it to that 2b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's 3a. 3 times a will look like that. So this vector is 3a. And then we add it to that 2b. Oh, no. We subtracted 2b from that. So minus b looks like this."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we add it to that 2b. Oh, no. We subtracted 2b from that. So minus b looks like this. Minus b looks like this. Minus 2b looks like this. This is minus 2b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So minus b looks like this. Minus b looks like this. Minus 2b looks like this. This is minus 2b. All the way in standard form, standard position. Minus 2b. So if you add 3a to minus 2b, we get to this vector."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is minus 2b. All the way in standard form, standard position. Minus 2b. So if you add 3a to minus 2b, we get to this vector. 3a to minus 2b. You get this vector right here. And that's exactly what we did when we solved it mathematically."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you add 3a to minus 2b, we get to this vector. 3a to minus 2b. You get this vector right here. And that's exactly what we did when we solved it mathematically. You get the vector 3, 0. You get this vector right here, 3, 0. But this was just one combination, one linear combination of a and b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And that's exactly what we did when we solved it mathematically. You get the vector 3, 0. You get this vector right here, 3, 0. But this was just one combination, one linear combination of a and b. Instead of multiplying a times 3, I could have multiplied a times 1.5 and just gotten right here. So 1.5a minus 2b would still look the same. It would look something like this."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But this was just one combination, one linear combination of a and b. Instead of multiplying a times 3, I could have multiplied a times 1.5 and just gotten right here. So 1.5a minus 2b would still look the same. It would look something like this. It would look something like this. It would look something like this. So our new vector that we would find would be something like this."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It would look something like this. It would look something like this. It would look something like this. So our new vector that we would find would be something like this. I just showed you I could find this vector with a linear combination. I could find this vector with a linear combination. It actually turns out that you can represent any vector in R2 with some linear combination of these vectors right here, a and b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So our new vector that we would find would be something like this. I just showed you I could find this vector with a linear combination. I could find this vector with a linear combination. It actually turns out that you can represent any vector in R2 with some linear combination of these vectors right here, a and b. Let's just think of an example or maybe just try a mental visual example. Wherever we want to go, we could go arbitrarily. We could scale a up by some arbitrary value."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It actually turns out that you can represent any vector in R2 with some linear combination of these vectors right here, a and b. Let's just think of an example or maybe just try a mental visual example. Wherever we want to go, we could go arbitrarily. We could scale a up by some arbitrary value. So this is some scaled up. So this is some weight on a. And then we can add up arbitrary multiples of b. b goes straight up and down."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We could scale a up by some arbitrary value. So this is some scaled up. So this is some weight on a. And then we can add up arbitrary multiples of b. b goes straight up and down. So we could add up arbitrary multiples of b to that. So we could get any point on this line right there. Now if we scaled a up a little bit more, and then added any multiple b, we'd get anything on that line."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we can add up arbitrary multiples of b. b goes straight up and down. So we could add up arbitrary multiples of b to that. So we could get any point on this line right there. Now if we scaled a up a little bit more, and then added any multiple b, we'd get anything on that line. If we multiplied a times a negative number, and then added a b in either direction, we'd get anything on that line. We can keep doing that. And there's no reason why we can't pick an arbitrary a that can fill in any of these gaps."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now if we scaled a up a little bit more, and then added any multiple b, we'd get anything on that line. If we multiplied a times a negative number, and then added a b in either direction, we'd get anything on that line. We can keep doing that. And there's no reason why we can't pick an arbitrary a that can fill in any of these gaps. If we want a point here, we just take a little smaller a, and then we can add all the b's that fill up all of that line. So we can fill up any point in R2 with the combinations of a and b. So what we can write here is that the span of the vectors a and b equals R2, or equals all the vectors in R2, which is all the tuples."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And there's no reason why we can't pick an arbitrary a that can fill in any of these gaps. If we want a point here, we just take a little smaller a, and then we can add all the b's that fill up all of that line. So we can fill up any point in R2 with the combinations of a and b. So what we can write here is that the span of the vectors a and b equals R2, or equals all the vectors in R2, which is all the tuples. R2 is all the tuples made of two ordered tuples of two real numbers. So it equals all of R2. This just means that I can represent any vector in R2 with some linear combination of a and b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what we can write here is that the span of the vectors a and b equals R2, or equals all the vectors in R2, which is all the tuples. R2 is all the tuples made of two ordered tuples of two real numbers. So it equals all of R2. This just means that I can represent any vector in R2 with some linear combination of a and b. And you're like, hey, can't I do that with any two vectors? Well, what if a and b were the vector, let's say the vector 2, 2 was a. So a is equal to 2, 2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This just means that I can represent any vector in R2 with some linear combination of a and b. And you're like, hey, can't I do that with any two vectors? Well, what if a and b were the vector, let's say the vector 2, 2 was a. So a is equal to 2, 2. And let's say that b is the vector minus 2, minus 2. So b is that vector. So b is the vector minus 2, minus 2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So a is equal to 2, 2. And let's say that b is the vector minus 2, minus 2. So b is that vector. So b is the vector minus 2, minus 2. Now, can I represent any vector with these? Well, I can scale a up and down to get anywhere on this line. And then I can add b or anywhere to it."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So b is the vector minus 2, minus 2. Now, can I represent any vector with these? Well, I can scale a up and down to get anywhere on this line. And then I can add b or anywhere to it. And b is essentially going in the same direction, it's just in the opposite direction, but I can multiply it by negative and go anywhere in the line. So any combination of a and b will just end up on this line right here if I draw it in standard form. It'll be a vector with the same slope as either a or b, or same inclination, whatever you want to call it."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I can add b or anywhere to it. And b is essentially going in the same direction, it's just in the opposite direction, but I can multiply it by negative and go anywhere in the line. So any combination of a and b will just end up on this line right here if I draw it in standard form. It'll be a vector with the same slope as either a or b, or same inclination, whatever you want to call it. I could never, there's no combination of a and b that I could represent this vector. I could represent vector c. I just can't do it. I can add in standard form, I could just keep adding, scale up a, scale up b, put them heads to tails."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It'll be a vector with the same slope as either a or b, or same inclination, whatever you want to call it. I could never, there's no combination of a and b that I could represent this vector. I could represent vector c. I just can't do it. I can add in standard form, I could just keep adding, scale up a, scale up b, put them heads to tails. I'll just get to stuff on this line, I'll never get to this. So in this case, the span, and I want to be clear, this is for this particular a and b, not for the a and b. For this blue a and this yellow b, the span here is just this line."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can add in standard form, I could just keep adding, scale up a, scale up b, put them heads to tails. I'll just get to stuff on this line, I'll never get to this. So in this case, the span, and I want to be clear, this is for this particular a and b, not for the a and b. For this blue a and this yellow b, the span here is just this line. It's just this line. It's not all of R2. So this isn't just some kind of statement."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "For this blue a and this yellow b, the span here is just this line. It's just this line. It's not all of R2. So this isn't just some kind of statement. When I first did it with that example, I was like, oh, can't any two vectors represent anything in R2? Well, no. I just showed you two vectors, it can't represent that."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this isn't just some kind of statement. When I first did it with that example, I was like, oh, can't any two vectors represent anything in R2? Well, no. I just showed you two vectors, it can't represent that. So what is the span of the zero vector? I'll put even a cap over it. The zero vector, I make it really bold."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just showed you two vectors, it can't represent that. So what is the span of the zero vector? I'll put even a cap over it. The zero vector, I make it really bold. Well, the zero vector is just 0, 0. So I don't care what multiple I put on it, I could put the span of it is all of the linear combinations of this. So essentially, I could put arbitrary real numbers here, but I'm just going to end up with the 0, 0 vector."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The zero vector, I make it really bold. Well, the zero vector is just 0, 0. So I don't care what multiple I put on it, I could put the span of it is all of the linear combinations of this. So essentially, I could put arbitrary real numbers here, but I'm just going to end up with the 0, 0 vector. So the span of the zero vector is just the zero vector. The only vector I can get with a linear combination of this, the zero vector by itself, is just the zero vector itself. Likewise, if I take the span of just, let's say I go back to this example right here."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So essentially, I could put arbitrary real numbers here, but I'm just going to end up with the 0, 0 vector. So the span of the zero vector is just the zero vector. The only vector I can get with a linear combination of this, the zero vector by itself, is just the zero vector itself. Likewise, if I take the span of just, let's say I go back to this example right here. My a vector was right like that. Let me draw it in a better color. My a vector looked like that."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Likewise, if I take the span of just, let's say I go back to this example right here. My a vector was right like that. Let me draw it in a better color. My a vector looked like that. If I were to ask just what the span of a is, it's all the vectors you can get by creating a linear combination of just a. It's really just scaling. You can't even talk about combinations really."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My a vector looked like that. If I were to ask just what the span of a is, it's all the vectors you can get by creating a linear combination of just a. It's really just scaling. You can't even talk about combinations really. So it's just c times a, all of those vectors. We saw in the video where I parametrized or showed a parametric representation of a line, the span of just this vector a is the line that's formed when you just scale a up and down. So the span of a is just a line."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can't even talk about combinations really. So it's just c times a, all of those vectors. We saw in the video where I parametrized or showed a parametric representation of a line, the span of just this vector a is the line that's formed when you just scale a up and down. So the span of a is just a line. You have to have two vectors, and they can't be collinear in order to span all of R2. I haven't proven that to you yet, but we saw with this example, if you pick this a and this b, you can represent all of R2 with just these two vectors. Now, the two vectors that you're most familiar with that span R2 are, if you take a little physics class, you have your i and j unit vectors."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the span of a is just a line. You have to have two vectors, and they can't be collinear in order to span all of R2. I haven't proven that to you yet, but we saw with this example, if you pick this a and this b, you can represent all of R2 with just these two vectors. Now, the two vectors that you're most familiar with that span R2 are, if you take a little physics class, you have your i and j unit vectors. i and j. And in R notation, i, the unit vector i that you learn in physics class, would be the vector 1, 0. So this is i."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, the two vectors that you're most familiar with that span R2 are, if you take a little physics class, you have your i and j unit vectors. i and j. And in R notation, i, the unit vector i that you learn in physics class, would be the vector 1, 0. So this is i. That's the vector i. And then the vector j is the unit vector 0, 1. This is what you learn in physics class."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is i. That's the vector i. And then the vector j is the unit vector 0, 1. This is what you learn in physics class. Let me do it in a different color. This is j. j is that. And you learn that they're orthogonal."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is what you learn in physics class. Let me do it in a different color. This is j. j is that. And you learn that they're orthogonal. And we're going to talk a lot more about what orthogonality means, but in our traditional sense that we learn in high school, it means that they're 90 degrees. But you can clearly represent any angle or any vector in R2 by these two vectors. And the fact that they're orthogonal makes them extra nice."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you learn that they're orthogonal. And we're going to talk a lot more about what orthogonality means, but in our traditional sense that we learn in high school, it means that they're 90 degrees. But you can clearly represent any angle or any vector in R2 by these two vectors. And the fact that they're orthogonal makes them extra nice. And that's why these form, and I'm going to throw out a word here that I haven't defined yet, these form the basis. These form a basis for R2. In fact, you can represent anything in R2 by these two vectors alone."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the fact that they're orthogonal makes them extra nice. And that's why these form, and I'm going to throw out a word here that I haven't defined yet, these form the basis. These form a basis for R2. In fact, you can represent anything in R2 by these two vectors alone. And I'm not going to even define what basis is. That's going to be a future video. But let me just write the formal mathy definition of span, just so you're satisfied."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In fact, you can represent anything in R2 by these two vectors alone. And I'm not going to even define what basis is. That's going to be a future video. But let me just write the formal mathy definition of span, just so you're satisfied. So if I were to write the span of a set of vectors, v1, v2, all the way to vn, that just means the set of all of the vectors, where I have c1 times v1 plus c2 times v2, all the way to cn, let me scroll over, all the way to cn, vn. So this is a set of vectors, because I can pick my ci's to be any member of the real numbers, and that's true for i, so I should write 4i, being anywhere between 1 and n. All I'm saying is that, look, I can multiply each of these vectors by any value, any arbitrary value, real value, and then I can add them up. And now the set of all of the combinations, scaled up combinations I can get, that's the span of these vectors."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let me just write the formal mathy definition of span, just so you're satisfied. So if I were to write the span of a set of vectors, v1, v2, all the way to vn, that just means the set of all of the vectors, where I have c1 times v1 plus c2 times v2, all the way to cn, let me scroll over, all the way to cn, vn. So this is a set of vectors, because I can pick my ci's to be any member of the real numbers, and that's true for i, so I should write 4i, being anywhere between 1 and n. All I'm saying is that, look, I can multiply each of these vectors by any value, any arbitrary value, real value, and then I can add them up. And now the set of all of the combinations, scaled up combinations I can get, that's the span of these vectors. You can kind of view it as the space of all of the vectors that can be represented by a combination of these vectors right there. And so the word span, I think it does have an intuitive sense. If I say that my first example, I showed that those two vectors span, or a and b spans R2, I wrote it right here, that tells me that any vector in R2 can be represented by a linear combination of a and b."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now the set of all of the combinations, scaled up combinations I can get, that's the span of these vectors. You can kind of view it as the space of all of the vectors that can be represented by a combination of these vectors right there. And so the word span, I think it does have an intuitive sense. If I say that my first example, I showed that those two vectors span, or a and b spans R2, I wrote it right here, that tells me that any vector in R2 can be represented by a linear combination of a and b. And actually, just in case that visual kind of pseudo-proof doesn't do you justice, let me prove it to you algebraically. I'm telling you that I can take, let's say I want to represent, you know, I have some, let me rewrite my a's and b's again. So this was my vector a, it was 1, 2, and b was 0, 3."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I say that my first example, I showed that those two vectors span, or a and b spans R2, I wrote it right here, that tells me that any vector in R2 can be represented by a linear combination of a and b. And actually, just in case that visual kind of pseudo-proof doesn't do you justice, let me prove it to you algebraically. I'm telling you that I can take, let's say I want to represent, you know, I have some, let me rewrite my a's and b's again. So this was my vector a, it was 1, 2, and b was 0, 3. Let me remember that. So my vector a is 1, 2, and my vector b was 0, 3. Now my claim was that I can represent any point."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this was my vector a, it was 1, 2, and b was 0, 3. Let me remember that. So my vector a is 1, 2, and my vector b was 0, 3. Now my claim was that I can represent any point. Let's say I want to represent some arbitrary point x in R2. So its coordinates are x1 and x2. I need to be able to prove to you that I can get to any x1 and any x2 with some combination of these guys."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now my claim was that I can represent any point. Let's say I want to represent some arbitrary point x in R2. So its coordinates are x1 and x2. I need to be able to prove to you that I can get to any x1 and any x2 with some combination of these guys. So let's say that my combination, I say c1 times a plus c2 times b has to be equal to my vector x. So let me show you that I can always find a c1 or a c2 given that you give me some x's. So let's just write this right here with the actual vectors being represented in their kind of column form."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I need to be able to prove to you that I can get to any x1 and any x2 with some combination of these guys. So let's say that my combination, I say c1 times a plus c2 times b has to be equal to my vector x. So let me show you that I can always find a c1 or a c2 given that you give me some x's. So let's just write this right here with the actual vectors being represented in their kind of column form. So we have c1 times this vector plus c2 times the b vector, 0, 3, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's just write this right here with the actual vectors being represented in their kind of column form. So we have c1 times this vector plus c2 times the b vector, 0, 3, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2, c1 times 2 plus c2 times 3, 3c2, should be equal to x2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2, c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now if I can show you that I can always find c1's and c2's given any x1's and x2's, then I've proven that I can get to any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we also know that 2 times c2, c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now if I can show you that I can always find c1's and c2's given any x1's and x2's, then I've proven that I can get to any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns. This is just a 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is just a system of two unknowns. This is just a 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here. So we get minus 2c1. I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's multiply this equation up here by minus 2 and put it here. So we get minus 2c1. I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two, you get 3c2. These cancel out. You get 3."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two, you get 3c2. These cancel out. You get 3. Let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You get 3. Let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3. You get c2 is equal to 1 third x2 minus x1. Now we'd have to go substitute back in for c1. Well, we have this first equation right here that c1, this first equation just says c1 plus 0 is equal to x1."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or divide both sides by 3. You get c2 is equal to 1 third x2 minus x1. Now we'd have to go substitute back in for c1. Well, we have this first equation right here that c1, this first equation just says c1 plus 0 is equal to x1. So that one just gets us there. So c1 is equal to x1. So you give me any point in R2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we have this first equation right here that c1, this first equation just says c1 plus 0 is equal to x1. So that one just gets us there. So c1 is equal to x1. So you give me any point in R2. These are just two real numbers. And I can just perform this operation, and I'll tell you what weights to apply to a and b to get to that point. If you say, okay, what combination of a and b can get me to the point, let's say I want to get to the point, let me go back up here."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you give me any point in R2. These are just two real numbers. And I can just perform this operation, and I'll tell you what weights to apply to a and b to get to that point. If you say, okay, what combination of a and b can get me to the point, let's say I want to get to the point, let me go back up here. Well, it's way up there. Let's say I'm looking to get to the point 2, 2. So x1 is 2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you say, okay, what combination of a and b can get me to the point, let's say I want to get to the point, let me go back up here. Well, it's way up there. Let's say I'm looking to get to the point 2, 2. So x1 is 2. Let me write it down here. Say I'm trying to get to the point, the vector 2, 2. What combinations of a and b can be there?"}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So x1 is 2. Let me write it down here. Say I'm trying to get to the point, the vector 2, 2. What combinations of a and b can be there? Well, I know that c1 is equal to x1, so that's equal to 2. And c2 is equal to 1 third times 2 minus 2. So 2 minus 2 is 0."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What combinations of a and b can be there? Well, I know that c1 is equal to x1, so that's equal to 2. And c2 is equal to 1 third times 2 minus 2. So 2 minus 2 is 0. So c2 is equal to 0. So if I want to just get to the point 2, 2, I just multiply. Oh, I just realized this was looking suspicious."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 2 minus 2 is 0. So c2 is equal to 0. So if I want to just get to the point 2, 2, I just multiply. Oh, I just realized this was looking suspicious. I made a slight error here, and this was good that I actually tried it out with real numbers. Over here, when I had 3c2 is equal to x2 minus 2x1, I got rid of this 2 over here. There's a 2 over here."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Oh, I just realized this was looking suspicious. I made a slight error here, and this was good that I actually tried it out with real numbers. Over here, when I had 3c2 is equal to x2 minus 2x1, I got rid of this 2 over here. There's a 2 over here. I divide both sides by 3, I get 1 third times x2 minus 2x1. And that's why I was like, hey, this is looking strange. I had to take a moment of pause."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's a 2 over here. I divide both sides by 3, I get 1 third times x2 minus 2x1. And that's why I was like, hey, this is looking strange. I had to take a moment of pause. So let's go to my corrected definition of c2. c2 is equal to 1 third times x2, so 2 minus 2 times x1. So minus 2 times 2."}, {"video_title": "Linear combinations and span Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I had to take a moment of pause. So let's go to my corrected definition of c2. c2 is equal to 1 third times x2, so 2 minus 2 times x1. So minus 2 times 2. So it's equal to 1 third times 2 minus 4, which is equal to minus 2, so it's equal to minus 2 thirds. So if I multiply 2 times my vector a minus 2 thirds times my vector b, I will get to the vector 2, 2. And you can verify it for yourself."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So we could call this an m by n matrix. And what I want to do in this video is relate the linear independence, or linear dependence, of the column vectors of A to the null space of A. So first of all, what am I talking about, column vectors? Well, as you can see, there's n columns here, and we can view each of those as an m-dimensional vector. And so, let me do it this way. So you could view this one right over here. We could write that as v one, v one."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "Well, as you can see, there's n columns here, and we can view each of those as an m-dimensional vector. And so, let me do it this way. So you could view this one right over here. We could write that as v one, v one. This next one over here, this would be v two, v two. And you would have n of these, because we have n columns. And so this one right over here would be v n, v sub n. And so we could rewrite A, we could rewrite the matrix A, the m by n matrix A, I'm bolding it to show that that's a matrix."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "We could write that as v one, v one. This next one over here, this would be v two, v two. And you would have n of these, because we have n columns. And so this one right over here would be v n, v sub n. And so we could rewrite A, we could rewrite the matrix A, the m by n matrix A, I'm bolding it to show that that's a matrix. We could rewrite it as, so let me do it the same way. So, draw my little brackets there. We can write it, just express it, in terms of its column vectors."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "And so this one right over here would be v n, v sub n. And so we could rewrite A, we could rewrite the matrix A, the m by n matrix A, I'm bolding it to show that that's a matrix. We could rewrite it as, so let me do it the same way. So, draw my little brackets there. We can write it, just express it, in terms of its column vectors. And so we could just say, well, this is going to be v one for that column, v one for that column, v two for this column, all the way. We're gonna have n columns, so you're gonna have v n for the nth column. And remember, each of these are going to have m terms, or I should say m components in them."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "We can write it, just express it, in terms of its column vectors. And so we could just say, well, this is going to be v one for that column, v one for that column, v two for this column, all the way. We're gonna have n columns, so you're gonna have v n for the nth column. And remember, each of these are going to have m terms, or I should say m components in them. These are m dimensional column vectors. Now what I want to do, I said I want to relate the linear independence of these vectors to the null space of A. So let's remind ourselves what the null space of A even is."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "And remember, each of these are going to have m terms, or I should say m components in them. These are m dimensional column vectors. Now what I want to do, I said I want to relate the linear independence of these vectors to the null space of A. So let's remind ourselves what the null space of A even is. So the null space of A, the null space of A, is equal to, or I could say it's equal to the set, yes, the set of all vectors x that are members of R n, and I'm gonna double down on why I'm saying R n in a second, such that, such that, if I take my matrix A, if I take my matrix A and multiply it by one of those xs, by one of those xs, I am going to get, I am going to get the zero vector. So why does x have to be a member of R n? Well, just for the matrix multiplication to work, for this to be, if this is m by n, let me write this down, if this is m by n, well, in order to just make the matrix multiplication work, or you could say the matrix vector multiplication, this has to be an n by one, an n by one vector, and so it's going to have n components, so it's going to be a member of R n. If this was m by a, or let me use a different letter, if this was m by, I don't know, seven, then this would be R seven that we would be dealing with."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So let's remind ourselves what the null space of A even is. So the null space of A, the null space of A, is equal to, or I could say it's equal to the set, yes, the set of all vectors x that are members of R n, and I'm gonna double down on why I'm saying R n in a second, such that, such that, if I take my matrix A, if I take my matrix A and multiply it by one of those xs, by one of those xs, I am going to get, I am going to get the zero vector. So why does x have to be a member of R n? Well, just for the matrix multiplication to work, for this to be, if this is m by n, let me write this down, if this is m by n, well, in order to just make the matrix multiplication work, or you could say the matrix vector multiplication, this has to be an n by one, an n by one vector, and so it's going to have n components, so it's going to be a member of R n. If this was m by a, or let me use a different letter, if this was m by, I don't know, seven, then this would be R seven that we would be dealing with. So that is the null space. So another way of thinking about it is, well, if I take my matrix A and I multiply it by some vector x that's a member of this null space, I'm going to get the zero vector. So if I take my matrix A, which I've expressed here in terms of its column vectors, multiply it by some vector x, so some vector x, and actually let me make it clear that it doesn't have to have the same, so some vector x right over here."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "Well, just for the matrix multiplication to work, for this to be, if this is m by n, let me write this down, if this is m by n, well, in order to just make the matrix multiplication work, or you could say the matrix vector multiplication, this has to be an n by one, an n by one vector, and so it's going to have n components, so it's going to be a member of R n. If this was m by a, or let me use a different letter, if this was m by, I don't know, seven, then this would be R seven that we would be dealing with. So that is the null space. So another way of thinking about it is, well, if I take my matrix A and I multiply it by some vector x that's a member of this null space, I'm going to get the zero vector. So if I take my matrix A, which I've expressed here in terms of its column vectors, multiply it by some vector x, so some vector x, and actually let me make it clear that it doesn't have to have the same, so some vector x right over here. Let me draw the other bracket. So this is a vector x, and so it's going to have, it's a member of R n, so it's going to have n components. So you're going to have x one as a first component, x two, and go all the way to x n. If you multiply, so if we say that this x is a member of the null space of A, then this whole thing is going to be equal to the zero vector, is going to be equal to the zero vector, and once again, the zero vector, this is going to be an m by one vector."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So if I take my matrix A, which I've expressed here in terms of its column vectors, multiply it by some vector x, so some vector x, and actually let me make it clear that it doesn't have to have the same, so some vector x right over here. Let me draw the other bracket. So this is a vector x, and so it's going to have, it's a member of R n, so it's going to have n components. So you're going to have x one as a first component, x two, and go all the way to x n. If you multiply, so if we say that this x is a member of the null space of A, then this whole thing is going to be equal to the zero vector, is going to be equal to the zero vector, and once again, the zero vector, this is going to be an m by one vector. So it's going to look, actually let me write it like this. It's going to have the same number of rows as A. So I'll try to make it, the brackets roughly the same length."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So you're going to have x one as a first component, x two, and go all the way to x n. If you multiply, so if we say that this x is a member of the null space of A, then this whole thing is going to be equal to the zero vector, is going to be equal to the zero vector, and once again, the zero vector, this is going to be an m by one vector. So it's going to look, actually let me write it like this. It's going to have the same number of rows as A. So I'll try to make it, the brackets roughly the same length. So, and there we go. Try and draw my brackets neatly. So you're going to have m of these one, two, and then go all the way to the nth zero."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So I'll try to make it, the brackets roughly the same length. So, and there we go. Try and draw my brackets neatly. So you're going to have m of these one, two, and then go all the way to the nth zero. So let's actually just multiply this out using what we know of matrix multiplication. And by the definition of matrix multiplication, one way to view this, if you were to multiply our matrix A times our vector x here, you are going to get the first column vector, v one, v one, v one, times the first component here, x one. x one, plus the second component times the second column vector, x two, times v two, v two, and we're going to do that n times."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So you're going to have m of these one, two, and then go all the way to the nth zero. So let's actually just multiply this out using what we know of matrix multiplication. And by the definition of matrix multiplication, one way to view this, if you were to multiply our matrix A times our vector x here, you are going to get the first column vector, v one, v one, v one, times the first component here, x one. x one, plus the second component times the second column vector, x two, times v two, v two, and we're going to do that n times. So plus dot, dot, dot, x sub n times v sub n. V sub n, v sub n. And these all, when you add them together, are going to be equal to the zero vector. Now this should be, this is going to be equal to the zero vector. And now this should start ringing a bell to you."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "x one, plus the second component times the second column vector, x two, times v two, v two, and we're going to do that n times. So plus dot, dot, dot, x sub n times v sub n. V sub n, v sub n. And these all, when you add them together, are going to be equal to the zero vector. Now this should be, this is going to be equal to the zero vector. And now this should start ringing a bell to you. When we looked at linear independence, we saw something like this. In fact, we saw that these vectors v, v sub one, v sub two, these n vectors are linearly independent if and only if, any linear, if and only if the solution to this, or I guess you could say the weights on these vectors, the only way to get this to be true is if x one, x two, x n are all equal zero. So let me write this down."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "And now this should start ringing a bell to you. When we looked at linear independence, we saw something like this. In fact, we saw that these vectors v, v sub one, v sub two, these n vectors are linearly independent if and only if, any linear, if and only if the solution to this, or I guess you could say the weights on these vectors, the only way to get this to be true is if x one, x two, x n are all equal zero. So let me write this down. So v sub one, v sub two, v sub two all the way to v sub n are linearly independent, linearly independent if and only if, if and only if, only solution, so let me, only solution, or you could say weights on these vectors to this equation, the only solution is x one, x two, all the way to x n, are equal to zero. So if the only solution here, if the only way to get this sum to be equal to the zero vector is if x one, x two, and x, all the way through x n are equal to zero, well that means that our vectors v one, v two, all the way to v n are linearly independent, or vice versa. If they are linearly independent, the only solution to this, if we're solving for the weights on those vectors, is if for x one, x two, and x n to be equal to zero."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "So let me write this down. So v sub one, v sub two, v sub two all the way to v sub n are linearly independent, linearly independent if and only if, if and only if, only solution, so let me, only solution, or you could say weights on these vectors to this equation, the only solution is x one, x two, all the way to x n, are equal to zero. So if the only solution here, if the only way to get this sum to be equal to the zero vector is if x one, x two, and x, all the way through x n are equal to zero, well that means that our vectors v one, v two, all the way to v n are linearly independent, or vice versa. If they are linearly independent, the only solution to this, if we're solving for the weights on those vectors, is if for x one, x two, and x n to be equal to zero. Remember, linear independence, if you want to say it in a, it's still mathematical, but in a little bit more common language, is if these vectors are linearly independent, that means that none of these vectors can be constructed by linear combinations of the other vectors. Or, looking at it this way, this right over here is a, you could view this as a linear combination of all of the vectors, that the only way to get this linear combination of all the vectors to be equal to zero is if x one, x two, all the way through x n are equal to zero. And we proved that in other videos on linear independence."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "If they are linearly independent, the only solution to this, if we're solving for the weights on those vectors, is if for x one, x two, and x n to be equal to zero. Remember, linear independence, if you want to say it in a, it's still mathematical, but in a little bit more common language, is if these vectors are linearly independent, that means that none of these vectors can be constructed by linear combinations of the other vectors. Or, looking at it this way, this right over here is a, you could view this as a linear combination of all of the vectors, that the only way to get this linear combination of all the vectors to be equal to zero is if x one, x two, all the way through x n are equal to zero. And we proved that in other videos on linear independence. Well, if the only solution to this is all of the x ones through x ns are equal to zero, that means that the null space, this is only going to be true, you could say, if and only if the null space of A, the null space of A, let me make sure it looks like a matrix, I'm gonna bold it, the null space of A only contains one vector. It only contains the zero vector. Remember, this is, if all of these are going to be zero, well then the only solution here is going to be the zero vector, is going to be, is going to be the zero vector."}, {"video_title": "Relation of null space to linear independence of columns.mp3", "Sentence": "And we proved that in other videos on linear independence. Well, if the only solution to this is all of the x ones through x ns are equal to zero, that means that the null space, this is only going to be true, you could say, if and only if the null space of A, the null space of A, let me make sure it looks like a matrix, I'm gonna bold it, the null space of A only contains one vector. It only contains the zero vector. Remember, this is, if all of these are going to be zero, well then the only solution here is going to be the zero vector, is going to be, is going to be the zero vector. So the result that we're showing here is if the column vectors of a matrix are linearly independent, then the null space of that matrix is only going to consist of the zero vector. Or you could go the other way. If the null space of a matrix only contains the zero vector, well that means that the columns of that matrix are linearly independent."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're both in the set Rn, and they're non-zero. It turns out that the absolute value of their dot product of the two vectors is less than or equal to the product of their lengths. We've defined the dot product and we've defined lengths already. It's less than or equal to the product of their lengths. Just to push it a little even further, the only time that this is equal, the dot product of the two vectors is only going to be equal to the length, the equal and the less than or equal apply, only in the situation where one of these vectors is a scalar multiple of the other, or they're collinear. One's just a longer, shorter version of the other one. Only in the situation where x is equal to some scalar multiple of y."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's less than or equal to the product of their lengths. Just to push it a little even further, the only time that this is equal, the dot product of the two vectors is only going to be equal to the length, the equal and the less than or equal apply, only in the situation where one of these vectors is a scalar multiple of the other, or they're collinear. One's just a longer, shorter version of the other one. Only in the situation where x is equal to some scalar multiple of y. These inequalities, or I guess this inequality and this inequality, this is called the Cauchy-Schwarz inequality. Let's prove it, because you can't take something like this just at face value. You can just accept that."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Only in the situation where x is equal to some scalar multiple of y. These inequalities, or I guess this inequality and this inequality, this is called the Cauchy-Schwarz inequality. Let's prove it, because you can't take something like this just at face value. You can just accept that. Let me just construct a somewhat artificial function. Let me construct some function that's a function of some variable, some scalar t. Let me define P of t to be equal to the length of the vector t times the vector, some scalar t times the vector y minus the vector x. It's the length of this vector."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can just accept that. Let me just construct a somewhat artificial function. Let me construct some function that's a function of some variable, some scalar t. Let me define P of t to be equal to the length of the vector t times the vector, some scalar t times the vector y minus the vector x. It's the length of this vector. This is going to be a vector now. That's squared. Before I move forward, I want to make one little point here."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's the length of this vector. This is going to be a vector now. That's squared. Before I move forward, I want to make one little point here. If I take the length of any vector, of any real vector, the length of any real vector, I'll do it here. Let's say I take the length of some vector v. I want you to accept that this is going to be a positive number, or it's at least greater than or equal to zero, because this is just going to be each of its terms squared, v2 squared all the way to vn squared. All of these are real numbers."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Before I move forward, I want to make one little point here. If I take the length of any vector, of any real vector, the length of any real vector, I'll do it here. Let's say I take the length of some vector v. I want you to accept that this is going to be a positive number, or it's at least greater than or equal to zero, because this is just going to be each of its terms squared, v2 squared all the way to vn squared. All of these are real numbers. When you square a real number, you get something greater than or equal to zero. When you sum them up, you're going to have something greater than or equal to zero. If you take the square root of it, the principal square root, the positive square root, you're going to have something greater than or equal to zero."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All of these are real numbers. When you square a real number, you get something greater than or equal to zero. When you sum them up, you're going to have something greater than or equal to zero. If you take the square root of it, the principal square root, the positive square root, you're going to have something greater than or equal to zero. The length of any real vector is going to be greater than or equal to zero. This is the length of a real vector, so this is going to be greater than or equal to zero. previous video, I think it was two videos ago, I also showed that the length of a vector squared can also be rewritten as the dot product of that vector with itself."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you take the square root of it, the principal square root, the positive square root, you're going to have something greater than or equal to zero. The length of any real vector is going to be greater than or equal to zero. This is the length of a real vector, so this is going to be greater than or equal to zero. previous video, I think it was two videos ago, I also showed that the length of a vector squared can also be rewritten as the dot product of that vector with itself. So let's rewrite this vector that way. So this is equal to, the length of this vector squared is equal to the dot product of that vector with itself. So it's ty minus x dot ty minus x."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "previous video, I think it was two videos ago, I also showed that the length of a vector squared can also be rewritten as the dot product of that vector with itself. So let's rewrite this vector that way. So this is equal to, the length of this vector squared is equal to the dot product of that vector with itself. So it's ty minus x dot ty minus x. In the last video, I showed you that you can treat a multiplication, or you can treat the dot product very similar to regular multiplication when it comes to the associative, distributive, and commutative properties. So when you multiply these, you can kind of use this as multiplying these two binomials, you can do it the same way as you would just multiply two regular algebraic binomials. You're essentially just using the distributive property."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's ty minus x dot ty minus x. In the last video, I showed you that you can treat a multiplication, or you can treat the dot product very similar to regular multiplication when it comes to the associative, distributive, and commutative properties. So when you multiply these, you can kind of use this as multiplying these two binomials, you can do it the same way as you would just multiply two regular algebraic binomials. You're essentially just using the distributive property. But remember, this isn't just regular multiplication. This is the dot product we're doing. This is vector multiplication, or one version of vector multiplication."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're essentially just using the distributive property. But remember, this isn't just regular multiplication. This is the dot product we're doing. This is vector multiplication, or one version of vector multiplication. So if we distribute it out, this will become ty dot ty. So let me write that out. That'll be ty dot ty, and then we'll get a minus, let me do it this way, then we get the minus x times this ty."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is vector multiplication, or one version of vector multiplication. So if we distribute it out, this will become ty dot ty. So let me write that out. That'll be ty dot ty, and then we'll get a minus, let me do it this way, then we get the minus x times this ty. So you get, or instead of saying times, I should be very careful to say dot. So minus x dot ty. And then you end up, and then you have this ty times this minus x, so then you have minus ty dot x."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That'll be ty dot ty, and then we'll get a minus, let me do it this way, then we get the minus x times this ty. So you get, or instead of saying times, I should be very careful to say dot. So minus x dot ty. And then you end up, and then you have this ty times this minus x, so then you have minus ty dot x. And then finally, you have the x's dot with each other. And you can view them as minus 1x dot minus 1x. So you could say plus minus 1x, right?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you end up, and then you have this ty times this minus x, so then you have minus ty dot x. And then finally, you have the x's dot with each other. And you can view them as minus 1x dot minus 1x. So you could say plus minus 1x, right? I could just say, view this as plus minus 1 or plus minus 1. So this is minus 1x dot minus 1x. So let's see, so this is what my whole expression simplified to, or expanded to."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you could say plus minus 1x, right? I could just say, view this as plus minus 1 or plus minus 1. So this is minus 1x dot minus 1x. So let's see, so this is what my whole expression simplified to, or expanded to. I can't really call this a simplification. But we can use the fact that this is commutative and associative to rewrite this expression right here. This is equal to y dot y times t squared, right?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see, so this is what my whole expression simplified to, or expanded to. I can't really call this a simplification. But we can use the fact that this is commutative and associative to rewrite this expression right here. This is equal to y dot y times t squared, right? t is just a scalar. And actually, this is 2. These two things are equivalent, right?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to y dot y times t squared, right? t is just a scalar. And actually, this is 2. These two things are equivalent, right? They're just rearrangements of the same thing. And we saw that the dot product is associative. So this is just equal to 2 times x dot y times t. And I should do that in maybe a different color."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These two things are equivalent, right? They're just rearrangements of the same thing. And we saw that the dot product is associative. So this is just equal to 2 times x dot y times t. And I should do that in maybe a different color. So these two terms result in that term right there. And then if you just rearrange these, you have a minus 1 times a minus 1, they cancel out. So those would become plus."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is just equal to 2 times x dot y times t. And I should do that in maybe a different color. So these two terms result in that term right there. And then if you just rearrange these, you have a minus 1 times a minus 1, they cancel out. So those would become plus. And you're just left with plus x dot x. And I should do that in a different color as well. I'll do that orange color."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So those would become plus. And you're just left with plus x dot x. And I should do that in a different color as well. I'll do that orange color. So those terms end up with that term right there. And of course, that term results in that term. And remember, all I did is I rewrote this thing and said, look, this has got to be greater than or equal to 0."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'll do that orange color. So those terms end up with that term right there. And of course, that term results in that term. And remember, all I did is I rewrote this thing and said, look, this has got to be greater than or equal to 0. So I can rewrite that here. This thing is still just the same thing. I've just rewritten it."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And remember, all I did is I rewrote this thing and said, look, this has got to be greater than or equal to 0. So I can rewrite that here. This thing is still just the same thing. I've just rewritten it. So this is all going to be greater than or equal to 0. Now let's make a little bit of a substitution just to clean up our expression a little bit. And we'll later back substitute into this."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I've just rewritten it. So this is all going to be greater than or equal to 0. Now let's make a little bit of a substitution just to clean up our expression a little bit. And we'll later back substitute into this. Let's define this as a. Let's define this piece right here as b. So the whole thing, minus 2x dot y. I'll leave the t there."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we'll later back substitute into this. Let's define this as a. Let's define this piece right here as b. So the whole thing, minus 2x dot y. I'll leave the t there. And let's define this right here as c. x dot x as c. So then what does our expression become? It becomes a times t squared minus b times t plus c. And of course, we know that it's going to be greater than or equal to 0. It's the same thing as this up here."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the whole thing, minus 2x dot y. I'll leave the t there. And let's define this right here as c. x dot x as c. So then what does our expression become? It becomes a times t squared minus b times t plus c. And of course, we know that it's going to be greater than or equal to 0. It's the same thing as this up here. Greater than or equal to 0. I could write p of t here. Now this is greater than or equal to 0 for any t that I put in here, for any real t that I put in there."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's the same thing as this up here. Greater than or equal to 0. I could write p of t here. Now this is greater than or equal to 0 for any t that I put in here, for any real t that I put in there. And so I could just put in, let me just pick a t to be, let me evaluate our function at b over 2a. And I can definitely do this, because what was a? I just have to make sure I'm not dividing by 0 anyplace."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now this is greater than or equal to 0 for any t that I put in here, for any real t that I put in there. And so I could just put in, let me just pick a t to be, let me evaluate our function at b over 2a. And I can definitely do this, because what was a? I just have to make sure I'm not dividing by 0 anyplace. So a was this vector dotted with itself. And we said this is a non-zero vector. So this is the square of its length."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just have to make sure I'm not dividing by 0 anyplace. So a was this vector dotted with itself. And we said this is a non-zero vector. So this is the square of its length. It's a non-zero vector. So some of these terms up here would end up becoming positive when you take its length. So this thing right here is non-zero."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is the square of its length. It's a non-zero vector. So some of these terms up here would end up becoming positive when you take its length. So this thing right here is non-zero. If this is a non-zero vector, then 2 times the dot product with itself is also going to be non-zero. So we can do this. We don't have to worry about dividing by 0 or whatever else."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here is non-zero. If this is a non-zero vector, then 2 times the dot product with itself is also going to be non-zero. So we can do this. We don't have to worry about dividing by 0 or whatever else. But what will this be equal to? And I'll just stick to the green. It takes too long to keep switching between colors."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We don't have to worry about dividing by 0 or whatever else. But what will this be equal to? And I'll just stick to the green. It takes too long to keep switching between colors. This is equal to a times this expression squared. So it's b squared over 4a squared. I just squared 2a to get 4a squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It takes too long to keep switching between colors. This is equal to a times this expression squared. So it's b squared over 4a squared. I just squared 2a to get 4a squared. Minus b times this. So b times, this is just regular multiplication, b times b over 2a. I'll just write regular multiplication there."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just squared 2a to get 4a squared. Minus b times this. So b times, this is just regular multiplication, b times b over 2a. I'll just write regular multiplication there. Plus c. And we know all of that is greater than or equal to 0. Now if we simplify this a little bit, what do we get? Well, this a cancels out with this exponent there."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'll just write regular multiplication there. Plus c. And we know all of that is greater than or equal to 0. Now if we simplify this a little bit, what do we get? Well, this a cancels out with this exponent there. And you end up with a b squared right there. So we get b squared over 4a minus b squared over 2a. That's that term over there."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this a cancels out with this exponent there. And you end up with a b squared right there. So we get b squared over 4a minus b squared over 2a. That's that term over there. Plus c is greater than or equal to 0. And let me rewrite this. If I multiply the numerator and the denominator of this by 2, what do I get?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's that term over there. Plus c is greater than or equal to 0. And let me rewrite this. If I multiply the numerator and the denominator of this by 2, what do I get? I get 2b squared over 4a. And the whole reason I did that is because to get a common denominator here. So what do you get?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply the numerator and the denominator of this by 2, what do I get? I get 2b squared over 4a. And the whole reason I did that is because to get a common denominator here. So what do you get? You get b squared over 4a minus 2b squared over 4a. So what do these two terms simplify to? Well, the numerator is b squared minus 2b squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what do you get? You get b squared over 4a minus 2b squared over 4a. So what do these two terms simplify to? Well, the numerator is b squared minus 2b squared. So that just becomes minus b squared over 4a plus c is greater than or equal to 0. These two terms add up to this one right here. Now if we add this to both sides of the equation, we get c is greater than or equal to b squared over 4a."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, the numerator is b squared minus 2b squared. So that just becomes minus b squared over 4a plus c is greater than or equal to 0. These two terms add up to this one right here. Now if we add this to both sides of the equation, we get c is greater than or equal to b squared over 4a. It was a negative on the left-hand side. If I add it to both sides, it's going to be a positive on the right-hand side. So we're approaching something that looks like an inequality."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now if we add this to both sides of the equation, we get c is greater than or equal to b squared over 4a. It was a negative on the left-hand side. If I add it to both sides, it's going to be a positive on the right-hand side. So we're approaching something that looks like an inequality. So let's back substitute our original substitutions to see what we have now. So where is my original substitutions that I made? It was right here."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we're approaching something that looks like an inequality. So let's back substitute our original substitutions to see what we have now. So where is my original substitutions that I made? It was right here. So what was, and actually just to simplify it even more, let me multiply both sides by 4a. And once I said a, not only is it non-zero, it's going to be positive. This is the square of its length."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It was right here. So what was, and actually just to simplify it even more, let me multiply both sides by 4a. And once I said a, not only is it non-zero, it's going to be positive. This is the square of its length. And I already showed you that the length of any vector, any real vector is going to be positive. And the reason why I'm taking great pains to show that a is positive is because if I multiply both sides of it, I don't want to change the inequality sign. So let me multiply both sides of this by a before I substitute, so we get 4ac is greater than or equal to b squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is the square of its length. And I already showed you that the length of any vector, any real vector is going to be positive. And the reason why I'm taking great pains to show that a is positive is because if I multiply both sides of it, I don't want to change the inequality sign. So let me multiply both sides of this by a before I substitute, so we get 4ac is greater than or equal to b squared. There you go. And remember, I took great pains. I just said a is definitely a positive number because it is essentially the square of the length."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me multiply both sides of this by a before I substitute, so we get 4ac is greater than or equal to b squared. There you go. And remember, I took great pains. I just said a is definitely a positive number because it is essentially the square of the length. y dot y is the square of the length of y, and that's a positive value. It has to be positive if we're dealing with real vectors. Now let's back substitute this."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I just said a is definitely a positive number because it is essentially the square of the length. y dot y is the square of the length of y, and that's a positive value. It has to be positive if we're dealing with real vectors. Now let's back substitute this. So 4 times a, 4 times y dot y, y dot y is also, I might as well just write it there, y dot y is the same thing as the magnitude of y squared. That's y dot y. This is a. y dot y, I showed you that in the previous video."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's back substitute this. So 4 times a, 4 times y dot y, y dot y is also, I might as well just write it there, y dot y is the same thing as the magnitude of y squared. That's y dot y. This is a. y dot y, I showed you that in the previous video. Times c. c is x dot x. Well, x dot x is the same thing as the length of vector x squared. So this was c. So 4 times a times c is going to be greater than or equal to b squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a. y dot y, I showed you that in the previous video. Times c. c is x dot x. Well, x dot x is the same thing as the length of vector x squared. So this was c. So 4 times a times c is going to be greater than or equal to b squared. Now what was b? b was this thing here. So b squared would be 2 times x dot y squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this was c. So 4 times a times c is going to be greater than or equal to b squared. Now what was b? b was this thing here. So b squared would be 2 times x dot y squared. So we've gotten to this result so far. And so what can we do with this? Oh, sorry, this whole thing is squared."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So b squared would be 2 times x dot y squared. So we've gotten to this result so far. And so what can we do with this? Oh, sorry, this whole thing is squared. This whole thing right here was b. So let's see if we can simplify this. So we get, let me switch to a different color."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Oh, sorry, this whole thing is squared. This whole thing right here was b. So let's see if we can simplify this. So we get, let me switch to a different color. 4 times the length of y squared times the length of x squared is greater than or equal to, if we squared this quantity right here, we get 4 times x dot y. 4 times x dot y times x dot y. Actually, even better, let me just write it like this."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we get, let me switch to a different color. 4 times the length of y squared times the length of x squared is greater than or equal to, if we squared this quantity right here, we get 4 times x dot y. 4 times x dot y times x dot y. Actually, even better, let me just write it like this. Let me just write 4 times x dot y squared. Now we can divide both sides by 4. That won't change our inequality."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Actually, even better, let me just write it like this. Let me just write 4 times x dot y squared. Now we can divide both sides by 4. That won't change our inequality. So that just cancels out there. And now let's take the square root of both sides of this equation. So the square roots of both sides of this equation, these are positive values."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That won't change our inequality. So that just cancels out there. And now let's take the square root of both sides of this equation. So the square roots of both sides of this equation, these are positive values. So the square root of this height is the square root of each of its terms. That's just an exponent property. So if you take the square root of both sides, you get the length of y times the length of x is greater than or equal to the square root of this."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the square roots of both sides of this equation, these are positive values. So the square root of this height is the square root of each of its terms. That's just an exponent property. So if you take the square root of both sides, you get the length of y times the length of x is greater than or equal to the square root of this. And we're going to take the positive square root. We're going to take the positive square root on both sides of this equation. That keeps us from having to mess with anything on the inequality or anything like that."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you take the square root of both sides, you get the length of y times the length of x is greater than or equal to the square root of this. And we're going to take the positive square root. We're going to take the positive square root on both sides of this equation. That keeps us from having to mess with anything on the inequality or anything like that. So the positive square root is going to be the absolute value of x dot y. And I want to be very careful to say this is the absolute value, because it's possible that this thing right here is a negative value. But when you square it, you want to be careful that when you take the square root of it, that you stay a positive value, because otherwise when we take the principal square root, we might mess with the inequality."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That keeps us from having to mess with anything on the inequality or anything like that. So the positive square root is going to be the absolute value of x dot y. And I want to be very careful to say this is the absolute value, because it's possible that this thing right here is a negative value. But when you square it, you want to be careful that when you take the square root of it, that you stay a positive value, because otherwise when we take the principal square root, we might mess with the inequality. So we're taking the positive square root, which will be, so if you take the absolute value, you're ensuring that it's going to be positive. But this is our result. The absolute value of the dot product of our vectors is less than the product of the two vectors' length."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But when you square it, you want to be careful that when you take the square root of it, that you stay a positive value, because otherwise when we take the principal square root, we might mess with the inequality. So we're taking the positive square root, which will be, so if you take the absolute value, you're ensuring that it's going to be positive. But this is our result. The absolute value of the dot product of our vectors is less than the product of the two vectors' length. So we got our Cauchy-Schwarz inequality. Now, the last thing I said is, look, what happens if x is equal to some scalar multiple of y? Well, in that case, what's the absolute value?"}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The absolute value of the dot product of our vectors is less than the product of the two vectors' length. So we got our Cauchy-Schwarz inequality. Now, the last thing I said is, look, what happens if x is equal to some scalar multiple of y? Well, in that case, what's the absolute value? The absolute value of x dot y, well, that equals what? If we make the substitution, that equals the absolute value of c times y. That's just x dot y, which is equal to, just from the associative property, it's equal to the absolute value of c times, we want to make sure absolute value, stay everything, keep everything positive, y dot y."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, in that case, what's the absolute value? The absolute value of x dot y, well, that equals what? If we make the substitution, that equals the absolute value of c times y. That's just x dot y, which is equal to, just from the associative property, it's equal to the absolute value of c times, we want to make sure absolute value, stay everything, keep everything positive, y dot y. Well, this is just equal to c times the magnitude of y, or the length of y squared. Well, that just is equal to the magnitude of c times, or the absolute value of our scalar c, times our length of y, times our length of y. Well, this right here, this, I can rewrite this, I mean, you can prove this to yourself if you don't believe it, but this, we could put the c inside of the magnitude, and that could be a good exercise for you to prove, but it's pretty straightforward."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's just x dot y, which is equal to, just from the associative property, it's equal to the absolute value of c times, we want to make sure absolute value, stay everything, keep everything positive, y dot y. Well, this is just equal to c times the magnitude of y, or the length of y squared. Well, that just is equal to the magnitude of c times, or the absolute value of our scalar c, times our length of y, times our length of y. Well, this right here, this, I can rewrite this, I mean, you can prove this to yourself if you don't believe it, but this, we could put the c inside of the magnitude, and that could be a good exercise for you to prove, but it's pretty straightforward. You just do the definition of length, and you multiply it by c. This is equal to the magnitude of c y times, let me say, the length of c y times the length of y. Length of y, I've lost my vector notation someplace over here. I lost my, there you go."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this right here, this, I can rewrite this, I mean, you can prove this to yourself if you don't believe it, but this, we could put the c inside of the magnitude, and that could be a good exercise for you to prove, but it's pretty straightforward. You just do the definition of length, and you multiply it by c. This is equal to the magnitude of c y times, let me say, the length of c y times the length of y. Length of y, I've lost my vector notation someplace over here. I lost my, there you go. Now, this is x. So this is equal to the length of x times the length of y. So I showed you kind of the second part of the Cauchy- Schwarz inequality, that this is only equal to each other if one of them is a scalar multiple of the other."}, {"video_title": "Proof of the Cauchy-Schwarz inequality Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I lost my, there you go. Now, this is x. So this is equal to the length of x times the length of y. So I showed you kind of the second part of the Cauchy- Schwarz inequality, that this is only equal to each other if one of them is a scalar multiple of the other. And if you thought, if you're a little uncomfortable with some of these steps I took, it might be a good exercise to actually do it. For example, to prove that the absolute value of c times the length of the vector y is the same thing as the length of c times y. Anyway, hopefully you found this pretty useful."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "In the last video, we started off with two linear transformations. We had the linear transformation S that was a mapping from Rn to Rm. And then we had the linear transformation T that was also a mapping from Rn to Rm. And we defined the idea of the addition of these two transformations. So S plus T, this transformation of X, we defined as being equal to S of X, this vector, plus T of X. And of course, this input is still from Rn. And then each of these are vectors in Rm."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And we defined the idea of the addition of these two transformations. So S plus T, this transformation of X, we defined as being equal to S of X, this vector, plus T of X. And of course, this input is still from Rn. And then each of these are vectors in Rm. If we add two vectors in Rm to each other, we get another vector in Rm because Rm is a valid subspace. It's also closed under addition. So this is still a mapping."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And then each of these are vectors in Rm. If we add two vectors in Rm to each other, we get another vector in Rm because Rm is a valid subspace. It's also closed under addition. So this is still a mapping. So S plus T is still a mapping from Rn to Rm. And we also said that, look, every linear transformation we've shown in a previous video can be represented as a matrix. That we could say that S of X is equal to some matrix A times X."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "So this is still a mapping. So S plus T is still a mapping from Rn to Rm. And we also said that, look, every linear transformation we've shown in a previous video can be represented as a matrix. That we could say that S of X is equal to some matrix A times X. And we could also say that T of X is equal to some matrix B times X. And both of these would be m by n matrices. And let me write that, m by n, both of these."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "That we could say that S of X is equal to some matrix A times X. And we could also say that T of X is equal to some matrix B times X. And both of these would be m by n matrices. And let me write that, m by n, both of these. Because these are both mappings from Rn to Rm. And what we did is we made another definition. We defined, this was a definition right here."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And let me write that, m by n, both of these. Because these are both mappings from Rn to Rm. And what we did is we made another definition. We defined, this was a definition right here. Then we made another definition. We defined the addition of two matrices. We said the matrix A, this is any matrix A plus B."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "We defined, this was a definition right here. Then we made another definition. We defined the addition of two matrices. We said the matrix A, this is any matrix A plus B. They both have to have the same dimensions. So they're both m by n in this case. And we defined this addition to be a new matrix where each column of this matrix is the sum of the corresponding columns of these matrices."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "We said the matrix A, this is any matrix A plus B. They both have to have the same dimensions. So they're both m by n in this case. And we defined this addition to be a new matrix where each column of this matrix is the sum of the corresponding columns of these matrices. So this matrix's first column will be the sum of A's first column and B's first column. So A1 plus B1. The second column, I'll do a little line here, is A2 plus B2."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And we defined this addition to be a new matrix where each column of this matrix is the sum of the corresponding columns of these matrices. So this matrix's first column will be the sum of A's first column and B's first column. So A1 plus B1. The second column, I'll do a little line here, is A2 plus B2. And it goes all the way to An plus Bn. This was a definition. And the whole reason why we made this definition is because if you defined matrix addition in this way, then this thing, when you replace it with Ax plus Bx, you get to that this thing over here is equal to the corresponding matrices by this definition, A plus B times x."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "The second column, I'll do a little line here, is A2 plus B2. And it goes all the way to An plus Bn. This was a definition. And the whole reason why we made this definition is because if you defined matrix addition in this way, then this thing, when you replace it with Ax plus Bx, you get to that this thing over here is equal to the corresponding matrices by this definition, A plus B times x. This was the motivation to get to a nice expression like this for defining matrix addition in this way. Now this all seems very abstract, so let's actually add a matrix. Or let's add two matrices."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And the whole reason why we made this definition is because if you defined matrix addition in this way, then this thing, when you replace it with Ax plus Bx, you get to that this thing over here is equal to the corresponding matrices by this definition, A plus B times x. This was the motivation to get to a nice expression like this for defining matrix addition in this way. Now this all seems very abstract, so let's actually add a matrix. Or let's add two matrices. So we'll start off with a 2 by 2 case. So let's say I'm adding the matrix 1, 3, minus 2, 4 to the matrix. Remember, they have to have the same dimensions."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Or let's add two matrices. So we'll start off with a 2 by 2 case. So let's say I'm adding the matrix 1, 3, minus 2, 4 to the matrix. Remember, they have to have the same dimensions. To the matrix, 2, 7, minus 3, minus 1. What do we get? Well, by definition, you just add up their corresponding columns."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Remember, they have to have the same dimensions. To the matrix, 2, 7, minus 3, minus 1. What do we get? Well, by definition, you just add up their corresponding columns. You add up the first column. When you add up the corresponding columns, what happens when you add up two columns, two vectors? Well, you just add up their corresponding entries."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Well, by definition, you just add up their corresponding columns. You add up the first column. When you add up the corresponding columns, what happens when you add up two columns, two vectors? Well, you just add up their corresponding entries. So essentially, when you add up two matrices, you're just adding up all of the corresponding entries. I'll talk about it in this way just because that's how I defined it, but they're all equivalent. So the first thing, the first column in this matrix right here is going to be this column plus this column."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Well, you just add up their corresponding entries. So essentially, when you add up two matrices, you're just adding up all of the corresponding entries. I'll talk about it in this way just because that's how I defined it, but they're all equivalent. So the first thing, the first column in this matrix right here is going to be this column plus this column. So it's going to be 1 plus 2. Let me write it like this. 1 plus 2, and then minus 2 minus 3."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing, the first column in this matrix right here is going to be this column plus this column. So it's going to be 1 plus 2. Let me write it like this. 1 plus 2, and then minus 2 minus 3. And then the second column, that one right there, is going to be 3 plus 7, and then 4 minus 1. And so this will be equal to 3, 10, minus 5, and 3. Just like that."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "1 plus 2, and then minus 2 minus 3. And then the second column, that one right there, is going to be 3 plus 7, and then 4 minus 1. And so this will be equal to 3, 10, minus 5, and 3. Just like that. And notice, even though the definition is I'm adding up corresponding columns, but what in effect happened? Well, I'm just adding up the corresponding entries. I added the 1 to the 2, the 3 to the 7, the minus 2 to the minus 3, the 4 to the minus 1."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. And notice, even though the definition is I'm adding up corresponding columns, but what in effect happened? Well, I'm just adding up the corresponding entries. I added the 1 to the 2, the 3 to the 7, the minus 2 to the minus 3, the 4 to the minus 1. It's that straightforward. Nothing fancier than that. We could have rewritten this definition."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "I added the 1 to the 2, the 3 to the 7, the minus 2 to the minus 3, the 4 to the minus 1. It's that straightforward. Nothing fancier than that. We could have rewritten this definition. If we say that the vector or the matrix A is equal to A11, A12, all the way to A1n, and then if you go down, this is A21 all the way to An1, and then you go all the way down there to Ann. We've seen that before. And then the matrix B, just by the same argument or by a similar definition, this is B11."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "We could have rewritten this definition. If we say that the vector or the matrix A is equal to A11, A12, all the way to A1n, and then if you go down, this is A21 all the way to An1, and then you go all the way down there to Ann. We've seen that before. And then the matrix B, just by the same argument or by a similar definition, this is B11. That entry is B11. That's B12, all the way to B1n. This is B21, second row, all the way down to Bn."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And then the matrix B, just by the same argument or by a similar definition, this is B11. That entry is B11. That's B12, all the way to B1n. This is B21, second row, all the way down to Bn. Sorry, this is M. We have M rows, so this is Mn. So this right here is BM1. This would be BM2, all the way down to this is BMn right there."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "This is B21, second row, all the way down to Bn. Sorry, this is M. We have M rows, so this is Mn. So this right here is BM1. This would be BM2, all the way down to this is BMn right there. I want to be very careful. These are M by n matrices, so this row down here is the Mth row in both cases. But we could redefine our matrix or another way of stating this matrix addition definition is to say, look, if I'm going to add A plus B, I'm just going to add up the corresponding entries."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "This would be BM2, all the way down to this is BMn right there. I want to be very careful. These are M by n matrices, so this row down here is the Mth row in both cases. But we could redefine our matrix or another way of stating this matrix addition definition is to say, look, if I'm going to add A plus B, I'm just going to add up the corresponding entries. So this entry up here is going to be, I'll do it in a different color, is going to be A11 plus B11. This one's going to be A21 plus B21, all the way down to Am1 plus BM1. And then this is going to be, of course, A12 plus B12, all the way to A1n."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "But we could redefine our matrix or another way of stating this matrix addition definition is to say, look, if I'm going to add A plus B, I'm just going to add up the corresponding entries. So this entry up here is going to be, I'll do it in a different color, is going to be A11 plus B11. This one's going to be A21 plus B21, all the way down to Am1 plus BM1. And then this is going to be, of course, A12 plus B12, all the way to A1n. Let me scroll over a little bit, all the way over to A1n plus B1n, and then all the way down to Amn plus BMn. These are equivalent definitions. This takes a lot less space to write it, and I felt comfortable doing it because we've already defined vector addition."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "And then this is going to be, of course, A12 plus B12, all the way to A1n. Let me scroll over a little bit, all the way over to A1n plus B1n, and then all the way down to Amn plus BMn. These are equivalent definitions. This takes a lot less space to write it, and I felt comfortable doing it because we've already defined vector addition. But it essentially boils down to you just add up all of the corresponding entries. That's all matrix addition is. It's probably one of the simplest things that you've seen in your recent mathematical experience."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "This takes a lot less space to write it, and I felt comfortable doing it because we've already defined vector addition. But it essentially boils down to you just add up all of the corresponding entries. That's all matrix addition is. It's probably one of the simplest things that you've seen in your recent mathematical experience. Now, matrix scalar multiplication, very similar idea. We defined scalar multiplication times the transformation of x to be equal to a scalar times the transformation of x. This was a definition."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "It's probably one of the simplest things that you've seen in your recent mathematical experience. Now, matrix scalar multiplication, very similar idea. We defined scalar multiplication times the transformation of x to be equal to a scalar times the transformation of x. This was a definition. We also defined scalar times some matrix A to be equal to the scalar, a new matrix where each of its columns are the scalar times the column vectors of A. So A1, and then the next column is CA2, and then you go all the way to CAn. The whole motivation for this was because this could be simplified to, well, T I said was equal to Bx, the transformation."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "This was a definition. We also defined scalar times some matrix A to be equal to the scalar, a new matrix where each of its columns are the scalar times the column vectors of A. So A1, and then the next column is CA2, and then you go all the way to CAn. The whole motivation for this was because this could be simplified to, well, T I said was equal to Bx, the transformation. Sorry, this is A times transformation of x. The transformation T of x was equal to, so we still have our C, so it's going to be C times the matrix B times the vector x. That's what the transformation of x could be written as."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "The whole motivation for this was because this could be simplified to, well, T I said was equal to Bx, the transformation. Sorry, this is A times transformation of x. The transformation T of x was equal to, so we still have our C, so it's going to be C times the matrix B times the vector x. That's what the transformation of x could be written as. This would be equal to, by just manipulating, and we did this in the last video by actually breaking this up in the column vectors, multiplying them by each of the components of x, and then distributing the C and rearranging a little bit, we can now say using this definition that this is equal to some new matrix CB. We're using this definition, some new matrix CB, where it's essentially C times each of the column vectors of B times x. This right here was our motivation."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "That's what the transformation of x could be written as. This would be equal to, by just manipulating, and we did this in the last video by actually breaking this up in the column vectors, multiplying them by each of the components of x, and then distributing the C and rearranging a little bit, we can now say using this definition that this is equal to some new matrix CB. We're using this definition, some new matrix CB, where it's essentially C times each of the column vectors of B times x. This right here was our motivation. We wanted to be able to express this as a product of some new matrix and a vector because any linear transformation should be expressible in that way, and that's why we made this definition. Now let's apply it, and I wanted to show you that this is perhaps even simpler than matrix addition. If we want to multiply the scalar 5 times the matrix, I'll do a 3 by 2 matrix, so 1 minus 1, 2, 3, 7, 0."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "This right here was our motivation. We wanted to be able to express this as a product of some new matrix and a vector because any linear transformation should be expressible in that way, and that's why we made this definition. Now let's apply it, and I wanted to show you that this is perhaps even simpler than matrix addition. If we want to multiply the scalar 5 times the matrix, I'll do a 3 by 2 matrix, so 1 minus 1, 2, 3, 7, 0. This will just be equal to, by this definition I'm just saying, look, I'm multiplying the scalar times each of the column vectors, so it's 5 times 1, 2, 7, but what's that? That's just 5 times each of the entries, so it's 5 times 1, which is 5, 5 times 2, which is 10, 5 times 7, which is 35. Then the next column is going to be 5 times this column right here, which is just 5 times each of its entries, so 5 times minus 1 is minus 5, 5 times 3 is 15, 5 times 0 is 0."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "If we want to multiply the scalar 5 times the matrix, I'll do a 3 by 2 matrix, so 1 minus 1, 2, 3, 7, 0. This will just be equal to, by this definition I'm just saying, look, I'm multiplying the scalar times each of the column vectors, so it's 5 times 1, 2, 7, but what's that? That's just 5 times each of the entries, so it's 5 times 1, which is 5, 5 times 2, which is 10, 5 times 7, which is 35. Then the next column is going to be 5 times this column right here, which is just 5 times each of its entries, so 5 times minus 1 is minus 5, 5 times 3 is 15, 5 times 0 is 0. It's that simple. You literally, if we go back to this definition, we can define scalar multiplication of a matrix, so we could also define CA as being equal to, if we say this is a representation for A, of the scalar C times each of the entries of A. That's it."}, {"video_title": "More on matrix addition and scalar multiplication Linear Algebra Khan Academy.mp3", "Sentence": "Then the next column is going to be 5 times this column right here, which is just 5 times each of its entries, so 5 times minus 1 is minus 5, 5 times 3 is 15, 5 times 0 is 0. It's that simple. You literally, if we go back to this definition, we can define scalar multiplication of a matrix, so we could also define CA as being equal to, if we say this is a representation for A, of the scalar C times each of the entries of A. That's it. So it's C times A11, C times A12, all the way to C times A1N. You go down this way, C times A21, all the way down to C times AM1. Then if you go down the diagonal, it should be C times AMN."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I think by now we have a reasonable sense of what linear dependence means. So let's just do a slightly more formal definition of linear dependence. So we're going to say that a set of vectors, let me just define my set of vectors. So let me call my set S of vectors v1, v2, all the way to vn. So I'm going to say that they are linearly dependent if and only if, so sometimes it's written if with a lot of f's in there, so sometimes it's written if and only if. Sometimes it's shown like an arrow in two directions. If and only if I can satisfy this equation, I can find a set of constants, c1 times v1, so I can take a linear combination of my vectors all the way to cn vn that satisfy the equation that I can create this into the zero vector."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me call my set S of vectors v1, v2, all the way to vn. So I'm going to say that they are linearly dependent if and only if, so sometimes it's written if with a lot of f's in there, so sometimes it's written if and only if. Sometimes it's shown like an arrow in two directions. If and only if I can satisfy this equation, I can find a set of constants, c1 times v1, so I can take a linear combination of my vectors all the way to cn vn that satisfy the equation that I can create this into the zero vector. So sometimes it's just written as a bold zero, and sometimes you could just write it, I mean we don't know the dimensionality of this vector, it would be a bunch of zeros, we don't know how many actual elements are in each of these vectors, but you get the idea. So my set of vectors is linearly dependent, remember I'm saying dependent, not independent, is linearly dependent if and only if I can satisfy this equation for some ci's where not all of them are equal to zero, for some ci's where not all, and this is key, not all are zero. Or you could say it the other way, you could say at least one is non-zero."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If and only if I can satisfy this equation, I can find a set of constants, c1 times v1, so I can take a linear combination of my vectors all the way to cn vn that satisfy the equation that I can create this into the zero vector. So sometimes it's just written as a bold zero, and sometimes you could just write it, I mean we don't know the dimensionality of this vector, it would be a bunch of zeros, we don't know how many actual elements are in each of these vectors, but you get the idea. So my set of vectors is linearly dependent, remember I'm saying dependent, not independent, is linearly dependent if and only if I can satisfy this equation for some ci's where not all of them are equal to zero, for some ci's where not all, and this is key, not all are zero. Or you could say it the other way, you could say at least one is non-zero. So how does this gel with what we were talking about in the previous video where I said look, a set is linearly dependent if one of the vectors can be represented by the combination of the other vectors. So let me write that down. So in the last video I said look, one vector can be, well let me write it this way, one vector being represented by the sum of the other vectors, I can just write it like this."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or you could say it the other way, you could say at least one is non-zero. So how does this gel with what we were talking about in the previous video where I said look, a set is linearly dependent if one of the vectors can be represented by the combination of the other vectors. So let me write that down. So in the last video I said look, one vector can be, well let me write it this way, one vector being represented by the sum of the other vectors, I can just write it like this. I could write it a little bit more mathy. In the last video I said that linear dependence means that, let me just pick an arbitrary vector, v1, let's say that v1 is, you know, this is arbitrary, v1 could be represented by some combination of the other vectors, by, you know, let me call them a1 times v, let me be careful, a2 times v2 plus a3 times v3 plus all the way up to an times vn. This is what we said in the previous video."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in the last video I said look, one vector can be, well let me write it this way, one vector being represented by the sum of the other vectors, I can just write it like this. I could write it a little bit more mathy. In the last video I said that linear dependence means that, let me just pick an arbitrary vector, v1, let's say that v1 is, you know, this is arbitrary, v1 could be represented by some combination of the other vectors, by, you know, let me call them a1 times v, let me be careful, a2 times v2 plus a3 times v3 plus all the way up to an times vn. This is what we said in the previous video. Look, if this is linearly dependent, if any one of these guys can be represented as some combination of the other ones, so how does this imply that? And in order to show this if and only if, I have to show that this implies that and I have to show that that implies this. So this is almost a trivially easy proof because if I subtract v1 from both sides of this equation, I get 0 is equal to minus 1 v1 plus a2 v2 plus a3 v3 all the way to an vn, and clearly I've just said, well, this is linear dependent, that means that I can represent this vector as a sum of the other vectors, which means that minus 1 times v1 plus some combination of the other vectors is equal to 0, which means that I've been able to satisfy this equation and at least one of my constants is non-zero."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is what we said in the previous video. Look, if this is linearly dependent, if any one of these guys can be represented as some combination of the other ones, so how does this imply that? And in order to show this if and only if, I have to show that this implies that and I have to show that that implies this. So this is almost a trivially easy proof because if I subtract v1 from both sides of this equation, I get 0 is equal to minus 1 v1 plus a2 v2 plus a3 v3 all the way to an vn, and clearly I've just said, well, this is linear dependent, that means that I can represent this vector as a sum of the other vectors, which means that minus 1 times v1 plus some combination of the other vectors is equal to 0, which means that I've been able to satisfy this equation and at least one of my constants is non-zero. So I've shown you that if I can represent one of the vectors by a sum of the other ones, then this condition is definitely going to be true. Now let me go the other way. Let me show you if I have this situation that I can definitely represent one of the vectors as the sum of the others."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is almost a trivially easy proof because if I subtract v1 from both sides of this equation, I get 0 is equal to minus 1 v1 plus a2 v2 plus a3 v3 all the way to an vn, and clearly I've just said, well, this is linear dependent, that means that I can represent this vector as a sum of the other vectors, which means that minus 1 times v1 plus some combination of the other vectors is equal to 0, which means that I've been able to satisfy this equation and at least one of my constants is non-zero. So I've shown you that if I can represent one of the vectors by a sum of the other ones, then this condition is definitely going to be true. Now let me go the other way. Let me show you if I have this situation that I can definitely represent one of the vectors as the sum of the others. So let's say that this is true and one of these constants, remember, it's not just this, at least one is non-zero. So let me just assume, just for the sake of simplicity, I mean, these are all arbitrary, let me just assume, I'll do it in a new color, let me do it in the magenta, let me assume that c1 is not equal to 0. If c1 is not equal to 0, then I can divide both sides of this equation by c1."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you if I have this situation that I can definitely represent one of the vectors as the sum of the others. So let's say that this is true and one of these constants, remember, it's not just this, at least one is non-zero. So let me just assume, just for the sake of simplicity, I mean, these are all arbitrary, let me just assume, I'll do it in a new color, let me do it in the magenta, let me assume that c1 is not equal to 0. If c1 is not equal to 0, then I can divide both sides of this equation by c1. And what do I get? I get v1 plus c2 over c1 v2 plus all the way up to cn over c1 is equal to 0. And then I can multiply both sides of this by, or I could add negative v1 to both sides of this equation or subtract v1 from both sides."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If c1 is not equal to 0, then I can divide both sides of this equation by c1. And what do I get? I get v1 plus c2 over c1 v2 plus all the way up to cn over c1 is equal to 0. And then I can multiply both sides of this by, or I could add negative v1 to both sides of this equation or subtract v1 from both sides. Then I get c2 over c1 v2 plus all the way up to cn over c1 vn is equal to minus v1. Now if I just multiply both sides of this by negative 1, I get a minus, then all these become minuses, and this becomes a plus. And I just showed you that if at least one of these constants is non-zero, that I can represent my vector v1 as some combination of the other vectors."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I can multiply both sides of this by, or I could add negative v1 to both sides of this equation or subtract v1 from both sides. Then I get c2 over c1 v2 plus all the way up to cn over c1 vn is equal to minus v1. Now if I just multiply both sides of this by negative 1, I get a minus, then all these become minuses, and this becomes a plus. And I just showed you that if at least one of these constants is non-zero, that I can represent my vector v1 as some combination of the other vectors. So we're able to go this way too. If this condition is true, then I can represent one of the vectors as a combination of the others. If I can represent one of the vectors as a combination of the others, then this condition is true."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I just showed you that if at least one of these constants is non-zero, that I can represent my vector v1 as some combination of the other vectors. So we're able to go this way too. If this condition is true, then I can represent one of the vectors as a combination of the others. If I can represent one of the vectors as a combination of the others, then this condition is true. So hopefully that kind of proves that these two definitions are equivalent. Maybe it's a little bit of overkill. But let's apply that definition now to actually test, you might say, hey, Sal, why did you go through all of this effort?"}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I can represent one of the vectors as a combination of the others, then this condition is true. So hopefully that kind of proves that these two definitions are equivalent. Maybe it's a little bit of overkill. But let's apply that definition now to actually test, you might say, hey, Sal, why did you go through all of this effort? And I went through all of this effort because this is actually a really useful way to test whether things are linearly independent or dependent. Let's try it out. Let's use our newly found tool."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's apply that definition now to actually test, you might say, hey, Sal, why did you go through all of this effort? And I went through all of this effort because this is actually a really useful way to test whether things are linearly independent or dependent. Let's try it out. Let's use our newly found tool. So let's say I have the set of vectors. Let me do it up here. I want to be efficient with my space usage."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's use our newly found tool. So let's say I have the set of vectors. Let me do it up here. I want to be efficient with my space usage. So let's say I have the set of vectors 2, 1 and 3, 2. And my question to you is, are these linearly independent or are they linearly dependent? Well, in order for them to be linearly dependent, that means that there's some constant times 2, 1 plus some other constant times this second vector, 3, 2, where this should be equal to 0, where these both aren't necessarily 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I want to be efficient with my space usage. So let's say I have the set of vectors 2, 1 and 3, 2. And my question to you is, are these linearly independent or are they linearly dependent? Well, in order for them to be linearly dependent, that means that there's some constant times 2, 1 plus some other constant times this second vector, 3, 2, where this should be equal to 0, where these both aren't necessarily 0. So before I go forth with this problem, let's remember what we're going to find out. If either of these are non-zero, so if either C1 or C2 non-zero, then this implies that we are dealing with a dependent, linearly dependent set. Now, if C1 and C2 are both 0, if the only way to satisfy this equation, I mean, you can always satisfy it by setting everything equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, in order for them to be linearly dependent, that means that there's some constant times 2, 1 plus some other constant times this second vector, 3, 2, where this should be equal to 0, where these both aren't necessarily 0. So before I go forth with this problem, let's remember what we're going to find out. If either of these are non-zero, so if either C1 or C2 non-zero, then this implies that we are dealing with a dependent, linearly dependent set. Now, if C1 and C2 are both 0, if the only way to satisfy this equation, I mean, you can always satisfy it by setting everything equal to 0. But if the only way to satisfy it is by making both of these guys 0, then we're dealing with a linearly independent set. So let's try to do some math. And this will just take us back to our algebra one days."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, if C1 and C2 are both 0, if the only way to satisfy this equation, I mean, you can always satisfy it by setting everything equal to 0. But if the only way to satisfy it is by making both of these guys 0, then we're dealing with a linearly independent set. So let's try to do some math. And this will just take us back to our algebra one days. So in order for this to be true, that means 2 times C1 plus 3 times C2 is equal to, when I say this is equal to 0, it's really the 0 vector. So I can rewrite this as 0, 0. So it's equal to, so 2 times C1 plus 3 times C2 would be equal to that 0 there."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this will just take us back to our algebra one days. So in order for this to be true, that means 2 times C1 plus 3 times C2 is equal to, when I say this is equal to 0, it's really the 0 vector. So I can rewrite this as 0, 0. So it's equal to, so 2 times C1 plus 3 times C2 would be equal to that 0 there. And then we'd have 1 times C1 plus 2 times C2 is equal to that 0 right there. And now this is just a system, two equations, two unknowns. A couple of things we could do."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to, so 2 times C1 plus 3 times C2 would be equal to that 0 there. And then we'd have 1 times C1 plus 2 times C2 is equal to that 0 right there. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1 half. And then so if you multiply it by 1 half, you get C1 plus 3 halves C2 is equal to 0. And then if we subtract the green equation from the red equation, this becomes 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "A couple of things we could do. Let's just multiply this top equation by 1 half. And then so if you multiply it by 1 half, you get C1 plus 3 halves C2 is equal to 0. And then if we subtract the green equation from the red equation, this becomes 0. 2 minus 1 and 1 half. 3 halves is 1 and 1 half. So this is just 1 half."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then if we subtract the green equation from the red equation, this becomes 0. 2 minus 1 and 1 half. 3 halves is 1 and 1 half. So this is just 1 half. C2 is equal to 0. And this is easy to solve. C2 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is just 1 half. C2 is equal to 0. And this is easy to solve. C2 is equal to 0. So what's C1? Let's just substitute this back in. C2 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "C2 is equal to 0. So what's C1? Let's just substitute this back in. C2 is equal to 0. So this is equal to 0. So C1 plus 0 is equal to 0. So C1 is also equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "C2 is equal to 0. So this is equal to 0. So C1 plus 0 is equal to 0. So C1 is also equal to 0. We could have substituted back into that top equation as well. So the only solution to this equation involves both C1 and C2 being equal to 0. So they both have to be 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So C1 is also equal to 0. We could have substituted back into that top equation as well. So the only solution to this equation involves both C1 and C2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors. Which means that neither of them are redundant of the other one. You can't represent one as a combination of the other."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So they both have to be 0. So this is a linearly independent set of vectors. Which means that neither of them are redundant of the other one. You can't represent one as a combination of the other. And since we have two vectors here, and they're linearly independent, we can actually know that this will span R2. If one of these vectors was just some multiple of the other, then the span would have been some line within R2. Not all of them."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can't represent one as a combination of the other. And since we have two vectors here, and they're linearly independent, we can actually know that this will span R2. If one of these vectors was just some multiple of the other, then the span would have been some line within R2. Not all of them. But now I can represent any vector in R2 as some combination of those. Let's do another example. Let's say, let me scroll to the right, because sometimes this thing, when I go too far down, I haven't figured out why, when I go too far down it starts messing up."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Not all of them. But now I can represent any vector in R2 as some combination of those. Let's do another example. Let's say, let me scroll to the right, because sometimes this thing, when I go too far down, I haven't figured out why, when I go too far down it starts messing up. So my next example is the set of vectors. So I have the vector 2, 1. I have the vector 3, 2."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say, let me scroll to the right, because sometimes this thing, when I go too far down, I haven't figured out why, when I go too far down it starts messing up. So my next example is the set of vectors. So I have the vector 2, 1. I have the vector 3, 2. And I have the vector 1, 2. And I want to know, are these linearly dependent or linearly independent? So I do go through the same drill."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have the vector 3, 2. And I have the vector 1, 2. And I want to know, are these linearly dependent or linearly independent? So I do go through the same drill. I use that little theorem that I proved at the beginning of this video. So in order for them to be linearly dependent, there must be some set of weights that I can multiply these guys. So c1 times this vector plus c2 times this vector plus c3 times that vector."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I do go through the same drill. I use that little theorem that I proved at the beginning of this video. So in order for them to be linearly dependent, there must be some set of weights that I can multiply these guys. So c1 times this vector plus c2 times this vector plus c3 times that vector. That will equal the 0 vector. And if one of these is non-zero, then we're dealing with a linearly dependent set of vectors. And if all of them are 0, then it's independent."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So c1 times this vector plus c2 times this vector plus c3 times that vector. That will equal the 0 vector. And if one of these is non-zero, then we're dealing with a linearly dependent set of vectors. And if all of them are 0, then it's independent. So let's just do our linear algebra. Well, actually, I'll, well, let me just. So this means that 2 times c1 plus 3 times c2 plus c3 is equal to that 0 up there."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if all of them are 0, then it's independent. So let's just do our linear algebra. Well, actually, I'll, well, let me just. So this means that 2 times c1 plus 3 times c2 plus c3 is equal to that 0 up there. And then if we do the bottom rows, remember, when you multiply a scalar times a vector, you multiply it by each of these terms. So c1 times 1. So 1c1 plus 2c2 plus 2c3 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this means that 2 times c1 plus 3 times c2 plus c3 is equal to that 0 up there. And then if we do the bottom rows, remember, when you multiply a scalar times a vector, you multiply it by each of these terms. So c1 times 1. So 1c1 plus 2c2 plus 2c3 is equal to 0. Now, there's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because in the very best case, even if you assume that, let's say, that that vector and that vector are linearly independent, then these would span R2, which means that any vector in your two-dimensional space can be represented by some combination of those two."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 1c1 plus 2c2 plus 2c3 is equal to 0. Now, there's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because in the very best case, even if you assume that, let's say, that that vector and that vector are linearly independent, then these would span R2, which means that any vector in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them. Because this is just a vector in two-dimensional space. So it would be linearly dependent."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because in the very best case, even if you assume that, let's say, that that vector and that vector are linearly independent, then these would span R2, which means that any vector in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them. Because this is just a vector in two-dimensional space. So it would be linearly dependent. And then if you say, well, these aren't linearly independent, well, then if these aren't linearly independent, then they're just multiples of each other. In which case, this would definitely be a linearly dependent set. So when you see three vectors that are each only vectors in R2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it would be linearly dependent. And then if you say, well, these aren't linearly independent, well, then if these aren't linearly independent, then they're just multiples of each other. In which case, this would definitely be a linearly dependent set. So when you see three vectors that are each only vectors in R2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zeros, C3, C2s, and C1s, such that I can get a 0 here. If all of these had to be 0, I mean, you could always set them equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So when you see three vectors that are each only vectors in R2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zeros, C3, C2s, and C1s, such that I can get a 0 here. If all of these had to be 0, I mean, you could always set them equal to 0. But if they had to be equal to 0, then it would be linearly independent. But let me just show you. Look, I can just pick some random C3."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If all of these had to be 0, I mean, you could always set them equal to 0. But if they had to be equal to 0, then it would be linearly independent. But let me just show you. Look, I can just pick some random C3. Let me pick C3 to be equal to negative 1. So what would these two equations reduce to? I mean, if you have three unknowns and two equations, it means you don't have enough constraints on your system."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Look, I can just pick some random C3. Let me pick C3 to be equal to negative 1. So what would these two equations reduce to? I mean, if you have three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set C3, I just picked that out of a hat. I could have picked C3 to be anything. But if I set C3 to be equal to negative 1, what do these equations become?"}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I mean, if you have three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set C3, I just picked that out of a hat. I could have picked C3 to be anything. But if I set C3 to be equal to negative 1, what do these equations become? You get 2C1 plus 3C2 minus 1 is equal to 0. And you get C1 plus 2C2 minus 2 is equal to 0, right? 2 times minus 1."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if I set C3 to be equal to negative 1, what do these equations become? You get 2C1 plus 3C2 minus 1 is equal to 0. And you get C1 plus 2C2 minus 2 is equal to 0, right? 2 times minus 1. And then I can, what can I do here? If I multiply this second equation by 2, what do I get? I get 2 plus 4C2 minus 4 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times minus 1. And then I can, what can I do here? If I multiply this second equation by 2, what do I get? I get 2 plus 4C2 minus 4 is equal to 0. And now let's subtract this equation from that equation. So the C1's cancel out. 3C2 minus 4C2 is minus C2."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I get 2 plus 4C2 minus 4 is equal to 0. And now let's subtract this equation from that equation. So the C1's cancel out. 3C2 minus 4C2 is minus C2. And then minus 1 minus minus 4. So that's minus 1 plus 4. That's plus 3 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "3C2 minus 4C2 is minus C2. And then minus 1 minus minus 4. So that's minus 1 plus 4. That's plus 3 is equal to 0. And so we get our C, let me make sure I got that right. Minus, we have a minus 1 minus a minus 4. So plus 4."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's plus 3 is equal to 0. And so we get our C, let me make sure I got that right. Minus, we have a minus 1 minus a minus 4. So plus 4. So we have a plus 3. Right, so that is a minus 2. So we have our C2 is equal to minus 3 or C2 is equal to 3."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So plus 4. So we have a plus 3. Right, so that is a minus 2. So we have our C2 is equal to minus 3 or C2 is equal to 3. So we get C2 is equal to 3. And if C2 is equal to 3 and C3 is equal to minus 1, let's just substitute here. So we get C1 plus 2 times C2."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we have our C2 is equal to minus 3 or C2 is equal to 3. So we get C2 is equal to 3. And if C2 is equal to 3 and C3 is equal to minus 1, let's just substitute here. So we get C1 plus 2 times C2. So plus 6 plus 2 times C3. So minus 2 is equal to 0. C1 plus 4 is equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we get C1 plus 2 times C2. So plus 6 plus 2 times C3. So minus 2 is equal to 0. C1 plus 4 is equal to 0. C1 is equal to minus 4. So I'm giving you a combination of C's that will give us a 0 vector. If I multiply minus 4 times our first vector, 2, 1, that's C1, plus 3 times our second vector, 3, 2, minus 1 times our third vector, 1, 2, this should be equal to 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "C1 plus 4 is equal to 0. C1 is equal to minus 4. So I'm giving you a combination of C's that will give us a 0 vector. If I multiply minus 4 times our first vector, 2, 1, that's C1, plus 3 times our second vector, 3, 2, minus 1 times our third vector, 1, 2, this should be equal to 0. Let's verify it, just for fun. Minus 4 times 2 is minus 8 plus 9 minus 1. That's minus 9 plus 9."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply minus 4 times our first vector, 2, 1, that's C1, plus 3 times our second vector, 3, 2, minus 1 times our third vector, 1, 2, this should be equal to 0. Let's verify it, just for fun. Minus 4 times 2 is minus 8 plus 9 minus 1. That's minus 9 plus 9. That's 0. Minus 4 times minus 4 plus 6 minus 2. Minus 4 plus 6 minus 2, that's also 0."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's minus 9 plus 9. That's 0. Minus 4 times minus 4 plus 6 minus 2. Minus 4 plus 6 minus 2, that's also 0. So we've just shown a linear combination of these vectors where actually none of the constants are 0. But all we had to show is that at least one of the constants had to be non-zero. And we actually showed all three of them were."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 4 plus 6 minus 2, that's also 0. So we've just shown a linear combination of these vectors where actually none of the constants are 0. But all we had to show is that at least one of the constants had to be non-zero. And we actually showed all three of them were. But at least one of these had to be non-zero. And I was able to satisfy this equation. I was able to make them into the 0 vector."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And we actually showed all three of them were. But at least one of these had to be non-zero. And I was able to satisfy this equation. I was able to make them into the 0 vector. So this proves that this is a linearly dependent set of vectors, which means one of the vectors is redundant. And you can never just say, oh, this is the redundant vector because I can represent this as a combination of those two. You could just as easily pick this guy as the redundant vector and say, hey, I can represent this guy as a sum of those two."}, {"video_title": "More on linear independence Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I was able to make them into the 0 vector. So this proves that this is a linearly dependent set of vectors, which means one of the vectors is redundant. And you can never just say, oh, this is the redundant vector because I can represent this as a combination of those two. You could just as easily pick this guy as the redundant vector and say, hey, I can represent this guy as a sum of those two. There's not kind of one bad apple in the bunch. Any of them can be represented by the combination of some other, by all of the rest of them. So hopefully you have a better intuition of linear dependence and independence."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you're actually going to have an infinite number of solutions, but those infinite number of solutions could still be constrained within, well, let's say we're in four dimensions in this case, because we have four variables. Maybe we're constrained into a plane in four dimensions, or if we're in three dimensions, maybe we're constrained to a line. A line is an infinite number of solutions, but it's kind of a more constrained set. So let's solve this set of linear equations, and we've done this by elimination in the past, but what I want to do is I want to introduce the idea of matrices. And the matrices are really just arrays of numbers that are shorthand for this system of equations. So let me create a matrix here. So I could just create a coefficient matrix, where the coefficient matrix would just be, let me write it neatly, the coefficient matrix would just be the coefficients on the left-hand side of these linear equations."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's solve this set of linear equations, and we've done this by elimination in the past, but what I want to do is I want to introduce the idea of matrices. And the matrices are really just arrays of numbers that are shorthand for this system of equations. So let me create a matrix here. So I could just create a coefficient matrix, where the coefficient matrix would just be, let me write it neatly, the coefficient matrix would just be the coefficients on the left-hand side of these linear equations. So the coefficient there is 1, coefficient there is 2, and you have 2 to 4, 1 to 0, there's no coefficient on the x3 term here, because there's no x3 term there. So we'll say the coefficient on the x3 term there is 0, and then we have 1 minus 1, 1 minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I could just create a coefficient matrix, where the coefficient matrix would just be, let me write it neatly, the coefficient matrix would just be the coefficients on the left-hand side of these linear equations. So the coefficient there is 1, coefficient there is 2, and you have 2 to 4, 1 to 0, there's no coefficient on the x3 term here, because there's no x3 term there. So we'll say the coefficient on the x3 term there is 0, and then we have 1 minus 1, 1 minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. But what I want to do is I want to augment it. I want to augment it with what these equations need to be equal to. So let me augment it, and what I'm going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. But what I want to do is I want to augment it. I want to augment it with what these equations need to be equal to. So let me augment it, and what I'm going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. And I think you can see that whatever this right here is just another way of writing this. And just by the position, we know that these are the coefficients on the x1 terms, we know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me augment it, and what I'm going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. And I think you can see that whatever this right here is just another way of writing this. And just by the position, we know that these are the coefficients on the x1 terms, we know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. But we can essentially do the same operations on this that we otherwise would have done on that. So what we can do is we can replace any equation with that equation times some scalar multiple plus another equation. We can divide an equation or multiply an equation by a scalar."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what this does, it really just saves us from having to write x1 and x2 every time. But we can essentially do the same operations on this that we otherwise would have done on that. So what we can do is we can replace any equation with that equation times some scalar multiple plus another equation. We can divide an equation or multiply an equation by a scalar. We can subtract them from each other. We can swap them. So let's do that in an attempt to solve this equation."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We can divide an equation or multiply an equation by a scalar. We can subtract them from each other. We can swap them. So let's do that in an attempt to solve this equation. So the first thing I want to do, just like I've done in the past, is I want to get this equation to the form of, where if I can, I have a 1, my kind of leading coefficient in any of my rows is a 1, and that every other entry in that column is a 0. And in the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos when we tried to figure out if things were linearly independent or not."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that in an attempt to solve this equation. So the first thing I want to do, just like I've done in the past, is I want to get this equation to the form of, where if I can, I have a 1, my kind of leading coefficient in any of my rows is a 1, and that every other entry in that column is a 0. And in the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos when we tried to figure out if things were linearly independent or not. But now I'm going to make sure that if there's a 1, if there's a leading 1 in any of my rows, that everything else in that column is a 0. And that form I'm doing is called reduced row echelon form. Let me write that."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's what I was doing in some of the previous videos when we tried to figure out if things were linearly independent or not. But now I'm going to make sure that if there's a 1, if there's a leading 1 in any of my rows, that everything else in that column is a 0. And that form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. And if we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters instead of lowercase letters."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that. Reduced row echelon form. And if we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters instead of lowercase letters. And we'll talk more about how matrices relate to vectors in the future. But let's just solve this system of equations. So the first thing I want to do is, in an ideal world, I would get all of these guys right here to be 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters instead of lowercase letters. And we'll talk more about how matrices relate to vectors in the future. But let's just solve this system of equations. So the first thing I want to do is, in an ideal world, I would get all of these guys right here to be 0. So let me replace this guy with that guy with the first entry minus the second entry. So let me do that. So I am just going to, the first row isn't going to change."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing I want to do is, in an ideal world, I would get all of these guys right here to be 0. So let me replace this guy with that guy with the first entry minus the second entry. So let me do that. So I am just going to, the first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I am just going to, the first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get? 1 minus 1 is 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get? 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I don't want to turn it into a 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I don't want to turn it into a 0. So let's replace this row with this row minus 2 times that row. So what I'm going to do is this row minus 2 times the first row. And I'm going to replace this row with that."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to turn it into a 0. So let's replace this row with this row minus 2 times that row. So what I'm going to do is this row minus 2 times the first row. And I'm going to replace this row with that. So 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I'm going to replace this row with that. So 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. And then 4 minus 2 times 7 is 4 minus 14, which is minus 10."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. And then 4 minus 2 times 7 is 4 minus 14, which is minus 10. Now what can I do next? Well, ideally, you can see that this row is kind of... We'll talk more about what this row means when all of a sudden it's just all been zeroed out. There's nothing here."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then 4 minus 2 times 7 is 4 minus 14, which is minus 10. Now what can I do next? Well, ideally, you can see that this row is kind of... We'll talk more about what this row means when all of a sudden it's just all been zeroed out. There's nothing here. If I had a non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. So I'm just going to move over to this row. So the first thing I want to do is I want to make this leading coefficient here a 1."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's nothing here. If I had a non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. So I'm just going to move over to this row. So the first thing I want to do is I want to make this leading coefficient here a 1. So what I want to do is I'm going to multiply this entire row by minus 1. So if I multiply this entire row times minus 1... In fact, I don't even have to rewrite the matrix."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing I want to do is I want to make this leading coefficient here a 1. So what I want to do is I'm going to multiply this entire row by minus 1. So if I multiply this entire row times minus 1... In fact, I don't even have to rewrite the matrix. This becomes plus 1 minus 2 plus 5. I think you can accept that. All right, now what can we do?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "In fact, I don't even have to rewrite the matrix. This becomes plus 1 minus 2 plus 5. I think you can accept that. All right, now what can we do? Well, let's turn this right here into a 0. So let me rewrite my augmented matrix in the new form that I have. So I'm going to keep the middle row the same this time."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All right, now what can we do? Well, let's turn this right here into a 0. So let me rewrite my augmented matrix in the new form that I have. So I'm going to keep the middle row the same this time. So my middle row is 0, 0, 1, minus 2. And then it's augmented, and I get a 5 there. And what I want to do is I want to eliminate this minus 2 here."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep the middle row the same this time. So my middle row is 0, 0, 1, minus 2. And then it's augmented, and I get a 5 there. And what I want to do is I want to eliminate this minus 2 here. So why don't I add this row to 2 times that row? Then I would have minus 2 plus 2, and that will work out. So what do I get?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what I want to do is I want to eliminate this minus 2 here. So why don't I add this row to 2 times that row? Then I would have minus 2 plus 2, and that will work out. So what do I get? I get a...well, these are just leading 0s. And then I have minus 2 plus 2 times 1. Well, that's just 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what do I get? I get a...well, these are just leading 0s. And then I have minus 2 plus 2 times 1. Well, that's just 0. 4 plus 2 times minus 2. That's minus 4. So that's 4 plus minus 4."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's just 0. 4 plus 2 times minus 2. That's minus 4. So that's 4 plus minus 4. That's 0 as well. And then you have minus 10 plus 2 times 5. Well, that's just minus 10 plus 10, which is 0."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's 4 plus minus 4. That's 0 as well. And then you have minus 10 plus 2 times 5. Well, that's just minus 10 plus 10, which is 0. So that one just got zeroed out. And what I want to do, and normally when I just did regular elimination, I was happy just having this situation where I just had these leading 1s. Everything below it were 0s."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's just minus 10 plus 10, which is 0. So that one just got zeroed out. And what I want to do, and normally when I just did regular elimination, I was happy just having this situation where I just had these leading 1s. Everything below it were 0s. But I wasn't too concerned about what was above our 1s. But what I want to do is I want to make those into a 0 as well. So I want to make this guy a 0 as well."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Everything below it were 0s. But I wasn't too concerned about what was above our 1s. But what I want to do is I want to make those into a 0 as well. So I want to make this guy a 0 as well. So what I can do is I can replace this first row with that first row minus the second row. So what is 1 minus 0? Well, that's just 1."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I want to make this guy a 0 as well. So what I can do is I can replace this first row with that first row minus the second row. So what is 1 minus 0? Well, that's just 1. 2 minus 0 is 2. 1 minus 1 is 0. 1 minus minus 2 is 3."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's just 1. 2 minus 0 is 2. 1 minus 1 is 0. 1 minus minus 2 is 3. 7 minus 5 is 2. And there you have it. We have our matrix in reduced row echelon form."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus minus 2 is 3. 7 minus 5 is 2. And there you have it. We have our matrix in reduced row echelon form. This is the reduced row echelon form of our matrix. I'll write it in bold. Of our matrix A right there."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have our matrix in reduced row echelon form. This is the reduced row echelon form of our matrix. I'll write it in bold. Of our matrix A right there. And you know it's in reduced row echelon form because all of your leading 1s in each row So what are my leading 1s in each row? I have this one and I have that one. They're the only non-zero entry in their columns."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Of our matrix A right there. And you know it's in reduced row echelon form because all of your leading 1s in each row So what are my leading 1s in each row? I have this one and I have that one. They're the only non-zero entry in their columns. And these are called the pivot entries. Let me label that for you. That's called a pivot entry."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "They're the only non-zero entry in their columns. And these are called the pivot entries. Let me label that for you. That's called a pivot entry. They're the only non-zero entry in their respective columns. If I have any zeroed out rows, and I do have a zeroed out row, it's right there. So this is zeroed out row."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's called a pivot entry. They're the only non-zero entry in their respective columns. If I have any zeroed out rows, and I do have a zeroed out row, it's right there. So this is zeroed out row. Just the style or just the convention that for reduced row echelon form, that has to be your last row. So we have the leading entries are the only They're all 1. That's one case."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is zeroed out row. Just the style or just the convention that for reduced row echelon form, that has to be your last row. So we have the leading entries are the only They're all 1. That's one case. You can't have this a 5. You'd want to divide that equation by 5 if this was a 5. So your leading entries in each row are 1."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's one case. You can't have this a 5. You'd want to divide that equation by 5 if this was a 5. So your leading entries in each row are 1. The leading entry in each successive row is to the right of the leading entry of the row before it. This guy right here is to the right of that guy. This is just the style, the convention of reduced row echelon form."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So your leading entries in each row are 1. The leading entry in each successive row is to the right of the leading entry of the row before it. This guy right here is to the right of that guy. This is just the style, the convention of reduced row echelon form. If you have any zeroed out rows, it's in the last row. Then finally, of course, and I think I've said this multiple times, this is the only non-zero entry in the row. Now what does this do for me?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is just the style, the convention of reduced row echelon form. If you have any zeroed out rows, it's in the last row. Then finally, of course, and I think I've said this multiple times, this is the only non-zero entry in the row. Now what does this do for me? Now I can go back from this world, back to my linear equations. Because we remember that these were the coefficients on x1, these were the coefficients on x2, these were the coefficients on x3, on x4, and then these were my constants out here. So I can rewrite this system of equations in my reduced row echelon form as x1 plus 2x2."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what does this do for me? Now I can go back from this world, back to my linear equations. Because we remember that these were the coefficients on x1, these were the coefficients on x2, these were the coefficients on x3, on x4, and then these were my constants out here. So I can rewrite this system of equations in my reduced row echelon form as x1 plus 2x2. There's no x3 there, so plus 3x4 is equal to 2. And then this equation, no x1, no x2, I have an x3. I have x3 minus 2x4 is equal to 5."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I can rewrite this system of equations in my reduced row echelon form as x1 plus 2x2. There's no x3 there, so plus 3x4 is equal to 2. And then this equation, no x1, no x2, I have an x3. I have x3 minus 2x4 is equal to 5. And then I have no other equation here. This one got completely zeroed out, so I was able to reduce this system of equations to this system of equations. Now the variables that you associate with your pivot entries, we call these pivot variables."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have x3 minus 2x4 is equal to 5. And then I have no other equation here. This one got completely zeroed out, so I was able to reduce this system of equations to this system of equations. Now the variables that you associate with your pivot entries, we call these pivot variables. So x1 and x3 are pivot variables. And then the variables that aren't associated with a pivot entry, we call them free variables. So x2 and x4 are free variables."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now the variables that you associate with your pivot entries, we call these pivot variables. So x1 and x3 are pivot variables. And then the variables that aren't associated with a pivot entry, we call them free variables. So x2 and x4 are free variables. But now let's solve for, essentially, you can only solve for your pivot variables. The free variables we can set to any variable. And I said that in the beginning of this equation."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So x2 and x4 are free variables. But now let's solve for, essentially, you can only solve for your pivot variables. The free variables we can set to any variable. And I said that in the beginning of this equation. We have fewer equations than unknowns, so this is going to be a not well-constrained solution. You're not going to have just one point in R4 that solves this equation. You're going to have multiple points."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I said that in the beginning of this equation. We have fewer equations than unknowns, so this is going to be a not well-constrained solution. You're not going to have just one point in R4 that solves this equation. You're going to have multiple points. So let's solve for our pivot variables, because that's all we can solve for. So this equation tells us right here, it tells us x3, let me do it in a good color, x3 is equal to 5 plus 2x4. And then we get x1 is equal to 2 minus 2x2 plus minus 3x4."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're going to have multiple points. So let's solve for our pivot variables, because that's all we can solve for. So this equation tells us right here, it tells us x3, let me do it in a good color, x3 is equal to 5 plus 2x4. And then we get x1 is equal to 2 minus 2x2 plus minus 3x4. I just subtracted these from both sides of the equation. Now this right here is essentially as far as we can go to the solution of this system of equations. I can pick really any values for my free variables."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we get x1 is equal to 2 minus 2x2 plus minus 3x4. I just subtracted these from both sides of the equation. Now this right here is essentially as far as we can go to the solution of this system of equations. I can pick really any values for my free variables. I can pick any values for my x2's and my x4's, and I can solve for an x3. But what I want to do right now is write this in a slightly different form, so that we can visualize a little bit better. And of course, it's always hard to visualize things in four dimensions."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can pick really any values for my free variables. I can pick any values for my x2's and my x4's, and I can solve for an x3. But what I want to do right now is write this in a slightly different form, so that we can visualize a little bit better. And of course, it's always hard to visualize things in four dimensions. But so we can visualize things a little bit better as to kind of the set of this solution. So let's write it this way. If I were to write it in vector form, our solution is the vector x1, x2, x3, x4."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And of course, it's always hard to visualize things in four dimensions. But so we can visualize things a little bit better as to kind of the set of this solution. So let's write it this way. If I were to write it in vector form, our solution is the vector x1, x2, x3, x4. And what is it equal to? Well, it's equal to, let me write it like this. I'm just rewriting, I'm just essentially rewriting this solution set in vector form."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to write it in vector form, our solution is the vector x1, x2, x3, x4. And what is it equal to? Well, it's equal to, let me write it like this. I'm just rewriting, I'm just essentially rewriting this solution set in vector form. So x1 is equal to 2, let me write a little column there, plus x2, let me write it this way, plus x2 times something, plus x4 times something. So how do I, so x1 is equal to 2 minus 2 times x2, or plus x2 times minus 2. So I put a minus 2 there."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just rewriting, I'm just essentially rewriting this solution set in vector form. So x1 is equal to 2, let me write a little column there, plus x2, let me write it this way, plus x2 times something, plus x4 times something. So how do I, so x1 is equal to 2 minus 2 times x2, or plus x2 times minus 2. So I put a minus 2 there. And then I can say plus x4 times minus 3. So I can put a minus 3 there. This right here, these first entries of these vectors just literally represent that equation right there."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I put a minus 2 there. And then I can say plus x4 times minus 3. So I can put a minus 3 there. This right here, these first entries of these vectors just literally represent that equation right there. x1 is equal to 2 plus x2 times minus 2 plus x4 times minus 3. Now what does x3 equal? What does x3 equal?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This right here, these first entries of these vectors just literally represent that equation right there. x1 is equal to 2 plus x2 times minus 2 plus x4 times minus 3. Now what does x3 equal? What does x3 equal? x3 is equal to 5, put that 5 right there, plus x4 times 2. x2 doesn't apply to it, so we can just put a 0. 0 times x2 plus 2 times x4. Now what does x2 equal?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What does x3 equal? x3 is equal to 5, put that 5 right there, plus x4 times 2. x2 doesn't apply to it, so we can just put a 0. 0 times x2 plus 2 times x4. Now what does x2 equal? Well you could say x2 is equal to 0 plus 1 times x2, plus 0 times x4, right? x2 is just equal to x2, it's a free variable. And similarly, what is x4 equal to?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what does x2 equal? Well you could say x2 is equal to 0 plus 1 times x2, plus 0 times x4, right? x2 is just equal to x2, it's a free variable. And similarly, what is x4 equal to? x4 is equal to 0 plus 0 times x2 plus 1 times x4. Now what does this do for us? Well all of a sudden here, we've expressed our solution set as essentially the linear combination of 3 vectors."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And similarly, what is x4 equal to? x4 is equal to 0 plus 0 times x2 plus 1 times x4. Now what does this do for us? Well all of a sudden here, we've expressed our solution set as essentially the linear combination of 3 vectors. This is a vector in, you could view it as almost a coordinate, either a position vector. It is a vector in R4, you can view it as a position vector or a coordinate in R4. And then you could say, look, our solution set is essentially, this is in R4, each of these have 4 components, but you can imagine it in R3."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well all of a sudden here, we've expressed our solution set as essentially the linear combination of 3 vectors. This is a vector in, you could view it as almost a coordinate, either a position vector. It is a vector in R4, you can view it as a position vector or a coordinate in R4. And then you could say, look, our solution set is essentially, this is in R4, each of these have 4 components, but you can imagine it in R3. That I have my solution set is equal to some vector, some vector there, that's the vector, think of it as a position vector, it would be the coordinate 2, 0, 5, 0, which obviously, this is 4 dimensions right there. But it's equal to multiples of these 2 vectors. So let's call this vector right here, let's call this vector vector A, and let's call this vector right here vector B."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then you could say, look, our solution set is essentially, this is in R4, each of these have 4 components, but you can imagine it in R3. That I have my solution set is equal to some vector, some vector there, that's the vector, think of it as a position vector, it would be the coordinate 2, 0, 5, 0, which obviously, this is 4 dimensions right there. But it's equal to multiples of these 2 vectors. So let's call this vector right here, let's call this vector vector A, and let's call this vector right here vector B. So our solution set is all of this point, which is right there, or I guess we could call it that position vector, right? That position vector will look like that, where you're starting at the origin right there, plus multiples of these 2 guys. So if this is vector A, actually we'll do vector A in a different color, vector A looks like that, so let's say vector A looks like that, and then vector B looks like that."}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's call this vector right here, let's call this vector vector A, and let's call this vector right here vector B. So our solution set is all of this point, which is right there, or I guess we could call it that position vector, right? That position vector will look like that, where you're starting at the origin right there, plus multiples of these 2 guys. So if this is vector A, actually we'll do vector A in a different color, vector A looks like that, so let's say vector A looks like that, and then vector B looks like that. So this is vector B, and this is vector A. And I don't know if this is going to be easier or harder for you to visualize, because obviously we're dealing in 4 dimensions right here, and I'm kind of just drawing it on a 2-dimensional surface, but what you can imagine is, that the solution set is equal to this fixed point, this position vector, plus linear combinations of A and B. And we're dealing of course in R4, let me write that down, we're dealing in R4, but linear combinations of A and B are going to create a plane, right?"}, {"video_title": "Matrices Reduced row echelon form 1 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if this is vector A, actually we'll do vector A in a different color, vector A looks like that, so let's say vector A looks like that, and then vector B looks like that. So this is vector B, and this is vector A. And I don't know if this is going to be easier or harder for you to visualize, because obviously we're dealing in 4 dimensions right here, and I'm kind of just drawing it on a 2-dimensional surface, but what you can imagine is, that the solution set is equal to this fixed point, this position vector, plus linear combinations of A and B. And we're dealing of course in R4, let me write that down, we're dealing in R4, but linear combinations of A and B are going to create a plane, right? You can just take, you know, you can multiply A times 2, and B times 3, or A times minus 1, and B times minus 100, and you can just keep adding and subtracting these linear combinations of A and B, and they're going to construct a plane that contains the position vector, that contains the point 2, 0, 5, 0. So the solution for these 3 equations with 4 unknowns is a plane in R4. And I know that's really hard to visualize, and maybe I'll do another one in 3 dimensions, but hopefully this at least gives you a decent understanding of what an augmented matrix is, what reduced row echelon form is, and what are kind of the valid operations that you can perform on a matrix without kind of messing up the solution."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "And then you have some other vector x that we're going to take the projection of it onto l. So that's x. The projection of x onto l, this thing right here, is going to be some vector in l. So it's going to be some vector in l, such that when I take the difference between x and that vector, it's going to be orthogonal to l. So it's going to be some vector in l. This was our old definition, where we took the projection onto a line. Some vector in l, maybe it's there. And if I take the difference between that and that, this difference vector is going to be orthogonal to everything in l. It's going to be orthogonal to everything in l. Just like that. So this right here would be its difference vector. That would be x minus the projection of x onto l. And then, of course, this vector right here, this is the one we were defining it. That was the projection."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "And if I take the difference between that and that, this difference vector is going to be orthogonal to everything in l. It's going to be orthogonal to everything in l. Just like that. So this right here would be its difference vector. That would be x minus the projection of x onto l. And then, of course, this vector right here, this is the one we were defining it. That was the projection. Projection onto l of x. Now what's a different way that we could have written this? We could have written this exact same definition."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "That was the projection. Projection onto l of x. Now what's a different way that we could have written this? We could have written this exact same definition. We could have said it is the vector in l such that. So we could say is, let me write it here in purple, is the vector v in l such that v, let me write it this way, such that x minus v, right? x minus the projection of l is orthogonal."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "We could have written this exact same definition. We could have said it is the vector in l such that. So we could say is, let me write it here in purple, is the vector v in l such that v, let me write it this way, such that x minus v, right? x minus the projection of l is orthogonal. Let me write it. Is equal to w, which is orthogonal to everything in l. Being orthogonal to l literally means being orthogonal to every vector in l. So I just rewrote it a little bit different. Instead of just leaving it as a projection of x onto l, I said, hey, that's some vector v in l. So this is some vector v in l such that x minus v is equal to some other vector w. That's that other vector w, which is orthogonal to everything in l. Or we could write it, we could rewrite that statement right there, as x is equal to v plus w. So we could just say that the projection of x onto l is the unique vector v in l such that x is equal to v plus w, where w is a unique vector."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "x minus the projection of l is orthogonal. Let me write it. Is equal to w, which is orthogonal to everything in l. Being orthogonal to l literally means being orthogonal to every vector in l. So I just rewrote it a little bit different. Instead of just leaving it as a projection of x onto l, I said, hey, that's some vector v in l. So this is some vector v in l such that x minus v is equal to some other vector w. That's that other vector w, which is orthogonal to everything in l. Or we could write it, we could rewrite that statement right there, as x is equal to v plus w. So we could just say that the projection of x onto l is the unique vector v in l such that x is equal to v plus w, where w is a unique vector. I mean, it is going to be a unique vector in the orthogonal complement of l, right? This has got to be orthogonal to everything in l. So that's going to be a member of the orthogonal complement. It's going to be a member of the orthogonal complement of l. So this definition is actually completely consistent with our new subspace definition."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "Instead of just leaving it as a projection of x onto l, I said, hey, that's some vector v in l. So this is some vector v in l such that x minus v is equal to some other vector w. That's that other vector w, which is orthogonal to everything in l. Or we could write it, we could rewrite that statement right there, as x is equal to v plus w. So we could just say that the projection of x onto l is the unique vector v in l such that x is equal to v plus w, where w is a unique vector. I mean, it is going to be a unique vector in the orthogonal complement of l, right? This has got to be orthogonal to everything in l. So that's going to be a member of the orthogonal complement. It's going to be a member of the orthogonal complement of l. So this definition is actually completely consistent with our new subspace definition. And we could just extend it to arbitrary subspaces, not just lines. And let me help you visualize that. So let's say we're dealing with R3 here."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be a member of the orthogonal complement of l. So this definition is actually completely consistent with our new subspace definition. And we could just extend it to arbitrary subspaces, not just lines. And let me help you visualize that. So let's say we're dealing with R3 here. And I've got some subspace in R3. And let's say that subspace happens to be a plane. I'm going to make it a plane just so that it becomes clear that we don't have to take projections just onto lines."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we're dealing with R3 here. And I've got some subspace in R3. And let's say that subspace happens to be a plane. I'm going to make it a plane just so that it becomes clear that we don't have to take projections just onto lines. So this is my subspace v right there. And let me draw its orthogonal complement. Let's say its orthogonal complement looks something like that."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to make it a plane just so that it becomes clear that we don't have to take projections just onto lines. So this is my subspace v right there. And let me draw its orthogonal complement. Let's say its orthogonal complement looks something like that. Let's say it's a line. And then it intersects right there. Then it goes back."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "Let's say its orthogonal complement looks something like that. Let's say it's a line. And then it intersects right there. Then it goes back. And of course, it would have to intersect at the 0 vector. That's the only place where a subspace and its orthogonal complement overlap. And then it goes behind."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "Then it goes back. And of course, it would have to intersect at the 0 vector. That's the only place where a subspace and its orthogonal complement overlap. And then it goes behind. And then you can see it again. But obviously, you wouldn't be able to see it again because this plane would extend in every direction. But you get the idea."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "And then it goes behind. And then you can see it again. But obviously, you wouldn't be able to see it again because this plane would extend in every direction. But you get the idea. So this right here is the orthogonal complement of v, that line. Now, let's say I have some other arbitrary vector in R3 here. So let's say I have some vector that looks like that."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "But you get the idea. So this right here is the orthogonal complement of v, that line. Now, let's say I have some other arbitrary vector in R3 here. So let's say I have some vector that looks like that. Let's say that that is x. Now, our new definition for the projection of x onto v is equal to the unique vector v. This is a vector v. That's a subspace v. The unique vector v that is a member of v such that x is equal to v plus w, where w is a unique member of the orthogonal complement of v. This is our new definition. So if we say x is equal to some member of v and some member of its orthogonal complement, we can visually understand that here."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have some vector that looks like that. Let's say that that is x. Now, our new definition for the projection of x onto v is equal to the unique vector v. This is a vector v. That's a subspace v. The unique vector v that is a member of v such that x is equal to v plus w, where w is a unique member of the orthogonal complement of v. This is our new definition. So if we say x is equal to some member of v and some member of its orthogonal complement, we can visually understand that here. We could say, OK, it's going to be equal to, let's see, on v, it'll be equal to that vector right there. And then on v's orthogonal complement, you add that to it, so if you were to shift it over, you would get that vector just like that. This right here is v. That right there is v. And then this vector that goes up like this out of the plane, orthogonal to the plane, is w. And you can see if you take v plus w, you're going to get x."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "So if we say x is equal to some member of v and some member of its orthogonal complement, we can visually understand that here. We could say, OK, it's going to be equal to, let's see, on v, it'll be equal to that vector right there. And then on v's orthogonal complement, you add that to it, so if you were to shift it over, you would get that vector just like that. This right here is v. That right there is v. And then this vector that goes up like this out of the plane, orthogonal to the plane, is w. And you can see if you take v plus w, you're going to get x. And you can see that v is the projection onto the subspace capital V, so this is a vector v, is the projection onto the subspace capital V of the vector x. So the analogy to a shadow still holds. If you imagine kind of a light source coming straight down onto our subspace, kind of orthogonal to our subspace, the projection onto our subspace is kind of the shadow of our vector x."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "This right here is v. That right there is v. And then this vector that goes up like this out of the plane, orthogonal to the plane, is w. And you can see if you take v plus w, you're going to get x. And you can see that v is the projection onto the subspace capital V, so this is a vector v, is the projection onto the subspace capital V of the vector x. So the analogy to a shadow still holds. If you imagine kind of a light source coming straight down onto our subspace, kind of orthogonal to our subspace, the projection onto our subspace is kind of the shadow of our vector x. So hopefully that helps you visualize it a little better. But what we're doing here is we're going to generalize it. Here, earlier in this video I showed you a line."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "If you imagine kind of a light source coming straight down onto our subspace, kind of orthogonal to our subspace, the projection onto our subspace is kind of the shadow of our vector x. So hopefully that helps you visualize it a little better. But what we're doing here is we're going to generalize it. Here, earlier in this video I showed you a line. This is a plane. But we can generalize it to any subspace. This is an R3."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "Here, earlier in this video I showed you a line. This is a plane. But we can generalize it to any subspace. This is an R3. We could generalize it to Rn, to R100. And that's really the power of what we're doing here. It's easy to visualize it here, but it's not so easy to visualize it once you get to higher dimensions."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "This is an R3. We could generalize it to Rn, to R100. And that's really the power of what we're doing here. It's easy to visualize it here, but it's not so easy to visualize it once you get to higher dimensions. Actually, one other thing. Let me show you that this new definition is pretty much almost identical to exactly what we did with lines. This is identical to saying that the projection onto the subspace x is equal to some unique vector in V such that x minus the projection onto V of x is orthogonal to every member of V. Because this statement right here is saying, any vector that's orthogonal to any member of V says that it's a member of the orthogonal complement of V. So that statement could be written as x minus the projection onto V of x is a member of V's orthogonal complement, or we could call it some w. So if you call this your V, and if you call this whole thing your w, you get this exact definition right there."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "It's easy to visualize it here, but it's not so easy to visualize it once you get to higher dimensions. Actually, one other thing. Let me show you that this new definition is pretty much almost identical to exactly what we did with lines. This is identical to saying that the projection onto the subspace x is equal to some unique vector in V such that x minus the projection onto V of x is orthogonal to every member of V. Because this statement right here is saying, any vector that's orthogonal to any member of V says that it's a member of the orthogonal complement of V. So that statement could be written as x minus the projection onto V of x is a member of V's orthogonal complement, or we could call it some w. So if you call this your V, and if you call this whole thing your w, you get this exact definition right there. You would have w is equal to x minus v. And then if you add v to both sides, you get w plus v is equal to x. We defined v to be the projection of x onto V. w is a member of our orthogonal complement. And I don't want to get confused."}, {"video_title": "Visualizing a projection onto a plane Linear Algebra Khan Academy.mp3", "Sentence": "This is identical to saying that the projection onto the subspace x is equal to some unique vector in V such that x minus the projection onto V of x is orthogonal to every member of V. Because this statement right here is saying, any vector that's orthogonal to any member of V says that it's a member of the orthogonal complement of V. So that statement could be written as x minus the projection onto V of x is a member of V's orthogonal complement, or we could call it some w. So if you call this your V, and if you call this whole thing your w, you get this exact definition right there. You would have w is equal to x minus v. And then if you add v to both sides, you get w plus v is equal to x. We defined v to be the projection of x onto V. w is a member of our orthogonal complement. And I don't want to get confused. This is the orthogonal V, the vector V is the orthogonal projection of our vector x onto the subspace capital V. I probably should use different letters instead of using a lowercase and an uppercase V. It makes the language a little difficult. But I just wanted to give you another video to give you a visualization of projections onto subspaces other than lines, and to show you that our old definition with just a projection onto a line, which was a linear transformation, is essentially equivalent to this new definition. On the next video, I'll show you that this, for any subspace, is indeed a linear transformation."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you just have two vectors, say, a1, a2, all the way down to an. We define the addition of this vector and, let's say, some other vector, b1, b2, all the way down to bn, as a third vector. If you add these two, we define the addition operation to be a third. You'll result with a third vector, where each of its components are just the sum of the corresponding components of the two vectors you're adding. So it's going to be a1 plus b1, a2 plus b2, all the way down to an plus bn. We knew this, and we've done multiple videos where we use this definition of vector addition. We also know about scalar multiplication."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You'll result with a third vector, where each of its components are just the sum of the corresponding components of the two vectors you're adding. So it's going to be a1 plus b1, a2 plus b2, all the way down to an plus bn. We knew this, and we've done multiple videos where we use this definition of vector addition. We also know about scalar multiplication. That's the case. If I have some real number, c, and I multiply it times some vector, a1, a2, all the way down to an, we define scalar multiplication of a vector to be some scalar times this vector will result in, essentially, this vector where each of its components are multiplied by the scalar, c, a1, c, a2, all the way down to c, an. After seeing these two operations, you might be tempted to say, wouldn't it be nice if there was some way to multiply two vectors?"}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We also know about scalar multiplication. That's the case. If I have some real number, c, and I multiply it times some vector, a1, a2, all the way down to an, we define scalar multiplication of a vector to be some scalar times this vector will result in, essentially, this vector where each of its components are multiplied by the scalar, c, a1, c, a2, all the way down to c, an. After seeing these two operations, you might be tempted to say, wouldn't it be nice if there was some way to multiply two vectors? This is just a scalar times a vector. You're just scaling it up. That's actually the actual effect of what it's doing if you visualize it in three dimensions or less."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "After seeing these two operations, you might be tempted to say, wouldn't it be nice if there was some way to multiply two vectors? This is just a scalar times a vector. You're just scaling it up. That's actually the actual effect of what it's doing if you visualize it in three dimensions or less. It's actually scaling the size of the vector. We haven't even defined size very precisely just yet. You understand at least this operation."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's actually the actual effect of what it's doing if you visualize it in three dimensions or less. It's actually scaling the size of the vector. We haven't even defined size very precisely just yet. You understand at least this operation. For multiplying vectors or taking the product, there's actually two ways. I'm going to define one of them in this video. That's the dot product."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You understand at least this operation. For multiplying vectors or taking the product, there's actually two ways. I'm going to define one of them in this video. That's the dot product. You signify the dot product by saying a dot b. They borrowed one of the types of multiplication notations that you saw, but you can't write a cross here. That'll be actually a different type of vector multiplication."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's the dot product. You signify the dot product by saying a dot b. They borrowed one of the types of multiplication notations that you saw, but you can't write a cross here. That'll be actually a different type of vector multiplication. The dot product is almost fun to take because it's mathematically pretty straightforward, unlike the cross product. It's fun to take and it's interesting because it results. This is a1, a2, all the way down to an."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That'll be actually a different type of vector multiplication. The dot product is almost fun to take because it's mathematically pretty straightforward, unlike the cross product. It's fun to take and it's interesting because it results. This is a1, a2, all the way down to an. That vector dot my b vector, b1, b2, all the way down to bn, is going to be equal to the product of each of their corresponding components, so a1, b1, added together, plus a2, b2, plus a3, b3, plus all the way to an, bn. What is this? Is this a vector?"}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a1, a2, all the way down to an. That vector dot my b vector, b1, b2, all the way down to bn, is going to be equal to the product of each of their corresponding components, so a1, b1, added together, plus a2, b2, plus a3, b3, plus all the way to an, bn. What is this? Is this a vector? No, this is just a number. This is just going to be a real number. You're just taking the product and adding together a bunch of real numbers."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Is this a vector? No, this is just a number. This is just going to be a real number. You're just taking the product and adding together a bunch of real numbers. This is just going to be a scalar, a real scalar. In the dot product, you multiply two vectors and you end up with a scalar value. Let me show you a couple of examples, just in case this was a little bit too abstract."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're just taking the product and adding together a bunch of real numbers. This is just going to be a scalar, a real scalar. In the dot product, you multiply two vectors and you end up with a scalar value. Let me show you a couple of examples, just in case this was a little bit too abstract. Let's say that we take the dot product of the vector 2, 5, and we're going to dot that with the vector 7, 1. This is just going to be equal to 2 times 7, plus 5 times 1, or 14 plus 5, which is equal to 19. The dot product of this vector and this vector is 19."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you a couple of examples, just in case this was a little bit too abstract. Let's say that we take the dot product of the vector 2, 5, and we're going to dot that with the vector 7, 1. This is just going to be equal to 2 times 7, plus 5 times 1, or 14 plus 5, which is equal to 19. The dot product of this vector and this vector is 19. Let me do one more example, although I think this is a pretty straightforward idea. Let me do it in MoV. Say I have the vector 1, 2, 3, and I'm going to dot that with the vector minus 2, 0, 5."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The dot product of this vector and this vector is 19. Let me do one more example, although I think this is a pretty straightforward idea. Let me do it in MoV. Say I have the vector 1, 2, 3, and I'm going to dot that with the vector minus 2, 0, 5. It's 1 times minus 2, plus 2 times 0, plus 3 times 5. It's minus 2 plus 0, plus 15. Minus 2 plus 15 is equal to 13."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Say I have the vector 1, 2, 3, and I'm going to dot that with the vector minus 2, 0, 5. It's 1 times minus 2, plus 2 times 0, plus 3 times 5. It's minus 2 plus 0, plus 15. Minus 2 plus 15 is equal to 13. That's the dot product by this definition. Now I'm going to make another definition. I'm going to define the length of a vector."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 plus 15 is equal to 13. That's the dot product by this definition. Now I'm going to make another definition. I'm going to define the length of a vector. You might say, Sal, I know what the length of something is. I've been measuring things since I was a kid. Why do I have to wait until a college level, or hopefully you are taking this before college maybe, what is now considered a college level course, to have length defined for me?"}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to define the length of a vector. You might say, Sal, I know what the length of something is. I've been measuring things since I was a kid. Why do I have to wait until a college level, or hopefully you are taking this before college maybe, what is now considered a college level course, to have length defined for me? The answer is because we're abstracting things to well beyond just R3, or just three-dimensional space, what you're used to. We're abstracting these vectors could have 50 components. Our definition of length will work even for these 50 component vectors."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Why do I have to wait until a college level, or hopefully you are taking this before college maybe, what is now considered a college level course, to have length defined for me? The answer is because we're abstracting things to well beyond just R3, or just three-dimensional space, what you're used to. We're abstracting these vectors could have 50 components. Our definition of length will work even for these 50 component vectors. My definition of length, but it's also going to be consistent with what we know length to be. My definition, if I take the length of A, and the notation we use is just these double lines around the vector. The length of my vector A is equal to, and this is a definition, it equals the square root of each of the terms, each of my components squared, added up."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Our definition of length will work even for these 50 component vectors. My definition of length, but it's also going to be consistent with what we know length to be. My definition, if I take the length of A, and the notation we use is just these double lines around the vector. The length of my vector A is equal to, and this is a definition, it equals the square root of each of the terms, each of my components squared, added up. Plus A2 squared, plus all the way to plus AN squared. This is pretty straightforward. If I wanted to take, let's call this vector B."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The length of my vector A is equal to, and this is a definition, it equals the square root of each of the terms, each of my components squared, added up. Plus A2 squared, plus all the way to plus AN squared. This is pretty straightforward. If I wanted to take, let's call this vector B. If I wanted to take the magnitude of vector B, right here, this would be what? This would be the square root of 2 squared, 2 squared plus 5 squared, which is equal to the square root of, what is this? This is 4 plus 25, the square root of 29."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I wanted to take, let's call this vector B. If I wanted to take the magnitude of vector B, right here, this would be what? This would be the square root of 2 squared, 2 squared plus 5 squared, which is equal to the square root of, what is this? This is 4 plus 25, the square root of 29. That's the length of this vector. You might say, I already knew that. That's from the Pythagorean theorem."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is 4 plus 25, the square root of 29. That's the length of this vector. You might say, I already knew that. That's from the Pythagorean theorem. If I were to draw my vector B, let me draw it. Those are my axes. My vector B, if I draw it in standard form, looks like this."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's from the Pythagorean theorem. If I were to draw my vector B, let me draw it. Those are my axes. My vector B, if I draw it in standard form, looks like this. I go to the right, 2, 1, 2, and I go up 5, 1, 2, 3, 4, 5. It looks like this. My vector B looks like that."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My vector B, if I draw it in standard form, looks like this. I go to the right, 2, 1, 2, and I go up 5, 1, 2, 3, 4, 5. It looks like this. My vector B looks like that. From the Pythagorean theorem, look, if I wanted to figure out the length of this vector in R2, or if I'm drawing it in 2-dimensional space, I take this side, which is length 2, I take that side, which is length 5. This is going to be the square root from the Pythagorean theorem of 2 squared plus 5 squared, which is exactly what we did here. This definition of length is completely consistent with your idea of measuring things in 1, 2, or 3-dimensional space."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My vector B looks like that. From the Pythagorean theorem, look, if I wanted to figure out the length of this vector in R2, or if I'm drawing it in 2-dimensional space, I take this side, which is length 2, I take that side, which is length 5. This is going to be the square root from the Pythagorean theorem of 2 squared plus 5 squared, which is exactly what we did here. This definition of length is completely consistent with your idea of measuring things in 1, 2, or 3-dimensional space. What's neat about it is that now we can start thinking about the length of a vector that maybe has 50 components, which really, to visualize it in our traditional way, makes no sense. But we can still apply this notion of length and start to maybe abstract beyond what we traditionally associate length with. Now, can we somehow relate length with the dot product?"}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This definition of length is completely consistent with your idea of measuring things in 1, 2, or 3-dimensional space. What's neat about it is that now we can start thinking about the length of a vector that maybe has 50 components, which really, to visualize it in our traditional way, makes no sense. But we can still apply this notion of length and start to maybe abstract beyond what we traditionally associate length with. Now, can we somehow relate length with the dot product? What happens if I dot A with itself? What is A dot A? That's equal to, I'll just write it out again, A1 all the way down to An dotted with A1 all the way down to An."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, can we somehow relate length with the dot product? What happens if I dot A with itself? What is A dot A? That's equal to, I'll just write it out again, A1 all the way down to An dotted with A1 all the way down to An. That's equal to A1 times A1, which is A1 squared, plus A2 times A2, A2 squared, plus all the way down, keep doing that, all the way down to An times An, which is An squared. What's this? This is the same thing as the thing you see under the radical."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's equal to, I'll just write it out again, A1 all the way down to An dotted with A1 all the way down to An. That's equal to A1 times A1, which is A1 squared, plus A2 times A2, A2 squared, plus all the way down, keep doing that, all the way down to An times An, which is An squared. What's this? This is the same thing as the thing you see under the radical. These two things are equivalent. So we could write our definition of length, of vector length, we can write it in terms of the dot product, of our dot product definition. It equals the square root of the vector dotted with itself."}, {"video_title": "Vector dot product and vector length Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is the same thing as the thing you see under the radical. These two things are equivalent. So we could write our definition of length, of vector length, we can write it in terms of the dot product, of our dot product definition. It equals the square root of the vector dotted with itself. Or if we square both sides, we could say that our new length definition, our new length definition squared, is equal to the dot product of a vector with itself. This is a pretty neat, it's almost trivial to actually prove it, but this is a pretty neat outcome and we're going to use this in future videos. So this is an introduction to what the dot product is, what length is."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have a set of linearly independent vectors, v1, v2, all the way to vk, that are a basis for v. We've seen this many times before, and that's nice. It's a basis. But we've learned over the last few videos that it would be even better to have an orthonormal basis for v. So what we're going to address in this video is, is there some way that we can construct an orthonormal basis for v, given that we already have this basis, which I'm assuming isn't orthonormal, but it'll work whether or not it's orthonormal, and it'll just kind of generate another orthonormal basis. But can we somehow, just given any basis, generate an orthonormal basis for v and then be able to benefit from all of the properties of having an orthonormal basis? So let's see if we can make some headway here. So let's think about it, I guess, for simple cases. Let's say that I just have a one-dimensional subspace."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But can we somehow, just given any basis, generate an orthonormal basis for v and then be able to benefit from all of the properties of having an orthonormal basis? So let's see if we can make some headway here. So let's think about it, I guess, for simple cases. Let's say that I just have a one-dimensional subspace. Let's call that v1, for it's a one-dimensional subspace. And I'm just going to say it's the span of just the vector v1. Now, or you could say that v1 is the basis for, or the vector v1 is a basis for the subspace v1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that I just have a one-dimensional subspace. Let's call that v1, for it's a one-dimensional subspace. And I'm just going to say it's the span of just the vector v1. Now, or you could say that v1 is the basis for, or the vector v1 is a basis for the subspace v1. Now if I had this simple case, how could I ensure that this is orthonormal? Well, what I could do is, I could define some vector, let's call it u1. Let's call it u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, or you could say that v1 is the basis for, or the vector v1 is a basis for the subspace v1. Now if I had this simple case, how could I ensure that this is orthonormal? Well, what I could do is, I could define some vector, let's call it u1. Let's call it u1. That's essentially equal to, obviously this is orthogonal to all of the other guys. There aren't any other guys here. You could argue the zero vectors there."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's call it u1. That's essentially equal to, obviously this is orthogonal to all of the other guys. There aren't any other guys here. You could argue the zero vectors there. But there aren't any other members of this set. So it's orthogonal to everything else, because there isn't anything else. And then to make its length equal to 1, you can just take this vector and divide it by its length."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You could argue the zero vectors there. But there aren't any other members of this set. So it's orthogonal to everything else, because there isn't anything else. And then to make its length equal to 1, you can just take this vector and divide it by its length. So if we define some vector u1 to be equal to v1 divided by the length of v1, then the length of u1 is going to be what? It's going to be the length of v1 over, let me write it this way, over the length of v1 like this. This is just a constant right here."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then to make its length equal to 1, you can just take this vector and divide it by its length. So if we define some vector u1 to be equal to v1 divided by the length of v1, then the length of u1 is going to be what? It's going to be the length of v1 over, let me write it this way, over the length of v1 like this. This is just a constant right here. So this is just going to be 1 over the length of v1. That's just, that could be like 5 times the length of v1, which is just equal to 1. So this 1 will have a length of 1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is just a constant right here. So this is just going to be 1 over the length of v1. That's just, that could be like 5 times the length of v1, which is just equal to 1. So this 1 will have a length of 1. So just like that, if we define, if we say that, if we have the set of just u1, we could say that just the set of u1 is an orthonormal basis for v1, for the subspace v1. Now that was trivially easy. I mean, if k happens to be 1, we're done."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this 1 will have a length of 1. So just like that, if we define, if we say that, if we have the set of just u1, we could say that just the set of u1 is an orthonormal basis for v1, for the subspace v1. Now that was trivially easy. I mean, if k happens to be 1, we're done. That was super easy. We just have to divide by our length. You just have to essentially just normalize this vector and you have an orthonormal basis because there's nothing else to be orthogonal with."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I mean, if k happens to be 1, we're done. That was super easy. We just have to divide by our length. You just have to essentially just normalize this vector and you have an orthonormal basis because there's nothing else to be orthogonal with. But let's complicate the issue a little bit. Let's say we go to two dimensions. Let's say we have a subspace v2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You just have to essentially just normalize this vector and you have an orthonormal basis because there's nothing else to be orthogonal with. But let's complicate the issue a little bit. Let's say we go to two dimensions. Let's say we have a subspace v2. That's the span of, let's say it's the first two of these vectors. It's the span of v1 and the vector v2. Now we know that v1 can easily be represented as, or something that v1 is a linear combination of u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have a subspace v2. That's the span of, let's say it's the first two of these vectors. It's the span of v1 and the vector v2. Now we know that v1 can easily be represented as, or something that v1 is a linear combination of u1. How do I know that? Well, I could just multiply both sides of this times the length of v1. We have u1 times the length of v1 is equal to v1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now we know that v1 can easily be represented as, or something that v1 is a linear combination of u1. How do I know that? Well, I could just multiply both sides of this times the length of v1. We have u1 times the length of v1 is equal to v1. So we could say that this is the same thing. This is equivalent to saying the span of the vector u1 and the vector v2, where u1 is the vector we got up here. How can I say this?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We have u1 times the length of v1 is equal to v1. So we could say that this is the same thing. This is equivalent to saying the span of the vector u1 and the vector v2, where u1 is the vector we got up here. How can I say this? Because anything that can be a linear combination of these guys can be a linear combination of those guys. Because wherever you had a v1, you can replace the v1 with the linear combination of u1 that gives you v1. So you just multiply u1 times that scalar and you get it."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "How can I say this? Because anything that can be a linear combination of these guys can be a linear combination of those guys. Because wherever you had a v1, you can replace the v1 with the linear combination of u1 that gives you v1. So you just multiply u1 times that scalar and you get it. I think you get the point. But how can we ensure that this is an orthonormal set? So what do we do?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you just multiply u1 times that scalar and you get it. I think you get the point. But how can we ensure that this is an orthonormal set? So what do we do? Let's graph it. So this is going to be a plane in Rn. So let's just make our chalkboard here, our video board here, the plane."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So what do we do? Let's graph it. So this is going to be a plane in Rn. So let's just make our chalkboard here, our video board here, the plane. So we have u1, which is a unit vector. It has length 1. So that is u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's just make our chalkboard here, our video board here, the plane. So we have u1, which is a unit vector. It has length 1. So that is u1. And v1 and v2 are linearly independent. That's by definition of a basis. So you can't represent v2 as a linear multiple or a linear combination of v1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that is u1. And v1 and v2 are linearly independent. That's by definition of a basis. So you can't represent v2 as a linear multiple or a linear combination of v1. And likewise, you can't represent v2 as a linear combination of u1. Because v1 is a linear combination of u1. So v1 is not going to be on the line spanned by u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you can't represent v2 as a linear multiple or a linear combination of v1. And likewise, you can't represent v2 as a linear combination of u1. Because v1 is a linear combination of u1. So v1 is not going to be on the line spanned by u1. So I can actually draw that. The line spanned by u1 is just like this. This is the line spanned by u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So v1 is not going to be on the line spanned by u1. So I can actually draw that. The line spanned by u1 is just like this. This is the line spanned by u1. Let me draw it a little bit better than that. I don't want to make it too dark. So this is the line."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is the line spanned by u1. Let me draw it a little bit better than that. I don't want to make it too dark. So this is the line. I'll do one last try. The line spanned by u1 looks like this. It looks like that."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is the line. I'll do one last try. The line spanned by u1 looks like this. It looks like that. It's just a line. And that is the subspace v1, right? With a span of u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It looks like that. It's just a line. And that is the subspace v1, right? With a span of u1. So this is equal to the span of u1. All we did is normalize v1 right there to get to u1. So the span of v1 is the same as the span of u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "With a span of u1. So this is equal to the span of u1. All we did is normalize v1 right there to get to u1. So the span of v1 is the same as the span of u1. That is this subspace. That's this line in Rn. Now we have vector v2, which is linearly independent from v1 and u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So the span of v1 is the same as the span of u1. That is this subspace. That's this line in Rn. Now we have vector v2, which is linearly independent from v1 and u1. So v2 would look, let's just say it looks something like that. v2. Now what we want to do is replace v2 with another vector that is definitely orthogonal to this."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now we have vector v2, which is linearly independent from v1 and u1. So v2 would look, let's just say it looks something like that. v2. Now what we want to do is replace v2 with another vector that is definitely orthogonal to this. And we'll still be able to construct v2 with some combination of this and our new vector. Well, that most obvious vector would be some vector that is orthogonal to v1. So it's orthogonal."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now what we want to do is replace v2 with another vector that is definitely orthogonal to this. And we'll still be able to construct v2 with some combination of this and our new vector. Well, that most obvious vector would be some vector that is orthogonal to v1. So it's orthogonal. So it's a member of the orthogonal complement of v1. So it's a member of the orthogonal complement of v1. And so if you look at it just visually here, if I add some member of v1 to that member of the orthogonal complement of v1, I'm going to get v2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's orthogonal. So it's a member of the orthogonal complement of v1. So it's a member of the orthogonal complement of v1. And so if you look at it just visually here, if I add some member of v1 to that member of the orthogonal complement of v1, I'm going to get v2. In fact, we've seen that multiple times. If you know this guy right here. Let's say that we know that any vector in Rn, let's say v2, can be represented as a sum of two vectors."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And so if you look at it just visually here, if I add some member of v1 to that member of the orthogonal complement of v1, I'm going to get v2. In fact, we've seen that multiple times. If you know this guy right here. Let's say that we know that any vector in Rn, let's say v2, can be represented as a sum of two vectors. I'll call them x and y. Where x is a member of v1 and y is a member of the orthogonal complement of v1. We've seen that multiple times."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that we know that any vector in Rn, let's say v2, can be represented as a sum of two vectors. I'll call them x and y. Where x is a member of v1 and y is a member of the orthogonal complement of v1. We've seen that multiple times. Now what was this thing by definition? We're looking for this thing. This is the x."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We've seen that multiple times. Now what was this thing by definition? We're looking for this thing. This is the x. So that is the x. And then this is the y. We're looking for the y."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is the x. So that is the x. And then this is the y. We're looking for the y. Because if we can find this vector y, then if we replace v with that vector y, we'll still be able to generate v. Because you can take a multiple of u and add it to y and get v. And so anything that you were able to generate with v2 before, you can now generate with our u1, multiples of that, plus multiples of our new vector. So we're trying to figure out what this vector y is right there. So how do we do it?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We're looking for the y. Because if we can find this vector y, then if we replace v with that vector y, we'll still be able to generate v. Because you can take a multiple of u and add it to y and get v. And so anything that you were able to generate with v2 before, you can now generate with our u1, multiples of that, plus multiples of our new vector. So we're trying to figure out what this vector y is right there. So how do we do it? Well, it's going to be v2 minus this vector x. And what is this vector x by definition? This vector x is, by definition, the projection of v2 onto the subspace v1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So how do we do it? Well, it's going to be v2 minus this vector x. And what is this vector x by definition? This vector x is, by definition, the projection of v2 onto the subspace v1. So the vector that we're trying to find, the vector y, that if we find it, and that we can replace v2 with, y, the vector y, is just equal to v2. It's just equal to v2, let me write it this way, minus the projection of v2 onto v1. That's what y is."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This vector x is, by definition, the projection of v2 onto the subspace v1. So the vector that we're trying to find, the vector y, that if we find it, and that we can replace v2 with, y, the vector y, is just equal to v2. It's just equal to v2, let me write it this way, minus the projection of v2 onto v1. That's what y is. And remember, if we can replace v2, the reason why the span of u1, v2 is the same thing as the span of u1. Let me call this y2 here. Let me just call that y2, because we're probably going to have to use y's in the future."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's what y is. And remember, if we can replace v2, the reason why the span of u1, v2 is the same thing as the span of u1. Let me call this y2 here. Let me just call that y2, because we're probably going to have to use y's in the future. u1, let me, y2. So the reason why this span is the same thing as the span of u1 and y2 is because I can generate v2 with a linear combination of u1 and y2, right? I can scale up u1 and then add y2, and I'll get to v2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me just call that y2, because we're probably going to have to use y's in the future. u1, let me, y2. So the reason why this span is the same thing as the span of u1 and y2 is because I can generate v2 with a linear combination of u1 and y2, right? I can scale up u1 and then add y2, and I'll get to v2. So anything that I could have generated with v2, I can get with a linear combination of these guys. That's why these are equivalent. And what's neat about this now is that these guys are orthogonal relative to each other, right?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I can scale up u1 and then add y2, and I'll get to v2. So anything that I could have generated with v2, I can get with a linear combination of these guys. That's why these are equivalent. And what's neat about this now is that these guys are orthogonal relative to each other, right? By definition, y was a member of the orthogonal complement. So if you dot these two guys, you're going to get 0. And how can we actually solve for it?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And what's neat about this now is that these guys are orthogonal relative to each other, right? By definition, y was a member of the orthogonal complement. So if you dot these two guys, you're going to get 0. And how can we actually solve for it? Well, this is useful as well, because v1 has an orthonormal basis, right? v1 has an orthonormal basis. Has ortho, I'll just write ortho basis."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And how can we actually solve for it? Well, this is useful as well, because v1 has an orthonormal basis, right? v1 has an orthonormal basis. Has ortho, I'll just write ortho basis. I'm tired of writing orthonormal. And we saw, I think it was two or three videos ago, that there's a very neat thing about orthonormal basis is it's very easy to determine the projection onto those basis. It's essentially, let me write it over here, the projection of the vector v2 onto the subspace v1 is equal to v2 dotted with the subspace v1's first basis vector, which is the vector u1, right?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Has ortho, I'll just write ortho basis. I'm tired of writing orthonormal. And we saw, I think it was two or three videos ago, that there's a very neat thing about orthonormal basis is it's very easy to determine the projection onto those basis. It's essentially, let me write it over here, the projection of the vector v2 onto the subspace v1 is equal to v2 dotted with the subspace v1's first basis vector, which is the vector u1, right? That's its first basis vector. We're dealing with an orthonormal basis, times u1. And then if we had more basis vectors, we would say plus v2 dotted with our next basis vector times that basis vector, so on and so forth."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's essentially, let me write it over here, the projection of the vector v2 onto the subspace v1 is equal to v2 dotted with the subspace v1's first basis vector, which is the vector u1, right? That's its first basis vector. We're dealing with an orthonormal basis, times u1. And then if we had more basis vectors, we would say plus v2 dotted with our next basis vector times that basis vector, so on and so forth. But v1 only has one basis vector. It only has the basis vector u1 right here, right? It was only spanned by that."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then if we had more basis vectors, we would say plus v2 dotted with our next basis vector times that basis vector, so on and so forth. But v1 only has one basis vector. It only has the basis vector u1 right here, right? It was only spanned by that. So we can rewrite this thing right here. y2, this vector right here that I'm going to replace v2 with, is equal to v2 minus the projection of v2 onto the subspace v1, onto that line, which is just this. v2 dot u1 times the vector u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It was only spanned by that. So we can rewrite this thing right here. y2, this vector right here that I'm going to replace v2 with, is equal to v2 minus the projection of v2 onto the subspace v1, onto that line, which is just this. v2 dot u1 times the vector u1. And just like that, we've solved for y. So we have a basis for v2 where this guy and this guy are orthogonal. This guy is a unit vector."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "v2 dot u1 times the vector u1. And just like that, we've solved for y. So we have a basis for v2 where this guy and this guy are orthogonal. This guy is a unit vector. He's been normalized. But this guy hasn't been normalized yet. So to normalize it, let's just define some other vector as u2 and do the same thing."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This guy is a unit vector. He's been normalized. But this guy hasn't been normalized yet. So to normalize it, let's just define some other vector as u2 and do the same thing. Let's just normalize it. So u2 is equal to y2 divided by the length of y2. So now we can say that the subspace v2 is equal to the span of u1."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So to normalize it, let's just define some other vector as u2 and do the same thing. Let's just normalize it. So u2 is equal to y2 divided by the length of y2. So now we can say that the subspace v2 is equal to the span of u1. And instead of y2, I'll put u2 there. Because I can generate y2 by just scaling up u2. So u2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So now we can say that the subspace v2 is equal to the span of u1. And instead of y2, I'll put u2 there. Because I can generate y2 by just scaling up u2. So u2. And what's neat about this now is I have two unit vectors, or two normalized vectors. And they're orthogonal with respect to each other. And they span what v1 and v2 spanned originally."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So u2. And what's neat about this now is I have two unit vectors, or two normalized vectors. And they're orthogonal with respect to each other. And they span what v1 and v2 spanned originally. Now, we need to keep going. What about if we want to go to, we're going to have a v3 here, what do we do? Well, we do the same idea."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And they span what v1 and v2 spanned originally. Now, we need to keep going. What about if we want to go to, we're going to have a v3 here, what do we do? Well, we do the same idea. So let's define a subspace v3. This is going to be a three-dimensional subspace. Once we get beyond v3, it becomes a little hard to visualize."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, we do the same idea. So let's define a subspace v3. This is going to be a three-dimensional subspace. Once we get beyond v3, it becomes a little hard to visualize. But I think you're going to see the pattern after this step. If we define v3 is equal to the span of these guys, u1, u2, and then in our original basis, v3. I didn't write it there, but there's a v3 there."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Once we get beyond v3, it becomes a little hard to visualize. But I think you're going to see the pattern after this step. If we define v3 is equal to the span of these guys, u1, u2, and then in our original basis, v3. I didn't write it there, but there's a v3 there. And v3. So in our original non-orthonormal basis, we have v3. Now, what is this going to look like?"}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I didn't write it there, but there's a v3 there. And v3. So in our original non-orthonormal basis, we have v3. Now, what is this going to look like? Or how can we turn this into an orthonormal basis? So if you visualize all of these, so the span of u1 and u2, our subspace v2 is just going to be a plane. It's just going to be a plane in R3."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this going to look like? Or how can we turn this into an orthonormal basis? So if you visualize all of these, so the span of u1 and u2, our subspace v2 is just going to be a plane. It's just going to be a plane in R3. It's going to look like that right there. And so we have our new span is going to be everything in that plane, all linear combinations of things in that plane, plus linear combinations of our vector v3, which is linearly independent from these guys, because it was linearly independent from the stuff we used to construct these guys. So v3 is going to pop out of this plane."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's just going to be a plane in R3. It's going to look like that right there. And so we have our new span is going to be everything in that plane, all linear combinations of things in that plane, plus linear combinations of our vector v3, which is linearly independent from these guys, because it was linearly independent from the stuff we used to construct these guys. So v3 is going to pop out of this plane. It can't be represented by a linear combination of those guys. So let's say v3 pops out of the plane. v3 like that."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So v3 is going to pop out of this plane. It can't be represented by a linear combination of those guys. So let's say v3 pops out of the plane. v3 like that. Now, we want another vector that can represent everything of this span, but it is orthogonal to these dudes right here. Or another way to think about it, it's orthogonal to the plane. So we want another vector that is orthogonal to the plane."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "v3 like that. Now, we want another vector that can represent everything of this span, but it is orthogonal to these dudes right here. Or another way to think about it, it's orthogonal to the plane. So we want another vector that is orthogonal to the plane. So let's call that vector y sub 3, not y to the third power. And if we figure out our y sub 3, we can replace that. We can replace v3 with it, because v3 can be represented as a linear combination of u1 and u2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we want another vector that is orthogonal to the plane. So let's call that vector y sub 3, not y to the third power. And if we figure out our y sub 3, we can replace that. We can replace v3 with it, because v3 can be represented as a linear combination of u1 and u2. That's going to be a linear combination of u1, u2 is going to be some vector in this plane, plus y3. I can represent this guy with this green vector plus this vector right here. So if we replace them with y3, we can still get to v3, and we can still get to all the linear combinations that v3 can help construct."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We can replace v3 with it, because v3 can be represented as a linear combination of u1 and u2. That's going to be a linear combination of u1, u2 is going to be some vector in this plane, plus y3. I can represent this guy with this green vector plus this vector right here. So if we replace them with y3, we can still get to v3, and we can still get to all the linear combinations that v3 can help construct. So what is this y3? Well, by the same exact logic, this green vector right here is the projection of my vector v3 onto the subspace v2. And your vector y3 is just equal to the vector v3 minus the projection of v3 onto v2."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if we replace them with y3, we can still get to v3, and we can still get to all the linear combinations that v3 can help construct. So what is this y3? Well, by the same exact logic, this green vector right here is the projection of my vector v3 onto the subspace v2. And your vector y3 is just equal to the vector v3 minus the projection of v3 onto v2. So what is that going to be? Well, the projection, I'll write it here, the projection onto the subspace v2 of v3, of the vector v3, is going to be equal to, we're going to use the same exact logic we did before. We saw this two or three videos ago."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And your vector y3 is just equal to the vector v3 minus the projection of v3 onto v2. So what is that going to be? Well, the projection, I'll write it here, the projection onto the subspace v2 of v3, of the vector v3, is going to be equal to, we're going to use the same exact logic we did before. We saw this two or three videos ago. Because v2 is defined with an orthonormal basis, we can say that the projection of v3 onto that subspace is v3 dot our first basis vector, dot u1, times our first basis vector, plus v3 dot our second basis vector, our second orthonormal basis vector, times our second orthonormal basis vector. It's that easy. That's what one of the neat things about having an orthonormal basis was."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We saw this two or three videos ago. Because v2 is defined with an orthonormal basis, we can say that the projection of v3 onto that subspace is v3 dot our first basis vector, dot u1, times our first basis vector, plus v3 dot our second basis vector, our second orthonormal basis vector, times our second orthonormal basis vector. It's that easy. That's what one of the neat things about having an orthonormal basis was. This is an orthonormal basis. So we can define the projection in this way. So y is just going to be v3 minus this business right here, and I'll just write it out."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's what one of the neat things about having an orthonormal basis was. This is an orthonormal basis. So we can define the projection in this way. So y is just going to be v3 minus this business right here, and I'll just write it out. So y3 is going to be v3 minus the projection of v3 onto v2. So minus that business right there. Let me copy it and then let me paste it."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So y is just going to be v3 minus this business right here, and I'll just write it out. So y3 is going to be v3 minus the projection of v3 onto v2. So minus that business right there. Let me copy it and then let me paste it. And then you get y3. And y3 is nice. So if we replace this with y3, which we can, because we can now get v3 as a linear combination of these guys and y3, that's nice because all of these guys are orthogonal with respect to each other."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me copy it and then let me paste it. And then you get y3. And y3 is nice. So if we replace this with y3, which we can, because we can now get v3 as a linear combination of these guys and y3, that's nice because all of these guys are orthogonal with respect to each other. But y3 does not have length 1 yet. It hasn't been normalized. So we can just replace y3 with another unit vector."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if we replace this with y3, which we can, because we can now get v3 as a linear combination of these guys and y3, that's nice because all of these guys are orthogonal with respect to each other. But y3 does not have length 1 yet. It hasn't been normalized. So we can just replace y3 with another unit vector. We'll just say u3 is equal to y3 divided by the length of y3. Whatever that might be. But if we have what y3 is, we can figure out its length."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we can just replace y3 with another unit vector. We'll just say u3 is equal to y3 divided by the length of y3. Whatever that might be. But if we have what y3 is, we can figure out its length. Divide by that. And then if you put, so this is equal to the span of u1, u2, and u3. So u3 might be a scaled down version of y3."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But if we have what y3 is, we can figure out its length. Divide by that. And then if you put, so this is equal to the span of u1, u2, and u3. So u3 might be a scaled down version of y3. But we're talking about linear combinations when we talk about spans. So you can scale them up and add it to the projection. You're still going to get v3."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So u3 might be a scaled down version of y3. But we're talking about linear combinations when we talk about spans. So you can scale them up and add it to the projection. You're still going to get v3. And now these are all normalized vectors. And so you now have an orthonormal basis for the subspace v3. I think you get the pattern now."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You're still going to get v3. And now these are all normalized vectors. And so you now have an orthonormal basis for the subspace v3. I think you get the pattern now. You can keep doing this. Now if you're curious about, you can define v4. And when I defined the problem, I just said that we're dealing with a k-dimensional subspace."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I think you get the pattern now. You can keep doing this. Now if you're curious about, you can define v4. And when I defined the problem, I just said that we're dealing with a k-dimensional subspace. We go up to vk. But you just keep going until you get to your k. If k was 3, we'd be done. If k is 4, you say you define a subspace v4 is equal to the span of u1, u2, u3."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And when I defined the problem, I just said that we're dealing with a k-dimensional subspace. We go up to vk. But you just keep going until you get to your k. If k was 3, we'd be done. If k is 4, you say you define a subspace v4 is equal to the span of u1, u2, u3. And then throw our non-orthonormal vector, v4, into the mix. And so you can replace v4 with y4 is equal to v4 minus the projection, it becomes a little hard to visualize. You're not projecting onto a three-dimensional space."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "If k is 4, you say you define a subspace v4 is equal to the span of u1, u2, u3. And then throw our non-orthonormal vector, v4, into the mix. And so you can replace v4 with y4 is equal to v4 minus the projection, it becomes a little hard to visualize. You're not projecting onto a three-dimensional space. The projection of v4 onto our subspace v3. Completely analogous to this, it's just v3 is now a three-dimensional space, not a plane. You're finding its projection onto this."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You're not projecting onto a three-dimensional space. The projection of v4 onto our subspace v3. Completely analogous to this, it's just v3 is now a three-dimensional space, not a plane. You're finding its projection onto this. This is definitely going to be orthogonal to everything else in to our subspace v3. And you can construct v4 with y4, because v4 by definition is equal to, we can just rearrange this, y4 plus the projection of v4 onto v3. So you can construct v4 with y4 and some linear combination of these guys."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You're finding its projection onto this. This is definitely going to be orthogonal to everything else in to our subspace v3. And you can construct v4 with y4, because v4 by definition is equal to, we can just rearrange this, y4 plus the projection of v4 onto v3. So you can construct v4 with y4 and some linear combination of these guys. So I can replace this guy with y4. And then I would normalize y4. I would divide it by its length and I get u4."}, {"video_title": "The Gram-Schmidt process Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you can construct v4 with y4 and some linear combination of these guys. So I can replace this guy with y4. And then I would normalize y4. I would divide it by its length and I get u4. And I just keep doing that until I get up to k. If I do v5, I do the process over and over and over again. And this process of creating an orthonormal basis is called the Gram-Schmidt process. It might seem a little abstract the way I did it here, but in the next video I'm actually going to find orthonormal bases for subspaces."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That's my matrix right there. And I could just write it as a series of n column vectors. So it could be A1, A2, all the way to An. Now let's say that I have some other vector B. Let's say B is a member of the column space of A. And remember, the column space is just a set of all of the vectors that can be represented as a linear combination of the columns of A. So that means that B can be represented as a linear combination of the columns of A."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now let's say that I have some other vector B. Let's say B is a member of the column space of A. And remember, the column space is just a set of all of the vectors that can be represented as a linear combination of the columns of A. So that means that B can be represented as a linear combination of the columns of A. So I'll just write the constant factors as x1 times A1 plus x2 times A1 plus x2 times A2, all the way to plus xn times An. Where x1, x2, xn, they're all just arbitrary real numbers. Or another way to state this is that that means that A, which I could write as A1, A2, all the way to An, times some vector x1, x2, all the way to xn, is equal to B."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So that means that B can be represented as a linear combination of the columns of A. So I'll just write the constant factors as x1 times A1 plus x2 times A1 plus x2 times A2, all the way to plus xn times An. Where x1, x2, xn, they're all just arbitrary real numbers. Or another way to state this is that that means that A, which I could write as A1, A2, all the way to An, times some vector x1, x2, all the way to xn, is equal to B. These two statements are equivalent. We know that B is a member of the column space. That means that B can be represented as a linear combination of the columns of A."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to state this is that that means that A, which I could write as A1, A2, all the way to An, times some vector x1, x2, all the way to xn, is equal to B. These two statements are equivalent. We know that B is a member of the column space. That means that B can be represented as a linear combination of the columns of A. And then this statement right here can be rewritten this way. So you can write that the equation Ax equals B has at least one solution x that is a member of Rn. And the entries of x would represent the weights on the column vectors of A to get your linear combination B."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That means that B can be represented as a linear combination of the columns of A. And then this statement right here can be rewritten this way. So you can write that the equation Ax equals B has at least one solution x that is a member of Rn. And the entries of x would represent the weights on the column vectors of A to get your linear combination B. This is all a bit of review. Now, let's draw Rn. Any solution to this equation right here is going to be a member of Rn."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And the entries of x would represent the weights on the column vectors of A to get your linear combination B. This is all a bit of review. Now, let's draw Rn. Any solution to this equation right here is going to be a member of Rn. Remember, this was an m by n matrix. We had n columns. This has to be a member of Rn right there."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Any solution to this equation right here is going to be a member of Rn. Remember, this was an m by n matrix. We had n columns. This has to be a member of Rn right there. So let's draw Rn. So Rn maybe looks like that. So that is Rn."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "This has to be a member of Rn right there. So let's draw Rn. So Rn maybe looks like that. So that is Rn. And let's look at some of the subspaces that we have in Rn. We have the null space. That's going to be an Rn."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So that is Rn. And let's look at some of the subspaces that we have in Rn. We have the null space. That's going to be an Rn. The null space is all of the solutions to the equation Ax is equal to 0. That's going to be an Rn. It's all of the x's that satisfy that equation."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be an Rn. The null space is all of the solutions to the equation Ax is equal to 0. That's going to be an Rn. It's all of the x's that satisfy that equation. So let me draw that right here. So let's say I have the null space right there. So that is the null space of A."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "It's all of the x's that satisfy that equation. So let me draw that right here. So let's say I have the null space right there. So that is the null space of A. And then what else do we have in Rn? Well, we have the orthogonal complement of the null space of A. Let me draw that."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So that is the null space of A. And then what else do we have in Rn? Well, we have the orthogonal complement of the null space of A. Let me draw that. So we have the orthogonal complement. Let me do that in a different color. We have the orthogonal complement of the null space of A, which we could also call, we learned this in the last video, this is also going to be equal to the row space of A."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw that. So we have the orthogonal complement. Let me do that in a different color. We have the orthogonal complement of the null space of A, which we could also call, we learned this in the last video, this is also going to be equal to the row space of A. The row space of A is the column space of A transpose. So we have two spaces here. This is the row space of A."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "We have the orthogonal complement of the null space of A, which we could also call, we learned this in the last video, this is also going to be equal to the row space of A. The row space of A is the column space of A transpose. So we have two spaces here. This is the row space of A. So I have two subsets of Rn. I have the null space, and then I have the null space's complement, orthogonal complement, which is the row space of A. Now, we've seen in several videos now, and I proved it I think two videos ago, that any vector in Rn can be represented as a sum of a member of our null space."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "This is the row space of A. So I have two subsets of Rn. I have the null space, and then I have the null space's complement, orthogonal complement, which is the row space of A. Now, we've seen in several videos now, and I proved it I think two videos ago, that any vector in Rn can be represented as a sum of a member of our null space. Let's call that vector n. And let's say some vector in our row space. Let's call that vector r. Any vector in Rn can be represented as a sum of some vector in our null space and some vector in our row space. So any solution to this equation is a member of Rn."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now, we've seen in several videos now, and I proved it I think two videos ago, that any vector in Rn can be represented as a sum of a member of our null space. Let's call that vector n. And let's say some vector in our row space. Let's call that vector r. Any vector in Rn can be represented as a sum of some vector in our null space and some vector in our row space. So any solution to this equation is a member of Rn. So it must be able to be represented by some member of our null space and some member of our row space. So let's write that down. So x is a solution."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So any solution to this equation is a member of Rn. So it must be able to be represented by some member of our null space and some member of our row space. So let's write that down. So x is a solution. So let's say that x is a solution to Ax equals b, which also means that x is a member of Rn. x is a member of Rn. So because it's a member of Rn, we can represent it as a combination of one vector here and one vector there."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So x is a solution. So let's say that x is a solution to Ax equals b, which also means that x is a member of Rn. x is a member of Rn. So because it's a member of Rn, we can represent it as a combination of one vector here and one vector there. So let's say that x is equal to some vector r, let's call it r naught, plus n naught. Where r naught is a member of our row space and n naught is a member of the row space's orthogonal complement. They're the orthogonal complement of each other."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So because it's a member of Rn, we can represent it as a combination of one vector here and one vector there. So let's say that x is equal to some vector r, let's call it r naught, plus n naught. Where r naught is a member of our row space and n naught is a member of the row space's orthogonal complement. They're the orthogonal complement of each other. So n naught is a member of our null space. Fair enough. Now, one thing we might wonder is, clearly this vector isn't a solution to Ax equals b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "They're the orthogonal complement of each other. So n naught is a member of our null space. Fair enough. Now, one thing we might wonder is, clearly this vector isn't a solution to Ax equals b. This vector is a solution to Ax is equal to 0. But we might be curious as to whether this solution right here, this member of our row space, is a solution to Ax is equal to b. This is kind of what we're focused on in this."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now, one thing we might wonder is, clearly this vector isn't a solution to Ax equals b. This vector is a solution to Ax is equal to 0. But we might be curious as to whether this solution right here, this member of our row space, is a solution to Ax is equal to b. This is kind of what we're focused on in this. So let's solve for r naught right here. So if we solve for r naught, if we subtract n naught from both sides, we get r naught is equal to x minus n naught. All I do is subtract n naught from both sides and then I switch things around."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "This is kind of what we're focused on in this. So let's solve for r naught right here. So if we solve for r naught, if we subtract n naught from both sides, we get r naught is equal to x minus n naught. All I do is subtract n naught from both sides and then I switch things around. I solve for r naught. Now, if we multiply A times r naught is equal to A times this whole thing, let me switch colors, that's equal to A times x naught minus n naught, which is equal to Ax minus A n naught. What is this equal to?"}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "All I do is subtract n naught from both sides and then I switch things around. I solve for r naught. Now, if we multiply A times r naught is equal to A times this whole thing, let me switch colors, that's equal to A times x naught minus n naught, which is equal to Ax minus A n naught. What is this equal to? Well, A times x, we already said that x is a solution to Ax equals b. So this right here is going to be equal to b. And n naught is a member of our null space, which means it satisfies this solution right here, that A times any member of our null space is going to be equal to the 0 vector."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "What is this equal to? Well, A times x, we already said that x is a solution to Ax equals b. So this right here is going to be equal to b. And n naught is a member of our null space, which means it satisfies this solution right here, that A times any member of our null space is going to be equal to the 0 vector. So that's going to be equal to the 0 vector. So you have the vector b minus the 0 vector. And you're just going to have the vector b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And n naught is a member of our null space, which means it satisfies this solution right here, that A times any member of our null space is going to be equal to the 0 vector. So that's going to be equal to the 0 vector. So you have the vector b minus the 0 vector. And you're just going to have the vector b. So we just found out that A times this member of our row space, let's call that r naught, that guy right there maybe. A times r naught is equal to b. So this is a solution."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And you're just going to have the vector b. So we just found out that A times this member of our row space, let's call that r naught, that guy right there maybe. A times r naught is equal to b. So this is a solution. So r naught is a solution to Ax is equal to b. So so far, it's kind of an interesting result that we have already. If you give me any vector here, b, that is a member of our column space, then there is going to be some member of our row space, there's going to be some member of our row space right here, that is a solution to Ax is equal to b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So this is a solution. So r naught is a solution to Ax is equal to b. So so far, it's kind of an interesting result that we have already. If you give me any vector here, b, that is a member of our column space, then there is going to be some member of our row space, there's going to be some member of our row space right here, that is a solution to Ax is equal to b. Now the next question you might wonder is, is this the only guy in our row space that is a solution to Ax is equal to b? And to prove that, let's assume that there's another guy here. Let's say that r1 is a member of our row space and a solution to Ax is equal to b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "If you give me any vector here, b, that is a member of our column space, then there is going to be some member of our row space, there's going to be some member of our row space right here, that is a solution to Ax is equal to b. Now the next question you might wonder is, is this the only guy in our row space that is a solution to Ax is equal to b? And to prove that, let's assume that there's another guy here. Let's say that r1 is a member of our row space and a solution to Ax is equal to b. Now the row space is a valid subspace. So if I take the sum or the difference of any two vectors in the row space, I'll get another member of the row space. That's one of the requirements for being a valid subspace."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that r1 is a member of our row space and a solution to Ax is equal to b. Now the row space is a valid subspace. So if I take the sum or the difference of any two vectors in the row space, I'll get another member of the row space. That's one of the requirements for being a valid subspace. So let's see this. So if I take two members of our subspace, so if I take r1 minus r0 and I take their difference, which is just the sum of, well you multiply 1 times a negative and that has to be a member of the subspace and then you're summing them. So this has to be a member of our subspace."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That's one of the requirements for being a valid subspace. So let's see this. So if I take two members of our subspace, so if I take r1 minus r0 and I take their difference, which is just the sum of, well you multiply 1 times a negative and that has to be a member of the subspace and then you're summing them. So this has to be a member of our subspace. So this must also be a member of our row space. That's because our row space is a valid subspace. You get two of its members, you take its difference, that also has to be a member."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So this has to be a member of our subspace. So this must also be a member of our row space. That's because our row space is a valid subspace. You get two of its members, you take its difference, that also has to be a member. Fair enough. Now let's see what happens when you multiply this guy by a. So if I take a times r1 minus r0, what do I get?"}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "You get two of its members, you take its difference, that also has to be a member. Fair enough. Now let's see what happens when you multiply this guy by a. So if I take a times r1 minus r0, what do I get? I get a times r1 minus a times r0. We already figured out, or for r1 we assumed that it is a solution to ax is equal to b. And r0 we already found out it is a solution to ax is equal to b, so either of these, when you multiply them by a, it equals b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So if I take a times r1 minus r0, what do I get? I get a times r1 minus a times r0. We already figured out, or for r1 we assumed that it is a solution to ax is equal to b. And r0 we already found out it is a solution to ax is equal to b, so either of these, when you multiply them by a, it equals b. So this equals b and that equals b. So you get b minus b, which is the 0 vector. Now this is interesting."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And r0 we already found out it is a solution to ax is equal to b, so either of these, when you multiply them by a, it equals b. So this equals b and that equals b. So you get b minus b, which is the 0 vector. Now this is interesting. This tells us that r minus r0 is a solution to the equation ax is equal to 0. When I put r1 minus r0 in the place of x right there and I multiplied it times a, I got 0. Which implies that r1 minus r0, that this vector is a member of our null space."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now this is interesting. This tells us that r minus r0 is a solution to the equation ax is equal to 0. When I put r1 minus r0 in the place of x right there and I multiplied it times a, I got 0. Which implies that r1 minus r0, that this vector is a member of our null space. So I have a vector here that's a member of my row space. And we got that from the fact that both of these are members of our row space and the row space is closed under addition and subtraction. And the vector r1 minus r0 is a member of my null space."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Which implies that r1 minus r0, that this vector is a member of our null space. So I have a vector here that's a member of my row space. And we got that from the fact that both of these are members of our row space and the row space is closed under addition and subtraction. And the vector r1 minus r0 is a member of my null space. And we've seen this several times already. If I have a vector that is in a subspace and it's also in the orthogonal complement of the subspace, the null space is also the orthogonal complement of the row space, then the only possible vector that that can be is the 0 vector. That's the only vector that's inside of a subspace and its orthogonal complement or a subspace and its orthogonal complement."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And the vector r1 minus r0 is a member of my null space. And we've seen this several times already. If I have a vector that is in a subspace and it's also in the orthogonal complement of the subspace, the null space is also the orthogonal complement of the row space, then the only possible vector that that can be is the 0 vector. That's the only vector that's inside of a subspace and its orthogonal complement or a subspace and its orthogonal complement. These two guys are the orthogonal complement of each other. We drew it up here. So we get that r1 minus r0 must be equal to the 0 vector."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That's the only vector that's inside of a subspace and its orthogonal complement or a subspace and its orthogonal complement. These two guys are the orthogonal complement of each other. We drew it up here. So we get that r1 minus r0 must be equal to the 0 vector. That's the only vector that's in a subspace and its orthogonal complement. Which implies that r1 must be equal to r0. If we take the difference, we get the 0 vector."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So we get that r1 minus r0 must be equal to the 0 vector. That's the only vector that's in a subspace and its orthogonal complement. Which implies that r1 must be equal to r0. If we take the difference, we get the 0 vector. So we have a couple of neat results here. If I have some, what do we know so far? We know that if we have some vector b that is a member of our column space of A, then there exists a unique member."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "If we take the difference, we get the 0 vector. So we have a couple of neat results here. If I have some, what do we know so far? We know that if we have some vector b that is a member of our column space of A, then there exists a unique member. We just proved the uniqueness. There exists a unique member of the row space of A. Let me write it of the row space of A such that a unique member of the row space of A, let me call that r0."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "We know that if we have some vector b that is a member of our column space of A, then there exists a unique member. We just proved the uniqueness. There exists a unique member of the row space of A. Let me write it of the row space of A such that a unique member of the row space of A, let me call that r0. Let me do that in a different color. I want to make this really stand out in your brain. So we know that r0 is a member of the row space of A such that r0 is a solution to Ax is equal to b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it of the row space of A such that a unique member of the row space of A, let me call that r0. Let me do that in a different color. I want to make this really stand out in your brain. So we know that r0 is a member of the row space of A such that r0 is a solution to Ax is equal to b. Which is kind of a little bit of a complex statement here, but it's interesting. You give me any b that's a member of the column space of A, then there will exist a unique member of the row space of A, that's my unique member of the row space of A, that is a solution to Ax is equal to b. Now, we can go further with this."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So we know that r0 is a member of the row space of A such that r0 is a solution to Ax is equal to b. Which is kind of a little bit of a complex statement here, but it's interesting. You give me any b that's a member of the column space of A, then there will exist a unique member of the row space of A, that's my unique member of the row space of A, that is a solution to Ax is equal to b. Now, we can go further with this. We wrote up here that any solution to this equation, Ax is equal to b, can be written as a sum of r0 plus n0, where r0 is a member of our row space and n0 is a member of our null space. And that's because we have a subspace and it's orthogonal complement. So any member of our n can be represented as a sum of a subspace and a member of the subspace is orthogonal complement."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now, we can go further with this. We wrote up here that any solution to this equation, Ax is equal to b, can be written as a sum of r0 plus n0, where r0 is a member of our row space and n0 is a member of our null space. And that's because we have a subspace and it's orthogonal complement. So any member of our n can be represented as a sum of a subspace and a member of the subspace is orthogonal complement. So let me rewrite that down here. So we already said that any solution x to Ax is equal to b can be written as a combination of r0 plus n0, fair enough. Now, what happens if I wanted to take the square of the length of x on both sides of that?"}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So any member of our n can be represented as a sum of a subspace and a member of the subspace is orthogonal complement. So let me rewrite that down here. So we already said that any solution x to Ax is equal to b can be written as a combination of r0 plus n0, fair enough. Now, what happens if I wanted to take the square of the length of x on both sides of that? So let me write this down and you'll see why I'm writing this, because I have another interesting result to show you. So if I were to take the square of any solution to this equation right here, well that's going to be the same thing as x dot x, which is the same thing as this thing dot itself, same thing as r0 plus n0 dot r0 plus n0. And what is this equal to?"}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Now, what happens if I wanted to take the square of the length of x on both sides of that? So let me write this down and you'll see why I'm writing this, because I have another interesting result to show you. So if I were to take the square of any solution to this equation right here, well that's going to be the same thing as x dot x, which is the same thing as this thing dot itself, same thing as r0 plus n0 dot r0 plus n0. And what is this equal to? This is equal to r0 dot r0 plus n0 dot r0 plus n0 dot r0 again plus n0 dot n0. I just kind of foiled it out and we can do that because we know the dot product exhibits the distributive property. So this thing right here is equal to the length of r0 squared."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? This is equal to r0 dot r0 plus n0 dot r0 plus n0 dot r0 again plus n0 dot n0. I just kind of foiled it out and we can do that because we know the dot product exhibits the distributive property. So this thing right here is equal to the length of r0 squared. Now we're going to have plus, well, what is n0 dot r0? What is n0 dot r0? We don't even have to simplify this much more."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here is equal to the length of r0 squared. Now we're going to have plus, well, what is n0 dot r0? What is n0 dot r0? We don't even have to simplify this much more. n0 is a member of our null space. r0 is a member of our row space. Each of them is in a subspace that is the orthogonal complement of the other, which means that everything here dotted with anything in here is equal to 0."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "We don't even have to simplify this much more. n0 is a member of our null space. r0 is a member of our row space. Each of them is in a subspace that is the orthogonal complement of the other, which means that everything here dotted with anything in here is equal to 0. So r0 dot n0 is going to be equal to 0. These guys are orthogonal to each other. So that's going to be equal to 0."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Each of them is in a subspace that is the orthogonal complement of the other, which means that everything here dotted with anything in here is equal to 0. So r0 dot n0 is going to be equal to 0. These guys are orthogonal to each other. So that's going to be equal to 0. And then you get plus, what's this? n0 dot n0 is just the length of the vector n0 squared. These are all vectors."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be equal to 0. And then you get plus, what's this? n0 dot n0 is just the length of the vector n0 squared. These are all vectors. And so we get the length of the vector x squared is equal to the length of our member of our row space squared, our unique member of our row space squared, plus that member of our null space squared. Now, this is definitely going to be a positive number. That's definitely at minimum 0, but it has to be something larger than 0."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "These are all vectors. And so we get the length of the vector x squared is equal to the length of our member of our row space squared, our unique member of our row space squared, plus that member of our null space squared. Now, this is definitely going to be a positive number. That's definitely at minimum 0, but it has to be something larger than 0. So we can say that this quantity right here is definitely greater than or equal to just r0 squared. Or another way to think about it is, you give me any solution to the equation Ax is equal to b, and the square of its length is going to be greater than or equal to the square of r0's length. Or, since both of the lengths are always positive, you can take the positive square root, and you know you won't have to switch sides there, that the length of any solution to Ax equal to b is going to be greater than or equal to the length of r0."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "That's definitely at minimum 0, but it has to be something larger than 0. So we can say that this quantity right here is definitely greater than or equal to just r0 squared. Or another way to think about it is, you give me any solution to the equation Ax is equal to b, and the square of its length is going to be greater than or equal to the square of r0's length. Or, since both of the lengths are always positive, you can take the positive square root, and you know you won't have to switch sides there, that the length of any solution to Ax equal to b is going to be greater than or equal to the length of r0. So that makes r0 kind of a special solution. So now let's write our entire statement, everything that we've learned in this video. So if b is a member of the column space of A, then there exists a unique r0 that is a member of the row space of A such that r0 is a solution to Ax is equal to b."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "Or, since both of the lengths are always positive, you can take the positive square root, and you know you won't have to switch sides there, that the length of any solution to Ax equal to b is going to be greater than or equal to the length of r0. So that makes r0 kind of a special solution. So now let's write our entire statement, everything that we've learned in this video. So if b is a member of the column space of A, then there exists a unique r0 that is a member of the row space of A such that r0 is a solution to Ax is equal to b. And not only is it a solution, it's a special solution. r0 is the solution with the least, or no solution has a smaller length than r0. Let me write it that way."}, {"video_title": "Unique rowspace solution to Ax = b Linear Algebra Khan Academy.mp3", "Sentence": "So if b is a member of the column space of A, then there exists a unique r0 that is a member of the row space of A such that r0 is a solution to Ax is equal to b. And not only is it a solution, it's a special solution. r0 is the solution with the least, or no solution has a smaller length than r0. Let me write it that way. Maybe some other solution could have the same length. So and no other solution can have a smaller length. Maybe we could write that if you give me any vector b that's a member of the column space of A, then there exists a unique member of the row space that is essentially the smallest solution."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I have this 4 by 4 matrix A here. And let's see if we can figure out its determinant, the determinant of A. And before just doing it the way we've done it in the past, where you go down one of the rows or one of the columns, and you notice there's no zeros here. So there's no easy row or easy column to take the determinant by. You know, we could have gone down this row and do all the sub-matrices. But this becomes pretty involved. You end up doing 4 3 by 3 determinants."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So there's no easy row or easy column to take the determinant by. You know, we could have gone down this row and do all the sub-matrices. But this becomes pretty involved. You end up doing 4 3 by 3 determinants. And then each of those are composed of 3 2 by 2 determinants, it becomes a pretty hairy process. Let's see if we can use some of the realizations we've discovered in the last few videos to simplify this process. Well, one of the realizations is that row operations, or if you subtract, let me write it this way, if you replace row j with, let's say, row j minus some scalar multiple times row i, it does not change the determinant."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You end up doing 4 3 by 3 determinants. And then each of those are composed of 3 2 by 2 determinants, it becomes a pretty hairy process. Let's see if we can use some of the realizations we've discovered in the last few videos to simplify this process. Well, one of the realizations is that row operations, or if you subtract, let me write it this way, if you replace row j with, let's say, row j minus some scalar multiple times row i, it does not change the determinant. Does not change the determinant of A. We saw that, I think, it was two videos ago. So this was a pretty big realization."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, one of the realizations is that row operations, or if you subtract, let me write it this way, if you replace row j with, let's say, row j minus some scalar multiple times row i, it does not change the determinant. Does not change the determinant of A. We saw that, I think, it was two videos ago. So this was a pretty big realization. We can do these type of row operations and it won't change the determinant. The other realization we had was that these upper triangular matrices, you can figure out their determinant. So what does upper triangular look like?"}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this was a pretty big realization. We can do these type of row operations and it won't change the determinant. The other realization we had was that these upper triangular matrices, you can figure out their determinant. So what does upper triangular look like? Let me just review it. The upper triangular, everything below the diagonal, so let's say the diagonal has, let me just draw its terms like that. These are some non-zero terms."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what does upper triangular look like? Let me just review it. The upper triangular, everything below the diagonal, so let's say the diagonal has, let me just draw its terms like that. These are some non-zero terms. Or they don't have to be. Then upper triangular, everything below the diagonal is a 0. And everything above the diagonal probably isn't a 0, but you never know."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are some non-zero terms. Or they don't have to be. Then upper triangular, everything below the diagonal is a 0. And everything above the diagonal probably isn't a 0, but you never know. But they're non-zero terms. So all the red stuff here is non-zero. All the stuff in green is 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And everything above the diagonal probably isn't a 0, but you never know. But they're non-zero terms. So all the red stuff here is non-zero. All the stuff in green is 0. I didn't touch on it in that video, but there is also such a thing as a lower triangular that you might have guessed how it looks. Everything above the main diagonal is 0. So this is the main diagonal right here, all the way down like that."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All the stuff in green is 0. I didn't touch on it in that video, but there is also such a thing as a lower triangular that you might have guessed how it looks. Everything above the main diagonal is 0. So this is the main diagonal right here, all the way down like that. All of these guys are going to be non-zero. All of that's going to be non-zero. And then the 0's are going to be above the diagonal."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the main diagonal right here, all the way down like that. All of these guys are going to be non-zero. All of that's going to be non-zero. And then the 0's are going to be above the diagonal. Like that. We saw in the last video that the determinant of this guy is just equal to the product of the diagonal entries, which is a very good or it's a very simple way of finding a determinant. And you could use the same argument we made in the last video to say that the same is true of the lower triangular matrix, that its determinant is also just the product of those entries."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the 0's are going to be above the diagonal. Like that. We saw in the last video that the determinant of this guy is just equal to the product of the diagonal entries, which is a very good or it's a very simple way of finding a determinant. And you could use the same argument we made in the last video to say that the same is true of the lower triangular matrix, that its determinant is also just the product of those entries. I won't prove it here, but you can use the exact same argument you used in the video that I just did on the upper triangular. So given this, that the determinant of this is just the product of those guys, and that I can perform row operations on this guy and not change the determinant, maybe a simpler way to calculate this determinant is to get this guy into an upper triangular form and then just multiply the entries down the diagonal. So let's do that."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you could use the same argument we made in the last video to say that the same is true of the lower triangular matrix, that its determinant is also just the product of those entries. I won't prove it here, but you can use the exact same argument you used in the video that I just did on the upper triangular. So given this, that the determinant of this is just the product of those guys, and that I can perform row operations on this guy and not change the determinant, maybe a simpler way to calculate this determinant is to get this guy into an upper triangular form and then just multiply the entries down the diagonal. So let's do that. So we want to find the determinant of A. So the determinant of A, let me rewrite A right here. It's 1, 2, 2, 1, 1, 1, 2, 4, 2, 2, 7, 5, 2, minus 1, 4, minus 6, 3."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So we want to find the determinant of A. So the determinant of A, let me rewrite A right here. It's 1, 2, 2, 1, 1, 1, 2, 4, 2, 2, 7, 5, 2, minus 1, 4, minus 6, 3. Now let's put this in, let's try to get this into upper triangular form. So let's replace the second row with the second. So I'm just going to keep the first row the same."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's 1, 2, 2, 1, 1, 1, 2, 4, 2, 2, 7, 5, 2, minus 1, 4, minus 6, 3. Now let's put this in, let's try to get this into upper triangular form. So let's replace the second row with the second. So I'm just going to keep the first row the same. 1, 2, 2, 1. And let's replace the second row with the second row minus the first row. The second row minus the first row is going to be equal to 1 minus 1 is 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just going to keep the first row the same. 1, 2, 2, 1. And let's replace the second row with the second row minus the first row. The second row minus the first row is going to be equal to 1 minus 1 is 0. So in this case, the constant is just 1. So 1 minus 1 is 0. 2 minus 2 is 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The second row minus the first row is going to be equal to 1 minus 1 is 0. So in this case, the constant is just 1. So 1 minus 1 is 0. 2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row with the third row minus 2 times the second row."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's replace the third row with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. And then let's, let me get a good color here, do pink. Let's replace the last row with the last row essentially plus the first row."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. And then let's, let me get a good color here, do pink. Let's replace the last row with the last row essentially plus the first row. You could say minus minus 1 times the first row, which is the same thing as the last row plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's replace the last row with the last row essentially plus the first row. You could say minus minus 1 times the first row, which is the same thing as the last row plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6. Minus 6 plus 2 is minus 4. And then 3 plus 1 is 4. So there we have it like that."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "4 plus 2 is 6. Minus 6 plus 2 is minus 4. And then 3 plus 1 is 4. So there we have it like that. And this guy has two 0's here, so maybe I want to swap some rows. So let me swap some rows. If we swap rows, what happens?"}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So there we have it like that. And this guy has two 0's here, so maybe I want to swap some rows. So let me swap some rows. If we swap rows, what happens? So let me, I'm going to swap the middle two rows just for fun. Well, not just for fun. Because I want a pivot entry right here, or I shouldn't say pivot entry, I want to do it in upper triangular form."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we swap rows, what happens? So let me, I'm going to swap the middle two rows just for fun. Well, not just for fun. Because I want a pivot entry right here, or I shouldn't say pivot entry, I want to do it in upper triangular form. So I want a non-zero entry here. This is a 0, so I'm going to move this guy down. So I'm going to keep the top row the same, 1, 2, 2, 1."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because I want a pivot entry right here, or I shouldn't say pivot entry, I want to do it in upper triangular form. So I want a non-zero entry here. This is a 0, so I'm going to move this guy down. So I'm going to keep the top row the same, 1, 2, 2, 1. I'm going to keep the bottom row the same, 0, 0, 6, minus 4, 4. And I'm going to swap these guys right here. So this is going to be 0, 3, 1, 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep the top row the same, 1, 2, 2, 1. I'm going to keep the bottom row the same, 0, 0, 6, minus 4, 4. And I'm going to swap these guys right here. So this is going to be 0, 3, 1, 0. And then 0, 0, 2, 1. Now, can I just swap entries like that? Well, I can, but you have to remember that when you swap entries, your resulting determinant is going to be the negative of your original determinant."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be 0, 3, 1, 0. And then 0, 0, 2, 1. Now, can I just swap entries like that? Well, I can, but you have to remember that when you swap entries, your resulting determinant is going to be the negative of your original determinant. So if we swap these two guys, the determinant of this is going to be the negative of this determinant. When you swap two rows, you just flip the sign of the determinant. We saw that."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, I can, but you have to remember that when you swap entries, your resulting determinant is going to be the negative of your original determinant. So if we swap these two guys, the determinant of this is going to be the negative of this determinant. When you swap two rows, you just flip the sign of the determinant. We saw that. That was one of the first videos we did on these kind of messing with the determinants. Now, what do we want to do here? To get this guy into upper triangular form, it would be nice to get this to be a 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We saw that. That was one of the first videos we did on these kind of messing with the determinants. Now, what do we want to do here? To get this guy into upper triangular form, it would be nice to get this to be a 0. So to get that to be a 0, let me keep everything else the same, so I have a 1, 2, 2, 1. I have a 0, 3, 1, 0. The third row is 0, 0, 2, 1."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "To get this guy into upper triangular form, it would be nice to get this to be a 0. So to get that to be a 0, let me keep everything else the same, so I have a 1, 2, 2, 1. I have a 0, 3, 1, 0. The third row is 0, 0, 2, 1. And now this last row, let me replace it with the last row minus 3 times this row. So let me write it like this. Well, I have to carry that negative sign as well."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The third row is 0, 0, 2, 1. And now this last row, let me replace it with the last row minus 3 times this row. So let me write it like this. Well, I have to carry that negative sign as well. So I'm going to replace this last row with the last row minus 3 times, or this last row minus 2 times the second row, I want to zero it out. So 0 minus 2 times 0 is 0. 6 minus 2 times 3 is 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, I have to carry that negative sign as well. So I'm going to replace this last row with the last row minus 3 times, or this last row minus 2 times the second row, I want to zero it out. So 0 minus 2 times 0 is 0. 6 minus 2 times 3 is 0. Minus 4 minus 2 times 1 is minus 6. And then 4 minus 2 times 0 is just 4. We're almost there."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "6 minus 2 times 3 is 0. Minus 4 minus 2 times 1 is minus 6. And then 4 minus 2 times 0 is just 4. We're almost there. Now we want to zero this guy out. So let's replace this one. So I'm going to keep my top three rows the same again."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're almost there. Now we want to zero this guy out. So let's replace this one. So I'm going to keep my top three rows the same again. And let me see if I can write it a little bit neater. So my first row is 1, 2, 2, 1. My second row is 0, 3, 1, 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep my top three rows the same again. And let me see if I can write it a little bit neater. So my first row is 1, 2, 2, 1. My second row is 0, 3, 1, 0. Fourth row is 0, 0, 2, 1. And I'm going to take the matrix. I haven't written the fourth row yet."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "My second row is 0, 3, 1, 0. Fourth row is 0, 0, 2, 1. And I'm going to take the matrix. I haven't written the fourth row yet. And of course, the negative of this is going to be the determinant of our original matrix, because we had swapped those rows. But let's replace this last row with the last row plus 3 times 3 times the third row. So we get 0 plus 3 times 0 is 0."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I haven't written the fourth row yet. And of course, the negative of this is going to be the determinant of our original matrix, because we had swapped those rows. But let's replace this last row with the last row plus 3 times 3 times the third row. So we get 0 plus 3 times 0 is 0. 0 plus 3 times 0 is 0. Minus 6 plus 3 times 2 is 0. 4 plus 3 times 1 is 7."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we get 0 plus 3 times 0 is 0. 0 plus 3 times 0 is 0. Minus 6 plus 3 times 2 is 0. 4 plus 3 times 1 is 7. And just like that, we have a determinant of a matrix in upper triangular form. So this is going to be equal to the product of these guys. So this is going to be equal to, can't forget our negative sign, let's throw our negative sign out there and put a parentheses just like that."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "4 plus 3 times 1 is 7. And just like that, we have a determinant of a matrix in upper triangular form. So this is going to be equal to the product of these guys. So this is going to be equal to, can't forget our negative sign, let's throw our negative sign out there and put a parentheses just like that. This is going to be the product of that diagonal entry. 1 times 3 times 3 times 2 times 7, which is 6 times 7, which is 42. So the determinant of this matrix is minus 42."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to, can't forget our negative sign, let's throw our negative sign out there and put a parentheses just like that. This is going to be the product of that diagonal entry. 1 times 3 times 3 times 2 times 7, which is 6 times 7, which is 42. So the determinant of this matrix is minus 42. Which was pretty fast. This was a pretty fast shortcut. And it actually turns out it tends to be computationally more efficient to use these takeaways to put things into upper triangular form first."}, {"video_title": "Simpler 4x4 determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of this matrix is minus 42. Which was pretty fast. This was a pretty fast shortcut. And it actually turns out it tends to be computationally more efficient to use these takeaways to put things into upper triangular form first. And then if you do swaps, you have to remember to make the determinant negative. And then just multiply down the diagonal. And we did that there."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So the first one is 2x minus y is equal to 2. The second one is x plus 2y is equal to 1. And the third one is x plus y is equal to 4. So let's first just graph these, just to have a visual representation of what we're trying to do. So I like writing my lines in y equals mx plus b form. So this top line becomes what? Minus y is equal to minus 2x plus 2."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's first just graph these, just to have a visual representation of what we're trying to do. So I like writing my lines in y equals mx plus b form. So this top line becomes what? Minus y is equal to minus 2x plus 2. I just subtracted 2x from both sides. Or we could write that y is equal to 2x minus 2. That's that first line here."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus y is equal to minus 2x plus 2. I just subtracted 2x from both sides. Or we could write that y is equal to 2x minus 2. That's that first line here. The second line, I'll do it in green. We could write this as 2y is equal to minus x plus 1. Or we could write that y is equal to minus 1 half x plus 1 half."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's that first line here. The second line, I'll do it in green. We could write this as 2y is equal to minus x plus 1. Or we could write that y is equal to minus 1 half x plus 1 half. I just divided both sides by 2. And then this last line right here, we could write this as y is equal to minus x plus 4. I could go straight here."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or we could write that y is equal to minus 1 half x plus 1 half. I just divided both sides by 2. And then this last line right here, we could write this as y is equal to minus x plus 4. I could go straight here. y is equal to minus x plus 4. Now let me graph these. Let me draw an axis."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I could go straight here. y is equal to minus x plus 4. Now let me graph these. Let me draw an axis. That is my y-axis. I can call it my y-axis, since we're actually dealing with x and y's now. And let's say that this is my, I'll do it in this gray as well, and say that is my x-axis."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw an axis. That is my y-axis. I can call it my y-axis, since we're actually dealing with x and y's now. And let's say that this is my, I'll do it in this gray as well, and say that is my x-axis. Just like that. And this first guy is going to be 2x minus 2. So its y-intercept is going to be at minus 2."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that this is my, I'll do it in this gray as well, and say that is my x-axis. Just like that. And this first guy is going to be 2x minus 2. So its y-intercept is going to be at minus 2. It's going to have a slope of 2, so it's going to be a pretty steep line. Just like that. So that's that first line, just like that."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So its y-intercept is going to be at minus 2. It's going to have a slope of 2, so it's going to be a pretty steep line. Just like that. So that's that first line, just like that. 2x minus 2. The next line is minus 1 half plus 1 half. So if we go plus 1 half, that's right there."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that's that first line, just like that. 2x minus 2. The next line is minus 1 half plus 1 half. So if we go plus 1 half, that's right there. And then the slope is minus 1 half. So for every 2 we go over, we go down 1. So it's going to be like that."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if we go plus 1 half, that's right there. And then the slope is minus 1 half. So for every 2 we go over, we go down 1. So it's going to be like that. It's actually going to be orthogonal, right? Because it's the negative inverse of this guy. So it's going to look something like that."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be like that. It's actually going to be orthogonal, right? Because it's the negative inverse of this guy. So it's going to look something like that. Let me draw it like that. Just like that. And this guy is minus x plus 4."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to look something like that. Let me draw it like that. Just like that. And this guy is minus x plus 4. So we go, this is 1, 2, 3, 4. And you go minus x. So for every 1 you go over, you go down 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And this guy is minus x plus 4. So we go, this is 1, 2, 3, 4. And you go minus x. So for every 1 you go over, you go down 1. So this other line is going to look something like this. This last line is going to look something like that. Just like that."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So for every 1 you go over, you go down 1. So this other line is going to look something like this. This last line is going to look something like that. Just like that. Now, I said at the beginning of this video that I want to find the intersection of these three lines. But notice, there is no intersection of these three lines. They all intersect the other two, but they don't all intersect each other in one point."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. Now, I said at the beginning of this video that I want to find the intersection of these three lines. But notice, there is no intersection of these three lines. They all intersect the other two, but they don't all intersect each other in one point. We can kind of call the system as being over-determined. We've over-constrained it. There is no intersection of all three of these points."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "They all intersect the other two, but they don't all intersect each other in one point. We can kind of call the system as being over-determined. We've over-constrained it. There is no intersection of all three of these points. So if I were to actually try to solve the system, I would find no solution. And to say that this has no solution is equivalent to saying that this matrix has no solution, or this equation has no solution. Let me write this."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "There is no intersection of all three of these points. So if I were to actually try to solve the system, I would find no solution. And to say that this has no solution is equivalent to saying that this matrix has no solution, or this equation has no solution. Let me write this. I'm just going to rewrite the system like this. This is equivalent to the matrix times the vector x, y is equal to 2, 1, and 4. And so this first equation is 2 times x minus 1 times y."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this. I'm just going to rewrite the system like this. This is equivalent to the matrix times the vector x, y is equal to 2, 1, and 4. And so this first equation is 2 times x minus 1 times y. So it's 2 and minus 1. 2 times x minus 1 times y is equal to 2. That's that first equation over there."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And so this first equation is 2 times x minus 1 times y. So it's 2 and minus 1. 2 times x minus 1 times y is equal to 2. That's that first equation over there. The second equation, actually I could even, well, I won't color code it. That'll take forever. That's 1 times x plus 2 times y is equal to 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's that first equation over there. The second equation, actually I could even, well, I won't color code it. That'll take forever. That's 1 times x plus 2 times y is equal to 1. And then we have x plus y is equal to 4. This system and this equation, this system right here, these are equivalent. Now, this isn't going to have any solution."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's 1 times x plus 2 times y is equal to 1. And then we have x plus y is equal to 4. This system and this equation, this system right here, these are equivalent. Now, this isn't going to have any solution. You could try to find a solution to this. You could create an augmented matrix, put it in reduced row echelon form, but there is no intersection to these three things. So you're not going to find a solution to A times some vector, we could call this some vector x, is equal to B."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, this isn't going to have any solution. You could try to find a solution to this. You could create an augmented matrix, put it in reduced row echelon form, but there is no intersection to these three things. So you're not going to find a solution to A times some vector, we could call this some vector x, is equal to B. Or another way to say it is that B is not in the column space of this matrix right here. Now, we learned in the last video that sure, we can't find a solution to Ax equals B. Ax equals B has no solution."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you're not going to find a solution to A times some vector, we could call this some vector x, is equal to B. Or another way to say it is that B is not in the column space of this matrix right here. Now, we learned in the last video that sure, we can't find a solution to Ax equals B. Ax equals B has no solution. We see it graphically here. These lines don't intersect with each other. And you could prove it for yourself algebraically by trying to find a solution here."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Ax equals B has no solution. We see it graphically here. These lines don't intersect with each other. And you could prove it for yourself algebraically by trying to find a solution here. You'll end up with a 0 equals 1. But we can almost get there by finding a least squares solution. And we find a least squares solution if we multiply both sides by A transpose."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And you could prove it for yourself algebraically by trying to find a solution here. You'll end up with a 0 equals 1. But we can almost get there by finding a least squares solution. And we find a least squares solution if we multiply both sides by A transpose. We know that A transpose times A times our least squares solution is going to be equal to A transpose times B. So at least we can find the closest fit for our solution. So let's find the vector x that is our least squares solution."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And we find a least squares solution if we multiply both sides by A transpose. We know that A transpose times A times our least squares solution is going to be equal to A transpose times B. So at least we can find the closest fit for our solution. So let's find the vector x that is our least squares solution. So what is A transpose times A? So A transpose looks like this. You'll have 2 minus 1, 1, 2, 1, 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's find the vector x that is our least squares solution. So what is A transpose times A? So A transpose looks like this. You'll have 2 minus 1, 1, 2, 1, 1. That is A transpose. And then of course, A is just this thing. 2 minus 1, 1, 2, 1, 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You'll have 2 minus 1, 1, 2, 1, 1. That is A transpose. And then of course, A is just this thing. 2 minus 1, 1, 2, 1, 1. So A transpose A is going to be equal to, we have a 2 by 3 times a 3 by 2 matrix. So it's going to be a 2 by 2 matrix. So what do we get?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 1, 1, 2, 1, 1. So A transpose A is going to be equal to, we have a 2 by 3 times a 3 by 2 matrix. So it's going to be a 2 by 2 matrix. So what do we get? We get 2 times 2, which is 4, plus 1 times 1, plus 1 times 1, so it's 4 plus 1 plus 1. So that's equal to 6. And then we have 2 times minus 1, which is minus 2, plus 1 times 2, so those cancel out."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So what do we get? We get 2 times 2, which is 4, plus 1 times 1, plus 1 times 1, so it's 4 plus 1 plus 1. So that's equal to 6. And then we have 2 times minus 1, which is minus 2, plus 1 times 2, so those cancel out. You get a minus 2 plus 2 is 0, plus 1 times 1. So that's just going to be 1. And then we get minus 1 times 2, which is minus 2, plus 2 times 1, which is 2, so the minus 2 plus 2 is 0, plus 1 times 1, so we get a 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then we have 2 times minus 1, which is minus 2, plus 1 times 2, so those cancel out. You get a minus 2 plus 2 is 0, plus 1 times 1. So that's just going to be 1. And then we get minus 1 times 2, which is minus 2, plus 2 times 1, which is 2, so the minus 2 plus 2 is 0, plus 1 times 1, so we get a 1. And then finally, we get minus 1 times minus 1, which is positive 1, plus 2 times 2, which is 4, so we're now at 5, plus 1 times 1, so this is going to be 6. So this is A transpose A. Now, what is A transpose times B?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then we get minus 1 times 2, which is minus 2, plus 2 times 1, which is 2, so the minus 2 plus 2 is 0, plus 1 times 1, so we get a 1. And then finally, we get minus 1 times minus 1, which is positive 1, plus 2 times 2, which is 4, so we're now at 5, plus 1 times 1, so this is going to be 6. So this is A transpose A. Now, what is A transpose times B? A transpose is 2, 1, 1, minus 1, 2, 1. And then B is just the 3 by 1 vector, or the vector that's a member of R3, 2, 1, 4. So what is this going to be equal to?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is A transpose times B? A transpose is 2, 1, 1, minus 1, 2, 1. And then B is just the 3 by 1 vector, or the vector that's a member of R3, 2, 1, 4. So what is this going to be equal to? This is going to be equal to, let's see, we have a 3. This is a 2 by 3 times a 3 by 1. We're going to get a 2 by 1 vector."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So what is this going to be equal to? This is going to be equal to, let's see, we have a 3. This is a 2 by 3 times a 3 by 1. We're going to get a 2 by 1 vector. We're going to get a 2 by 1 vector here. So 2 times 2 is 4, plus 1 times 1, so that's plus 1, so that's 5, plus, let me actually write it down. I'm going to make a careless mistake, 2 times 2, which is 4, plus 1 times 1, which is 1, plus 1 times 4, which is 4."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We're going to get a 2 by 1 vector. We're going to get a 2 by 1 vector here. So 2 times 2 is 4, plus 1 times 1, so that's plus 1, so that's 5, plus, let me actually write it down. I'm going to make a careless mistake, 2 times 2, which is 4, plus 1 times 1, which is 1, plus 1 times 4, which is 4. And then here you have minus 1 times 2, which is minus 2. 2 times 1, which is 2, plus 1 times 4, which is 4. So A transpose times B is equal to 9, and this is going to be 4."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to make a careless mistake, 2 times 2, which is 4, plus 1 times 1, which is 1, plus 1 times 4, which is 4. And then here you have minus 1 times 2, which is minus 2. 2 times 1, which is 2, plus 1 times 4, which is 4. So A transpose times B is equal to 9, and this is going to be 4. So we can rewrite this guy right here as the matrix A transpose A, which is just 6, 1, 1, 6, times my least squares solution. So this is actually going to be in the column space of A, is equal to A transpose times B, which is just the vector 9, 4. And this will be a little bit more straightforward to find a solution for."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So A transpose times B is equal to 9, and this is going to be 4. So we can rewrite this guy right here as the matrix A transpose A, which is just 6, 1, 1, 6, times my least squares solution. So this is actually going to be in the column space of A, is equal to A transpose times B, which is just the vector 9, 4. And this will be a little bit more straightforward to find a solution for. In fact, there will be a solution. We've proved it in the last video. So to find a solution, let's create our little augmented matrix."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And this will be a little bit more straightforward to find a solution for. In fact, there will be a solution. We've proved it in the last video. So to find a solution, let's create our little augmented matrix. 6, 1, augmented with a 9, put a 4 there, and you get a 1 and a 6, just like that. Let's put the left-hand side in reduced row echelon form. Actually, first I'm going to swap these two rows."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So to find a solution, let's create our little augmented matrix. 6, 1, augmented with a 9, put a 4 there, and you get a 1 and a 6, just like that. Let's put the left-hand side in reduced row echelon form. Actually, first I'm going to swap these two rows. That's my first row operation that I choose to do, just because I like to have that 1 there. It's a nice pivot entry. So then I get, it goes to 1, 6, 4, and 6, 1, 9."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Actually, first I'm going to swap these two rows. That's my first row operation that I choose to do, just because I like to have that 1 there. It's a nice pivot entry. So then I get, it goes to 1, 6, 4, and 6, 1, 9. And then let me replace my second row with the second row minus 6 times the first row. So I'm going to keep my first row the same. So I have 1, 6, and 4."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So then I get, it goes to 1, 6, 4, and 6, 1, 9. And then let me replace my second row with the second row minus 6 times the first row. So I'm going to keep my first row the same. So I have 1, 6, and 4. And then my second row, I'm going to replace my second row with the second row minus 6 times the first row. So 6 minus 6 is 0. 1 minus 6 times 6, that's 1 minus 36, so that's minus 35."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I have 1, 6, and 4. And then my second row, I'm going to replace my second row with the second row minus 6 times the first row. So 6 minus 6 is 0. 1 minus 6 times 6, that's 1 minus 36, so that's minus 35. And then 9 minus 6 times 4 is 9 minus 24. This one always gets me in trouble. So 9 minus 24, that's the negative of 24 minus 9."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 6 times 6, that's 1 minus 36, so that's minus 35. And then 9 minus 6 times 4 is 9 minus 24. This one always gets me in trouble. So 9 minus 24, that's the negative of 24 minus 9. So that is minus 15. Let me make sure I didn't make a careless mistake. 1 minus 36 is minus 35."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 9 minus 24, that's the negative of 24 minus 9. So that is minus 15. Let me make sure I didn't make a careless mistake. 1 minus 36 is minus 35. 9 minus 24 is minus 15. So that's what I get right there. And then let me go to the right now."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 36 is minus 35. 9 minus 24 is minus 15. So that's what I get right there. And then let me go to the right now. So let me divide this row right here. Let me divide it by minus 35. So I'm going to keep my first row the same, 1, 6, and 4."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then let me go to the right now. So let me divide this row right here. Let me divide it by minus 35. So I'm going to keep my first row the same, 1, 6, and 4. And then this guy right here, I'm going to divide by minus 35. So I'm going to get 0, 1, and then minus 15 over minus 35. That's 15 over 35."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep my first row the same, 1, 6, and 4. And then this guy right here, I'm going to divide by minus 35. So I'm going to get 0, 1, and then minus 15 over minus 35. That's 15 over 35. Or that's 3 over 7. So that is equal to 3 sevenths. And then if I really want to, well, let me just put this in complete reduced row echelon form."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's 15 over 35. Or that's 3 over 7. So that is equal to 3 sevenths. And then if I really want to, well, let me just put this in complete reduced row echelon form. That'll be nice. Let me keep my second row the same. So my second row is 0, 1, and 3 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then if I really want to, well, let me just put this in complete reduced row echelon form. That'll be nice. Let me keep my second row the same. So my second row is 0, 1, and 3 sevenths. And then my first row, I'm going to replace it with my first row minus 6 times my second row. So 1 minus 6 times 0 is 1. 6 minus 6 times 1 is 0."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So my second row is 0, 1, and 3 sevenths. And then my first row, I'm going to replace it with my first row minus 6 times my second row. So 1 minus 6 times 0 is 1. 6 minus 6 times 1 is 0. And then we have 4 minus 6 times 3 sevenths. So 4 is 28 sevenths. Let me write it up here."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "6 minus 6 times 1 is 0. And then we have 4 minus 6 times 3 sevenths. So 4 is 28 sevenths. Let me write it up here. So we have 4, which is 28 sevenths, minus 6 times 3 sevenths, so minus 18 sevenths. That's 6 times 3 sevenths. So this is going to be equal to 10 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it up here. So we have 4, which is 28 sevenths, minus 6 times 3 sevenths, so minus 18 sevenths. That's 6 times 3 sevenths. So this is going to be equal to 10 sevenths. And just like that, I've solved this new equation. So you could say that, let me write it this way. We could write that x star, this is going to the first entry of x star, which we could call x, is going to be 10 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to 10 sevenths. And just like that, I've solved this new equation. So you could say that, let me write it this way. We could write that x star, this is going to the first entry of x star, which we could call x, is going to be 10 sevenths. Let me write this. x star, the solution, is going to be 10 sevenths and 3 sevenths. So I'm saying that if you take x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to get as close to a solution as possible."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We could write that x star, this is going to the first entry of x star, which we could call x, is going to be 10 sevenths. Let me write this. x star, the solution, is going to be 10 sevenths and 3 sevenths. So I'm saying that if you take x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to get as close to a solution as possible. So let's see what that looks like visually. What is 10 sevenths? Let me write this down."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I'm saying that if you take x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to get as close to a solution as possible. So let's see what that looks like visually. What is 10 sevenths? Let me write this down. x star is equal to 10 sevenths and 3 sevenths. Or we're saying the closest, our least square solution is x is equal to 10 sevenths. So x is a little over 1."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this down. x star is equal to 10 sevenths and 3 sevenths. Or we're saying the closest, our least square solution is x is equal to 10 sevenths. So x is a little over 1. And then y is going to be 3 sevenths, a little less than 1 half. So our least square solution is going to be this one right there. And so when you put this value for x, when you put x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to minimize the collective squares of the distances between all of these guys."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So x is a little over 1. And then y is going to be 3 sevenths, a little less than 1 half. So our least square solution is going to be this one right there. And so when you put this value for x, when you put x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to minimize the collective squares of the distances between all of these guys. And I drew this a little bit too small to show that. But let's actually figure out what our minimized difference is. So remember, the whole point of this is to minimize the distance between a x star and b."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And so when you put this value for x, when you put x is equal to 10 sevenths and y is equal to 3 sevenths, you're going to minimize the collective squares of the distances between all of these guys. And I drew this a little bit too small to show that. But let's actually figure out what our minimized difference is. So remember, the whole point of this is to minimize the distance between a x star and b. Or between b and a x star. Now, what was a x star equal to? a x star was equal to 9 fourths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So remember, the whole point of this is to minimize the distance between a x star and b. Or between b and a x star. Now, what was a x star equal to? a x star was equal to 9 fourths. So this right here, sorry, a x star, that's not equal to 9 fourths, that's a transpose a x star is equal to 9 fourths. a x star is our original matrix A, which is this one right here. So 2, let me write it down here."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "a x star was equal to 9 fourths. So this right here, sorry, a x star, that's not equal to 9 fourths, that's a transpose a x star is equal to 9 fourths. a x star is our original matrix A, which is this one right here. So 2, let me write it down here. I know it's off the page right now. So our original matrix A was 2 minus 1, 1, 2, 1, 1. That was our original matrix A right there."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 2, let me write it down here. I know it's off the page right now. So our original matrix A was 2 minus 1, 1, 2, 1, 1. That was our original matrix A right there. And then our x star, we were able to determine, is 10 sevenths and 3 sevenths. So a x star is going to be this product. So what is it equal to?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That was our original matrix A right there. And then our x star, we were able to determine, is 10 sevenths and 3 sevenths. So a x star is going to be this product. So what is it equal to? It is equal to, it's going to be a 3 by 1 matrix. So we get 2 times 10 sevenths, which is 20 sevenths, minus 1 times 3 sevenths. So minus 3 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So what is it equal to? It is equal to, it's going to be a 3 by 1 matrix. So we get 2 times 10 sevenths, which is 20 sevenths, minus 1 times 3 sevenths. So minus 3 sevenths. And then we get, let me see, yep, that's 20 minus 3 sevenths, and we have 10 sevenths. And we have 10 sevenths minus 2 times 3 sevenths. So minus 6 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So minus 3 sevenths. And then we get, let me see, yep, that's 20 minus 3 sevenths, and we have 10 sevenths. And we have 10 sevenths minus 2 times 3 sevenths. So minus 6 sevenths. Or plus, sorry, this is a plus. 2 times 10 sevenths is 20 sevenths, minus 1 times 3 sevenths. Then we have 1 times 10 sevenths plus 2 times 3 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So minus 6 sevenths. Or plus, sorry, this is a plus. 2 times 10 sevenths is 20 sevenths, minus 1 times 3 sevenths. Then we have 1 times 10 sevenths plus 2 times 3 sevenths. And then we have 10 sevenths plus 3 sevenths. 10 sevenths plus 3 sevenths. So a x, so this is a, and x star, or least squares approximation for x, is equal to, what is this?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Then we have 1 times 10 sevenths plus 2 times 3 sevenths. And then we have 10 sevenths plus 3 sevenths. 10 sevenths plus 3 sevenths. So a x, so this is a, and x star, or least squares approximation for x, is equal to, what is this? This is 17 sevenths, this is 16 sevenths, and this is 13 sevenths. And we want to find out what this minimum distance is. So let's see, so this is going to be this thing."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So a x, so this is a, and x star, or least squares approximation for x, is equal to, what is this? This is 17 sevenths, this is 16 sevenths, and this is 13 sevenths. And we want to find out what this minimum distance is. So let's see, so this is going to be this thing. So 17 sevenths, 16 sevenths, and 13 sevenths, minus our original b. Now our original b was 2, 1, and 4. So I'm claiming that my solution that we've just found, this minimizes this distance."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's see, so this is going to be this thing. So 17 sevenths, 16 sevenths, and 13 sevenths, minus our original b. Now our original b was 2, 1, and 4. So I'm claiming that my solution that we've just found, this minimizes this distance. Because this is the projection of b onto the column space of a, we saw that before. So our b, we see all the way up here, is 2, 1, 4. Just like that."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I'm claiming that my solution that we've just found, this minimizes this distance. Because this is the projection of b onto the column space of a, we saw that before. So our b, we see all the way up here, is 2, 1, 4. Just like that. So if we take the length of this, let me switch colors. This is equal to the length, let me write all of this in sevenths, I'll just do it in my head. I don't want to waste too much time."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. So if we take the length of this, let me switch colors. This is equal to the length, let me write all of this in sevenths, I'll just do it in my head. I don't want to waste too much time. So this is 17 sevenths minus 14 sevenths. 2 is 14 sevenths, so this is going to be 3 sevenths. And we have 16 sevenths minus 7 sevenths."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I don't want to waste too much time. So this is 17 sevenths minus 14 sevenths. 2 is 14 sevenths, so this is going to be 3 sevenths. And we have 16 sevenths minus 7 sevenths. So that's 9 sevenths. And then we have 13 sevenths minus 28 sevenths. So that is minus 15 over 7."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And we have 16 sevenths minus 7 sevenths. So that's 9 sevenths. And then we have 13 sevenths minus 28 sevenths. So that is minus 15 over 7. So this is the vector that separates the b that was not in my column space of a from my best, the projection of b. And if we find its length, its length is going to be equal to let's find the square of its length first. The square of its length is going to be 3 sevenths squared."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that is minus 15 over 7. So this is the vector that separates the b that was not in my column space of a from my best, the projection of b. And if we find its length, its length is going to be equal to let's find the square of its length first. The square of its length is going to be 3 sevenths squared. So that is 9 49ths plus 9 sevenths squared, which is 81 49ths, plus minus 15 sevenths squared. So that's what's 15 squared. 15 squared is 225, I think."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The square of its length is going to be 3 sevenths squared. So that is 9 49ths plus 9 sevenths squared, which is 81 49ths, plus minus 15 sevenths squared. So that's what's 15 squared. 15 squared is 225, I think. Let me make sure. I'm prone to careless mistakes. 5 times 5 is 25."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "15 squared is 225, I think. Let me make sure. I'm prone to careless mistakes. 5 times 5 is 25. 1 times 5, this is 75. And then I have a 150. Yep, 225."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "5 times 5 is 25. 1 times 5, this is 75. And then I have a 150. Yep, 225. So plus 225 over 49, which is equal to 9 plus 81 is 90. And then so 225 plus 90, we get a 5. What is it?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Yep, 225. So plus 225 over 49, which is equal to 9 plus 81 is 90. And then so 225 plus 90, we get a 5. What is it? 315. So this is equal to 315 over 49. Or if we actually wanted the difference, that's going to be the square root of this."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "What is it? 315. So this is equal to 315 over 49. Or if we actually wanted the difference, that's going to be the square root of this. So if we take just the regular distance, that's equal to the square root of that. So it's equal to the square root of 315 over 7. And square root of 315, it looks like that is simplifiable."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or if we actually wanted the difference, that's going to be the square root of this. So if we take just the regular distance, that's equal to the square root of that. So it's equal to the square root of 315 over 7. And square root of 315, it looks like that is simplifiable. Does 9 go into it? Looks like 9 goes into it maybe 35 times. So it would be what?"}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And square root of 315, it looks like that is simplifiable. Does 9 go into it? Looks like 9 goes into it maybe 35 times. So it would be what? 3 square roots of 35 over 7. So that is just a measure. You're not going to find any, let me put it this way, you're not going to be able to find any member of R2, any values of x and y, that's going to give you a smaller value than this when you find the distance between its solution and the solution you were trying to get to."}, {"video_title": "Least squares examples Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it would be what? 3 square roots of 35 over 7. So that is just a measure. You're not going to find any, let me put it this way, you're not going to be able to find any member of R2, any values of x and y, that's going to give you a smaller value than this when you find the distance between its solution and the solution you were trying to get to. So this, based on our least square solution, is the best estimate you're going to get. x is equal to 10 sevenths, y is equal to 3 sevenths. A little bit right, just like that."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And I know for a fact that it's a basis for the subspace v. What I want to show you in this video is that if this guy has n elements right here, that any set that spans v has to have at least n elements. So any spanning set must have at least n elements or n members or cardinality of n. There's all just different ways of saying you've got n vectors in this set. So let's see if we could, I'm saying that every set that spans v must have at least n elements if the sum basis set has n elements for v. Let's see if we can kind of run with a set that has less than n elements and see if we reach any contradictions. So let's say that I have some set B here. And it's equal to the vectors b1, b2, all the way to bm. And m is less than n. So I have some set of vectors here that have fewer elements than my set A. And let's say that B spans, you come to me one day and you say, look, I found you this set of vectors right here."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that I have some set B here. And it's equal to the vectors b1, b2, all the way to bm. And m is less than n. So I have some set of vectors here that have fewer elements than my set A. And let's say that B spans, you come to me one day and you say, look, I found you this set of vectors right here. And not only does it have fewer elements than A, but it spans v. And I look at you very suspiciously because I always thought that this green statement was true. So we start a little bit of a thought experiment. And I say, OK, you claim that your set spans v, so let's do something."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that B spans, you come to me one day and you say, look, I found you this set of vectors right here. And not only does it have fewer elements than A, but it spans v. And I look at you very suspiciously because I always thought that this green statement was true. So we start a little bit of a thought experiment. And I say, OK, you claim that your set spans v, so let's do something. Let me define a new set. Let me call this new set b1 prime. And you'll see why I'm doing this kind of strange notation."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And I say, OK, you claim that your set spans v, so let's do something. Let me define a new set. Let me call this new set b1 prime. And you'll see why I'm doing this kind of strange notation. What's essentially going to be is the set B plus my vector a1. So it's a1, and then I have all of my elements of B. So b1, b2, all the way to bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And you'll see why I'm doing this kind of strange notation. What's essentially going to be is the set B plus my vector a1. So it's a1, and then I have all of my elements of B. So b1, b2, all the way to bm. Now, I think you and I could both agree that this set is linearly dependent. How do I know that? Linear dependence means that at least one of the elements of the set can be represented as a linear combination of the others."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So b1, b2, all the way to bm. Now, I think you and I could both agree that this set is linearly dependent. How do I know that? Linear dependence means that at least one of the elements of the set can be represented as a linear combination of the others. Well, we know that a1, I mean, it's one of the basis vectors for v, for this definition of a basis. But all of the basis vectors are members of v, right? You can obviously represent, if this set is a basis for v, then this means that this set spans v, or that every member of v can be represented as a linear combination of these guys, or another way is every linear combination of these guys is in v. And one of the linear combinations of these guys is you just set the coefficient on a1 to be 1 and the coefficients on everyone else to be 0."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Linear dependence means that at least one of the elements of the set can be represented as a linear combination of the others. Well, we know that a1, I mean, it's one of the basis vectors for v, for this definition of a basis. But all of the basis vectors are members of v, right? You can obviously represent, if this set is a basis for v, then this means that this set spans v, or that every member of v can be represented as a linear combination of these guys, or another way is every linear combination of these guys is in v. And one of the linear combinations of these guys is you just set the coefficient on a1 to be 1 and the coefficients on everyone else to be 0. So obviously, a1 is also in the set. So if a1 is in v and all of these guys span v, by definition, if these guys span v, some linear combination of these guys can be used to construct any member of v. So you can take some linear combination of these guys to construct a1. So you could say a1 is equal to d1, where the d's are the constants, d1, b1, plus d2, b2, all the way to dm, bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "You can obviously represent, if this set is a basis for v, then this means that this set spans v, or that every member of v can be represented as a linear combination of these guys, or another way is every linear combination of these guys is in v. And one of the linear combinations of these guys is you just set the coefficient on a1 to be 1 and the coefficients on everyone else to be 0. So obviously, a1 is also in the set. So if a1 is in v and all of these guys span v, by definition, if these guys span v, some linear combination of these guys can be used to construct any member of v. So you can take some linear combination of these guys to construct a1. So you could say a1 is equal to d1, where the d's are the constants, d1, b1, plus d2, b2, all the way to dm, bm. And at least one of these have to be non-zero. We know that a is a non-zero vector. If it was a zero vector, this couldn't be a basis because it wouldn't be linearly independent because you can always represent a zero vector as really just a zero times any other vector."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So you could say a1 is equal to d1, where the d's are the constants, d1, b1, plus d2, b2, all the way to dm, bm. And at least one of these have to be non-zero. We know that a is a non-zero vector. If it was a zero vector, this couldn't be a basis because it wouldn't be linearly independent because you can always represent a zero vector as really just a zero times any other vector. So this won't be a zero vector. So at least one of these are non-zero. So let's just say, for the sake of argument, that dj, so the coefficient on bj, is non-zero."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "If it was a zero vector, this couldn't be a basis because it wouldn't be linearly independent because you can always represent a zero vector as really just a zero times any other vector. So this won't be a zero vector. So at least one of these are non-zero. So let's just say, for the sake of argument, that dj, so the coefficient on bj, is non-zero. So dj does not equal 0. So what we could actually do is we could solve for that term. So over here someplace you have the term dj, bj, and it's plus a bunch of other stuff."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So let's just say, for the sake of argument, that dj, so the coefficient on bj, is non-zero. So dj does not equal 0. So what we could actually do is we could solve for that term. So over here someplace you have the term dj, bj, and it's plus a bunch of other stuff. We can solve for this term. If we subtract it from both sides of the equation and then divide both sides by minus dj and put this minus a1 on the other side, what do we get? I know that was a lot of operations, but that's just straight up algebra."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So over here someplace you have the term dj, bj, and it's plus a bunch of other stuff. We can solve for this term. If we subtract it from both sides of the equation and then divide both sides by minus dj and put this minus a1 on the other side, what do we get? I know that was a lot of operations, but that's just straight up algebra. I think you can say that we could rewrite this right here. We can solve for bj. We could solve for our bj term and say that is equal to 1."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I know that was a lot of operations, but that's just straight up algebra. I think you can say that we could rewrite this right here. We can solve for bj. We could solve for our bj term and say that is equal to 1. Sorry, that should be equal to minus 1 over its coefficient times, and if we subtract the a1 from both sides, minus a1 and then plus all of these guys, plus d1, b1, plus all the way, you're going to have a little bit of a gap here. I'll just draw it like that. It's very unconventional notation, where this guy sat, all the way to dm, bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "We could solve for our bj term and say that is equal to 1. Sorry, that should be equal to minus 1 over its coefficient times, and if we subtract the a1 from both sides, minus a1 and then plus all of these guys, plus d1, b1, plus all the way, you're going to have a little bit of a gap here. I'll just draw it like that. It's very unconventional notation, where this guy sat, all the way to dm, bm. And I'm doing all of this just to show you that by definition you can write a1 as a linear combination of these other guys, but you can just rearrange things. You can rearrange it so that you can write one of the other guys as a linear combination of the rest of the other guys and a1. And you say, you know what?"}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "It's very unconventional notation, where this guy sat, all the way to dm, bm. And I'm doing all of this just to show you that by definition you can write a1 as a linear combination of these other guys, but you can just rearrange things. You can rearrange it so that you can write one of the other guys as a linear combination of the rest of the other guys and a1. And you say, you know what? This guy is now redundant. I don't need this guy any longer to continue to span v. Clearly this set still spanned v. I mean, I added an extra vector here, but I can remove this guy right here, I can remove him from my set b1 prime and still span v. And how do I know that? Because I can achieve him, by removing him I don't lose anything."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And you say, you know what? This guy is now redundant. I don't need this guy any longer to continue to span v. Clearly this set still spanned v. I mean, I added an extra vector here, but I can remove this guy right here, I can remove him from my set b1 prime and still span v. And how do I know that? Because I can achieve him, by removing him I don't lose anything. Because if I needed this vector to create some other vector, I can construct him with a linear combination of the rest of the b's plus my a1. So let's get rid of him and let's call that set b1. And actually, just for the sake of notation, let me just change his name."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Because I can achieve him, by removing him I don't lose anything. Because if I needed this vector to create some other vector, I can construct him with a linear combination of the rest of the b's plus my a1. So let's get rid of him and let's call that set b1. And actually, just for the sake of notation, let me just change his name. Let me just say, this is a little unconventional. You won't see it done like this in any textbook, but I think it's a little bit easier instead of having to keep talking about these little guys that are embedded someplace in the middle of the stream. I mean, these names, b1, b2, bm, they're arbitrary names."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And actually, just for the sake of notation, let me just change his name. Let me just say, this is a little unconventional. You won't see it done like this in any textbook, but I think it's a little bit easier instead of having to keep talking about these little guys that are embedded someplace in the middle of the stream. I mean, these names, b1, b2, bm, they're arbitrary names. So let me just rename, let me swap the labels. Let me just say that bj is equal to b1 and that b1 is equal to bj. I'm just swapping their names."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I mean, these names, b1, b2, bm, they're arbitrary names. So let me just rename, let me swap the labels. Let me just say that bj is equal to b1 and that b1 is equal to bj. I'm just swapping their names. So I'm essentially just going to remove, I took that guy, I renamed him b1, I renamed b1 bj so that I could swap them. So I'm essentially just going to remove b1 from the vector just to make my notation easier. You could just keep saying, oh, I'm going to remove this bj from the middle, but it becomes very confusing then."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I'm just swapping their names. So I'm essentially just going to remove, I took that guy, I renamed him b1, I renamed b1 bj so that I could swap them. So I'm essentially just going to remove b1 from the vector just to make my notation easier. You could just keep saying, oh, I'm going to remove this bj from the middle, but it becomes very confusing then. So let me call my new set, after removing the bj that I've renamed as b1, let me just call that straight up b1. So my straight up set b1 is equal to a1, and then remember, I removed the bj and I renamed that as b1 and then I renamed b1 as bj. So now my set looks like this."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "You could just keep saying, oh, I'm going to remove this bj from the middle, but it becomes very confusing then. So let me call my new set, after removing the bj that I've renamed as b1, let me just call that straight up b1. So my straight up set b1 is equal to a1, and then remember, I removed the bj and I renamed that as b1 and then I renamed b1 as bj. So now my set looks like this. Let me go to the other color. b2, and for all we know, bj might have been b1. We don't know."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So now my set looks like this. Let me go to the other color. b2, and for all we know, bj might have been b1. We don't know. And it doesn't, I mean, there's probably multiple of these that are non-zero. So we could have picked any of those to be our bj. But anyway, we took our bj, renamed it b1, and removed the b1."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "We don't know. And it doesn't, I mean, there's probably multiple of these that are non-zero. So we could have picked any of those to be our bj. But anyway, we took our bj, renamed it b1, and removed the b1. And so now our set looks like this. b3 all the way to bm. And this will still span, this set still spans v. And we know that because the guy we removed can be constructed with any linear combination of these guys."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "But anyway, we took our bj, renamed it b1, and removed the b1. And so now our set looks like this. b3 all the way to bm. And this will still span, this set still spans v. And we know that because the guy we removed can be constructed with any linear combination of these guys. So we haven't lost our ability to construct all of the vectors in v. Now let me create another vector. Let me create the vector b, let me do it in a new color. Let's say I have the vector b2 prime."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And this will still span, this set still spans v. And we know that because the guy we removed can be constructed with any linear combination of these guys. So we haven't lost our ability to construct all of the vectors in v. Now let me create another vector. Let me create the vector b, let me do it in a new color. Let's say I have the vector b2 prime. And what I'm going to do here is now I'm going to take another element from our spanning, from our basis of v. I'll take the second element. I'll take a2. I'll take a2 and throw it on this guy."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have the vector b2 prime. And what I'm going to do here is now I'm going to take another element from our spanning, from our basis of v. I'll take the second element. I'll take a2. I'll take a2 and throw it on this guy. So now we have the set, let me write it this way, b2 prime is equal to, I'm just going to add a2 to this guy. So you have a1, a2, and then you have all the rest of these guys. b2, b3, all the way to bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I'll take a2 and throw it on this guy. So now we have the set, let me write it this way, b2 prime is equal to, I'm just going to add a2 to this guy. So you have a1, a2, and then you have all the rest of these guys. b2, b3, all the way to bm. And of course, this still spans v. I just added something here. But this is definitely linearly dependent. Remember, I didn't say in the beginning whether this was linear dependent or not."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "b2, b3, all the way to bm. And of course, this still spans v. I just added something here. But this is definitely linearly dependent. Remember, I didn't say in the beginning whether this was linear dependent or not. It may or may not be. But when you add this other vector that's in v, you definitely know that you're linear dependent because these guys can construct that guy. Similarly, we know that this vector, b1, this spans v. So when we add this new element here, we know that it can be written as a linear combination of the other ones."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Remember, I didn't say in the beginning whether this was linear dependent or not. It may or may not be. But when you add this other vector that's in v, you definitely know that you're linear dependent because these guys can construct that guy. Similarly, we know that this vector, b1, this spans v. So when we add this new element here, we know that it can be written as a linear combination of the other ones. So we know that this right here is linearly dependent. And we could say that a2 is equal to some constant times c1 times a1 plus, let me put some constants, so plus d2. Actually, let me just write it this way."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Similarly, we know that this vector, b1, this spans v. So when we add this new element here, we know that it can be written as a linear combination of the other ones. So we know that this right here is linearly dependent. And we could say that a2 is equal to some constant times c1 times a1 plus, let me put some constants, so plus d2. Actually, let me just write it this way. Since plus c2, b2, plus c3, b3, all the way to cm, bm. Now, what I'm going to claim is that at least one of these coefficients is non-zero. So at least one of the ci's does not equal to 0."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me just write it this way. Since plus c2, b2, plus c3, b3, all the way to cm, bm. Now, what I'm going to claim is that at least one of these coefficients is non-zero. So at least one of the ci's does not equal to 0. And I'll make the further claim that there's at least one that's outside of this one. There's at least one that's outside of that one. That it has to be that at least one of the coefficients on these b's, on these b terms, has to be non-zero."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So at least one of the ci's does not equal to 0. And I'll make the further claim that there's at least one that's outside of this one. There's at least one that's outside of that one. That it has to be that at least one of the coefficients on these b's, on these b terms, has to be non-zero. And the way you can kind of think about it is, what if all of these guys were 0? If all of these guys were 0, then that means that a2 is a linear combination of a1, right? All of these guys would cancel out and you'd have a2 is equal to some non-zero constant times a1."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "That it has to be that at least one of the coefficients on these b's, on these b terms, has to be non-zero. And the way you can kind of think about it is, what if all of these guys were 0? If all of these guys were 0, then that means that a2 is a linear combination of a1, right? All of these guys would cancel out and you'd have a2 is equal to some non-zero constant times a1. Well, we know that's not the case because these two guys come from the same linearly independent set. They both come from that spanning basis. The word spanning basis, I shouldn't say it like that because it's redundant."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "All of these guys would cancel out and you'd have a2 is equal to some non-zero constant times a1. Well, we know that's not the case because these two guys come from the same linearly independent set. They both come from that spanning basis. The word spanning basis, I shouldn't say it like that because it's redundant. A basis is a spanning set that is linearly independent. If they're linearly independent, we know that a2 cannot be represented as some linear combination of the rest of these guys. So we know that one of the coefficients on the b terms has to be non-zero."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "The word spanning basis, I shouldn't say it like that because it's redundant. A basis is a spanning set that is linearly independent. If they're linearly independent, we know that a2 cannot be represented as some linear combination of the rest of these guys. So we know that one of the coefficients on the b terms has to be non-zero. And once again, let's just say that the coefficient, someplace here, you're going to have a cj, bj. This is a different one than we had before. And we know that this guy, one of them, at least one of them, has to be non-zero."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So we know that one of the coefficients on the b terms has to be non-zero. And once again, let's just say that the coefficient, someplace here, you're going to have a cj, bj. This is a different one than we had before. And we know that this guy, one of them, at least one of them, has to be non-zero. Because if all of these guys were non-zero, then you wouldn't be able to say that this vector and that vector are linearly independent because they would be scalar multiples of each other. So we're going to do the same exercise. This guy right here, that's someplace cj, bj right here."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And we know that this guy, one of them, at least one of them, has to be non-zero. Because if all of these guys were non-zero, then you wouldn't be able to say that this vector and that vector are linearly independent because they would be scalar multiples of each other. So we're going to do the same exercise. This guy right here, that's someplace cj, bj right here. Obviously, this coefficient is non-zero. So we can solve for our bj. Once again, we can say that bj is equal to minus 1 over cj times, now it's minus a2 plus c1, a1, all the way to cm, bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "This guy right here, that's someplace cj, bj right here. Obviously, this coefficient is non-zero. So we can solve for our bj. Once again, we can say that bj is equal to minus 1 over cj times, now it's minus a2 plus c1, a1, all the way to cm, bm. So we have some bj here that can be represented as a linear combination of the rest of the people, including our new a2. And so, just like we did before, let's remove him. Let's take him out of the set."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Once again, we can say that bj is equal to minus 1 over cj times, now it's minus a2 plus c1, a1, all the way to cm, bm. So we have some bj here that can be represented as a linear combination of the rest of the people, including our new a2. And so, just like we did before, let's remove him. Let's take him out of the set. And before I take him out of the set, I'm going to rename him, just solely for the purpose of notational simplicity, I'm just going to rename our bj b2 and our b2 is equal to bj. So I'm just rearranging the names. And I'm going to remove our b2."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Let's take him out of the set. And before I take him out of the set, I'm going to rename him, just solely for the purpose of notational simplicity, I'm just going to rename our bj b2 and our b2 is equal to bj. So I'm just rearranging the names. And I'm going to remove our b2. Or I'm going to remove what I now call our b2. It was whatever was out here that could be represented as a linear combination of everything else, including our new a2. So let me call that set."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And I'm going to remove our b2. Or I'm going to remove what I now call our b2. It was whatever was out here that could be represented as a linear combination of everything else, including our new a2. So let me call that set. When I remove one of those terms right here, and now I renamed it b2, I call this set b2. And now it's equal to a1, a2. And now I have the leftovers of my b's."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So let me call that set. When I remove one of those terms right here, and now I renamed it b2, I call this set b2. And now it's equal to a1, a2. And now I have the leftovers of my b's. So I have b3, b4, all the way to bm. I still notice, I still have exactly m elements. And this still spans, so this spans v. It spans v because the element that I took out of it can be represented as a linear combination of these guys."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And now I have the leftovers of my b's. So I have b3, b4, all the way to bm. I still notice, I still have exactly m elements. And this still spans, so this spans v. It spans v because the element that I took out of it can be represented as a linear combination of these guys. So if I ever want to construct anything that needed that, I can construct that vector with some combination of these guys. So it wasn't necessary. So it still spans v. So this process I'm doing, I can just keep repeating it."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And this still spans, so this spans v. It spans v because the element that I took out of it can be represented as a linear combination of these guys. So if I ever want to construct anything that needed that, I can construct that vector with some combination of these guys. So it wasn't necessary. So it still spans v. So this process I'm doing, I can just keep repeating it. I can add an a3. I can define b3 prime. I can just add a3 to this set right here, a2, a3."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So it still spans v. So this process I'm doing, I can just keep repeating it. I can add an a3. I can define b3 prime. I can just add a3 to this set right here, a2, a3. And then I have my b3, b4, all the way to bm. And I'll say, oh, this is linearly dependent because this guy spans v. So everything but this guy spans v. So obviously, you can construct this guy with a linear combination of the rest of them. So you could say a3 is equal to sum a1 plus sum a2 plus c3, b3, all the way to cm, bm."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I can just add a3 to this set right here, a2, a3. And then I have my b3, b4, all the way to bm. And I'll say, oh, this is linearly dependent because this guy spans v. So everything but this guy spans v. So obviously, you can construct this guy with a linear combination of the rest of them. So you could say a3 is equal to sum a1 plus sum a2 plus c3, b3, all the way to cm, bm. And we know that at least one of the coefficients on the b terms has to be non-zero. Because if all of those were zero, then you would be saying that a3 could be a linear combination of the a terms. And we know that a3 can't be represented as a linear combination of the a terms because all these a terms come from a linearly independent set."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So you could say a3 is equal to sum a1 plus sum a2 plus c3, b3, all the way to cm, bm. And we know that at least one of the coefficients on the b terms has to be non-zero. Because if all of those were zero, then you would be saying that a3 could be a linear combination of the a terms. And we know that a3 can't be represented as a linear combination of the a terms because all these a terms come from a linearly independent set. So you do the same operation. You find you can solve, let's say that cj, some term right here, the cj is non-zero. Then you can solve for that bj."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And we know that a3 can't be represented as a linear combination of the a terms because all these a terms come from a linearly independent set. So you do the same operation. You find you can solve, let's say that cj, some term right here, the cj is non-zero. Then you can solve for that bj. And then I do that little renaming thing I do where I rename the bj b3 and rename b3 bj. And then I remove b3. And I get the set b3 is equal to a1, a2, a3."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Then you can solve for that bj. And then I do that little renaming thing I do where I rename the bj b3 and rename b3 bj. And then I remove b3. And I get the set b3 is equal to a1, a2, a3. And then I have b4 all the way to bm. And this still spans v. And I keep doing it. So what's going to happen eventually?"}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And I get the set b3 is equal to a1, a2, a3. And then I have b4 all the way to bm. And this still spans v. And I keep doing it. So what's going to happen eventually? If I keep doing this process over and over and over again, eventually I'll essentially replace all of the bm's. Or I'll replace all of the n terms. So eventually I'll have a set that looks like this."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So what's going to happen eventually? If I keep doing this process over and over and over again, eventually I'll essentially replace all of the bm's. Or I'll replace all of the n terms. So eventually I'll have a set that looks like this. I'll have a set that looks bm. Well, I will have replaced each of these guys with an a. Or I would have replaced each of these guys with an a."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So eventually I'll have a set that looks like this. I'll have a set that looks bm. Well, I will have replaced each of these guys with an a. Or I would have replaced each of these guys with an a. So I'll have a1, a2, a3, all the way to am. You can always do this by definition if you started with that initial set b that is a spanning set. And once you do this process, you'll get the same result that this also spans v. Now let me write this."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "Or I would have replaced each of these guys with an a. So I'll have a1, a2, a3, all the way to am. You can always do this by definition if you started with that initial set b that is a spanning set. And once you do this process, you'll get the same result that this also spans v. Now let me write this. This is the result we got by starting off with a spanning set b that has m elements where we said that m is less than n. So we always have enough a elements to do this because we have more a elements than there were b elements to begin with. And we get a result that this spans v. But we already said that the set a, which is equal to a1, a2, all the way to am, and then am plus 1. I don't know how many more terms there are between m and n, but then you go all the way to an."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And once you do this process, you'll get the same result that this also spans v. Now let me write this. This is the result we got by starting off with a spanning set b that has m elements where we said that m is less than n. So we always have enough a elements to do this because we have more a elements than there were b elements to begin with. And we get a result that this spans v. But we already said that the set a, which is equal to a1, a2, all the way to am, and then am plus 1. I don't know how many more terms there are between m and n, but then you go all the way to an. Remember, we said that n is greater than m. Or when we defined b, we said that m is less than n. Same thing, that this was a smaller set. Now we're saying that this spans v, but at the same time we said that this was a basis. This was just our starting fact, that this is a basis for v. Basis means two things."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "I don't know how many more terms there are between m and n, but then you go all the way to an. Remember, we said that n is greater than m. Or when we defined b, we said that m is less than n. Same thing, that this was a smaller set. Now we're saying that this spans v, but at the same time we said that this was a basis. This was just our starting fact, that this is a basis for v. Basis means two things. It means it spans v, and it means it's linearly independent. Now, we just got this result by assuming that we had some set b that's smaller than this set here that spans v. We were able to construct this by saying that a1 through am also spans v. So the result we got is that this spans v. But if this subset of a spans v, then a becomes linearly independent, because if this subset spans v, that means that an can be represented as some linear combination of these guys, which would imply that you're linearly dependent, which is a contradiction with our original statement that set A is a basis for v, because that means that it's linearly independent. If you're able to do this, then this means that this is, if there is some smaller spanning set, you get the result that A has to be linearly dependent, even though we said it was linearly independent."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "This was just our starting fact, that this is a basis for v. Basis means two things. It means it spans v, and it means it's linearly independent. Now, we just got this result by assuming that we had some set b that's smaller than this set here that spans v. We were able to construct this by saying that a1 through am also spans v. So the result we got is that this spans v. But if this subset of a spans v, then a becomes linearly independent, because if this subset spans v, that means that an can be represented as some linear combination of these guys, which would imply that you're linearly dependent, which is a contradiction with our original statement that set A is a basis for v, because that means that it's linearly independent. If you're able to do this, then this means that this is, if there is some smaller spanning set, you get the result that A has to be linearly dependent, even though we said it was linearly independent. So we now know, we get our contradiction. We say that there cannot be. So this leads us to there cannot be a set, let me write this, a spanning set B that has fewer elements than A."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "If you're able to do this, then this means that this is, if there is some smaller spanning set, you get the result that A has to be linearly dependent, even though we said it was linearly independent. So we now know, we get our contradiction. We say that there cannot be. So this leads us to there cannot be a set, let me write this, a spanning set B that has fewer elements than A. And this is a pretty neat outcome, because now if anyone tells you that, if I come up to you and I say, hey, I found some set x, so set x spans the subspace, I don't know, let's just call that v again, then you know that x has 5 elements, you now know that no set that spans the subspace v can have fewer than 5 elements. Even better, if I told you that x is a basis for v, and it has 5 elements, and y is a basis for v, y is another basis, my handwriting's degrading, y is also a basis for v, you know that y also has to have exactly 5 elements. y has 5 elements."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So this leads us to there cannot be a set, let me write this, a spanning set B that has fewer elements than A. And this is a pretty neat outcome, because now if anyone tells you that, if I come up to you and I say, hey, I found some set x, so set x spans the subspace, I don't know, let's just call that v again, then you know that x has 5 elements, you now know that no set that spans the subspace v can have fewer than 5 elements. Even better, if I told you that x is a basis for v, and it has 5 elements, and y is a basis for v, y is another basis, my handwriting's degrading, y is also a basis for v, you know that y also has to have exactly 5 elements. y has 5 elements. How do I know that? Well, if y is a basis, then that means that it spans v. And we know it can't have anything less than 5 elements, we just proved that. So one way we know that y has to have greater than or equal to 5 elements."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "y has 5 elements. How do I know that? Well, if y is a basis, then that means that it spans v. And we know it can't have anything less than 5 elements, we just proved that. So one way we know that y has to have greater than or equal to 5 elements. But on the other hand, we know that if y is a basis for v, and x is a basis, x also spans v, so we know that x has to have fewer elements than y. So we know that y has to have greater than x, or let me just call it y's elements, have to be greater than x's elements, because any spanning set has to have more elements, or at least as many elements, as a basis set, x's elements. And then since x is a spanning set, x's elements have to be greater than or equal to y's elements, because y is a basis."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "So one way we know that y has to have greater than or equal to 5 elements. But on the other hand, we know that if y is a basis for v, and x is a basis, x also spans v, so we know that x has to have fewer elements than y. So we know that y has to have greater than x, or let me just call it y's elements, have to be greater than x's elements, because any spanning set has to have more elements, or at least as many elements, as a basis set, x's elements. And then since x is a spanning set, x's elements have to be greater than or equal to y's elements, because y is a basis. I know you can't read that, but if this guy's elements are less than this guy's elements, but it's also greater than or equal to, we know that the number of elements that x has, so x's elements, or the cardinality, or the number of elements in it, is equal to y's elements. And so now that we know that any basis for a vector space, so let's say that x, though let me just go back to our set A, A is equal to A1, A2, all the way to An, we can now say that any basis for some vector, for some subspace V, they all have the same number of elements. And so we can define a new term called the dimension of V, sometimes it's written just dimension of V, is equal to the number of elements, sometimes called the cardinality, of any basis of V. And I went through great brains in this video to show that any basis of V all has the same number of elements, so this is well defined."}, {"video_title": "Proof Any subspace basis has same number of elements Linear Algebra Khan Academy.mp3", "Sentence": "And then since x is a spanning set, x's elements have to be greater than or equal to y's elements, because y is a basis. I know you can't read that, but if this guy's elements are less than this guy's elements, but it's also greater than or equal to, we know that the number of elements that x has, so x's elements, or the cardinality, or the number of elements in it, is equal to y's elements. And so now that we know that any basis for a vector space, so let's say that x, though let me just go back to our set A, A is equal to A1, A2, all the way to An, we can now say that any basis for some vector, for some subspace V, they all have the same number of elements. And so we can define a new term called the dimension of V, sometimes it's written just dimension of V, is equal to the number of elements, sometimes called the cardinality, of any basis of V. And I went through great brains in this video to show that any basis of V all has the same number of elements, so this is well defined. You can't have one basis that has five elements and one that has six. By definition, they would either both have to have five or they would both have to have six. And so we can define our dimensionality."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw my axes. Those are my axes right there. Not perfectly drawn, but I think you get the idea. Let me draw a line that goes through the origin here. So that is my line there. And we know that a line in any Rn, we're doing it in R2, can be defined as just all of the possible scalar multiples of some vector. So let's say that this is some vector right here that's on the line."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw a line that goes through the origin here. So that is my line there. And we know that a line in any Rn, we're doing it in R2, can be defined as just all of the possible scalar multiples of some vector. So let's say that this is some vector right here that's on the line. We can define our line. We could say L is equal to the set of all the scalar multiples. Let me say that this is vector."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that this is some vector right here that's on the line. We can define our line. We could say L is equal to the set of all the scalar multiples. Let me say that this is vector. Let's say that that is v right there. So it's all the possible scalar multiples of our vector v, where the scalar multiples, by definition, they're just any real number. So obviously, if you take all of the possible multiples of v, both positive multiples and negative multiples and less than 1 multiples, fraction multiples, you'll have a set of vectors that will essentially define, or I guess specify, every point on that line that goes through the origin."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me say that this is vector. Let's say that that is v right there. So it's all the possible scalar multiples of our vector v, where the scalar multiples, by definition, they're just any real number. So obviously, if you take all of the possible multiples of v, both positive multiples and negative multiples and less than 1 multiples, fraction multiples, you'll have a set of vectors that will essentially define, or I guess specify, every point on that line that goes through the origin. And we know, of course, if this wasn't a line that went through the origin, you would have to shift it by some vector, it would have to be some other vector plus cv. But anyway, we're starting off with this line definition that goes through the origin. What I want to do in this video is to define the idea of a projection onto L of some other vector x."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So obviously, if you take all of the possible multiples of v, both positive multiples and negative multiples and less than 1 multiples, fraction multiples, you'll have a set of vectors that will essentially define, or I guess specify, every point on that line that goes through the origin. And we know, of course, if this wasn't a line that went through the origin, you would have to shift it by some vector, it would have to be some other vector plus cv. But anyway, we're starting off with this line definition that goes through the origin. What I want to do in this video is to define the idea of a projection onto L of some other vector x. So let me draw my some other vector x. Let's say that this right here is my other vector x. Now, a projection, I'm going to give you just a sense of it, and then we'll define it a little bit more precisely."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What I want to do in this video is to define the idea of a projection onto L of some other vector x. So let me draw my some other vector x. Let's say that this right here is my other vector x. Now, a projection, I'm going to give you just a sense of it, and then we'll define it a little bit more precisely. A projection I always imagine is if you had some light source that were perpendicular somehow, or orthogonal to our line, so let's say our light source, the light was shining down like this, and I'm doing that direction because that is perpendicular to my line. I imagine the projection of x onto this line is kind of the shadow of x. So if this light was coming down, I would just draw a perpendicular like that, and the shadow of x onto L would be that vector right there."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, a projection, I'm going to give you just a sense of it, and then we'll define it a little bit more precisely. A projection I always imagine is if you had some light source that were perpendicular somehow, or orthogonal to our line, so let's say our light source, the light was shining down like this, and I'm doing that direction because that is perpendicular to my line. I imagine the projection of x onto this line is kind of the shadow of x. So if this light was coming down, I would just draw a perpendicular like that, and the shadow of x onto L would be that vector right there. So we can kind of view it as the shadow of x on our line L. That's one way to think of it. Another way to think of it, and you can think of it however you like, is how much of x goes in the L direction? But the technique would be the same."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this light was coming down, I would just draw a perpendicular like that, and the shadow of x onto L would be that vector right there. So we can kind of view it as the shadow of x on our line L. That's one way to think of it. Another way to think of it, and you can think of it however you like, is how much of x goes in the L direction? But the technique would be the same. You draw a perpendicular from x to L, and you say, OK, then how much of L do I have to go in that direction to get to my perpendicular? Either of those are how I think of the idea of a projection. I think the shadow is part of the motivation for why it's even called a projection."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But the technique would be the same. You draw a perpendicular from x to L, and you say, OK, then how much of L do I have to go in that direction to get to my perpendicular? Either of those are how I think of the idea of a projection. I think the shadow is part of the motivation for why it's even called a projection. When you project something, you're kind of beaming light and seeing where the light hits on a wall, and you're kind of doing that here. You're beaming light, and you're seeing where that light hits on a line in this case. But you can't do anything with this definition."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think the shadow is part of the motivation for why it's even called a projection. When you project something, you're kind of beaming light and seeing where the light hits on a wall, and you're kind of doing that here. You're beaming light, and you're seeing where that light hits on a line in this case. But you can't do anything with this definition. This is just kind of an intuitive sense of what a projection is. So we need to figure out some way to calculate this, or a more mathematically precise definition. And one thing we can do is, when I created this projection, let me actually draw another projection of another line, or another vector, just so you get the idea, if I had some other vector over here that looked like that, the projection of this onto the line would look something like this."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But you can't do anything with this definition. This is just kind of an intuitive sense of what a projection is. So we need to figure out some way to calculate this, or a more mathematically precise definition. And one thing we can do is, when I created this projection, let me actually draw another projection of another line, or another vector, just so you get the idea, if I had some other vector over here that looked like that, the projection of this onto the line would look something like this. You just draw a perpendicular, and its projection would be like that. So I don't want to draw it for just this case. I want to give you the sense that it's the shadow of any vector onto this line."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And one thing we can do is, when I created this projection, let me actually draw another projection of another line, or another vector, just so you get the idea, if I had some other vector over here that looked like that, the projection of this onto the line would look something like this. You just draw a perpendicular, and its projection would be like that. So I don't want to draw it for just this case. I want to give you the sense that it's the shadow of any vector onto this line. So how can we think about it with our original example? Well, in every case, no matter how I perceive it, I dropped a perpendicular down here. And so if we can construct a vector right here, we could say, hey, that vector is always going to be perpendicular to the line."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I want to give you the sense that it's the shadow of any vector onto this line. So how can we think about it with our original example? Well, in every case, no matter how I perceive it, I dropped a perpendicular down here. And so if we can construct a vector right here, we could say, hey, that vector is always going to be perpendicular to the line. And we can do that. I wouldn't have been talking about it if we couldn't. So let me define this vector."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so if we can construct a vector right here, we could say, hey, that vector is always going to be perpendicular to the line. And we can do that. I wouldn't have been talking about it if we couldn't. So let me define this vector. Actually, I'm not even going to define it. What is this vector going to be? If this vector, let me not use all these this's, let me say that this vector, we know we want to somehow get to this blue vector."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me define this vector. Actually, I'm not even going to define it. What is this vector going to be? If this vector, let me not use all these this's, let me say that this vector, we know we want to somehow get to this blue vector. Let me keep it in blue. That blue vector is the projection of x onto l. That's what we want to get to. Now, one thing we can look at is this pink vector right there."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If this vector, let me not use all these this's, let me say that this vector, we know we want to somehow get to this blue vector. Let me keep it in blue. That blue vector is the projection of x onto l. That's what we want to get to. Now, one thing we can look at is this pink vector right there. What is that pink vector? That pink vector that I just drew, this is x. That's the vector x minus the projection, minus this blue vector over here, minus the projection of x onto l. If you add the projection to the pink vector, you get x."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, one thing we can look at is this pink vector right there. What is that pink vector? That pink vector that I just drew, this is x. That's the vector x minus the projection, minus this blue vector over here, minus the projection of x onto l. If you add the projection to the pink vector, you get x. If you add this blue projection of x to x minus the projection of x, you're of course going to get x. And we also know that this pink vector is orthogonal to the line itself, which means it's orthogonal to every vector on the line, which also means that its dot product is going to be 0. So let me define the projection this way."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the vector x minus the projection, minus this blue vector over here, minus the projection of x onto l. If you add the projection to the pink vector, you get x. If you add this blue projection of x to x minus the projection of x, you're of course going to get x. And we also know that this pink vector is orthogonal to the line itself, which means it's orthogonal to every vector on the line, which also means that its dot product is going to be 0. So let me define the projection this way. This is going to be my slightly more mathematical definition. The projection onto l of some vector x is going to be some vector that's in l. I drew it right here, this blue vector. I'll trace it with white right here."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me define the projection this way. This is going to be my slightly more mathematical definition. The projection onto l of some vector x is going to be some vector that's in l. I drew it right here, this blue vector. I'll trace it with white right here. Some vector in l where, and this might be a little bit unintuitive, where x minus the projection onto l of x is orthogonal to my line. So I'm saying the projection, this is my definition. I'm defining the projection of x onto l to some vector in l where x minus that projection is orthogonal to l. This is my definition."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll trace it with white right here. Some vector in l where, and this might be a little bit unintuitive, where x minus the projection onto l of x is orthogonal to my line. So I'm saying the projection, this is my definition. I'm defining the projection of x onto l to some vector in l where x minus that projection is orthogonal to l. This is my definition. That is a little bit more precise, and I think it makes a bit of sense why it connects to the idea of the shadow or projection. But how can we deal with this? I mean, this is still just in words."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm defining the projection of x onto l to some vector in l where x minus that projection is orthogonal to l. This is my definition. That is a little bit more precise, and I think it makes a bit of sense why it connects to the idea of the shadow or projection. But how can we deal with this? I mean, this is still just in words. How can I actually calculate the projection of x onto l? Well, the key clue here is this notion that x minus the projection of x is orthogonal to l. So let's see if we can use that somehow. So the first thing we need to realize is by definition, because the projection of x onto l is some vector in l, that means it's some scalar multiple of v, some scalar multiple of our defining vector, of our v right there."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I mean, this is still just in words. How can I actually calculate the projection of x onto l? Well, the key clue here is this notion that x minus the projection of x is orthogonal to l. So let's see if we can use that somehow. So the first thing we need to realize is by definition, because the projection of x onto l is some vector in l, that means it's some scalar multiple of v, some scalar multiple of our defining vector, of our v right there. So we could also say, hey, look, we could rewrite our projection of x onto l. We could write it as some scalar multiple times our vector v. We can say that this is equivalent to our projection. Now we also know that x minus our projection is orthogonal to l. So we also know that x minus our projection, and I just said that I could rewrite my projection as some multiple of this vector right there. You could see it the way I drew it here."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing we need to realize is by definition, because the projection of x onto l is some vector in l, that means it's some scalar multiple of v, some scalar multiple of our defining vector, of our v right there. So we could also say, hey, look, we could rewrite our projection of x onto l. We could write it as some scalar multiple times our vector v. We can say that this is equivalent to our projection. Now we also know that x minus our projection is orthogonal to l. So we also know that x minus our projection, and I just said that I could rewrite my projection as some multiple of this vector right there. You could see it the way I drew it here. It almost looks like it's 2 times this vector. So we know that x minus our projection, this is our projection right here, is orthogonal to l. Orthogonality, by definition, means its dot product with any vector in l is 0. So let's dot it with some vector in l. Well, we could dot it with this vector v. That's what we used to define l. So let's dot it with v. And we know that that must be equal to 0."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could see it the way I drew it here. It almost looks like it's 2 times this vector. So we know that x minus our projection, this is our projection right here, is orthogonal to l. Orthogonality, by definition, means its dot product with any vector in l is 0. So let's dot it with some vector in l. Well, we could dot it with this vector v. That's what we used to define l. So let's dot it with v. And we know that that must be equal to 0. We're taking this vector right here, dotting it with v, and we know that this has to be equal to 0. That has to be equal to 0. So let's use our properties of dot products to see if we can calculate a particular value of c. Because once we know a particular value of c, then we can just always multiply that times the vector v, which we are given, and we will have our projection."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's dot it with some vector in l. Well, we could dot it with this vector v. That's what we used to define l. So let's dot it with v. And we know that that must be equal to 0. We're taking this vector right here, dotting it with v, and we know that this has to be equal to 0. That has to be equal to 0. So let's use our properties of dot products to see if we can calculate a particular value of c. Because once we know a particular value of c, then we can just always multiply that times the vector v, which we are given, and we will have our projection. And then I'll show it to you with some actual numbers. So let's see if we can calculate a c. So if we distribute the c, or sorry, if we distribute the v, we know the dot product exhibits the distributive property. This expression can be rewritten as x dot v minus c times v dot v. I rearranged things."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's use our properties of dot products to see if we can calculate a particular value of c. Because once we know a particular value of c, then we can just always multiply that times the vector v, which we are given, and we will have our projection. And then I'll show it to you with some actual numbers. So let's see if we can calculate a c. So if we distribute the c, or sorry, if we distribute the v, we know the dot product exhibits the distributive property. This expression can be rewritten as x dot v minus c times v dot v. I rearranged things. We know that the scalar, we know that c minus cv dot v is the same thing. We could write it as minus c times v dot v. And all of this, of course, is equal to 0. And then if we want to solve for c, let's add cv dot v to both sides of the equation."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This expression can be rewritten as x dot v minus c times v dot v. I rearranged things. We know that the scalar, we know that c minus cv dot v is the same thing. We could write it as minus c times v dot v. And all of this, of course, is equal to 0. And then if we want to solve for c, let's add cv dot v to both sides of the equation. And you get x dot v is equal to c times v dot v. Solving for c, let's divide both sides of this equation by v dot v. You get, there's a different color. c is equal to this, x dot v divided by v dot v. Now what was c? c was, we're saying the projection of x, let me write it here, the projection of x onto l is equal to some scalar multiple, right?"}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then if we want to solve for c, let's add cv dot v to both sides of the equation. And you get x dot v is equal to c times v dot v. Solving for c, let's divide both sides of this equation by v dot v. You get, there's a different color. c is equal to this, x dot v divided by v dot v. Now what was c? c was, we're saying the projection of x, let me write it here, the projection of x onto l is equal to some scalar multiple, right? We know it's in the line. So it's some scalar multiple of this defining vector. Some scalar multiple of the vector v. And we just figured out what that scalar multiple is going to be."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "c was, we're saying the projection of x, let me write it here, the projection of x onto l is equal to some scalar multiple, right? We know it's in the line. So it's some scalar multiple of this defining vector. Some scalar multiple of the vector v. And we just figured out what that scalar multiple is going to be. It's going to be x dot v over v dot v. And this, of course, is just going to be a number, right? This is a scalar still. Even though we have all these vectors here, when you take their dot products, you just end up with a number."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Some scalar multiple of the vector v. And we just figured out what that scalar multiple is going to be. It's going to be x dot v over v dot v. And this, of course, is just going to be a number, right? This is a scalar still. Even though we have all these vectors here, when you take their dot products, you just end up with a number. And you multiply that number times v. You just kind of scale v, and you get your projection. So in this case, the way I drew it up here, my dot product should end up with some scaling factor that's maybe close to 2. So that if I start with a v and I scale it up by 2, this value would be 2, and I'd get a projection that looks something like that."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Even though we have all these vectors here, when you take their dot products, you just end up with a number. And you multiply that number times v. You just kind of scale v, and you get your projection. So in this case, the way I drew it up here, my dot product should end up with some scaling factor that's maybe close to 2. So that if I start with a v and I scale it up by 2, this value would be 2, and I'd get a projection that looks something like that. Now this looks a little abstract to you, so let's do it with some real vectors. And I think it'll make a little bit more sense. And nothing I did here only applies to R2."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that if I start with a v and I scale it up by 2, this value would be 2, and I'd get a projection that looks something like that. Now this looks a little abstract to you, so let's do it with some real vectors. And I think it'll make a little bit more sense. And nothing I did here only applies to R2. Everything I did here can be extended to an arbitrarily high dimension. So even though we're doing it in R2, and R2 and R3 is where we tend to deal with projections the most, this could apply to Rn. So let me do it in this particular case."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And nothing I did here only applies to R2. Everything I did here can be extended to an arbitrarily high dimension. So even though we're doing it in R2, and R2 and R3 is where we tend to deal with projections the most, this could apply to Rn. So let me do it in this particular case. Let me define my line L to be the set of all scalar multiples of the vector, I don't know, let's say the vector 2, 1, such that all of them are, the c is any real number. So let me draw my axes here. That's my vertical axes."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me do it in this particular case. Let me define my line L to be the set of all scalar multiples of the vector, I don't know, let's say the vector 2, 1, such that all of them are, the c is any real number. So let me draw my axes here. That's my vertical axes. This is my horizontal axis right there. And so my line is all the scalar multiples of the vector 2.1. And actually, let me just call my vector 2.1, let me call that right there, the vector v. So let me draw that."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's my vertical axes. This is my horizontal axis right there. And so my line is all the scalar multiples of the vector 2.1. And actually, let me just call my vector 2.1, let me call that right there, the vector v. So let me draw that. So I go 1, 2, go up 1. That right there is my vector v. Vector v right there. And the line is all of the possible scalar multiples of that, so let me draw that."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And actually, let me just call my vector 2.1, let me call that right there, the vector v. So let me draw that. So I go 1, 2, go up 1. That right there is my vector v. Vector v right there. And the line is all of the possible scalar multiples of that, so let me draw that. So all the possible scalar multiples of that, you just keep going in that direction, or you keep going backwards in that direction, or anything in between. That's what my line is. If we just, all of the scalar multiples of my vector v. Now let's say I have another vector x."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the line is all of the possible scalar multiples of that, so let me draw that. So all the possible scalar multiples of that, you just keep going in that direction, or you keep going backwards in that direction, or anything in between. That's what my line is. If we just, all of the scalar multiples of my vector v. Now let's say I have another vector x. Let's say I have a vector x, and let's say that x is equal to 2, 3. So let me draw x. x is 2, and then you go 1, 2, 3. So x will look like this."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we just, all of the scalar multiples of my vector v. Now let's say I have another vector x. Let's say I have a vector x, and let's say that x is equal to 2, 3. So let me draw x. x is 2, and then you go 1, 2, 3. So x will look like this. Vector x will look like that. Let me draw a little bit better than that. Vector x will look like that."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So x will look like this. Vector x will look like that. Let me draw a little bit better than that. Vector x will look like that. That is vector x. What we want to do is figure out the projection of x onto l. We can use this definition right here. So let me write it down."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Vector x will look like that. That is vector x. What we want to do is figure out the projection of x onto l. We can use this definition right here. So let me write it down. The projection of x onto l is equal to what? It's equal to x dot v, where v is kind of the defining vector for our line. So it's equal to x, which is 2, 3, dot v, which is 2, 1, all of that over v dot v. So all of that over 2, 1, dot 2, 1, times our original defining vector v. So what's our original defining vector?"}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me write it down. The projection of x onto l is equal to what? It's equal to x dot v, where v is kind of the defining vector for our line. So it's equal to x, which is 2, 3, dot v, which is 2, 1, all of that over v dot v. So all of that over 2, 1, dot 2, 1, times our original defining vector v. So what's our original defining vector? It's this one right here, 2, 1. So times the vector 2, 1. And what does this equal?"}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to x, which is 2, 3, dot v, which is 2, 1, all of that over v dot v. So all of that over 2, 1, dot 2, 1, times our original defining vector v. So what's our original defining vector? It's this one right here, 2, 1. So times the vector 2, 1. And what does this equal? See, when you take these two dotted each other, you have 2 times 2 plus 3 times 1. So 4 plus 3, so you get 7. This all simplified to 7."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what does this equal? See, when you take these two dotted each other, you have 2 times 2 plus 3 times 1. So 4 plus 3, so you get 7. This all simplified to 7. And then this, you get 2 times 2 plus 1 times 1. So 4 plus 1 is 5. So you get 7 fifths."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This all simplified to 7. And then this, you get 2 times 2 plus 1 times 1. So 4 plus 1 is 5. So you get 7 fifths. That all simplified to 5. That was a very fast simplification. You might have been daunted by this kind of strange looking expression, but when you take dot products, they actually tend to simplify quite very quickly."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you get 7 fifths. That all simplified to 5. That was a very fast simplification. You might have been daunted by this kind of strange looking expression, but when you take dot products, they actually tend to simplify quite very quickly. And then you just multiply that times your defining vector for the line. So we're scaling it up by a factor of 7 fifths. So multiply it times the vector 2, 1."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You might have been daunted by this kind of strange looking expression, but when you take dot products, they actually tend to simplify quite very quickly. And then you just multiply that times your defining vector for the line. So we're scaling it up by a factor of 7 fifths. So multiply it times the vector 2, 1. And what do you get? You get the vector, we do it in a new color. You get the vector 14 over 5 and the vector 7 over 5."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So multiply it times the vector 2, 1. And what do you get? You get the vector, we do it in a new color. You get the vector 14 over 5 and the vector 7 over 5. And just so we can visualize this or plot it a little better, let me write it as decimals. 14 over 5 is 2 and 4 fifths, which is 2.8. And this is 1 and 2 fifths, which is 1.4."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get the vector 14 over 5 and the vector 7 over 5. And just so we can visualize this or plot it a little better, let me write it as decimals. 14 over 5 is 2 and 4 fifths, which is 2.8. And this is 1 and 2 fifths, which is 1.4. And so the projection of x onto L is 2.8, 1.4. So let's see, 1 to 2.8 is right about there. And I go 1.4 is right about there."}, {"video_title": "Introduction to projections Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is 1 and 2 fifths, which is 1.4. And so the projection of x onto L is 2.8, 1.4. So let's see, 1 to 2.8 is right about there. And I go 1.4 is right about there. So the vector is going to be right about there. I haven't even drawn this too precisely, but you get the idea. This is the projection."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "A couple of videos ago, we learned that a function that is a mapping from the set X to the set Y is invertible if and only if, and I'll write that as an if with two F's, if, if and only if for every Y, so let me write this down, for every, I'll do this in yellow, for every Y that is a member of our codomain, there exists a unique, and I'll make that a little bit bold, a unique X that is a member of our domain such that F of this X is equal to this Y. So that's just saying that if I take my domain right here, that's X, and then I take a codomain here, that is Y, we say that the function F is invertible, we say that the function F is invertible, and we know what invertibility means, it means that there's this other function called the inverse that can essentially, if you apply, if you take that in composition with F, it's like taking the identity on X, or if you take F in composition with it, it's like taking the identity on Y, and we've done that multiple times, so I won't repeat that there. We know what invertibility means, but we learn that it's invertible if and only if for every Y here, so you take any Y here, any Y that's a member of your codomain, there exists a unique X such that F of X is equal to that Y. F of X is equal to, let me write it this way, if this is an X, let's say that's an X naught, F of X naught would be equal to Y. So this Y would be equal to F of X naught. You apply the function here, it's going to map it to this point here. It wouldn't be invertible if you had this, if you had two members of X mapping here. That would break invertibility if you had this situation, because then you wouldn't have the unique condition."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "So this Y would be equal to F of X naught. You apply the function here, it's going to map it to this point here. It wouldn't be invertible if you had this, if you had two members of X mapping here. That would break invertibility if you had this situation, because then you wouldn't have the unique condition. You have to have a unique X that maps to this thing, and what I just drew here with this other pink mapping, we don't have a unique X that maps to Y, we have two X's that map to Y. Now based on what I just told you on that last video, what does this mean? If we have a unique X that maps to each Y, that means that we have to have a one-to-one mapping, that F has to be one-to-one."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "That would break invertibility if you had this situation, because then you wouldn't have the unique condition. You have to have a unique X that maps to this thing, and what I just drew here with this other pink mapping, we don't have a unique X that maps to Y, we have two X's that map to Y. Now based on what I just told you on that last video, what does this mean? If we have a unique X that maps to each Y, that means that we have to have a one-to-one mapping, that F has to be one-to-one. So let me write that. So another way of saying this is that F is one-to-one, or injective. So if we have two guys mapping to the same Y, that would break down this condition."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "If we have a unique X that maps to each Y, that means that we have to have a one-to-one mapping, that F has to be one-to-one. So let me write that. So another way of saying this is that F is one-to-one, or injective. So if we have two guys mapping to the same Y, that would break down this condition. We wouldn't be one-to-one, and we couldn't say that there exists a unique X solution to this equation right here. Now the other part of this is that for every Y, you could pick any Y here, and there exists a unique X that maps to that. So I don't care, there cannot be some Y here, let's say that there's some Y here, and no one maps to that."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "So if we have two guys mapping to the same Y, that would break down this condition. We wouldn't be one-to-one, and we couldn't say that there exists a unique X solution to this equation right here. Now the other part of this is that for every Y, you could pick any Y here, and there exists a unique X that maps to that. So I don't care, there cannot be some Y here, let's say that there's some Y here, and no one maps to that. No one maps. If that's the case, then we don't have our conditions for invertibility. So that would be not invertible."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "So I don't care, there cannot be some Y here, let's say that there's some Y here, and no one maps to that. No one maps. If that's the case, then we don't have our conditions for invertibility. So that would be not invertible. So everything in Y, every element of Y has to be mapped to, all of these guys have to be mapped to, and they can only be mapped to, by one of the elements of X. So everything here has to be mapped to by a unique guy. Now, in the last video, what did we call it when a function maps to every element of your codomain?"}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "So that would be not invertible. So everything in Y, every element of Y has to be mapped to, all of these guys have to be mapped to, and they can only be mapped to, by one of the elements of X. So everything here has to be mapped to by a unique guy. Now, in the last video, what did we call it when a function maps to every element of your codomain? So this every here, what is another way of saying that? That a function maps to every element of your codomain? Well in the last video, I explained that that notion is called a surjective."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "Now, in the last video, what did we call it when a function maps to every element of your codomain? So this every here, what is another way of saying that? That a function maps to every element of your codomain? Well in the last video, I explained that that notion is called a surjective. Surjective, or onto function. So the whole reason why I'm doing this video is because I really just want to restate this condition for invertibility in using the vocabulary that I introduced to you in the last video. So given that the statement for every Y that's a member of our codomain, there exists a X that maps to it, we could just say that F is surjective."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "Well in the last video, I explained that that notion is called a surjective. Surjective, or onto function. So the whole reason why I'm doing this video is because I really just want to restate this condition for invertibility in using the vocabulary that I introduced to you in the last video. So given that the statement for every Y that's a member of our codomain, there exists a X that maps to it, we could just say that F is surjective. And then the fact is, if we just said that F is surjective, that means that everything here is mapped to, but it could be mapped to, maybe this person right here could be mapped to by more than one. Surjective doesn't by itself make the condition that there's a unique mapping from a member of X to that element of Y. So in order to get that, in order to satisfy the unique condition of this condition for invertibility, we have to say that F is also injective."}, {"video_title": "Relating invertibility to being onto and one-to-one Linear Algebra Khan Academy.mp3", "Sentence": "So given that the statement for every Y that's a member of our codomain, there exists a X that maps to it, we could just say that F is surjective. And then the fact is, if we just said that F is surjective, that means that everything here is mapped to, but it could be mapped to, maybe this person right here could be mapped to by more than one. Surjective doesn't by itself make the condition that there's a unique mapping from a member of X to that element of Y. So in order to get that, in order to satisfy the unique condition of this condition for invertibility, we have to say that F is also injective. And obviously maybe the less formal terms for each of these, you could call this onto, and you could call this one to one. So using the terminology that we learned in the last video, we can restate this condition for invertibility. We can say that a function that is a mapping from the domain X to the codomain Y is invertible if, and only if, I'll write it out, and only if, F is both surjective and injective, or we could have said that F is invertible if and only if F is onto, I could write that here, and one to one, and these are really just fancy ways of saying for every Y in our codomain, there's a unique X that F maps to it."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And what it does is, it takes any vector in R2 and it maps it to a rotated version of that vector. Or another way of saying it is that the rotation, let me apply, let me, it's not one of those double Rs. The rotation of some vector x is going to be equal to some counterclockwise theta degree rotation of x. So this is what we want to construct using our new linear transformation tools. And just to make sure that we can actually even do this, we need to make sure there's an actual linear transformation. I'll just do that visually, because I actually don't even have a mathematical definition for this yet. This is about as good as I've given you."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So this is what we want to construct using our new linear transformation tools. And just to make sure that we can actually even do this, we need to make sure there's an actual linear transformation. I'll just do that visually, because I actually don't even have a mathematical definition for this yet. This is about as good as I've given you. So let me just draw some really fast, some axes right here. I have to draw them a little bit neater than that. See that's my vertical axis, that's my horizontal axis."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This is about as good as I've given you. So let me just draw some really fast, some axes right here. I have to draw them a little bit neater than that. See that's my vertical axis, that's my horizontal axis. I could call this, I'll call this one the x1 and I'll call this the x2 axis. In the last video I called them the x and the y, but you know this is the first component of our vectors. This is our second component of our vectors."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "See that's my vertical axis, that's my horizontal axis. I could call this, I'll call this one the x1 and I'll call this the x2 axis. In the last video I called them the x and the y, but you know this is the first component of our vectors. This is our second component of our vectors. And so if I have some vector x like that, we know that a rotation of this, a counterclockwise rotation of this, will look like this. I'll do the rotations in blue. It will look like this."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This is our second component of our vectors. And so if I have some vector x like that, we know that a rotation of this, a counterclockwise rotation of this, will look like this. I'll do the rotations in blue. It will look like this. Where this angle right here is theta. So this right here is the rotation for an angle of theta of x. That's what this vector right here is."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "It will look like this. Where this angle right here is theta. So this right here is the rotation for an angle of theta of x. That's what this vector right here is. So what do we have to do to make sure that this is a linear combination? We have to show two things. We have to show that the transformation, so the rotation through theta of the sum of two vectors, it's equivalent to the sum of each of their individual rotations."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "That's what this vector right here is. So what do we have to do to make sure that this is a linear combination? We have to show two things. We have to show that the transformation, so the rotation through theta of the sum of two vectors, it's equivalent to the sum of each of their individual rotations. The rotation of the vector x plus the rotation of the vector y. And I'll just show that to you visually. This is the vector x."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "We have to show that the transformation, so the rotation through theta of the sum of two vectors, it's equivalent to the sum of each of their individual rotations. The rotation of the vector x plus the rotation of the vector y. And I'll just show that to you visually. This is the vector x. Let's say that the vector y looks something like, let me draw, well I'll draw both of the original vectors in yellow. So let's say the vector y looks like that. So that's y."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This is the vector x. Let's say that the vector y looks something like, let me draw, well I'll draw both of the original vectors in yellow. So let's say the vector y looks like that. So that's y. So what's x plus y? So let's put heads to tails. If we just shift y up here, if you just shift y up there, that's also vector y."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So that's y. So what's x plus y? So let's put heads to tails. If we just shift y up here, if you just shift y up there, that's also vector y. Not drawn in standard position, but x plus y would then, well, it would look pretty close to this. I'm not drawing it. Let me draw it a little bit neater."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "If we just shift y up here, if you just shift y up there, that's also vector y. Not drawn in standard position, but x plus y would then, well, it would look pretty close to this. I'm not drawing it. Let me draw it a little bit neater. x plus y would look like that. That would be the vector x plus y. And what would its rotation look like through an angle of theta?"}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw it a little bit neater. x plus y would look like that. That would be the vector x plus y. And what would its rotation look like through an angle of theta? What's rotation if you just rotate this guy through an angle of theta? I'm just approximating. It would look something like this."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And what would its rotation look like through an angle of theta? What's rotation if you just rotate this guy through an angle of theta? I'm just approximating. It would look something like this. It would look something like that. So this right here, this would be the rotation through an angle of theta of x plus y. Now let's see if that's the same thing as if we rotate x and rotate y and then add them together."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "It would look something like this. It would look something like that. So this right here, this would be the rotation through an angle of theta of x plus y. Now let's see if that's the same thing as if we rotate x and rotate y and then add them together. So what's y if we rotate it through an angle of theta? If we rotate y through an angle of theta, it's going to look something like, it's going to look something like, this is all approximation, I should be doing it with a ruler and a protractor, maybe it looks something like this. And the rotation, so this is right here, that's the rotation of y through an angle of theta."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Now let's see if that's the same thing as if we rotate x and rotate y and then add them together. So what's y if we rotate it through an angle of theta? If we rotate y through an angle of theta, it's going to look something like, it's going to look something like, this is all approximation, I should be doing it with a ruler and a protractor, maybe it looks something like this. And the rotation, so this is right here, that's the rotation of y through an angle of theta. That's the same theta that I've been doing the whole time. Let me make it in a color that you can actually see. So that's that vector right there."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And the rotation, so this is right here, that's the rotation of y through an angle of theta. That's the same theta that I've been doing the whole time. Let me make it in a color that you can actually see. So that's that vector right there. The rotation of x was right there, so if we add the rotation of x plus the rotation of y, I'm kind of fudging it a little bit, but I think you get the idea. You actually get, so this is the rotation of x plus the rotation of y, you actually get the rotation of x plus y. So at least visually it satisfied that first condition."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So that's that vector right there. The rotation of x was right there, so if we add the rotation of x plus the rotation of y, I'm kind of fudging it a little bit, but I think you get the idea. You actually get, so this is the rotation of x plus the rotation of y, you actually get the rotation of x plus y. So at least visually it satisfied that first condition. Now the second condition that we need for this to be a valid linear transformation is that the rotation through an angle of theta of a scaled up version of a vector, it should be equal to a scaled up version of the rotated vector. I'll just do another visual example here. So let's say that's my vertical axis, that's my horizontal axis, and let me say that this is my vector x."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So at least visually it satisfied that first condition. Now the second condition that we need for this to be a valid linear transformation is that the rotation through an angle of theta of a scaled up version of a vector, it should be equal to a scaled up version of the rotated vector. I'll just do another visual example here. So let's say that's my vertical axis, that's my horizontal axis, and let me say that this is my vector x. Now let's draw a scaled up version of it. So a scaled up version of x, maybe it's just like x, but it gets scaled up a little bit. So it goes all the way out here."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that's my vertical axis, that's my horizontal axis, and let me say that this is my vector x. Now let's draw a scaled up version of it. So a scaled up version of x, maybe it's just like x, but it gets scaled up a little bit. So it goes all the way out here. So this is my c times x, and now we're going to rotate that through an angle of theta. So if we rotate that through an angle of theta, you get a vector that looks something like this, rotating it counterclockwise. So this vector right here is the rotation by an angle of theta counterclockwise of cx."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So it goes all the way out here. So this is my c times x, and now we're going to rotate that through an angle of theta. So if we rotate that through an angle of theta, you get a vector that looks something like this, rotating it counterclockwise. So this vector right here is the rotation by an angle of theta counterclockwise of cx. So that's what this is right there. Now what happens if we take the rotation of x first? So if we just rotate x first, we're going to get this vector right here."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So this vector right here is the rotation by an angle of theta counterclockwise of cx. So that's what this is right there. Now what happens if we take the rotation of x first? So if we just rotate x first, we're going to get this vector right here. So this right here is just the rotation through an angle of theta of x, and then we scale it up. Well, we see it's the same thing. If this scaled up to that when you multiply it by c, then this thing is going to scale up to that when you multiply it by c. So at least visually, I've shown you that this is satisfied."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So if we just rotate x first, we're going to get this vector right here. So this right here is just the rotation through an angle of theta of x, and then we scale it up. Well, we see it's the same thing. If this scaled up to that when you multiply it by c, then this thing is going to scale up to that when you multiply it by c. So at least visually, I've shown you that this is satisfied. So rotation definitely is a linear transformation, at least the way I've shown you. Now let's actually construct a mathematical definition for it. Let's actually construct a matrix that will perform the transformation."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "If this scaled up to that when you multiply it by c, then this thing is going to scale up to that when you multiply it by c. So at least visually, I've shown you that this is satisfied. So rotation definitely is a linear transformation, at least the way I've shown you. Now let's actually construct a mathematical definition for it. Let's actually construct a matrix that will perform the transformation. So I'm saying that my rotation transformation from R2 to R2 of some vector x can be defined as some 2 by 2 matrix. And it's 2 by 2 because it's a mapping from R2 to R2, times any vector x. And I'm saying I can do this because I've at least shown you visually that it is indeed a linear transformation."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Let's actually construct a matrix that will perform the transformation. So I'm saying that my rotation transformation from R2 to R2 of some vector x can be defined as some 2 by 2 matrix. And it's 2 by 2 because it's a mapping from R2 to R2, times any vector x. And I'm saying I can do this because I've at least shown you visually that it is indeed a linear transformation. And how do I find A? Well, I start with, since this is mapping from R2, I start with the identity matrix in R2, which is 1, 0, 0, 1. Its columns are the basis vectors for R2."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And I'm saying I can do this because I've at least shown you visually that it is indeed a linear transformation. And how do I find A? Well, I start with, since this is mapping from R2, I start with the identity matrix in R2, which is 1, 0, 0, 1. Its columns are the basis vectors for R2. We refer this one as E1 and this column vector as E2. And to figure out A, we essentially just perform the transformation on each of these columns. So let me write that."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Its columns are the basis vectors for R2. We refer this one as E1 and this column vector as E2. And to figure out A, we essentially just perform the transformation on each of these columns. So let me write that. So A, our matrix A is going to be, the first column of it is going to be the rotation transformation performed on the vector 1, 0. And our second column is going to be the rotation transformation, there's a little data here that I'm forgetting to write, times the second column vector, times the transformation of that one, 0, 1. This is what our A is going to look like."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that. So A, our matrix A is going to be, the first column of it is going to be the rotation transformation performed on the vector 1, 0. And our second column is going to be the rotation transformation, there's a little data here that I'm forgetting to write, times the second column vector, times the transformation of that one, 0, 1. This is what our A is going to look like. So how do we figure out what these are? I'm trying to get to some numbers and this doesn't get me there. So let's try to do that."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This is what our A is going to look like. So how do we figure out what these are? I'm trying to get to some numbers and this doesn't get me there. So let's try to do that. Let me draw some more axes here. Let me pick a different color, I'll do it in gray. So that is my vertical axes, that is my horizontal axes."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So let's try to do that. Let me draw some more axes here. Let me pick a different color, I'll do it in gray. So that is my vertical axes, that is my horizontal axes. And I could call this the X1 axis and this is the X2 axis. Now this basis vector E1, what does it look like? Well, it's 1 in the horizontal, X1 is 1 and X2 is 0."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So that is my vertical axes, that is my horizontal axes. And I could call this the X1 axis and this is the X2 axis. Now this basis vector E1, what does it look like? Well, it's 1 in the horizontal, X1 is 1 and X2 is 0. So if this is 1 out here, E1 will look like that. Let me do it in a more vibrant color. E1 will look like that right there."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's 1 in the horizontal, X1 is 1 and X2 is 0. So if this is 1 out here, E1 will look like that. Let me do it in a more vibrant color. E1 will look like that right there. Now, if I'm rotating, let me write what E2 looks like. E2 would look like, I'll do it in yellow, E2 would look like this right here. E2, that's that vector, 0, 1."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "E1 will look like that right there. Now, if I'm rotating, let me write what E2 looks like. E2 would look like, I'll do it in yellow, E2 would look like this right here. E2, that's that vector, 0, 1. This is 1 in our X2 direction. Now if I rotate E1 by an angle theta, what will it look like? So if I rotate E1 by an angle theta, I'll do it in this color right here, it will still have a length of 1, but it will be rotated like that."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "E2, that's that vector, 0, 1. This is 1 in our X2 direction. Now if I rotate E1 by an angle theta, what will it look like? So if I rotate E1 by an angle theta, I'll do it in this color right here, it will still have a length of 1, but it will be rotated like that. And that angle right there is theta. So this right here is the rotation of E1 by theta. These are all vectors, of course."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So if I rotate E1 by an angle theta, I'll do it in this color right here, it will still have a length of 1, but it will be rotated like that. And that angle right there is theta. So this right here is the rotation of E1 by theta. These are all vectors, of course. That's what that is. So what are the coordinates for this, or how do we specify this new vector? We can break out a little bit of our trigonometry."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "These are all vectors, of course. That's what that is. So what are the coordinates for this, or how do we specify this new vector? We can break out a little bit of our trigonometry. Its new X1 coordinate, we could call it, or its X1 entry, is going to be this length right here. So if we draw a right triangle, it is the side that is adjacent to theta. This side is the hypotenuse, which is length 1."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "We can break out a little bit of our trigonometry. Its new X1 coordinate, we could call it, or its X1 entry, is going to be this length right here. So if we draw a right triangle, it is the side that is adjacent to theta. This side is the hypotenuse, which is length 1. So how do we figure out this side? If we call this side the adjacent side, the adjacent side over the hypotenuse, which is just 1, is equal to the cosine of theta. That comes from Soh Cah Toa."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This side is the hypotenuse, which is length 1. So how do we figure out this side? If we call this side the adjacent side, the adjacent side over the hypotenuse, which is just 1, is equal to the cosine of theta. That comes from Soh Cah Toa. Let me write that. Cosine is adjacent over hypotenuse. And the adjacent side is going to be our new X1 coordinate."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "That comes from Soh Cah Toa. Let me write that. Cosine is adjacent over hypotenuse. And the adjacent side is going to be our new X1 coordinate. We can obviously ignore that 1. A divided by 1 is equal to cosine theta, which means that A is equal to cosine theta, which means that this length of our rotated vector is equal to cosine theta. Its horizontal component, or its horizontal coordinate, is equal to cosine of theta."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And the adjacent side is going to be our new X1 coordinate. We can obviously ignore that 1. A divided by 1 is equal to cosine theta, which means that A is equal to cosine theta, which means that this length of our rotated vector is equal to cosine theta. Its horizontal component, or its horizontal coordinate, is equal to cosine of theta. Now, if we take what's its vertical component going to be, what's the vertical component? Well, its vertical component is going to be this height right here. It's going to be that height right there, which is the same thing as that height right there."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Its horizontal component, or its horizontal coordinate, is equal to cosine of theta. Now, if we take what's its vertical component going to be, what's the vertical component? Well, its vertical component is going to be this height right here. It's going to be that height right there, which is the same thing as that height right there. Or we could say sine of theta. Let me call this the opposite. Sine of theta is equal to the opposite over 1."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be that height right there, which is the same thing as that height right there. Or we could say sine of theta. Let me call this the opposite. Sine of theta is equal to the opposite over 1. So this is going to be equal to sine of theta. This over 1, which is just this, is equal to sine of theta from Soh Cah Toa. So this vertical component is equal to sine of theta."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Sine of theta is equal to the opposite over 1. So this is going to be equal to sine of theta. This over 1, which is just this, is equal to sine of theta from Soh Cah Toa. So this vertical component is equal to sine of theta. So the new rotated basis vector can be written as cosine of theta for its x component, or for its horizontal component, and sine of theta for its vertical component. This is the new rotated basis vector. Now, what about E2?"}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So this vertical component is equal to sine of theta. So the new rotated basis vector can be written as cosine of theta for its x component, or for its horizontal component, and sine of theta for its vertical component. This is the new rotated basis vector. Now, what about E2? We could do something very similar there. E2 is going to end up looking like this when you rotate it by an angle of theta. It's going to look like that."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Now, what about E2? We could do something very similar there. E2 is going to end up looking like this when you rotate it by an angle of theta. It's going to look like that. That angle right there is theta. We can create a little right triangle right there. So if we want to know its x coordinate, so now we're concerned with the rotation through an angle of theta of E2, which is that right there, of E2."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "It's going to look like that. That angle right there is theta. We can create a little right triangle right there. So if we want to know its x coordinate, so now we're concerned with the rotation through an angle of theta of E2, which is that right there, of E2. This is E2 right there. This is going to be equal to what? Its new x coordinate, or its first entry in this vector, if we wanted to draw it in standard position, or the point that it is specifying, is going to be equal to this distance, which is equal to this distance on this triangle, but the coordinate is going to be the negative of this."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So if we want to know its x coordinate, so now we're concerned with the rotation through an angle of theta of E2, which is that right there, of E2. This is E2 right there. This is going to be equal to what? Its new x coordinate, or its first entry in this vector, if we wanted to draw it in standard position, or the point that it is specifying, is going to be equal to this distance, which is equal to this distance on this triangle, but the coordinate is going to be the negative of this. If this is a distance of 2, this coordinate is going to be minus 2. So what's this? We have an angle."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Its new x coordinate, or its first entry in this vector, if we wanted to draw it in standard position, or the point that it is specifying, is going to be equal to this distance, which is equal to this distance on this triangle, but the coordinate is going to be the negative of this. If this is a distance of 2, this coordinate is going to be minus 2. So what's this? We have an angle. It's a right triangle. This is opposite to the angle. Opposite over 1, opposite over a hypotenuse, is equal to cosine of theta."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "We have an angle. It's a right triangle. This is opposite to the angle. Opposite over 1, opposite over a hypotenuse, is equal to cosine of theta. So this opposite side is equal to the cosine of theta. So the x coordinate right here, oh, sorry, my trigonometry is messing up. This is the opposite side."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Opposite over 1, opposite over a hypotenuse, is equal to cosine of theta. So this opposite side is equal to the cosine of theta. So the x coordinate right here, oh, sorry, my trigonometry is messing up. This is the opposite side. So, Katoa, sine is equal to opposite. Let me write it, because I was, sine of theta is equal to opposite over hypotenuse. So the sine of theta, sine of this angle, is equal to this opposite over the hypotenuse."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This is the opposite side. So, Katoa, sine is equal to opposite. Let me write it, because I was, sine of theta is equal to opposite over hypotenuse. So the sine of theta, sine of this angle, is equal to this opposite over the hypotenuse. The hypotenuse is 1, has length 1, because these are the standard basis vectors. So this is equal to the sine of theta. Now, this distance is equal to the sine of theta, but it's going in the negative direction."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So the sine of theta, sine of this angle, is equal to this opposite over the hypotenuse. The hypotenuse is 1, has length 1, because these are the standard basis vectors. So this is equal to the sine of theta. Now, this distance is equal to the sine of theta, but it's going in the negative direction. So it's going to be equal to the minus sine of theta. And then what's its new y component going to be, of this rotated version of E2? Well, we just look right here."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Now, this distance is equal to the sine of theta, but it's going in the negative direction. So it's going to be equal to the minus sine of theta. And then what's its new y component going to be, of this rotated version of E2? Well, we just look right here. We have our angle. This is adjacent to the angle. This adjacent side over the hypotenuse, so adjacent over 1, which is just this adjacent right here, is just going to be equal to cosine of theta."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Well, we just look right here. We have our angle. This is adjacent to the angle. This adjacent side over the hypotenuse, so adjacent over 1, which is just this adjacent right here, is just going to be equal to cosine of theta. So its new y coordinate is going to be cosine of theta. So when we apply the transformation to each of our basis vectors, we get A is equal to the transformation applied to E1, which is cosine of theta and sine of theta, and the transformation applied to E2, which is minus sine of theta times the cosine of theta. So now this is a big result."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "This adjacent side over the hypotenuse, so adjacent over 1, which is just this adjacent right here, is just going to be equal to cosine of theta. So its new y coordinate is going to be cosine of theta. So when we apply the transformation to each of our basis vectors, we get A is equal to the transformation applied to E1, which is cosine of theta and sine of theta, and the transformation applied to E2, which is minus sine of theta times the cosine of theta. So now this is a big result. We've now been able to mathematically specify our rotation transformation using a matrix. So we can now say that the rotation transformation, and it's a transformation from R2 to R2. It's a function."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So now this is a big result. We've now been able to mathematically specify our rotation transformation using a matrix. So we can now say that the rotation transformation, and it's a transformation from R2 to R2. It's a function. We can say that the rotation through an angle of theta of any vector x in our domain is equal to the matrix cosine of theta, sine of theta, minus sine of theta, cosine of theta, times your vector in your domain, times x1 and x2. And you might be saying, oh, Sal, we did all this work and that's kind of neat, but how do I apply this? I still have all these cosines of thetas and sines of thetas there."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "It's a function. We can say that the rotation through an angle of theta of any vector x in our domain is equal to the matrix cosine of theta, sine of theta, minus sine of theta, cosine of theta, times your vector in your domain, times x1 and x2. And you might be saying, oh, Sal, we did all this work and that's kind of neat, but how do I apply this? I still have all these cosines of thetas and sines of thetas there. How do I do it? Well, what you do is you pick an angle you want to rotate to and just evaluate these, and you'll just have a normal matrix with numbers in it. So let's say that we wanted to rotate through an angle of 45 degrees, some vector."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "I still have all these cosines of thetas and sines of thetas there. How do I do it? Well, what you do is you pick an angle you want to rotate to and just evaluate these, and you'll just have a normal matrix with numbers in it. So let's say that we wanted to rotate through an angle of 45 degrees, some vector. Well, this is going to be equal to what? We just apply or we evaluate each of these ratios at 45 degrees. The cosine of 45 degrees is square root of 2 over 2."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that we wanted to rotate through an angle of 45 degrees, some vector. Well, this is going to be equal to what? We just apply or we evaluate each of these ratios at 45 degrees. The cosine of 45 degrees is square root of 2 over 2. Sine of 45 degrees is square root of 2 over 2. Sine of 45 is square root of 2 over 2. We have a minus there, so minus square root of 2 over 2."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "The cosine of 45 degrees is square root of 2 over 2. Sine of 45 degrees is square root of 2 over 2. Sine of 45 is square root of 2 over 2. We have a minus there, so minus square root of 2 over 2. And then cosine is just another square root of 2 over 2. And we multiply it times our vector x. So this matrix, if we multiply it times any vector x, literally, so if we have some coordinates right here, and let's say that I were to, I don't know, let's say I were to have a bunch of vectors that specify some square here."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "We have a minus there, so minus square root of 2 over 2. And then cosine is just another square root of 2 over 2. And we multiply it times our vector x. So this matrix, if we multiply it times any vector x, literally, so if we have some coordinates right here, and let's say that I were to, I don't know, let's say I were to have a bunch of vectors that specify some square here. Let me see if I can do it properly. Well, maybe it has some triangle here. Maybe that will be a little easier for me to draw."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "So this matrix, if we multiply it times any vector x, literally, so if we have some coordinates right here, and let's say that I were to, I don't know, let's say I were to have a bunch of vectors that specify some square here. Let me see if I can do it properly. Well, maybe it has some triangle here. Maybe that will be a little easier for me to draw. I'll do a square. Let's say it has some square here. Let's say I have some square here in my domain."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Maybe that will be a little easier for me to draw. I'll do a square. Let's say it has some square here. Let's say I have some square here in my domain. So this is an R2. Now when I apply, if I literally multiply this times each of the basis vectors, or actually all of the vectors that specify this set here, I will get, when I transform it, I'll get a rotated version of this by 45 degrees. And just to draw it, I'll actually draw a little 45 degree angle right there."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have some square here in my domain. So this is an R2. Now when I apply, if I literally multiply this times each of the basis vectors, or actually all of the vectors that specify this set here, I will get, when I transform it, I'll get a rotated version of this by 45 degrees. And just to draw it, I'll actually draw a little 45 degree angle right there. And then it will map it to this image right there, which is a pretty neat result. And you can all, if you ever attempted to write any computer game that involves marbles or pinballs going around, this is a very useful thing to know, how do you rotate things, and in the future we'll talk about other types of transformations. But this is a super useful one, and this is super hard to do."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And just to draw it, I'll actually draw a little 45 degree angle right there. And then it will map it to this image right there, which is a pretty neat result. And you can all, if you ever attempted to write any computer game that involves marbles or pinballs going around, this is a very useful thing to know, how do you rotate things, and in the future we'll talk about other types of transformations. But this is a super useful one, and this is super hard to do. And I remember the first time I wrote a computer program that tried to do this type of thing, I just did it by hand. But when you have this tool at your disposal, all you have to do is evaluate this matrix at the angle you want to rotate it by, and then multiply it times your position vectors. And so obviously you have a bunch of position vectors here, but here you can just do it times the vertices, and then you can say, okay, and everything else is just connect the dots between them, and then you have your rotated image."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "But this is a super useful one, and this is super hard to do. And I remember the first time I wrote a computer program that tried to do this type of thing, I just did it by hand. But when you have this tool at your disposal, all you have to do is evaluate this matrix at the angle you want to rotate it by, and then multiply it times your position vectors. And so obviously you have a bunch of position vectors here, but here you can just do it times the vertices, and then you can say, okay, and everything else is just connect the dots between them, and then you have your rotated image. And just to be clear, these are the points specified by a set of vectors, and I always want to make this clear, right? This point right here is specified by some position vector that looks like that. When you apply the rotation on 45 degrees of that vector, this vector then looks like this."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "And so obviously you have a bunch of position vectors here, but here you can just do it times the vertices, and then you can say, okay, and everything else is just connect the dots between them, and then you have your rotated image. And just to be clear, these are the points specified by a set of vectors, and I always want to make this clear, right? This point right here is specified by some position vector that looks like that. When you apply the rotation on 45 degrees of that vector, this vector then looks like this. And the vector that specified this corner right here, let me do it in a different color, that specified this corner right here when you rotate it by 45 degrees, then becomes this vector. And the vector that specified that corner over there, that now becomes this vector. That's what's actually being mapped or actually being transformed."}, {"video_title": "Linear transformation examples Rotations in R2 Linear Algebra Khan Academy.mp3", "Sentence": "When you apply the rotation on 45 degrees of that vector, this vector then looks like this. And the vector that specified this corner right here, let me do it in a different color, that specified this corner right here when you rotate it by 45 degrees, then becomes this vector. And the vector that specified that corner over there, that now becomes this vector. That's what's actually being mapped or actually being transformed. Anyway, hopefully you found this pretty neat. I thought this was a, at least for me, this is kind of the neat, the first really neat transformation. And you can already start thinking about how to extend this into multiple dimensions, especially three-dimensionals."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "We know what an orthonormal basis is, but the next obvious question is, what are they good for? And one of the many answers to that question is that they make for good coordinate systems. For example, the standard basis or the standard coordinates, let me write the standard basis in Rn. So if we're dealing with Rn, so the standard basis for Rn is equal to, well, I could write it as e1, e2, and all of that, but I'll actually write out the vectors. e1 is just 1 with a bunch of 0's all the way, e2, and this is going to be n 0's right there. e2 is going to be 0, 1 with a bunch of 0's all the way, and then you're going to go all the way to en, which is going to have a bunch of 0's, and then you're going to have a 1. The standard basis that we've been dealing with throughout this playlist is an orthonormal set."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So if we're dealing with Rn, so the standard basis for Rn is equal to, well, I could write it as e1, e2, and all of that, but I'll actually write out the vectors. e1 is just 1 with a bunch of 0's all the way, e2, and this is going to be n 0's right there. e2 is going to be 0, 1 with a bunch of 0's all the way, and then you're going to go all the way to en, which is going to have a bunch of 0's, and then you're going to have a 1. The standard basis that we've been dealing with throughout this playlist is an orthonormal set. So this is an orthonormal basis. Clearly the length of any of these guys is 1. If you were to take this guy dotted with yourself, you're going to get 1 times 1 plus a bunch of 0's times each other, so it's going to be 1 squared, so it's going to be 1."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "The standard basis that we've been dealing with throughout this playlist is an orthonormal set. So this is an orthonormal basis. Clearly the length of any of these guys is 1. If you were to take this guy dotted with yourself, you're going to get 1 times 1 plus a bunch of 0's times each other, so it's going to be 1 squared, so it's going to be 1. And that's true of any of these guys. And clearly they're orthogonal. You take the dot product of any of these guys with any of the other ones, you're going to get a 1 times a 0, and a 1 times a 0, and a bunch of 0's, so you're going to get 0's."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "If you were to take this guy dotted with yourself, you're going to get 1 times 1 plus a bunch of 0's times each other, so it's going to be 1 squared, so it's going to be 1. And that's true of any of these guys. And clearly they're orthogonal. You take the dot product of any of these guys with any of the other ones, you're going to get a 1 times a 0, and a 1 times a 0, and a bunch of 0's, so you're going to get 0's. So they clearly each have lengths of 1, and they're all orthogonal. And clearly this is a good coordinate system. But what about other orthonormal bases?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "You take the dot product of any of these guys with any of the other ones, you're going to get a 1 times a 0, and a 1 times a 0, and a bunch of 0's, so you're going to get 0's. So they clearly each have lengths of 1, and they're all orthogonal. And clearly this is a good coordinate system. But what about other orthonormal bases? Obviously this is one specific example. I need to show you that all orthonormal bases make for good coordinate systems. So let's say I have some set, some orthonormal set of vectors, such as v1, v2, all the way to vk."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But what about other orthonormal bases? Obviously this is one specific example. I need to show you that all orthonormal bases make for good coordinate systems. So let's say I have some set, some orthonormal set of vectors, such as v1, v2, all the way to vk. And it is an orthonormal basis for some subspace v. And this is a k-dimensional subspace, because you have k basis vectors in your basis. Or you have k vectors in your basis. Now, let's experiment with this a little bit."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have some set, some orthonormal set of vectors, such as v1, v2, all the way to vk. And it is an orthonormal basis for some subspace v. And this is a k-dimensional subspace, because you have k basis vectors in your basis. Or you have k vectors in your basis. Now, let's experiment with this a little bit. I'm claiming that this would make, that the coordinate system with respect to this is good. But what does it mean to be good? I mean, the standard basis is good, but that's just because we use it, and it seems to be easy to deal with."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's experiment with this a little bit. I'm claiming that this would make, that the coordinate system with respect to this is good. But what does it mean to be good? I mean, the standard basis is good, but that's just because we use it, and it seems to be easy to deal with. But let's see what I, when I say good in this context, what do I mean? So let's experiment. If I say that some vector x is a member of v, that means that x can be represented as a linear combination of these characters up here."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "I mean, the standard basis is good, but that's just because we use it, and it seems to be easy to deal with. But let's see what I, when I say good in this context, what do I mean? So let's experiment. If I say that some vector x is a member of v, that means that x can be represented as a linear combination of these characters up here. So x can be represented as some constant times v1 plus some constant times v2 plus the ith constant times vi, all the way, if you just keep going, all the way to the kth constant times vk. That's what being a member of the subspace means. The subspace is spanned by these guys, so this guy can be represented as a linear combination of those guys."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "If I say that some vector x is a member of v, that means that x can be represented as a linear combination of these characters up here. So x can be represented as some constant times v1 plus some constant times v2 plus the ith constant times vi, all the way, if you just keep going, all the way to the kth constant times vk. That's what being a member of the subspace means. The subspace is spanned by these guys, so this guy can be represented as a linear combination of those guys. Now, what happens if we take the dot product of both sides of this equation with vi? So I'm going to take vi, I'm going to dot both of these sides with vi. So I get vi dotted with x is going to be equal to what?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "The subspace is spanned by these guys, so this guy can be represented as a linear combination of those guys. Now, what happens if we take the dot product of both sides of this equation with vi? So I'm going to take vi, I'm going to dot both of these sides with vi. So I get vi dotted with x is going to be equal to what? Well, it's going to be, we could just put the constants out. It's going to be c1 times vi dot v1 plus c2 times vi dot v2 plus all the way to ci times vi dot vi. And then you keep going plus all the way to ck times vi dot vk."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So I get vi dotted with x is going to be equal to what? Well, it's going to be, we could just put the constants out. It's going to be c1 times vi dot v1 plus c2 times vi dot v2 plus all the way to ci times vi dot vi. And then you keep going plus all the way to ck times vi dot vk. Now, this is an orthonormal set. That means if I take two vectors that are different than each other in our basis right here, then if you take their dot product, you're going to get 0. They're orthogonal to each other."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then you keep going plus all the way to ck times vi dot vk. Now, this is an orthonormal set. That means if I take two vectors that are different than each other in our basis right here, then if you take their dot product, you're going to get 0. They're orthogonal to each other. So these are two different vectors in our set. They're going to be orthogonal, so this term is going to be 0. It's going to be 0 times c1, so it's going to be 0."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "They're orthogonal to each other. So these are two different vectors in our set. They're going to be orthogonal, so this term is going to be 0. It's going to be 0 times c1, so it's going to be 0. This term is also going to be 0, assuming that i isn't 2. Let's just assume that. This term over here, let's assume that i isn't k. It's also going to be equal to 0."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be 0 times c1, so it's going to be 0. This term is also going to be 0, assuming that i isn't 2. Let's just assume that. This term over here, let's assume that i isn't k. It's also going to be equal to 0. So all of the terms are going to be 0, except for the case where v sub i is equal to, well, in this case, v sub i. Except for the case where this subscript is equal to that subscript. And then what is v sub i dot v sub i?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This term over here, let's assume that i isn't k. It's also going to be equal to 0. So all of the terms are going to be 0, except for the case where v sub i is equal to, well, in this case, v sub i. Except for the case where this subscript is equal to that subscript. And then what is v sub i dot v sub i? Well, this is, you know, orthonormal has two parts. They're orthogonal to each other, and they're each normalized, or they each have length 1. So v sub i dot v sub i dot v sub i is going to be equal to 1."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And then what is v sub i dot v sub i? Well, this is, you know, orthonormal has two parts. They're orthogonal to each other, and they're each normalized, or they each have length 1. So v sub i dot v sub i dot v sub i is going to be equal to 1. So this whole equation has simplified to v sub i, which is one of these guys, it's the i-th member of our basis set, dot x, where x is just any member of the subspace, is equal to, the only thing that's left over is 1 times ci. So it's just equal to ci. Now why is this useful?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So v sub i dot v sub i dot v sub i is going to be equal to 1. So this whole equation has simplified to v sub i, which is one of these guys, it's the i-th member of our basis set, dot x, where x is just any member of the subspace, is equal to, the only thing that's left over is 1 times ci. So it's just equal to ci. Now why is this useful? You know, we were just experimenting around, and we got this nice little result here. Why is this useful in terms of having a coordinate system with respect to this basis? So let's remind ourselves what a coordinate system is here."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now why is this useful? You know, we were just experimenting around, and we got this nice little result here. Why is this useful in terms of having a coordinate system with respect to this basis? So let's remind ourselves what a coordinate system is here. So if we wanted to represent the vector x, which is a member of our subspace, with coordinates that are with respect to this basis of the subspace, right? The subspace can have many bases, but this is the basis that we're choosing. So we want to write x with respect to the basis b."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let's remind ourselves what a coordinate system is here. So if we wanted to represent the vector x, which is a member of our subspace, with coordinates that are with respect to this basis of the subspace, right? The subspace can have many bases, but this is the basis that we're choosing. So we want to write x with respect to the basis b. What do we do? The coordinates are going to just be the coefficients on the different basis vectors. This is all a bit of review, right?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So we want to write x with respect to the basis b. What do we do? The coordinates are going to just be the coefficients on the different basis vectors. This is all a bit of review, right? It's going to be c1, c2, you're going to go down to ci, and then you're going to go down all the way to ck. You're going to have k terms, because this is a k dimensional subspace. Now, normally this is not such an easy thing to figure out."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This is all a bit of review, right? It's going to be c1, c2, you're going to go down to ci, and then you're going to go down all the way to ck. You're going to have k terms, because this is a k dimensional subspace. Now, normally this is not such an easy thing to figure out. If I give you some vector x, I mean, we've seen it before. If you have x represented in b coordinate system, then you can multiply it times the change of basis matrix, and you can just get regular x. But if you have regular x and you need to find this, one, if c is invertible, then you can apply this equation right here, which isn't always the case."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now, normally this is not such an easy thing to figure out. If I give you some vector x, I mean, we've seen it before. If you have x represented in b coordinate system, then you can multiply it times the change of basis matrix, and you can just get regular x. But if you have regular x and you need to find this, one, if c is invertible, then you can apply this equation right here, which isn't always the case. This is only if c is invertible. And first of all, c will not always be invertible if this isn't a square matrix and this isn't going to apply. So this is one way that if I give you your x to get your b representation of x, but if c isn't invertible, then you're just going to have to solve this equation."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But if you have regular x and you need to find this, one, if c is invertible, then you can apply this equation right here, which isn't always the case. This is only if c is invertible. And first of all, c will not always be invertible if this isn't a square matrix and this isn't going to apply. So this is one way that if I give you your x to get your b representation of x, but if c isn't invertible, then you're just going to have to solve this equation. You're going to have something on the right hand side here. You're going to have a change of basis matrix, and then you're going to have to solve that equation. So it's kind of a, for an arbitrary basis, that can be pretty painful."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So this is one way that if I give you your x to get your b representation of x, but if c isn't invertible, then you're just going to have to solve this equation. You're going to have something on the right hand side here. You're going to have a change of basis matrix, and then you're going to have to solve that equation. So it's kind of a, for an arbitrary basis, that can be pretty painful. But what do we have here? We have a very simple solution for finding the different coordinates of x. So this is the same thing as being equal to c1 is just going to be equal to my first basis vector dotted with x."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So it's kind of a, for an arbitrary basis, that can be pretty painful. But what do we have here? We have a very simple solution for finding the different coordinates of x. So this is the same thing as being equal to c1 is just going to be equal to my first basis vector dotted with x. We say ci is just the ith basis vector dotted with x, so c1 is going to be the first basis vector dotted with x. c2 is going to be my second basis vector dotted with x. And you're going to go all the way down to ck is going to be my kth basis vector dotted with x. Let me show you that this is actually easier."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So this is the same thing as being equal to c1 is just going to be equal to my first basis vector dotted with x. We say ci is just the ith basis vector dotted with x, so c1 is going to be the first basis vector dotted with x. c2 is going to be my second basis vector dotted with x. And you're going to go all the way down to ck is going to be my kth basis vector dotted with x. Let me show you that this is actually easier. So let's do a concrete example. So let's say I want to leave this result up here. Let's say that I have two vectors, let's say that v1 is the vector 3 fifths."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let me show you that this is actually easier. So let's do a concrete example. So let's say I want to leave this result up here. Let's say that I have two vectors, let's say that v1 is the vector 3 fifths. Let me write it this way. Let's say it's 3 over 5 and 4 over 5. And that v2 is equal to minus 4 fifths, 3 fifths."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that I have two vectors, let's say that v1 is the vector 3 fifths. Let me write it this way. Let's say it's 3 over 5 and 4 over 5. And that v2 is equal to minus 4 fifths, 3 fifths. And let's say that the set B is equal to, it's comprised of just those two vectors, v1 and v2. Now I'm claiming, or I'm about to claim, that this is an orthonormal set. Let's just prove it to ourselves."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And that v2 is equal to minus 4 fifths, 3 fifths. And let's say that the set B is equal to, it's comprised of just those two vectors, v1 and v2. Now I'm claiming, or I'm about to claim, that this is an orthonormal set. Let's just prove it to ourselves. So what is the length of v1 squared? Well that's just v1 dotted with itself, so that's 3 fifths squared, which is 9 25ths, plus 4 fifths squared, which is 16 25ths, which is equal to 25 over 25, which is equal to 1. So this guy definitely has length 1."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let's just prove it to ourselves. So what is the length of v1 squared? Well that's just v1 dotted with itself, so that's 3 fifths squared, which is 9 25ths, plus 4 fifths squared, which is 16 25ths, which is equal to 25 over 25, which is equal to 1. So this guy definitely has length 1. What is the length of v2 squared? Well it's going to be this guy squared. Negative 4 fifths squared is 9 25ths, plus 3, sorry, minus 4 fifths squared is plus 16 25ths."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So this guy definitely has length 1. What is the length of v2 squared? Well it's going to be this guy squared. Negative 4 fifths squared is 9 25ths, plus 3, sorry, minus 4 fifths squared is plus 16 25ths. And then 3 fifths squared is 9 25ths. And once again, the length squared is going to be 1, or the length is going to be 1. So both of these guys definitely have length 1."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Negative 4 fifths squared is 9 25ths, plus 3, sorry, minus 4 fifths squared is plus 16 25ths. And then 3 fifths squared is 9 25ths. And once again, the length squared is going to be 1, or the length is going to be 1. So both of these guys definitely have length 1. And now we just have to verify that they're orthogonal with respect to each other. So what is v1 dot v2? It's going to be 3 fifths times minus 4 fifths, so it's going to be minus 12 25ths, plus 4 fifths times 3 fifths, which is going to be plus 12 25ths, which is equal to 0."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So both of these guys definitely have length 1. And now we just have to verify that they're orthogonal with respect to each other. So what is v1 dot v2? It's going to be 3 fifths times minus 4 fifths, so it's going to be minus 12 25ths, plus 4 fifths times 3 fifths, which is going to be plus 12 25ths, which is equal to 0. So these guys are definitely orthogonal with respect to each other, and their lengths are 1. So this is definitely an orthonormal set. And so that also tells us that they're linearly independent."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be 3 fifths times minus 4 fifths, so it's going to be minus 12 25ths, plus 4 fifths times 3 fifths, which is going to be plus 12 25ths, which is equal to 0. So these guys are definitely orthogonal with respect to each other, and their lengths are 1. So this is definitely an orthonormal set. And so that also tells us that they're linearly independent. So let's say that my set B is the basis for some subspace v. And let's say, and actually, we don't have to say that. It's the basis for R2. How do we know it's the basis for R2?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And so that also tells us that they're linearly independent. So let's say that my set B is the basis for some subspace v. And let's say, and actually, we don't have to say that. It's the basis for R2. How do we know it's the basis for R2? I have two linearly independent vectors in my basis, and it's spanning a two-dimensional space, R2. So this can be a basis for all of R2. Now, given what we've seen already, let's pick some random member of R2."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "How do we know it's the basis for R2? I have two linearly independent vectors in my basis, and it's spanning a two-dimensional space, R2. So this can be a basis for all of R2. Now, given what we've seen already, let's pick some random member of R2. So if we pick some random member of R2, let's say that x is equal to, I don't know, I'm just going to pick some random numbers, 9 and minus 2. 9 and minus 2. If we didn't know this was an orthonormal basis, and we wanted to figure out x in B's coordinates, what we would have to do is we would have to create the change of basis matrix."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Now, given what we've seen already, let's pick some random member of R2. So if we pick some random member of R2, let's say that x is equal to, I don't know, I'm just going to pick some random numbers, 9 and minus 2. 9 and minus 2. If we didn't know this was an orthonormal basis, and we wanted to figure out x in B's coordinates, what we would have to do is we would have to create the change of basis matrix. So the change of basis matrix would be 3 5ths, 4 5ths, minus 4 5ths, and then 3 5ths. We would say that times my B coordinate representation of x is going to be equal to my regular representation of x, or my standard coordinates of x. And I would have to solve this 2 by 2 system."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "If we didn't know this was an orthonormal basis, and we wanted to figure out x in B's coordinates, what we would have to do is we would have to create the change of basis matrix. So the change of basis matrix would be 3 5ths, 4 5ths, minus 4 5ths, and then 3 5ths. We would say that times my B coordinate representation of x is going to be equal to my regular representation of x, or my standard coordinates of x. And I would have to solve this 2 by 2 system. And in the 2 by 2 case, it's not so bad. But we have this neat tool here for orthonormal sets or orthonormal bases. So instead of solving this equation, we can just say that x represented in B coordinates is going to be equal to, let me scroll down a little bit, it's going to be equal to v1, which is this guy right here, dotted with x."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And I would have to solve this 2 by 2 system. And in the 2 by 2 case, it's not so bad. But we have this neat tool here for orthonormal sets or orthonormal bases. So instead of solving this equation, we can just say that x represented in B coordinates is going to be equal to, let me scroll down a little bit, it's going to be equal to v1, which is this guy right here, dotted with x. So it's going to be v1 dot x dot x. And then this guy right here is just going to be v2 dot x. And I can do this because this is an orthonormal basis."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So instead of solving this equation, we can just say that x represented in B coordinates is going to be equal to, let me scroll down a little bit, it's going to be equal to v1, which is this guy right here, dotted with x. So it's going to be v1 dot x dot x. And then this guy right here is just going to be v2 dot x. And I can do this because this is an orthonormal basis. And what is this equal to? x is 9 minus 2. If I dot that with v1, I get 9 times 3 5ths, which is 27 5ths, right?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And I can do this because this is an orthonormal basis. And what is this equal to? x is 9 minus 2. If I dot that with v1, I get 9 times 3 5ths, which is 27 5ths, right? 27 over 5. 9 times 3 is 27 over 5, plus minus 2 times 4 5ths. So that's minus 8 5ths, right?"}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "If I dot that with v1, I get 9 times 3 5ths, which is 27 5ths, right? 27 over 5. 9 times 3 is 27 over 5, plus minus 2 times 4 5ths. So that's minus 8 5ths, right? Minus 2 times 4 5ths is minus 8 5ths. And then the second term is v2 dot x. So v2 dot x, I get 9 times, let me scroll up a little bit, 9 times minus 4 5ths, that's minus 36 over 5, plus minus 2 times 3 5ths."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So that's minus 8 5ths, right? Minus 2 times 4 5ths is minus 8 5ths. And then the second term is v2 dot x. So v2 dot x, I get 9 times, let me scroll up a little bit, 9 times minus 4 5ths, that's minus 36 over 5, plus minus 2 times 3 5ths. So that's plus minus 2 times 3 5ths is minus 6 5ths. So the b-coordinate representation of x, just being able to use this property right here of orthonormal basis, is equal to, what is this? 27 minus 8 is 19 over 5."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So v2 dot x, I get 9 times, let me scroll up a little bit, 9 times minus 4 5ths, that's minus 36 over 5, plus minus 2 times 3 5ths. So that's plus minus 2 times 3 5ths is minus 6 5ths. So the b-coordinate representation of x, just being able to use this property right here of orthonormal basis, is equal to, what is this? 27 minus 8 is 19 over 5. And then minus 36 minus 6 is minus 42 over 5. Not a pretty answer, but we would have had this ugly answer either way we solved it. But hopefully you see that when we have an orthonormal basis, solving for the coordinates with respect to that basis becomes a lot easier."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "27 minus 8 is 19 over 5. And then minus 36 minus 6 is minus 42 over 5. Not a pretty answer, but we would have had this ugly answer either way we solved it. But hopefully you see that when we have an orthonormal basis, solving for the coordinates with respect to that basis becomes a lot easier. And this is just an example in R2. You can imagine how difficult it could be if you start dealing with R4 or R100. Then all of a sudden solving these systems aren't so easy, but taking dot products are always fairly straightforward."}, {"video_title": "Coordinates with respect to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "But hopefully you see that when we have an orthonormal basis, solving for the coordinates with respect to that basis becomes a lot easier. And this is just an example in R2. You can imagine how difficult it could be if you start dealing with R4 or R100. Then all of a sudden solving these systems aren't so easy, but taking dot products are always fairly straightforward. So earlier in this video, when I said orthonormal basis, what are they good for? And I said, well, the standard basis is good, so these are good coordinate systems. We've used it before."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have a linear transformation T that's a mapping between Rn and Rm. We know that we can represent this linear transformation as a matrix product. So we can say that T of x, so the transformation T, let me write a little higher to save space. So we can write that the transformation T applied to some vector x is equal to some matrix times x. And this matrix, since it's a mapping from Rn to Rm, this is going to be a m by n matrix. Because each of these entries are going to have n components. They're going to be members of Rn."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So we can write that the transformation T applied to some vector x is equal to some matrix times x. And this matrix, since it's a mapping from Rn to Rm, this is going to be a m by n matrix. Because each of these entries are going to have n components. They're going to be members of Rn. So this guy has to have n columns in order to be able to have this matrix vector product well defined. So we can go back to what we've been talking about now. In the last couple of videos we've been talking about invertibility of functions."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "They're going to be members of Rn. So this guy has to have n columns in order to be able to have this matrix vector product well defined. So we can go back to what we've been talking about now. In the last couple of videos we've been talking about invertibility of functions. And we can just as easily apply it to this transformation. Because transformations are just functions. And we just use the word transformations when we start talking about maps between vector spaces or between sets of vectors."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "In the last couple of videos we've been talking about invertibility of functions. And we can just as easily apply it to this transformation. Because transformations are just functions. And we just use the word transformations when we start talking about maps between vector spaces or between sets of vectors. But they're essentially the same thing. Everything we've done in the last few videos have been very general. I never told you what the set of our domain and the set of our codomains were made up of."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And we just use the word transformations when we start talking about maps between vector spaces or between sets of vectors. But they're essentially the same thing. Everything we've done in the last few videos have been very general. I never told you what the set of our domain and the set of our codomains were made up of. Now we're dealing with vectors. So we can apply the same ideas. So T is a mapping right here from Rn to Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "I never told you what the set of our domain and the set of our codomains were made up of. Now we're dealing with vectors. So we can apply the same ideas. So T is a mapping right here from Rn to Rm. So if you take some vector here, x, T will map it to some other vector in Rm. Call that Ax. If we take this matrix vector product right there."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So T is a mapping right here from Rn to Rm. So if you take some vector here, x, T will map it to some other vector in Rm. Call that Ax. If we take this matrix vector product right there. And this is the mapping T right there. So let's ask the same questions about T that we've been asking in general about functions. Is T invertible?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "If we take this matrix vector product right there. And this is the mapping T right there. So let's ask the same questions about T that we've been asking in general about functions. Is T invertible? And we learned in the last video that there's two conditions for invertibility. T has to be onto, or the other word was surjective. Surjective."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Is T invertible? And we learned in the last video that there's two conditions for invertibility. T has to be onto, or the other word was surjective. Surjective. That's one condition for invertibility. And then T also has to be one-to-one. And the fancy word for that was injective."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Surjective. That's one condition for invertibility. And then T also has to be one-to-one. And the fancy word for that was injective. Injective. So in this video I'm going to just focus on this first one. So I'm not going to prove to you whether T is invertible, but we'll at least be able to try to figure out whether T is onto or whether it's surjective."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And the fancy word for that was injective. Injective. So in this video I'm going to just focus on this first one. So I'm not going to prove to you whether T is invertible, but we'll at least be able to try to figure out whether T is onto or whether it's surjective. So just as a reminder, what does onto or surjective mean? It means that you take any element in Rm, you take any element in your codomain. So you give me any element in your codomain."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So I'm not going to prove to you whether T is invertible, but we'll at least be able to try to figure out whether T is onto or whether it's surjective. So just as a reminder, what does onto or surjective mean? It means that you take any element in Rm, you take any element in your codomain. So you give me any element in your codomain. Let's call that element b. It's going to be a vector. The fact, if T is surjective or if T is onto, that means that any b that you pick in our codomain, there's always going to be some vector, at least one vector in our domain, that if you apply the transformation to it, you're going to get b."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So you give me any element in your codomain. Let's call that element b. It's going to be a vector. The fact, if T is surjective or if T is onto, that means that any b that you pick in our codomain, there's always going to be some vector, at least one vector in our domain, that if you apply the transformation to it, you're going to get b. Or another way to think about it is the image of our transformation is all of Rm. All of these guys can be reached. So let's think about what that means."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "The fact, if T is surjective or if T is onto, that means that any b that you pick in our codomain, there's always going to be some vector, at least one vector in our domain, that if you apply the transformation to it, you're going to get b. Or another way to think about it is the image of our transformation is all of Rm. All of these guys can be reached. So let's think about what that means. So we know that the transformation is Ax. It's some matrix A. So the transformation of x, I'll just rewrite it, is equal to some matrix A, that's an m by n matrix, times the vector x."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So let's think about what that means. So we know that the transformation is Ax. It's some matrix A. So the transformation of x, I'll just rewrite it, is equal to some matrix A, that's an m by n matrix, times the vector x. Now, if T has to be onto, that means that Ax, this matrix vector product, has to be equal to any member. Any member of our codomain can be reached by multiplying our matrix A by some member of our domain. So what's another way of thinking about it?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So the transformation of x, I'll just rewrite it, is equal to some matrix A, that's an m by n matrix, times the vector x. Now, if T has to be onto, that means that Ax, this matrix vector product, has to be equal to any member. Any member of our codomain can be reached by multiplying our matrix A by some member of our domain. So what's another way of thinking about it? Another way of thinking about it is, for any b, so onto implies that for any vector b that is a member of Rm, so any b here, there exists at least one solution to A times x is equal to b. Where, of course, x is, the vector x is a member of Rn. It's just another way of saying exactly what I've been saying the first part of this video."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So what's another way of thinking about it? Another way of thinking about it is, for any b, so onto implies that for any vector b that is a member of Rm, so any b here, there exists at least one solution to A times x is equal to b. Where, of course, x is, the vector x is a member of Rn. It's just another way of saying exactly what I've been saying the first part of this video. You give me any b in this set, and then there has to be, if we assume that T is onto, or for T to be onto, there has to be at least one solution to Ax is equal to b. There has to be at least one x out here that if I multiply it by a, I get to my b. And this has to be true for every, maybe I should write for every, instead of any, but there's the same idea."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "It's just another way of saying exactly what I've been saying the first part of this video. You give me any b in this set, and then there has to be, if we assume that T is onto, or for T to be onto, there has to be at least one solution to Ax is equal to b. There has to be at least one x out here that if I multiply it by a, I get to my b. And this has to be true for every, maybe I should write for every, instead of any, but there's the same idea. For every b in Rm, we have to be able to find at least one x that makes this true. So what does that mean? That means that A times x has to be equal to, you can construct any member of Rm by taking a product of A and x where x is a member of Rn, where x is a member of here."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And this has to be true for every, maybe I should write for every, instead of any, but there's the same idea. For every b in Rm, we have to be able to find at least one x that makes this true. So what does that mean? That means that A times x has to be equal to, you can construct any member of Rm by taking a product of A and x where x is a member of Rn, where x is a member of here. Now what is this? If x is an arbitrary member of Rn, let me write it like this. We know the matrix A will look like this."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "That means that A times x has to be equal to, you can construct any member of Rm by taking a product of A and x where x is a member of Rn, where x is a member of here. Now what is this? If x is an arbitrary member of Rn, let me write it like this. We know the matrix A will look like this. It'll be a bunch of column vectors. A1, A2, and it has n columns, so it looks like that. That's what our matrix A looks like."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "We know the matrix A will look like this. It'll be a bunch of column vectors. A1, A2, and it has n columns, so it looks like that. That's what our matrix A looks like. So we're saying if you take this product, you have to be able to reach any guy, any member of Rm. But what does this product look like if we're taking this product, instead of writing an x there, I could write x like this. x1, x2, all the way to xn."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "That's what our matrix A looks like. So we're saying if you take this product, you have to be able to reach any guy, any member of Rm. But what does this product look like if we're taking this product, instead of writing an x there, I could write x like this. x1, x2, all the way to xn. So this product is going to be x1 times the first column vector of A plus x2 times the second column vector of A, all the way to plus xn times the nth column vector of A. That's what this product is. And in order for t to be onto, this combination has to be equal to any vector in Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "x1, x2, all the way to xn. So this product is going to be x1 times the first column vector of A plus x2 times the second column vector of A, all the way to plus xn times the nth column vector of A. That's what this product is. And in order for t to be onto, this combination has to be equal to any vector in Rm. Well, what does this mean? These are just linear combinations of the column vectors of A. So another way to say that is for t to be onto, so for t to be surjective or onto, the column vectors of A have to span Rm, have to span our codomain."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And in order for t to be onto, this combination has to be equal to any vector in Rm. Well, what does this mean? These are just linear combinations of the column vectors of A. So another way to say that is for t to be onto, so for t to be surjective or onto, the column vectors of A have to span Rm, have to span our codomain. They have to span this right here. You have to be able to get any vector here with a linear combination of these guys. And the linear combination is set up because the weights are just arbitrary members of the real numbers."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So another way to say that is for t to be onto, so for t to be surjective or onto, the column vectors of A have to span Rm, have to span our codomain. They have to span this right here. You have to be able to get any vector here with a linear combination of these guys. And the linear combination is set up because the weights are just arbitrary members of the real numbers. This vector is just a bunch of arbitrary real numbers. So for t to be onto, the span of A1, A2, all the way to An has to be equal to Rm, has to be equal to your codomain. That just means that you can achieve any vector in your codomain with the linear combinations of the column vectors of this."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And the linear combination is set up because the weights are just arbitrary members of the real numbers. This vector is just a bunch of arbitrary real numbers. So for t to be onto, the span of A1, A2, all the way to An has to be equal to Rm, has to be equal to your codomain. That just means that you can achieve any vector in your codomain with the linear combinations of the column vectors of this. Well, what's the span of a matrix's column vectors? By definition, that is the matrix's column space. So we could say, so that means that the span of these guys have to be Rn, or another way is that the column space of A, let me switch colors, the column space of A has to be equal to Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "That just means that you can achieve any vector in your codomain with the linear combinations of the column vectors of this. Well, what's the span of a matrix's column vectors? By definition, that is the matrix's column space. So we could say, so that means that the span of these guys have to be Rn, or another way is that the column space of A, let me switch colors, the column space of A has to be equal to Rm. So how do we know if a vector's column space is equal to Rm? And so here, maybe it might be instructive to think about when can we not find solutions to the equation Ax is equal to b. So whenever you're faced with this type of an equation, what do we do?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So we could say, so that means that the span of these guys have to be Rn, or another way is that the column space of A, let me switch colors, the column space of A has to be equal to Rm. So how do we know if a vector's column space is equal to Rm? And so here, maybe it might be instructive to think about when can we not find solutions to the equation Ax is equal to b. So whenever you're faced with this type of an equation, what do we do? We can set up an augmented matrix that looks like this, where you put A on this side, and then you put the vector b on the right-hand side. And then you essentially perform a bunch of row operations, and you have to perform on the entire rows, in both cases, on both sides. And we've done this multiple times."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So whenever you're faced with this type of an equation, what do we do? We can set up an augmented matrix that looks like this, where you put A on this side, and then you put the vector b on the right-hand side. And then you essentially perform a bunch of row operations, and you have to perform on the entire rows, in both cases, on both sides. And we've done this multiple times. And your goal is to get the left-hand side into reduced row echelon form. So what you want to do is eventually get your augmented matrix to look like this, where the left-handed side is, let me define R, big capital R, to be the reduced row echelon form of A. And we've done many videos on that."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And we've done this multiple times. And your goal is to get the left-hand side into reduced row echelon form. So what you want to do is eventually get your augmented matrix to look like this, where the left-handed side is, let me define R, big capital R, to be the reduced row echelon form of A. And we've done many videos on that. That's just, you have a matrix where you have your pivots, and the pivot will be the only non-zero entry in its column. But not every column has to have a pivot in it. Maybe you have a non-pivot column, and then it could have a bunch of 0's, and maybe this has a pivot."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And we've done many videos on that. That's just, you have a matrix where you have your pivots, and the pivot will be the only non-zero entry in its column. But not every column has to have a pivot in it. Maybe you have a non-pivot column, and then it could have a bunch of 0's, and maybe this has a pivot. This would have to be 0 if there's a pivot there. These would have to be 0, and so on and so forth. And maybe the next pivot is right there."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Maybe you have a non-pivot column, and then it could have a bunch of 0's, and maybe this has a pivot. This would have to be 0 if there's a pivot there. These would have to be 0, and so on and so forth. And maybe the next pivot is right there. These would have to be 0. And you get the idea. You could have some columns that don't have pivots, but whenever you do have a pivot, they have to be the only non-zero entry in their column."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And maybe the next pivot is right there. These would have to be 0. And you get the idea. You could have some columns that don't have pivots, but whenever you do have a pivot, they have to be the only non-zero entry in their column. This is reduced row echelon form. So what we do with any matrix is we keep performing those row operations so that we eventually get it into a reduced row echelon form. And as we do that, we're performing those same operations on the right-hand side."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "You could have some columns that don't have pivots, but whenever you do have a pivot, they have to be the only non-zero entry in their column. This is reduced row echelon form. So what we do with any matrix is we keep performing those row operations so that we eventually get it into a reduced row echelon form. And as we do that, we're performing those same operations on the right-hand side. We're performing on the entire row of these augmented matrices. So this guy, this vector b right here, I guess I could write it as a vector. It's eventually going to be some other vector c right here."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And as we do that, we're performing those same operations on the right-hand side. We're performing on the entire row of these augmented matrices. So this guy, this vector b right here, I guess I could write it as a vector. It's eventually going to be some other vector c right here. If this is 1, 2, 3, maybe I perform a bunch of operations, this will be 3, 2, 1, or something of that nature. Now, when does this not have a solution? And we went on this, we reviewed this early on."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "It's eventually going to be some other vector c right here. If this is 1, 2, 3, maybe I perform a bunch of operations, this will be 3, 2, 1, or something of that nature. Now, when does this not have a solution? And we went on this, we reviewed this early on. The only time where you don't have a solution, remember, there's three cases where you have many solutions, and that's the situation where you have three variables. We've talked about that before. You have the case where you have only one unique solution, one solution, that's the other case."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And we went on this, we reviewed this early on. The only time where you don't have a solution, remember, there's three cases where you have many solutions, and that's the situation where you have three variables. We've talked about that before. You have the case where you have only one unique solution, one solution, that's the other case. And then you have the final case where you have no solutions. And when do you have no solutions? What has to happen to have no solutions?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "You have the case where you have only one unique solution, one solution, that's the other case. And then you have the final case where you have no solutions. And when do you have no solutions? What has to happen to have no solutions? To have no solutions when you perform these row operations, you have to eventually get to a point where your matrix looks something like this. I don't know what all of this stuff looks like. Maybe there's a 1 here, a bunch of stuff."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "What has to happen to have no solutions? To have no solutions when you perform these row operations, you have to eventually get to a point where your matrix looks something like this. I don't know what all of this stuff looks like. Maybe there's a 1 here, a bunch of stuff. There's a 1 here, and a 0. But if you have a whole row, at least one whole row of 0's, so you just have a bunch of 0's like that, and over here you have something that is non-zero. This is the only time that you have no solution."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Maybe there's a 1 here, a bunch of stuff. There's a 1 here, and a 0. But if you have a whole row, at least one whole row of 0's, so you just have a bunch of 0's like that, and over here you have something that is non-zero. This is the only time that you have no solution. So let's remember what I'm even talking about this stuff for. We're saying that our transformation is onto, if it's column vectors or if it's column space, is Rm. If it's column vectors, span Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "This is the only time that you have no solution. So let's remember what I'm even talking about this stuff for. We're saying that our transformation is onto, if it's column vectors or if it's column space, is Rm. If it's column vectors, span Rm. And what I'm trying to figure out is how do I know that it spans Rm? And so essentially, for it to span Rm, you can give me any b here, any b that's a member of Rm, and I should be able to get a solution. So we asked ourselves the question, when do we not get a solution?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "If it's column vectors, span Rm. And what I'm trying to figure out is how do I know that it spans Rm? And so essentially, for it to span Rm, you can give me any b here, any b that's a member of Rm, and I should be able to get a solution. So we asked ourselves the question, when do we not get a solution? So we don't get a solution in two situations. Well, we definitely don't get a solution if we have a bunch of 0's in a row, and then we have something non-zero here. That's definitely not going to be a solution."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So we asked ourselves the question, when do we not get a solution? So we don't get a solution in two situations. Well, we definitely don't get a solution if we have a bunch of 0's in a row, and then we have something non-zero here. That's definitely not going to be a solution. Now, there's the other case where we have a bunch of 0's. So there's the other case where we do have some solutions, where it's only valid for certain b's. So that's the case."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "That's definitely not going to be a solution. Now, there's the other case where we have a bunch of 0's. So there's the other case where we do have some solutions, where it's only valid for certain b's. So that's the case. Let me draw it like this. Let me start it like this. So let's say I have my matrix A, and I have my b1, b2."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So that's the case. Let me draw it like this. Let me start it like this. So let's say I have my matrix A, and I have my b1, b2. Let me write it all the way. And then you have bm. Remember, this is a member of Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have my matrix A, and I have my b1, b2. Let me write it all the way. And then you have bm. Remember, this is a member of Rm. And we do our reduced row echelon form with this augmented matrix. And A goes to its reduced row echelon form. And let's say its reduced row echelon form has a row of 0's at the end of it."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Remember, this is a member of Rm. And we do our reduced row echelon form with this augmented matrix. And A goes to its reduced row echelon form. And let's say its reduced row echelon form has a row of 0's at the end of it. So it just has a row of 0's right there. Everything else looks like your standard stuff. 1's and 0's."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And let's say its reduced row echelon form has a row of 0's at the end of it. So it just has a row of 0's right there. Everything else looks like your standard stuff. 1's and 0's. But the last row, let's say it's a bunch of 0's. And when we perform the row operations here on this generalized member of Rm, this last row has some function. Maybe it just looks like 2b1 plus 3b2."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "1's and 0's. But the last row, let's say it's a bunch of 0's. And when we perform the row operations here on this generalized member of Rm, this last row has some function. Maybe it just looks like 2b1 plus 3b2. I'm just writing a particular case. It won't always be this. Minus b3."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Maybe it just looks like 2b1 plus 3b2. I'm just writing a particular case. It won't always be this. Minus b3. And it'll be a bunch of, it'll essentially be some function of all of the b's. So let me write it this way. I'm writing a particular case in here."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Minus b3. And it'll be a bunch of, it'll essentially be some function of all of the b's. So let me write it this way. I'm writing a particular case in here. Maybe I shouldn't have written a particular case. This will be some function of b1, b2, all the way to bm. Now, clearly if this is non-zero, we don't have a solution."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "I'm writing a particular case in here. Maybe I shouldn't have written a particular case. This will be some function of b1, b2, all the way to bm. Now, clearly if this is non-zero, we don't have a solution. And so if we don't have a solution for some cases of b, then we are definitely not spanning Rm. So let me write that down. If we don't have a solution for some cases of b, then we don't span Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Now, clearly if this is non-zero, we don't have a solution. And so if we don't have a solution for some cases of b, then we are definitely not spanning Rm. So let me write that down. If we don't have a solution for some cases of b, then we don't span Rm. Let me, I don't know if I'm overstating something that is maybe obvious to you, but I really want to make sure you understand this. Anytime you just want to solve the equation Ax is equal to b, and remember, we want to make sure that this can be true for any b we chose. What we could do is we just set up this augmented matrix like this, and we perform a bunch of row operations until we get this matrix A into reduced row echelon form."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "If we don't have a solution for some cases of b, then we don't span Rm. Let me, I don't know if I'm overstating something that is maybe obvious to you, but I really want to make sure you understand this. Anytime you just want to solve the equation Ax is equal to b, and remember, we want to make sure that this can be true for any b we chose. What we could do is we just set up this augmented matrix like this, and we perform a bunch of row operations until we get this matrix A into reduced row echelon form. As we do this, the right-hand side is going to be a bunch of functions of b. So maybe the first row is b1 minus b2 plus b4, or something, and the next row will be something like that. We've seen examples of this in the past."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "What we could do is we just set up this augmented matrix like this, and we perform a bunch of row operations until we get this matrix A into reduced row echelon form. As we do this, the right-hand side is going to be a bunch of functions of b. So maybe the first row is b1 minus b2 plus b4, or something, and the next row will be something like that. We've seen examples of this in the past. If you end up by doing the reduced row echelon form with a row of 0's here, the only way that you're going to have a solution is if your vector b, if its entries satisfy this function here on the right so that this thing equals 0. So it's only going to be true for certain b's. And if this only has a solution for the certain b's that make this equal to 0, then we definitely are not spanning all of Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "We've seen examples of this in the past. If you end up by doing the reduced row echelon form with a row of 0's here, the only way that you're going to have a solution is if your vector b, if its entries satisfy this function here on the right so that this thing equals 0. So it's only going to be true for certain b's. And if this only has a solution for the certain b's that make this equal to 0, then we definitely are not spanning all of Rm. Let me do it visually. So if that is Rm, and if this is only 0 for a couple of b's, for let's say for some handful of b's, then these are the only guys that we can reach by multiplying a times some vector in Rn. And we definitely won't be spanning all of Rm."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And if this only has a solution for the certain b's that make this equal to 0, then we definitely are not spanning all of Rm. Let me do it visually. So if that is Rm, and if this is only 0 for a couple of b's, for let's say for some handful of b's, then these are the only guys that we can reach by multiplying a times some vector in Rn. And we definitely won't be spanning all of Rm. In order to span all of Rm, when we put this in a reduced row echelon form, we have to always find a solution. And the only way we're always going to be finding a solution is if we don't run into this condition where we have a row of 0's, because when you have a row of 0's, then you have to put the constraint that this right hand side has to be 0. So what's the only reduced row echelon form where you don't have a row of 0's at the end?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And we definitely won't be spanning all of Rm. In order to span all of Rm, when we put this in a reduced row echelon form, we have to always find a solution. And the only way we're always going to be finding a solution is if we don't run into this condition where we have a row of 0's, because when you have a row of 0's, then you have to put the constraint that this right hand side has to be 0. So what's the only reduced row echelon form where you don't have a row of 0's at the end? Well, any row in reduced row echelon form either has to have all 0's, or it has to have a pivot entry in every row. So the only way that you span, so t is onto, if and only if, it's the column space of its transformation vector is equal to Rm, its column vectors span all of Rm. And the only way that that's happening is if the reduced row echelon form of A has a pivot entry in every row."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So what's the only reduced row echelon form where you don't have a row of 0's at the end? Well, any row in reduced row echelon form either has to have all 0's, or it has to have a pivot entry in every row. So the only way that you span, so t is onto, if and only if, it's the column space of its transformation vector is equal to Rm, its column vectors span all of Rm. And the only way that that's happening is if the reduced row echelon form of A has a pivot entry in every row. And how many rows does it have? This is an m by n matrix. It has m rows and n columns."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And the only way that that's happening is if the reduced row echelon form of A has a pivot entry in every row. And how many rows does it have? This is an m by n matrix. It has m rows and n columns. So it has a pivot entry in every row. That means that it has to have m pivot entries. Now what's another way of thinking about that?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "It has m rows and n columns. So it has a pivot entry in every row. That means that it has to have m pivot entries. Now what's another way of thinking about that? Remember, earlier on, several videos ago, we thought about how do you figure out, and this might confuse things a little bit, how do you figure out the basis for your column space? So how do you figure out the basis for your column space? So the basis for the column space of a matrix."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Now what's another way of thinking about that? Remember, earlier on, several videos ago, we thought about how do you figure out, and this might confuse things a little bit, how do you figure out the basis for your column space? So how do you figure out the basis for your column space? So the basis for the column space of a matrix. And this is a bit of a review. What we did is we say, look, you take your matrix and you put it in reduced row echelon form. So you put it in reduced row echelon form of your matrix."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So the basis for the column space of a matrix. And this is a bit of a review. What we did is we say, look, you take your matrix and you put it in reduced row echelon form. So you put it in reduced row echelon form of your matrix. And then essentially, let me draw it a little bit different here. Well, you put it in reduced row echelon form. So let's say that's its reduced row echelon form."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So you put it in reduced row echelon form of your matrix. And then essentially, let me draw it a little bit different here. Well, you put it in reduced row echelon form. So let's say that's its reduced row echelon form. And you look at which columns have pivot entries. And the corresponding columns in your original matrix forms the basis for your column space. So let me draw that out."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that's its reduced row echelon form. And you look at which columns have pivot entries. And the corresponding columns in your original matrix forms the basis for your column space. So let me draw that out. So I'll do a particular instance. So let's say that it has its column vectors a1, let's say a2, all the way to an. That's what a looks like."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw that out. So I'll do a particular instance. So let's say that it has its column vectors a1, let's say a2, all the way to an. That's what a looks like. And when you put it in reduced row echelon form, let's say that this column right here has a pivot entry. Let's say that this one doesn't. Let's say there's like a 2 there."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "That's what a looks like. And when you put it in reduced row echelon form, let's say that this column right here has a pivot entry. Let's say that this one doesn't. Let's say there's like a 2 there. I'm just picking particular numbers. Let's say that none of the other ones have any, let's say there's a 3. Let's say all of these are non-pivot entries."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Let's say there's like a 2 there. I'm just picking particular numbers. Let's say that none of the other ones have any, let's say there's a 3. Let's say all of these are non-pivot entries. And then our last entry, n, is a pivot entry. So it just has a bunch of 0's and then a 1 like that. How do you determine what are the basis vectors for the column space?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Let's say all of these are non-pivot entries. And then our last entry, n, is a pivot entry. So it just has a bunch of 0's and then a 1 like that. How do you determine what are the basis vectors for the column space? Well, obviously the column space is everything that's spanned by all of these guys. But what's the minimum set you need to have that same span? Well, you look at which one has a corresponding pivot entries or pivot columns."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "How do you determine what are the basis vectors for the column space? Well, obviously the column space is everything that's spanned by all of these guys. But what's the minimum set you need to have that same span? Well, you look at which one has a corresponding pivot entries or pivot columns. You say, I have a pivot column here and I have a pivot column there. So the basis for my column space must be this column in my original matrix and that column in my original matrix. And then we said, well, how do you define the dimension of its column space and you just essentially count the number of vectors you need for your basis?"}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Well, you look at which one has a corresponding pivot entries or pivot columns. You say, I have a pivot column here and I have a pivot column there. So the basis for my column space must be this column in my original matrix and that column in my original matrix. And then we said, well, how do you define the dimension of its column space and you just essentially count the number of vectors you need for your basis? And we call that the rank of A. So the rank of A, this is all review, the rank of A was equal to the dimension of the column space of A, which is equal to number of basis vectors for the column space. And this is how you determine it."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And then we said, well, how do you define the dimension of its column space and you just essentially count the number of vectors you need for your basis? And we call that the rank of A. So the rank of A, this is all review, the rank of A was equal to the dimension of the column space of A, which is equal to number of basis vectors for the column space. And this is how you determine it. You essentially figure out how many pivot columns you have. The number of pivot columns you have is the number of basis vectors you have. And so that's going to be the rank of A."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And this is how you determine it. You essentially figure out how many pivot columns you have. The number of pivot columns you have is the number of basis vectors you have. And so that's going to be the rank of A. Now the whole reason why I talked about this is we just said that our transformation T is onto if and only if its column space is Rm, which is the case if it has a pivot entry in every row in its reduced row echelon form, or since it has m rows, it has to have m pivot entries. So for every row you have a pivot entry, but every pivot entry corresponds to a pivot column. Every pivot entry corresponds to a pivot column."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And so that's going to be the rank of A. Now the whole reason why I talked about this is we just said that our transformation T is onto if and only if its column space is Rm, which is the case if it has a pivot entry in every row in its reduced row echelon form, or since it has m rows, it has to have m pivot entries. So for every row you have a pivot entry, but every pivot entry corresponds to a pivot column. Every pivot entry corresponds to a pivot column. So if you have m pivot entries, you also have m pivot columns, which means that if you were to do this exercise right here, you would have m basis vectors for your column space, or that you would have a rank of m. So this whole video was just a big, long way of saying that T is onto. And another way of saying that is that if you have your domain here, which was Rn, and you have your codomain here, that is Rm, that every member of Rm can be reached by T by some member of Rn. Any guy here, there's always at least one guy here that if you apply T to it, you get right there."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Every pivot entry corresponds to a pivot column. So if you have m pivot entries, you also have m pivot columns, which means that if you were to do this exercise right here, you would have m basis vectors for your column space, or that you would have a rank of m. So this whole video was just a big, long way of saying that T is onto. And another way of saying that is that if you have your domain here, which was Rn, and you have your codomain here, that is Rm, that every member of Rm can be reached by T by some member of Rn. Any guy here, there's always at least one guy here that if you apply T to it, you get right there. There might be more than one. We're not even talking about one to one yet. So we say that T is onto if and only if the rank of the rank of its transformation matrix, A, is equal to m. So that was the big takeaway of this video."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Any guy here, there's always at least one guy here that if you apply T to it, you get right there. There might be more than one. We're not even talking about one to one yet. So we say that T is onto if and only if the rank of the rank of its transformation matrix, A, is equal to m. So that was the big takeaway of this video. So actually, let's just actually do an example. Because sometimes when you do things really abstract, it seems a little bit confusing. When you see something particular."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So we say that T is onto if and only if the rank of the rank of its transformation matrix, A, is equal to m. So that was the big takeaway of this video. So actually, let's just actually do an example. Because sometimes when you do things really abstract, it seems a little bit confusing. When you see something particular. Let's see. Let me define some transformation, S. Let's say the transformation, S, is a mapping from R2 to R3. And let's say that S applied to some vector, x, is equal to the matrix 1, 2, 3, 4, 5, 6 times the vector, x."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "When you see something particular. Let's see. Let me define some transformation, S. Let's say the transformation, S, is a mapping from R2 to R3. And let's say that S applied to some vector, x, is equal to the matrix 1, 2, 3, 4, 5, 6 times the vector, x. So this is clearly a 3 by 2 matrix. Let's see if S is S onto. Well, based on what we just did, we just have to go and put this guy in reduced row echelon form."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that S applied to some vector, x, is equal to the matrix 1, 2, 3, 4, 5, 6 times the vector, x. So this is clearly a 3 by 2 matrix. Let's see if S is S onto. Well, based on what we just did, we just have to go and put this guy in reduced row echelon form. So let's do that. So if you put this guy into reduced row echelon form. So let's keep 1, 2, 3, 4, 5, 6."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Well, based on what we just did, we just have to go and put this guy in reduced row echelon form. So let's do that. So if you put this guy into reduced row echelon form. So let's keep 1, 2, 3, 4, 5, 6. Now let's keep the first row the same. It's 1, 2. Let's replace the second row with the second row minus 3 times the first row."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So let's keep 1, 2, 3, 4, 5, 6. Now let's keep the first row the same. It's 1, 2. Let's replace the second row with the second row minus 3 times the first row. Actually, let's replace it with 3 times the first row minus the second row. So 3 times 1 minus 3 is 0. 3 times 2 minus 4, that's 6 minus 4, is 2."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Let's replace the second row with the second row minus 3 times the first row. Actually, let's replace it with 3 times the first row minus the second row. So 3 times 1 minus 3 is 0. 3 times 2 minus 4, that's 6 minus 4, is 2. Now let's replace the third row with 5 times the first row minus the third row. So 5 times 1 minus 5 is 0. 5 times 2 is 10 minus 6 is 4."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "3 times 2 minus 4, that's 6 minus 4, is 2. Now let's replace the third row with 5 times the first row minus the third row. So 5 times 1 minus 5 is 0. 5 times 2 is 10 minus 6 is 4. Now let's see if we can get a 1 right here. So I'm going to keep my middle row the same. Actually, let's just divide the middle row by 2, or multiply it by 1 half."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "5 times 2 is 10 minus 6 is 4. Now let's see if we can get a 1 right here. So I'm going to keep my middle row the same. Actually, let's just divide the middle row by 2, or multiply it by 1 half. So you get 0, 1, and we have a 0, 4, 1, 2. And now let's try to make these 0. Again, reduced row echelon form."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let's just divide the middle row by 2, or multiply it by 1 half. So you get 0, 1, and we have a 0, 4, 1, 2. And now let's try to make these 0. Again, reduced row echelon form. So I'm going to keep my middle row the same, 0, 1. And let's replace the top row with the top row minus 2 times the second row. So 1 minus 2 times 0 is 1."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "Again, reduced row echelon form. So I'm going to keep my middle row the same, 0, 1. And let's replace the top row with the top row minus 2 times the second row. So 1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. Let's replace this last row with the last row minus 4 times this row. So we get 0 minus 4 times that is 0."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So 1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. Let's replace this last row with the last row minus 4 times this row. So we get 0 minus 4 times that is 0. 4 minus 4 times 1 is 0. So notice, we have a row with 0's here. We have 2 pivot entries, or 2 rows with pivot entries."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "So we get 0 minus 4 times that is 0. 4 minus 4 times 1 is 0. So notice, we have a row with 0's here. We have 2 pivot entries, or 2 rows with pivot entries. And we also have 2 pivot columns right there. So the rank of this guy right here, the rank of 1, 2, 3, 4, 5, 6, the rank of that is equal to 2, which is not equal to our codomain. It is not equal to R3."}, {"video_title": "Determining whether a transformation is onto Linear Algebra Khan Academy.mp3", "Sentence": "We have 2 pivot entries, or 2 rows with pivot entries. And we also have 2 pivot columns right there. So the rank of this guy right here, the rank of 1, 2, 3, 4, 5, 6, the rank of that is equal to 2, which is not equal to our codomain. It is not equal to R3. So S is not onto, or not surjective. And that's one of the two conditions for invertibility. So we definitely know that S is not invertible."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "We've defined the notion of a projection onto a subspace, but I haven't shown you yet that it's definitely a linear transformation, nor have I shown you that if you know a basis for its subspace, how do you actually find the projection onto it? So let's see if we can make any progress on that here. So let's say I have some subspace. Let's say v is a subspace of Rn. And let's say I've got some basis vectors for v. So let's say these are my basis vectors. Basis vector 1, 2, 2, and I have k of them. I don't know what v's dimension is, but let's say it's k. It's got k basis vectors."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Let's say v is a subspace of Rn. And let's say I've got some basis vectors for v. So let's say these are my basis vectors. Basis vector 1, 2, 2, and I have k of them. I don't know what v's dimension is, but let's say it's k. It's got k basis vectors. So it is a basis for v. And that means that any vector, that's not v vector, that's v subspace. Now that means that any vector, let me call some vector, I don't know, let's say any vector a that is a member of my subspace can be represented. That means that a can be represented as a linear combination of these guys."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "I don't know what v's dimension is, but let's say it's k. It's got k basis vectors. So it is a basis for v. And that means that any vector, that's not v vector, that's v subspace. Now that means that any vector, let me call some vector, I don't know, let's say any vector a that is a member of my subspace can be represented. That means that a can be represented as a linear combination of these guys. So I'll make my linear combination, let's say it is y1 times b1 plus y2 times b2 all the way to plus yk times bk. That's what the definition of a basis is. The span of these guys is your subspace v. So any member of v can be represented as a linear combination of my basis vectors."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "That means that a can be represented as a linear combination of these guys. So I'll make my linear combination, let's say it is y1 times b1 plus y2 times b2 all the way to plus yk times bk. That's what the definition of a basis is. The span of these guys is your subspace v. So any member of v can be represented as a linear combination of my basis vectors. Now if I were to construct a matrix, let's make it an n by k matrix, whose columns are essentially the basis vectors of my subspace. So a looks like this. The first column is my first basis vector."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "The span of these guys is your subspace v. So any member of v can be represented as a linear combination of my basis vectors. Now if I were to construct a matrix, let's make it an n by k matrix, whose columns are essentially the basis vectors of my subspace. So a looks like this. The first column is my first basis vector. My second column is my second basis vector. And I go all the way to my kth column, and I have k columns, is going to be my kth basis vector. If I'm going to have my kth basis vector, let me make the closing bracket the same color as my opening bracket, just like that, it's going to have n rows because each of these basis vectors are members of Rn."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "The first column is my first basis vector. My second column is my second basis vector. And I go all the way to my kth column, and I have k columns, is going to be my kth basis vector. If I'm going to have my kth basis vector, let me make the closing bracket the same color as my opening bracket, just like that, it's going to have n rows because each of these basis vectors are members of Rn. Remember, v is a subspace of Rn, so each of these guys are going to have n terms. So this matrix is going to have n rows. Now, saying that any member of the subspace v can be represented as a linear combination of these basis vectors is equivalent to saying that any member a of our subspace v can be represented as the product of our matrix a times some vector y, where somebody is equal to a for some y that is a member of Rk."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "If I'm going to have my kth basis vector, let me make the closing bracket the same color as my opening bracket, just like that, it's going to have n rows because each of these basis vectors are members of Rn. Remember, v is a subspace of Rn, so each of these guys are going to have n terms. So this matrix is going to have n rows. Now, saying that any member of the subspace v can be represented as a linear combination of these basis vectors is equivalent to saying that any member a of our subspace v can be represented as the product of our matrix a times some vector y, where somebody is equal to a for some y that is a member of Rk. Now, why is this statement and that statement equivalent? Well, you can imagine, if you were to just multiply this times some vector y in Rk, so it's y1, y2, all the way down to yk, this is going to be equal to y1 times b1 plus y2 times b2, all the way to plus yk times bk, which is the same thing as this. So you can always pick the right linear combination."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Now, saying that any member of the subspace v can be represented as a linear combination of these basis vectors is equivalent to saying that any member a of our subspace v can be represented as the product of our matrix a times some vector y, where somebody is equal to a for some y that is a member of Rk. Now, why is this statement and that statement equivalent? Well, you can imagine, if you were to just multiply this times some vector y in Rk, so it's y1, y2, all the way down to yk, this is going to be equal to y1 times b1 plus y2 times b2, all the way to plus yk times bk, which is the same thing as this. So you can always pick the right linear combination. You can always pick the right member of yk so that you get the right linear combination of your basis vectors to get any member of your subspace v. So any member of my subspace right there can be represented as the product of the matrix a with some vector in Rk. And we don't know much about this vector right here in Rk. Now, the projection, let's say that x is just some arbitrary member of Rn."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So you can always pick the right linear combination. You can always pick the right member of yk so that you get the right linear combination of your basis vectors to get any member of your subspace v. So any member of my subspace right there can be represented as the product of the matrix a with some vector in Rk. And we don't know much about this vector right here in Rk. Now, the projection, let's say that x is just some arbitrary member of Rn. The projection of x onto our subspace v, that is, by definition, it is going to be a member of your subspace. Or another way of saying it is that this guy, the projection onto v of x is going to be equal to my matrix A, I'll do it in blue, is going to be equal to A times some vector y or some vector y in Rk. If we knew what that vector y was, if we could always find it, then we would have a formula, so to speak, for figuring out the projection of x onto v. But we don't have that yet."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Now, the projection, let's say that x is just some arbitrary member of Rn. The projection of x onto our subspace v, that is, by definition, it is going to be a member of your subspace. Or another way of saying it is that this guy, the projection onto v of x is going to be equal to my matrix A, I'll do it in blue, is going to be equal to A times some vector y or some vector y in Rk. If we knew what that vector y was, if we could always find it, then we would have a formula, so to speak, for figuring out the projection of x onto v. But we don't have that yet. All I've said is any member of v can be represented as a product of our matrix A, which has the basis for v as columns, and some member of Rk. That just comes out of the fact that these guys span v. That any member of v is a linear combination of those guys. We know that the projection of x onto v is a member of our subspace v. It has to be inside of v. So it can also be represented this way."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "If we knew what that vector y was, if we could always find it, then we would have a formula, so to speak, for figuring out the projection of x onto v. But we don't have that yet. All I've said is any member of v can be represented as a product of our matrix A, which has the basis for v as columns, and some member of Rk. That just comes out of the fact that these guys span v. That any member of v is a linear combination of those guys. We know that the projection of x onto v is a member of our subspace v. It has to be inside of v. So it can also be represented this way. Now, what was our definition of our projection? Our definition of our projection, we say, well, let me write it this way. We know that x can be represented as the sum of the projection onto v of x plus some member of v complement."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "We know that the projection of x onto v is a member of our subspace v. It has to be inside of v. So it can also be represented this way. Now, what was our definition of our projection? Our definition of our projection, we say, well, let me write it this way. We know that x can be represented as the sum of the projection onto v of x plus some member of v complement. Or maybe I could even write plus the projection onto the orthogonal complement of v. You could write it this way. I could have also written this as w, where w is a member of v complement. Actually, let me write it that way."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "We know that x can be represented as the sum of the projection onto v of x plus some member of v complement. Or maybe I could even write plus the projection onto the orthogonal complement of v. You could write it this way. I could have also written this as w, where w is a member of v complement. Actually, let me write it that way. That might make it simpler. I don't want to write too many projections here. Plus w, where w is a unique member of the orthogonal complement of v. Or you could say it this way."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me write it that way. That might make it simpler. I don't want to write too many projections here. Plus w, where w is a unique member of the orthogonal complement of v. Or you could say it this way. If you subtract the projection of x onto v from both sides, you get x minus the projection of x onto v is equal to w. Or another way to say it is that this guy right here is going to be a member of the orthogonal complement of v. Because this is the same thing as w. Now, what's the orthogonal complement of v? If we go back to this matrix here, I have these basis vectors right here as the columns. So the column space of A is going to be equal to v. The column space of A is just the span of these basis vectors."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Plus w, where w is a unique member of the orthogonal complement of v. Or you could say it this way. If you subtract the projection of x onto v from both sides, you get x minus the projection of x onto v is equal to w. Or another way to say it is that this guy right here is going to be a member of the orthogonal complement of v. Because this is the same thing as w. Now, what's the orthogonal complement of v? If we go back to this matrix here, I have these basis vectors right here as the columns. So the column space of A is going to be equal to v. The column space of A is just the span of these basis vectors. And by definition, that is going to be equal to my subspace v. Now, what is the orthogonal complement of v? The orthogonal complement of v is going to be the orthogonal complement of my column space. And what's the orthogonal complement of a column space?"}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So the column space of A is going to be equal to v. The column space of A is just the span of these basis vectors. And by definition, that is going to be equal to my subspace v. Now, what is the orthogonal complement of v? The orthogonal complement of v is going to be the orthogonal complement of my column space. And what's the orthogonal complement of a column space? That's equal to the null space of A transpose. Or you could also call that the left null space of A. But that, we've seen that many, many videos ago."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "And what's the orthogonal complement of a column space? That's equal to the null space of A transpose. Or you could also call that the left null space of A. But that, we've seen that many, many videos ago. So we could say that x minus the projection of x onto v is a member of, let me write it this way, x minus the projection of x onto v. x minus the projection onto v of x is a member of the orthogonal complement of my column space of my matrix, which is equal to the null space of A transpose. That's the orthogonal complement of v. This is the same thing as the orthogonal complement of v. But what does this mean? What does this mean right here?"}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "But that, we've seen that many, many videos ago. So we could say that x minus the projection of x onto v is a member of, let me write it this way, x minus the projection of x onto v. x minus the projection onto v of x is a member of the orthogonal complement of my column space of my matrix, which is equal to the null space of A transpose. That's the orthogonal complement of v. This is the same thing as the orthogonal complement of v. But what does this mean? What does this mean right here? This means that if I take A transpose and I multiply it times this vector, because it's a member of A transpose's null space. So if I multiply it times that vector right there, so projection of x onto v, that I'm going to get 0. I'm going to get the 0 vector."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "What does this mean right here? This means that if I take A transpose and I multiply it times this vector, because it's a member of A transpose's null space. So if I multiply it times that vector right there, so projection of x onto v, that I'm going to get 0. I'm going to get the 0 vector. That's the definition of a null space. So let's write this out a little bit more. Let's see if we can algebraically manipulate it a bit."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to get the 0 vector. That's the definition of a null space. So let's write this out a little bit more. Let's see if we can algebraically manipulate it a bit. So if we distribute this matrix vector product, we get A transpose times the vector x minus A transpose times the projection, actually let me write this this way. Instead of keep writing the projection onto v of x, what did we say earlier in this video? We said that the projection of v onto x can be represented as the matrix product of the matrix A times some vector y in Rk."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if we can algebraically manipulate it a bit. So if we distribute this matrix vector product, we get A transpose times the vector x minus A transpose times the projection, actually let me write this this way. Instead of keep writing the projection onto v of x, what did we say earlier in this video? We said that the projection of v onto x can be represented as the matrix product of the matrix A times some vector y in Rk. That's where we started off the video. So let me write it that way, because that's going to simplify our work a little bit. So I'm going to distribute the A transpose, A transpose times x, and then A transpose minus A transpose times this thing."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "We said that the projection of v onto x can be represented as the matrix product of the matrix A times some vector y in Rk. That's where we started off the video. So let me write it that way, because that's going to simplify our work a little bit. So I'm going to distribute the A transpose, A transpose times x, and then A transpose minus A transpose times this thing. This thing I can write as A times some vector y. And this is just a byproduct of the notion that the projection is a member of our subspace. Because it's a member of our subspace, it's going to be some linear combination of the column vectors of A."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to distribute the A transpose, A transpose times x, and then A transpose minus A transpose times this thing. This thing I can write as A times some vector y. And this is just a byproduct of the notion that the projection is a member of our subspace. Because it's a member of our subspace, it's going to be some linear combination of the column vectors of A. We saw that up here. So it can be represented in this way. So instead of projection onto v of x, I can just write Ay."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Because it's a member of our subspace, it's going to be some linear combination of the column vectors of A. We saw that up here. So it can be represented in this way. So instead of projection onto v of x, I can just write Ay. This thing and this thing are equivalent, because this thing is a member of v. And all of that is going to be equal to 0. And then if we add this to both sides of this equation, we get that A transpose x is equal to A transpose A of y. Now this is interesting."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So instead of projection onto v of x, I can just write Ay. This thing and this thing are equivalent, because this thing is a member of v. And all of that is going to be equal to 0. And then if we add this to both sides of this equation, we get that A transpose x is equal to A transpose A of y. Now this is interesting. Now remember where we started off here. We said that the projection onto v of x is equal to Ay for some y that is a member of Rk. If we knew what that y was, if we could always solve for that y, then the projection of x would be well-defined."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Now this is interesting. Now remember where we started off here. We said that the projection onto v of x is equal to Ay for some y that is a member of Rk. If we knew what that y was, if we could always solve for that y, then the projection of x would be well-defined. And we could always just figure it out. Now, can we solve for y here? Well, we'll be able to solve for y if we can take the inverse of that matrix."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "If we knew what that y was, if we could always solve for that y, then the projection of x would be well-defined. And we could always just figure it out. Now, can we solve for y here? Well, we'll be able to solve for y if we can take the inverse of that matrix. If this matrix is always invertible, then we're always going to be able to solve for y here. Because we can just take the inverse of this matrix and multiply it times the left side of both sides of this equation. Now if you remember, three videos ago, I think it was three videos ago, I showed you that if I have a matrix A whose columns are linearly independent, then A transpose A is always invertible."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Well, we'll be able to solve for y if we can take the inverse of that matrix. If this matrix is always invertible, then we're always going to be able to solve for y here. Because we can just take the inverse of this matrix and multiply it times the left side of both sides of this equation. Now if you remember, three videos ago, I think it was three videos ago, I showed you that if I have a matrix A whose columns are linearly independent, then A transpose A is always invertible. The whole reason why I did that video is for this moment right here. Now, what about our matrix A? Well, our matrix A has column vectors that form the basis for a subspace."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Now if you remember, three videos ago, I think it was three videos ago, I showed you that if I have a matrix A whose columns are linearly independent, then A transpose A is always invertible. The whole reason why I did that video is for this moment right here. Now, what about our matrix A? Well, our matrix A has column vectors that form the basis for a subspace. By definition, basis vectors are linearly independent. So our matrix A has columns that are linearly independent. And if you watched that video and if you believed what I told you, then you know that A transpose A, in our case, is going to be invertible."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Well, our matrix A has column vectors that form the basis for a subspace. By definition, basis vectors are linearly independent. So our matrix A has columns that are linearly independent. And if you watched that video and if you believed what I told you, then you know that A transpose A, in our case, is going to be invertible. It has to be invertible. So let's take the inverse of it and multiply it times both sides. So if we take A transpose A inverse, we know that this exists because A has linearly independent columns."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "And if you watched that video and if you believed what I told you, then you know that A transpose A, in our case, is going to be invertible. It has to be invertible. So let's take the inverse of it and multiply it times both sides. So if we take A transpose A inverse, we know that this exists because A has linearly independent columns. And multiply it times this side right here, A transpose x. And then on this side, we get, we're going to do the same thing, A transpose A inverse times this thing right here, A transpose Ay. These two things, when you multiply them, when you multiply the inverse of a matrix times a matrix, you're just going to get the identity matrix."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So if we take A transpose A inverse, we know that this exists because A has linearly independent columns. And multiply it times this side right here, A transpose x. And then on this side, we get, we're going to do the same thing, A transpose A inverse times this thing right here, A transpose Ay. These two things, when you multiply them, when you multiply the inverse of a matrix times a matrix, you're just going to get the identity matrix. So that's just going to be equal to the identity matrix. And the identity matrix times y is just going to be y. So we get, and this is a vector, so if I flip them around, I get that y is equal to this expression right here, A transpose A inverse, which will always exist, times A transpose times x."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "These two things, when you multiply them, when you multiply the inverse of a matrix times a matrix, you're just going to get the identity matrix. So that's just going to be equal to the identity matrix. And the identity matrix times y is just going to be y. So we get, and this is a vector, so if I flip them around, I get that y is equal to this expression right here, A transpose A inverse, which will always exist, times A transpose times x. Now we said that the projection of x onto v is going to be equal to A times y for some y. Well, we just solved for the y using our definition of our projection. We just were able to solve for y."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So we get, and this is a vector, so if I flip them around, I get that y is equal to this expression right here, A transpose A inverse, which will always exist, times A transpose times x. Now we said that the projection of x onto v is going to be equal to A times y for some y. Well, we just solved for the y using our definition of our projection. We just were able to solve for y. So now we can define our projection of x onto v as a matrix vector product. So we can write the projection onto v of our vector x is equal to A times y. And y is just equal to that thing right there."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "We just were able to solve for y. So now we can define our projection of x onto v as a matrix vector product. So we can write the projection onto v of our vector x is equal to A times y. And y is just equal to that thing right there. So A times A transpose A inverse, which always exists because A has linearly independent columns, times A transpose times x. And this thing right here, this long, convoluted thing, that's just some matrix. Some matrix which always exists for any subspace that has some basis."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "And y is just equal to that thing right there. So A times A transpose A inverse, which always exists because A has linearly independent columns, times A transpose times x. And this thing right here, this long, convoluted thing, that's just some matrix. Some matrix which always exists for any subspace that has some basis. So we've just been able to express the projection of x onto a subspace as a matrix vector product. So this says, so anything that can be any matrix vector product transformation is a linear transformation. And not only did we show that it's a linear transformation, we showed that, look, if you can give me the basis for v, I'm going to throw that basis into some, I'm going to make those column vectors equal to the column of some matrix A."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "Some matrix which always exists for any subspace that has some basis. So we've just been able to express the projection of x onto a subspace as a matrix vector product. So this says, so anything that can be any matrix vector product transformation is a linear transformation. And not only did we show that it's a linear transformation, we showed that, look, if you can give me the basis for v, I'm going to throw that basis into some, I'm going to make those column vectors equal to the column of some matrix A. And then if I take matrix A, if I take its transpose, if I take A transpose times A and invert it, and if I multiply them all out in this way, I'm going to get the transformation matrix for the projection. Now this might seem really complicated, and it is hard to do by hand for many, many projections. But this is super useful if you're going to do some three-dimensional graphical programming."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "And not only did we show that it's a linear transformation, we showed that, look, if you can give me the basis for v, I'm going to throw that basis into some, I'm going to make those column vectors equal to the column of some matrix A. And then if I take matrix A, if I take its transpose, if I take A transpose times A and invert it, and if I multiply them all out in this way, I'm going to get the transformation matrix for the projection. Now this might seem really complicated, and it is hard to do by hand for many, many projections. But this is super useful if you're going to do some three-dimensional graphical programming. Let's say you have something in, you have these cubed, let's say you have some three-dimensional object. And you want to know what it looks like from the point of view of some observer. So let's say you have some observer."}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "But this is super useful if you're going to do some three-dimensional graphical programming. Let's say you have something in, you have these cubed, let's say you have some three-dimensional object. And you want to know what it looks like from the point of view of some observer. So let's say you have some observer. Some observer's point of view is essentially going to be some subspace. You want to see what the projection of this cube onto the subspace, how would it look to the person who's essentially onto this flat screen right there. How would that cube look from this point of view?"}, {"video_title": "A projection onto a subspace is a linear transformation Linear Algebra Khan Academy.mp3", "Sentence": "So let's say you have some observer. Some observer's point of view is essentially going to be some subspace. You want to see what the projection of this cube onto the subspace, how would it look to the person who's essentially onto this flat screen right there. How would that cube look from this point of view? Well, if you know the basis for this subspace, you can just apply this transformation. You can make a matrix whose columns are these basis vectors for this observer's point of view. And then you can apply this to every vector in this cube in R3 and you will know exactly how this cube should look from this person's point of view."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And we used the fact that lambda is an eigenvalue of A if and only if the determinant of lambda times the identity matrix, in this case it's the 2 by 2 identity matrix, minus A is equal to 0. This gave us a characteristic polynomial. And we solved for that. And we said, well, the eigenvalues for A are lambda is equal to 5 and lambda is equal to negative 1. That's what we saw in the last video. We said that if you were trying to solve A times some eigenvector is equal to lambda times that eigenvector, the 2 lambdas, which this equation can be solved for, are the lambdas 5 and minus 1. If we assume non-zero eigenvectors."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And we said, well, the eigenvalues for A are lambda is equal to 5 and lambda is equal to negative 1. That's what we saw in the last video. We said that if you were trying to solve A times some eigenvector is equal to lambda times that eigenvector, the 2 lambdas, which this equation can be solved for, are the lambdas 5 and minus 1. If we assume non-zero eigenvectors. So we have our eigenvalues, but I don't even call that half the battle. What we really want is our eigenvectors and our eigenvalues. So let's see if we can do that."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "If we assume non-zero eigenvectors. So we have our eigenvalues, but I don't even call that half the battle. What we really want is our eigenvectors and our eigenvalues. So let's see if we can do that. So if we manipulate this equation a little bit, and we manipulated it in the past, actually, to even come up with this statement over here, we can rewrite this over here as the zero vector is equal to lambda times my eigenvector minus A times my eigenvector, just subtracted Av from both sides, and we know lambda times some eigenvector is the same thing as lambda times the identity matrix times that eigenvector. So all I'm doing is rewriting this like that. You multiply the identity matrix times an eigenvector or times any vector, you're just going to get that vector."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if we can do that. So if we manipulate this equation a little bit, and we manipulated it in the past, actually, to even come up with this statement over here, we can rewrite this over here as the zero vector is equal to lambda times my eigenvector minus A times my eigenvector, just subtracted Av from both sides, and we know lambda times some eigenvector is the same thing as lambda times the identity matrix times that eigenvector. So all I'm doing is rewriting this like that. You multiply the identity matrix times an eigenvector or times any vector, you're just going to get that vector. So these two things are equivalent. Minus Av, that's still going to be equal to the zero vector. So far, all I've done is manipulated this thing, and this is really how we got to that thing up there."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "You multiply the identity matrix times an eigenvector or times any vector, you're just going to get that vector. So these two things are equivalent. Minus Av, that's still going to be equal to the zero vector. So far, all I've done is manipulated this thing, and this is really how we got to that thing up there. Then you factor out the v, so to speak, because we know that matrix vector products exhibit the distributive property. And we get lambda times the identity matrix minus A times my eigenvector have got to be equal to zero. Or another way to say it is for any lambda eigenvalue, let me write it for any eigenvalue lambda, the eigenvectors that correspond to that lambda, we can call that the eigenspace for lambda."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So far, all I've done is manipulated this thing, and this is really how we got to that thing up there. Then you factor out the v, so to speak, because we know that matrix vector products exhibit the distributive property. And we get lambda times the identity matrix minus A times my eigenvector have got to be equal to zero. Or another way to say it is for any lambda eigenvalue, let me write it for any eigenvalue lambda, the eigenvectors that correspond to that lambda, we can call that the eigenspace for lambda. So that's a new word. Eigenspace. Eigenspace just means all of the eigenvectors that correspond to some eigenvalue."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is for any lambda eigenvalue, let me write it for any eigenvalue lambda, the eigenvectors that correspond to that lambda, we can call that the eigenspace for lambda. So that's a new word. Eigenspace. Eigenspace just means all of the eigenvectors that correspond to some eigenvalue. The eigenspace for some particular eigenvalue is going to be equal to the set of vectors that satisfy this equation. Well, the set of vectors that satisfy this equation is just the null space of that right there. So it's equal to the null space of this matrix right there, the null space of lambda times the identity matrix, n by n identity matrix, minus A."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Eigenspace just means all of the eigenvectors that correspond to some eigenvalue. The eigenspace for some particular eigenvalue is going to be equal to the set of vectors that satisfy this equation. Well, the set of vectors that satisfy this equation is just the null space of that right there. So it's equal to the null space of this matrix right there, the null space of lambda times the identity matrix, n by n identity matrix, minus A. And so everything I've done here, this is true. This is the general case. But now we can apply this notion to this matrix A right here."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to the null space of this matrix right there, the null space of lambda times the identity matrix, n by n identity matrix, minus A. And so everything I've done here, this is true. This is the general case. But now we can apply this notion to this matrix A right here. So we know that 5 is an eigenvalue. So for, let's say, for lambda is equal to 5, the eigenspace that corresponds to 5 is equal to the null space of, well, what is 5 times the identity matrix? 5 is going to be the 2 by 2 identity matrix, right?"}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "But now we can apply this notion to this matrix A right here. So we know that 5 is an eigenvalue. So for, let's say, for lambda is equal to 5, the eigenspace that corresponds to 5 is equal to the null space of, well, what is 5 times the identity matrix? 5 is going to be the 2 by 2 identity matrix, right? 5 times the identity matrix is just 5, 0, 0, 5. Minus A, that's just 1, 2, 4, 3. So that is equal to the null space of the matrix."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "5 is going to be the 2 by 2 identity matrix, right? 5 times the identity matrix is just 5, 0, 0, 5. Minus A, that's just 1, 2, 4, 3. So that is equal to the null space of the matrix. 5 minus 1 is 4. 0 minus 2 is minus 2. 0 minus 4 is minus 4."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So that is equal to the null space of the matrix. 5 minus 1 is 4. 0 minus 2 is minus 2. 0 minus 4 is minus 4. And then 5 minus 3 is 2. So the null space of this matrix right here, this matrix is just a actual numerical representation of this matrix right here, the null space of this matrix is the set of all of the vectors that satisfy this, or all of the eigenvectors that correspond to this eigenvalue, or the eigenspace that corresponds to the eigenvalue 5. These are all equivalent statements."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 4 is minus 4. And then 5 minus 3 is 2. So the null space of this matrix right here, this matrix is just a actual numerical representation of this matrix right here, the null space of this matrix is the set of all of the vectors that satisfy this, or all of the eigenvectors that correspond to this eigenvalue, or the eigenspace that corresponds to the eigenvalue 5. These are all equivalent statements. So we just need to figure out the null space of this guy is all the vectors that satisfy the equation. 4 minus 2 minus 4, 2 times some eigenvector is equal to the 0 vector. And the null space of a matrix is equal to the null space of the reduced row echelon form of a matrix."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "These are all equivalent statements. So we just need to figure out the null space of this guy is all the vectors that satisfy the equation. 4 minus 2 minus 4, 2 times some eigenvector is equal to the 0 vector. And the null space of a matrix is equal to the null space of the reduced row echelon form of a matrix. So what's the reduced row echelon form of this guy? Well, let's, I guess a good starting point. Let me keep my first row the same, 4 minus 2."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And the null space of a matrix is equal to the null space of the reduced row echelon form of a matrix. So what's the reduced row echelon form of this guy? Well, let's, I guess a good starting point. Let me keep my first row the same, 4 minus 2. And let me replace my second row with my second row plus my first row. So minus 4 plus 4 is 0. 2 plus minus 2 is 0."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Let me keep my first row the same, 4 minus 2. And let me replace my second row with my second row plus my first row. So minus 4 plus 4 is 0. 2 plus minus 2 is 0. Now let me divide my first row by 4. And I get 1 minus 1 half, and then I get 0, 0. So what's the null space of this?"}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "2 plus minus 2 is 0. Now let me divide my first row by 4. And I get 1 minus 1 half, and then I get 0, 0. So what's the null space of this? This corresponds to v, this times v1, v2. That's just another way of writing my eigenvector v. It's got to be equal to the 0 vector. Or another way to say it is that my first entry, v1, which corresponds to this pivot column, plus or minus 1 half times my second entry has got to be equal to that 0 right there, or v1 is equal to 1 half v2."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So what's the null space of this? This corresponds to v, this times v1, v2. That's just another way of writing my eigenvector v. It's got to be equal to the 0 vector. Or another way to say it is that my first entry, v1, which corresponds to this pivot column, plus or minus 1 half times my second entry has got to be equal to that 0 right there, or v1 is equal to 1 half v2. And so if I wanted to write all of the eigenvectors that satisfy this, so I could write it this way, my eigenspace that corresponds to lambda equals 5, that corresponds to the eigenvalue 5, is equal to the set of all of the vectors v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1, and then v1 is going to be equal to 1 half times v2, or 1 half times t. 1 half times t, just like that."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is that my first entry, v1, which corresponds to this pivot column, plus or minus 1 half times my second entry has got to be equal to that 0 right there, or v1 is equal to 1 half v2. And so if I wanted to write all of the eigenvectors that satisfy this, so I could write it this way, my eigenspace that corresponds to lambda equals 5, that corresponds to the eigenvalue 5, is equal to the set of all of the vectors v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1, and then v1 is going to be equal to 1 half times v2, or 1 half times t. 1 half times t, just like that. For any t is a member of the real numbers. And if we wanted to, we could scale this up. We could say any real number times 1, 2, that would also be a, you know, the span."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "If we say that v2 is equal to t, so v2 is going to be equal to t times 1, and then v1 is going to be equal to 1 half times v2, or 1 half times t. 1 half times t, just like that. For any t is a member of the real numbers. And if we wanted to, we could scale this up. We could say any real number times 1, 2, that would also be a, you know, the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "We could say any real number times 1, 2, that would also be a, you know, the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1 half and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy, that work for the equation where the eigenvalue is equal to 5."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1 half and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy, that work for the equation where the eigenvalue is equal to 5. Now, what about when the eigenvalue is equal to minus 1? When the eigenvalue is equal to minus 1. So let's do that case."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Those are all of the eigenvectors that satisfy, that work for the equation where the eigenvalue is equal to 5. Now, what about when the eigenvalue is equal to minus 1? When the eigenvalue is equal to minus 1. So let's do that case. When lambda is equal to minus 1, then we're going to have the, it's going to be the null space. So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1, 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that case. When lambda is equal to minus 1, then we're going to have the, it's going to be the null space. So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1, 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus a, so minus 1, 2, 4, 3. And this is equal to the null space of minus 1, minus 1 is minus 2, 0, minus 2 is minus 2, 0, minus 4 is minus 4, and minus 1, minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus a, so minus 1, 2, 4, 3. And this is equal to the null space of minus 1, minus 1 is minus 2, 0, minus 2 is minus 2, 0, minus 4 is minus 4, and minus 1, minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row with my second row plus 2 times my first row."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row with my second row plus 2 times my first row. So I'll keep the first row the same. Minus 2, minus 2. And then my second row I'll replace it with it plus 2 times the first."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So if I replace my second row with my second row plus 2 times my first row. So I'll keep the first row the same. Minus 2, minus 2. And then my second row I'll replace it with it plus 2 times the first. Or even better, I'm going to replace it with it plus minus 2 times the first. So minus 4 plus 4 is 0. Minus 4 plus 4 is 0."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And then my second row I'll replace it with it plus 2 times the first. Or even better, I'm going to replace it with it plus minus 2 times the first. So minus 4 plus 4 is 0. Minus 4 plus 4 is 0. And then if I divide the top row by minus 2, the reduced row echelon form of this matrix right here is going to be 1, 1, 0. So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here. Or it equals the set of vectors that satisfy this equation."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Minus 4 plus 4 is 0. And then if I divide the top row by minus 2, the reduced row echelon form of this matrix right here is going to be 1, 1, 0. So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here. Or it equals the set of vectors that satisfy this equation. 1, 1, 0, 0. And then you have v1, v2 is equal to 0. Or you get v1 plus, these aren't vectors, these are just values."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Or it equals the set of vectors that satisfy this equation. 1, 1, 0, 0. And then you have v1, v2 is equal to 0. Or you get v1 plus, these aren't vectors, these are just values. v1 plus v2 is equal to 0. Sorry, is equal to 0. Because 0 is just equal to that thing right there."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Or you get v1 plus, these aren't vectors, these are just values. v1 plus v2 is equal to 0. Sorry, is equal to 0. Because 0 is just equal to that thing right there. So 1 times v1 plus 1 times v2 is going to be equal to that 0 right there. Or I could write v1 is equal to minus v2. Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Because 0 is just equal to that thing right there. So 1 times v1 plus 1 times v2 is going to be equal to that 0 right there. Or I could write v1 is equal to minus v2. Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1. So let's just graph this a little bit just to understand what we just did. We were able to find two eigenvalues for this. 5 and minus 1."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1. So let's just graph this a little bit just to understand what we just did. We were able to find two eigenvalues for this. 5 and minus 1. And we were able to find all of the vectors that are essentially the set of vectors that are the eigenvectors that correspond to each of these eigenvalues. So let's graph them. So if we go to R2."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "5 and minus 1. And we were able to find all of the vectors that are essentially the set of vectors that are the eigenvectors that correspond to each of these eigenvalues. So let's graph them. So if we go to R2. Let me draw my axis. That's my vertical axis. That's my horizontal axis."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So if we go to R2. Let me draw my axis. That's my vertical axis. That's my horizontal axis. So all of the vectors that correspond to lambda equal 5 are aligned to line 1 half 1. So the span of 1 half 1. So that is 1."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "That's my horizontal axis. So all of the vectors that correspond to lambda equal 5 are aligned to line 1 half 1. So the span of 1 half 1. So that is 1. So you go 1 half and 1 just like that. So that's that vector. The spanning vector."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So that is 1. So you go 1 half and 1 just like that. So that's that vector. The spanning vector. But anything along the span of this, all the multiples of this, are going to be valid eigenvectors. So anything along that line, all of the vectors, when you draw them in standard position, point to a point on that line, all of these vectors, any vector on there is going to be a valid eigenvector. And the corresponding eigenvalue is going to be equal to 5."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "The spanning vector. But anything along the span of this, all the multiples of this, are going to be valid eigenvectors. So anything along that line, all of the vectors, when you draw them in standard position, point to a point on that line, all of these vectors, any vector on there is going to be a valid eigenvector. And the corresponding eigenvalue is going to be equal to 5. So you give me this guy right here. When you apply the transformation, it's going to be 5 times this guy. If this guy is x, T of x is going to be 5 times this guy."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And the corresponding eigenvalue is going to be equal to 5. So you give me this guy right here. When you apply the transformation, it's going to be 5 times this guy. If this guy is x, T of x is going to be 5 times this guy. Whatever vector you give along this line, the transformation of that guy, and the transformation is literally multiplying it by the matrix A. Where did I have the matrix A? The matrix A right up there."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "If this guy is x, T of x is going to be 5 times this guy. Whatever vector you give along this line, the transformation of that guy, and the transformation is literally multiplying it by the matrix A. Where did I have the matrix A? The matrix A right up there. You're essentially just scaling this guy by 5 in either direction. This is for lambda equals 5. And for lambda equals 1, it's the span of this vector, which is minus 1, minus 1, 1, which looks like this."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "The matrix A right up there. You're essentially just scaling this guy by 5 in either direction. This is for lambda equals 5. And for lambda equals 1, it's the span of this vector, which is minus 1, minus 1, 1, which looks like this. So this vector looks like that. We care about the span of it. So any vector in this set, any vector that when you draw in standard position, lies or points to points on this line will be an eigenvector for the eigenvalue minus 1."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And for lambda equals 1, it's the span of this vector, which is minus 1, minus 1, 1, which looks like this. So this vector looks like that. We care about the span of it. So any vector in this set, any vector that when you draw in standard position, lies or points to points on this line will be an eigenvector for the eigenvalue minus 1. So lambda equals minus 1. So if you take, let's say you take the spanning vector here, you apply the transformation, you're going to get minus 1 times it. So if this is x, the transformation of x is going to be that right there."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So any vector in this set, any vector that when you draw in standard position, lies or points to points on this line will be an eigenvector for the eigenvalue minus 1. So lambda equals minus 1. So if you take, let's say you take the spanning vector here, you apply the transformation, you're going to get minus 1 times it. So if this is x, the transformation of x is going to be that right there. Same length, just in the opposite direction. If you have this guy right here, you apply the transformation, it's going to be in the same spanning line. Just like that."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "So if this is x, the transformation of x is going to be that right there. Same length, just in the opposite direction. If you have this guy right here, you apply the transformation, it's going to be in the same spanning line. Just like that. So the two eigenspaces for the matrix, where did I write it? I think it was the matrix 1, 2, 3, 1, 2, 4, 3. The two eigenvalues were 5 and minus 1."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. So the two eigenspaces for the matrix, where did I write it? I think it was the matrix 1, 2, 3, 1, 2, 4, 3. The two eigenvalues were 5 and minus 1. And then it has an infinite number of eigenvectors, so they actually create two eigenspaces. Each of them correspond to one of the eigenvalues. And these lines represent those two eigenspaces."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "The two eigenvalues were 5 and minus 1. And then it has an infinite number of eigenvectors, so they actually create two eigenspaces. Each of them correspond to one of the eigenvalues. And these lines represent those two eigenspaces. You give me any vector in either of these sets, and they're going to be an eigenvector. Or you give me any vector, I'm using the word vector too much, you give me any vector in either of these sets, and they will be an eigenvector for our matrix A. And then, depending which line it is, we know what their transformation is going to be."}, {"video_title": "Finding eigenvectors and eigenspaces example Linear Algebra Khan Academy.mp3", "Sentence": "And these lines represent those two eigenspaces. You give me any vector in either of these sets, and they're going to be an eigenvector. Or you give me any vector, I'm using the word vector too much, you give me any vector in either of these sets, and they will be an eigenvector for our matrix A. And then, depending which line it is, we know what their transformation is going to be. It's going to be on this guy. You take the transformation, the resulting vector is going to be 5 times the vector. If you take one of these eigenvectors and you transform it, the resulting transformation of the vector is going to be minus 1 times that vector."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I've got some basis B, and it's made up of k vectors, let's say it's v1, v2, all the way to vk. And let's say I have some vector A, and I know what A's coordinates are with respect to B. So this is the coordinates of A with respect to B are c1, c2, and I'm going to have k coordinates, because we have k basis vectors, or if this describes a subspace, this is a k-dimensional subspace, so I'm going to have k of these guys right there. And all this means, by our definition of coordinates with respect to a basis, this literally means that I can represent my vector A as a linear combination of these guys, where these coordinates are the weights. So A would be equal to c1 times v1 plus c2 times v2 plus all the way, keep adding them up, all the way to ck times vk. Now, another way to write this is that, let me write it this way, if I had a matrix where the column vectors were the basis vectors of B, so let me write it just like that. So let me say I have some matrix C that looks like this, where its column vectors are just these basis vectors."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And all this means, by our definition of coordinates with respect to a basis, this literally means that I can represent my vector A as a linear combination of these guys, where these coordinates are the weights. So A would be equal to c1 times v1 plus c2 times v2 plus all the way, keep adding them up, all the way to ck times vk. Now, another way to write this is that, let me write it this way, if I had a matrix where the column vectors were the basis vectors of B, so let me write it just like that. So let me say I have some matrix C that looks like this, where its column vectors are just these basis vectors. So we have v1, v2, all the way to vk. If we assume that these are, let's say that all of these are a member of Rn, then each of these are going to have n entries. So it's going to be an n by k matrix."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let me say I have some matrix C that looks like this, where its column vectors are just these basis vectors. So we have v1, v2, all the way to vk. If we assume that these are, let's say that all of these are a member of Rn, then each of these are going to have n entries. So it's going to be an n by k matrix. Each of these guys have n entries, so we're going to have n rows. And we have k columns. So let's imagine this matrix right there."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be an n by k matrix. Each of these guys have n entries, so we're going to have n rows. And we have k columns. So let's imagine this matrix right there. Another way to write this expression right there is to say that A is equal to the vector c1, c2, all the way to ck, multiplied by this matrix right there. This would be equal to A. This statement over here and this expression over here are completely identical."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's imagine this matrix right there. Another way to write this expression right there is to say that A is equal to the vector c1, c2, all the way to ck, multiplied by this matrix right there. This would be equal to A. This statement over here and this expression over here are completely identical. If I take this matrix vector product, what do I get? I get c1 times v1, plus c2 times v2, plus c3 times v3, all the way to ck times vk is equal to A. We've seen this multiple times in multiple different contexts."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This statement over here and this expression over here are completely identical. If I take this matrix vector product, what do I get? I get c1 times v1, plus c2 times v2, plus c3 times v3, all the way to ck times vk is equal to A. We've seen this multiple times in multiple different contexts. But what's interesting here is this expression is the same thing, and it's really I'm just applying new words to things that we've seen probably 100 times by now. We can rewrite this expression. This is c. And remember, c is just a matrix with our basis vectors as columns."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We've seen this multiple times in multiple different contexts. But what's interesting here is this expression is the same thing, and it's really I'm just applying new words to things that we've seen probably 100 times by now. We can rewrite this expression. This is c. And remember, c is just a matrix with our basis vectors as columns. c is equal to this guy. This is just the coordinates of A with respect to the basis B. So c times the vector that has the coordinates of the vector A with respect to the basis B."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is c. And remember, c is just a matrix with our basis vectors as columns. c is equal to this guy. This is just the coordinates of A with respect to the basis B. So c times the vector that has the coordinates of the vector A with respect to the basis B. That is going to be equal to A. Now, why did I go through the trouble of doing this? Because now you have a fairly straightforward way."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So c times the vector that has the coordinates of the vector A with respect to the basis B. That is going to be equal to A. Now, why did I go through the trouble of doing this? Because now you have a fairly straightforward way. If I were to give you this, if I were to give you that right there, and say, hey, what is A if I wanted to write it in the standard coordinates, or with respect to the standard basis, which is just kind of the way we've been writing vectors all along, then you just multiply it times this matrix c, this matrix that has the basis vectors as columns. The other way, if you have some matrix A, or sorry, if you have some vector A that you know can be represented as a linear combination of B, or it's in the span of this basis vectors, then you could solve for this guy right here to figure out A's coordinates with respect to B. And so this little matrix right here, what does it do?"}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Because now you have a fairly straightforward way. If I were to give you this, if I were to give you that right there, and say, hey, what is A if I wanted to write it in the standard coordinates, or with respect to the standard basis, which is just kind of the way we've been writing vectors all along, then you just multiply it times this matrix c, this matrix that has the basis vectors as columns. The other way, if you have some matrix A, or sorry, if you have some vector A that you know can be represented as a linear combination of B, or it's in the span of this basis vectors, then you could solve for this guy right here to figure out A's coordinates with respect to B. And so this little matrix right here, what does it do? It helps us change bases. If you multiply it times this guy, you're going from the vector represented by coordinates with respect to some basis, and you multiply it times this guy, you're going to get to the vector in just the standard, just with standard coordinates. So we call this matrix right here change of basis matrix."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And so this little matrix right here, what does it do? It helps us change bases. If you multiply it times this guy, you're going from the vector represented by coordinates with respect to some basis, and you multiply it times this guy, you're going to get to the vector in just the standard, just with standard coordinates. So we call this matrix right here change of basis matrix. Which sounds very fancy, but all it literally is is a matrix with the basis vectors as columns. So let's just apply this a little bit to see if we can do anything vaguely constructive with it. Let's say that I have some basis."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we call this matrix right here change of basis matrix. Which sounds very fancy, but all it literally is is a matrix with the basis vectors as columns. So let's just apply this a little bit to see if we can do anything vaguely constructive with it. Let's say that I have some basis. Let's say B for basis. Let's say I have two vectors. I'll define the vectors up here."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that I have some basis. Let's say B for basis. Let's say I have two vectors. I'll define the vectors up here. Let's say vector 1 is, so we're dealing with R3. So vector 1 is 1, 2, 3. And let's say that vector 2 is 1, 0, 1."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I'll define the vectors up here. Let's say vector 1 is, so we're dealing with R3. So vector 1 is 1, 2, 3. And let's say that vector 2 is 1, 0, 1. And let's say I'm going to define some basis B as being the set of the vectors v1 and v2. And I'll leave it to you to verify that these are not linear combinations of each other. So this is a valid basis that these aren't in any way linearly dependent."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that vector 2 is 1, 0, 1. And let's say I'm going to define some basis B as being the set of the vectors v1 and v2. And I'll leave it to you to verify that these are not linear combinations of each other. So this is a valid basis that these aren't in any way linearly dependent. Now let's say that I know some vector that's in the span of these guys. Let's say that I know some vector that's in the span of these. And all I know is how it happens to be represented in coordinates with respect to this basis."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is a valid basis that these aren't in any way linearly dependent. Now let's say that I know some vector that's in the span of these guys. Let's say that I know some vector that's in the span of these. And all I know is how it happens to be represented in coordinates with respect to this basis. So let's say I have some vector A. And when I represent the coordinates of A with respect to this basis, it's equal to 7, 7, minus 4. So how can we represent this guy in its standard coordinates?"}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And all I know is how it happens to be represented in coordinates with respect to this basis. So let's say I have some vector A. And when I represent the coordinates of A with respect to this basis, it's equal to 7, 7, minus 4. So how can we represent this guy in its standard coordinates? Or what is A equal to? Well, you could just say, oh, well, A is equal to 7 times v1 minus 4 times v2. And you'd be completely correct."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So how can we represent this guy in its standard coordinates? Or what is A equal to? Well, you could just say, oh, well, A is equal to 7 times v1 minus 4 times v2. And you'd be completely correct. But let's actually use this change of basis matrix that I've introduced you to in this video. So the change of basis matrix here is going to be just a matrix with v1 and v2 as its columns. So 1, 2, 3, and then 1, 0, 1."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And you'd be completely correct. But let's actually use this change of basis matrix that I've introduced you to in this video. So the change of basis matrix here is going to be just a matrix with v1 and v2 as its columns. So 1, 2, 3, and then 1, 0, 1. And then if we multiply our change of basis matrix times the vector representation with respect to that basis, so times 7 minus 4, we're going to get the vector represented in standard coordinates. So what is this going to be equal to? We have a 3 by 2 matrix times a 2 by 1."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 1, 2, 3, and then 1, 0, 1. And then if we multiply our change of basis matrix times the vector representation with respect to that basis, so times 7 minus 4, we're going to get the vector represented in standard coordinates. So what is this going to be equal to? We have a 3 by 2 matrix times a 2 by 1. We're going to get a 3 by 1 matrix, which makes complete sense, because we're dealing in R3. This vector right here, A is going to be a member of R3. So when we write it with standard coordinates, we should have three coordinates right there."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We have a 3 by 2 matrix times a 2 by 1. We're going to get a 3 by 1 matrix, which makes complete sense, because we're dealing in R3. This vector right here, A is going to be a member of R3. So when we write it with standard coordinates, we should have three coordinates right there. Now, when we represented A with respect to the basis, we only had two coordinates, because A was in the plane spanned by these two guys. Actually, this is a good excuse to draw this. So let me draw it in three dimensions."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So when we write it with standard coordinates, we should have three coordinates right there. Now, when we represented A with respect to the basis, we only had two coordinates, because A was in the plane spanned by these two guys. Actually, this is a good excuse to draw this. So let me draw it in three dimensions. Let's say the span of v1 and v2 looks like this. And let's say this is the 0 vector right there. So this right here is the span of v1 and v2."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw it in three dimensions. Let's say the span of v1 and v2 looks like this. And let's say this is the 0 vector right there. So this right here is the span of v1 and v2. Or another way, this is the subspace that B is the basis for. And so we know that A is in this guy. So let's say v1 looks like this."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is the span of v1 and v2. Or another way, this is the subspace that B is the basis for. And so we know that A is in this guy. So let's say v1 looks like this. Let's say v1 looks like this, and that v2, I'm not even looking at the numbers. I'm just doing it fairly abstract. Let's say v2 looks like this right here."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's say v1 looks like this. Let's say v1 looks like this, and that v2, I'm not even looking at the numbers. I'm just doing it fairly abstract. Let's say v2 looks like this right here. Now, the fact that A can be represented as a linear combination of v1 and v2 tells us that A is also going to be in this plane in R3. And in fact, it's 7 times v1. So 7 times v1."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say v2 looks like this right here. Now, the fact that A can be represented as a linear combination of v1 and v2 tells us that A is also going to be in this plane in R3. And in fact, it's 7 times v1. So 7 times v1. So it's 1v1, 2v1s, 3v1s, 4, 5, 6, 7. So it's 7 in that direction. And then it's minus 4 in the v2 direction."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 7 times v1. So it's 1v1, 2v1s, 3v1s, 4, 5, 6, 7. So it's 7 in that direction. And then it's minus 4 in the v2 direction. So that's 1 in the v2 direction. This is minus 1 in the v2 direction. Minus 2, minus 3, minus 4."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then it's minus 4 in the v2 direction. So that's 1 in the v2 direction. This is minus 1 in the v2 direction. Minus 2, minus 3, minus 4. Or we could do it here. 1, 2, 3, 4. So our vector A is going to look like this."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2, minus 3, minus 4. Or we could do it here. 1, 2, 3, 4. So our vector A is going to look like this. It's going to sit on the plane. So this is our vector A. It's going to sit on the plane."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So our vector A is going to look like this. It's going to sit on the plane. So this is our vector A. It's going to sit on the plane. And when we represent it with respect to this basis, when we represent its coordinates with respect to our basis B, we say OK, it's 7 of this guy. I'm just doing this abstractly. Don't pay attention to the numbers just now."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's going to sit on the plane. And when we represent it with respect to this basis, when we represent its coordinates with respect to our basis B, we say OK, it's 7 of this guy. I'm just doing this abstractly. Don't pay attention to the numbers just now. I just want to understand the idea. We say it's 7 of this guy minus 4 of this guy. So it takes you back here and you get this vector, which is in this plane."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Don't pay attention to the numbers just now. I just want to understand the idea. We say it's 7 of this guy minus 4 of this guy. So it takes you back here and you get this vector, which is in this plane. So we only needed 2 coordinates to specify it within this plane. Because this subspace was 2-dimensional. But we're dealing in R3."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it takes you back here and you get this vector, which is in this plane. So we only needed 2 coordinates to specify it within this plane. Because this subspace was 2-dimensional. But we're dealing in R3. We're dealing in R3. And if we just want the general version of A in standard coordinates, we'll have to essentially get 3 coordinates. And I want you to understand that A is sitting on this plane."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But we're dealing in R3. We're dealing in R3. And if we just want the general version of A in standard coordinates, we'll have to essentially get 3 coordinates. And I want you to understand that A is sitting on this plane. This plane just keeps going on and on and on in all of these directions. A actually sits on that plane. It's a linear combination of that guy and that guy."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And I want you to understand that A is sitting on this plane. This plane just keeps going on and on and on in all of these directions. A actually sits on that plane. It's a linear combination of that guy and that guy. But let's figure out what A looks like in standard coordinates. In standard coordinates, we get the first term is going to be 1 times 7 plus 1 times minus 4. So that's going to be 3."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's a linear combination of that guy and that guy. But let's figure out what A looks like in standard coordinates. In standard coordinates, we get the first term is going to be 1 times 7 plus 1 times minus 4. So that's going to be 3. Then you get 2 times 7 plus 0 times minus 4. That is 14. Then you're going to get 3 times 7 plus 1 times minus 4."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be 3. Then you get 2 times 7 plus 0 times minus 4. That is 14. Then you're going to get 3 times 7 plus 1 times minus 4. So 3 times 7 is 21. Minus 4 is 17. So A is the vector 3, 14, 17."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Then you're going to get 3 times 7 plus 1 times minus 4. So 3 times 7 is 21. Minus 4 is 17. So A is the vector 3, 14, 17. That is equal to A. Now let's say we wanted to go the other way. Let's say we have some vector."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So A is the vector 3, 14, 17. That is equal to A. Now let's say we wanted to go the other way. Let's say we have some vector. Let me pick a letter I haven't used recently. Let's say I have some vector d, which is 8 minus 6, 2. And let's say we know."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have some vector. Let me pick a letter I haven't used recently. Let's say I have some vector d, which is 8 minus 6, 2. And let's say we know. We know that d is a member of the span of our basis vectors. The span of v1 and v2. Which tells us that d can be represented as a linear combination of these guys, or that d is in this subspace, or that d can be represented as coordinates with respect to the basis B."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And let's say we know. We know that d is a member of the span of our basis vectors. The span of v1 and v2. Which tells us that d can be represented as a linear combination of these guys, or that d is in this subspace, or that d can be represented as coordinates with respect to the basis B. Remember, the basis B was just equal to the set of v1 and v2. That's all the basis B was. Now we know that if we have our change of basis matrix, times the vector made up of the coordinates of d with respect to B."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Which tells us that d can be represented as a linear combination of these guys, or that d is in this subspace, or that d can be represented as coordinates with respect to the basis B. Remember, the basis B was just equal to the set of v1 and v2. That's all the basis B was. Now we know that if we have our change of basis matrix, times the vector made up of the coordinates of d with respect to B. So let me write that down. d with respect to B is equal to d. We know that. We know if we have this guy's coordinates and we multiply by the change of basis matrix, we'll just get the regular standard coordinate representation of d. Now in this case, we have d. We're given this."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now we know that if we have our change of basis matrix, times the vector made up of the coordinates of d with respect to B. So let me write that down. d with respect to B is equal to d. We know that. We know if we have this guy's coordinates and we multiply by the change of basis matrix, we'll just get the regular standard coordinate representation of d. Now in this case, we have d. We're given this. We of course know what the change of basis matrix is. So if we wanted to represent d in coordinates with respect to B, we're going to have to solve this equation. So let's do that."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We know if we have this guy's coordinates and we multiply by the change of basis matrix, we'll just get the regular standard coordinate representation of d. Now in this case, we have d. We're given this. We of course know what the change of basis matrix is. So if we wanted to represent d in coordinates with respect to B, we're going to have to solve this equation. So let's do that. So our change of basis matrix is 1, 1, 2, 0, 3, 1. And we're going to have to multiply it sometimes some coordinates. This thing right here, we can represent it as, I'll do it in yellow, we're going to need two coordinates."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. So our change of basis matrix is 1, 1, 2, 0, 3, 1. And we're going to have to multiply it sometimes some coordinates. This thing right here, we can represent it as, I'll do it in yellow, we're going to need two coordinates. It's going to be some multiple of v1 plus some multiple of v2. So it's c1, c2. And we know it has to be two coordinates because this matrix vector product is only well defined if this is a member of R2."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This thing right here, we can represent it as, I'll do it in yellow, we're going to need two coordinates. It's going to be some multiple of v1 plus some multiple of v2. So it's c1, c2. And we know it has to be two coordinates because this matrix vector product is only well defined if this is a member of R2. Because this is a 3 by 2 matrix. We have two columns here, so we have to have two entries here. And then that's going to be equal to d. That's going to be equal to d. So we have 8 minus 6, 2."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And we know it has to be two coordinates because this matrix vector product is only well defined if this is a member of R2. Because this is a 3 by 2 matrix. We have two columns here, so we have to have two entries here. And then that's going to be equal to d. That's going to be equal to d. So we have 8 minus 6, 2. And so if we figure out what this vector is, we've figured out what the representation or the coordinates of d with respect to b are. So let's solve this. So to solve this, we can just set up an augmented matrix."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then that's going to be equal to d. That's going to be equal to d. So we have 8 minus 6, 2. And so if we figure out what this vector is, we've figured out what the representation or the coordinates of d with respect to b are. So let's solve this. So to solve this, we can just set up an augmented matrix. That's just our traditional way of solving a linear equation. So what we have, we have 1, 1, 2, 0, 3, 1. We augment it with this side right there."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So to solve this, we can just set up an augmented matrix. That's just our traditional way of solving a linear equation. So what we have, we have 1, 1, 2, 0, 3, 1. We augment it with this side right there. So we have 8 minus 6 and 2. And let's keep my first row the same. So I have 1, 1, augmented it with 8."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We augment it with this side right there. So we have 8 minus 6 and 2. And let's keep my first row the same. So I have 1, 1, augmented it with 8. And let's replace my second row with the second row minus 2 times the first row. So I'm going to get 2 minus 2 times 1 is, actually let me do it the other way. Let me replace my second row with 2 times my first row minus my second row."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I have 1, 1, augmented it with 8. And let's replace my second row with the second row minus 2 times the first row. So I'm going to get 2 minus 2 times 1 is, actually let me do it the other way. Let me replace my second row with 2 times my first row minus my second row. So 2 times 1 minus 2 is 0. 2 times 1 minus 0 is 2. 2 times 8 is 16."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me replace my second row with 2 times my first row minus my second row. So 2 times 1 minus 2 is 0. 2 times 1 minus 0 is 2. 2 times 8 is 16. Minus 6 is 10. Now let's replace the third row with 3 times the first row minus the third row. So 3 times 1 minus 3 is 0."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "2 times 8 is 16. Minus 6 is 10. Now let's replace the third row with 3 times the first row minus the third row. So 3 times 1 minus 3 is 0. 3 times 1 minus 1 is 2. And then 3 times 8 is 24. Minus 2."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 3 times 1 minus 3 is 0. 3 times 1 minus 1 is 2. And then 3 times 8 is 24. Minus 2. 3 times 8 is 24. Minus 2 is going to be 22. See, it looks like I must have made a mistake someplace because I have these two would lead to no solutions."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2. 3 times 8 is 24. Minus 2 is going to be 22. See, it looks like I must have made a mistake someplace because I have these two would lead to no solutions. Let me verify what I did, make sure I didn't make any strange errors. So the second row, I replace it with 2 times the first row minus the second row. So 2 times 1 minus 2 is 0."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "See, it looks like I must have made a mistake someplace because I have these two would lead to no solutions. Let me verify what I did, make sure I didn't make any strange errors. So the second row, I replace it with 2 times the first row minus the second row. So 2 times 1 minus 2 is 0. 2 times 1 minus 0 is 2. 2 times 8 minus minus 6. So there's my error."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 2 times 1 minus 2 is 0. 2 times 1 minus 0 is 2. 2 times 8 minus minus 6. So there's my error. That's equal to 22. That was my error. So these two things are equivalent."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So there's my error. That's equal to 22. That was my error. So these two things are equivalent. Now let me do, I'll do one step at a time. Let me divide, let me replace my third row with my third row minus my second row. Just get it out of the way."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So these two things are equivalent. Now let me do, I'll do one step at a time. Let me divide, let me replace my third row with my third row minus my second row. Just get it out of the way. So I'll keep this 1, 1, 8, 0, 2, 22. And then the third row is going to be my third row. I'm going to replace it with my third row minus my second row."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Just get it out of the way. So I'll keep this 1, 1, 8, 0, 2, 22. And then the third row is going to be my third row. I'm going to replace it with my third row minus my second row. So it's going to be 0, 0, 0. So that just gets zeroed out. Now let me divide my second row by 2."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to replace it with my third row minus my second row. So it's going to be 0, 0, 0. So that just gets zeroed out. Now let me divide my second row by 2. So I get 1, 1, and 8. And then this one becomes 0, 1, and 11. And then of course the third row is just a bunch of 0's."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now let me divide my second row by 2. So I get 1, 1, and 8. And then this one becomes 0, 1, and 11. And then of course the third row is just a bunch of 0's. And then let me keep my middle row the same. So it's 0, 1, and 11. And then let me replace my first row with my first row minus my middle row."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then of course the third row is just a bunch of 0's. And then let me keep my middle row the same. So it's 0, 1, and 11. And then let me replace my first row with my first row minus my middle row. So 1 minus 0 is 1. 1 minus 1 is 0. 8 minus 11 is minus 3."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then let me replace my first row with my first row minus my middle row. So 1 minus 0 is 1. 1 minus 1 is 0. 8 minus 11 is minus 3. And I'll keep my last row the same. So I've put the left-hand side in reduced row echelon form. So this right here is essentially telling me my solution."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "8 minus 11 is minus 3. And I'll keep my last row the same. So I've put the left-hand side in reduced row echelon form. So this right here is essentially telling me my solution. I know that c1, the solution, so I could write it this way. I could write that 1, 0, 0, 1, 0, 0 times c1, c2 is equal to minus 3, 11, 0. Or another way of writing this is that 1 times c1 plus 0 times c2, or c1 is equal to minus 3."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is essentially telling me my solution. I know that c1, the solution, so I could write it this way. I could write that 1, 0, 0, 1, 0, 0 times c1, c2 is equal to minus 3, 11, 0. Or another way of writing this is that 1 times c1 plus 0 times c2, or c1 is equal to minus 3. 1 times c1 plus 0 times c2 is equal to minus 3. And then we have 0 times c1 plus 1 times c2 is going to be equal to 11. So our solution to this equation is minus 3, 11."}, {"video_title": "Change of basis matrix Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or another way of writing this is that 1 times c1 plus 0 times c2, or c1 is equal to minus 3. 1 times c1 plus 0 times c2 is equal to minus 3. And then we have 0 times c1 plus 1 times c2 is going to be equal to 11. So our solution to this equation is minus 3, 11. Or another way of saying this is that if I wanted to write my vector d in coordinates with respect to my basis b, it would be the coordinates minus 3, 11. So if I want to write my vector d in coordinates with respect to my basis b, it's going to be equal to minus 3, 11, which implies that d is equal to minus 3 times vector 1 plus 11 times vector 2. And I'll leave that for you to verify."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I want to make a quick correction or clarification to the last video that you may or may not have found confusing. You may not have noticed it. But when I did the general case for multiplying a row by a scalar, I had this situation where I had the matrix A, and I defined it as an n by n matrix. So it was A11, A12, all the way to A1n. And then we went down this way, and then we picked a particular row i. So we call that AI1, AI2, all the way to AIN. And then we keep going down, assuming that this is the last row."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it was A11, A12, all the way to A1n. And then we went down this way, and then we picked a particular row i. So we call that AI1, AI2, all the way to AIN. And then we keep going down, assuming that this is the last row. So AN1 all the way to ANN. And when I wanted to find the determinant of A, and this is where I made a, I would call it a notational error. When I wanted to find the determinant of A, I wrote that it was equal to, well, we could go down in that video."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we keep going down, assuming that this is the last row. So AN1 all the way to ANN. And when I wanted to find the determinant of A, and this is where I made a, I would call it a notational error. When I wanted to find the determinant of A, I wrote that it was equal to, well, we could go down in that video. I went down this row. That's why I kind of highlighted it to begin with, and I wrote it down. So it's equal to, to do the checkerboard pattern, I said negative 1 to the i plus j, well, let's do the first term, i plus 1, times AI1 times its submatrix."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When I wanted to find the determinant of A, I wrote that it was equal to, well, we could go down in that video. I went down this row. That's why I kind of highlighted it to begin with, and I wrote it down. So it's equal to, to do the checkerboard pattern, I said negative 1 to the i plus j, well, let's do the first term, i plus 1, times AI1 times its submatrix. That's what I wrote in the last. So if you have AI1, you get rid of that row, that column. You have the submatrix right there, AI1."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to, to do the checkerboard pattern, I said negative 1 to the i plus j, well, let's do the first term, i plus 1, times AI1 times its submatrix. That's what I wrote in the last. So if you have AI1, you get rid of that row, that column. You have the submatrix right there, AI1. That's what I wrote in the last video. But that was incorrect. And I think when I did the 2 by 2 case and the 3 by 3 case, that's pretty clear."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have the submatrix right there, AI1. That's what I wrote in the last video. But that was incorrect. And I think when I did the 2 by 2 case and the 3 by 3 case, that's pretty clear. It's not times the matrix, it's times the determinant of the submatrix. So this right here is incorrect. And of course, you keep adding that to, and I wrote AI2 times its submatrix like that, AI2, all the way to AIN times its submatrix."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I think when I did the 2 by 2 case and the 3 by 3 case, that's pretty clear. It's not times the matrix, it's times the determinant of the submatrix. So this right here is incorrect. And of course, you keep adding that to, and I wrote AI2 times its submatrix like that, AI2, all the way to AIN times its submatrix. That's what I did in the video. That's incorrect. Let me do the incorrect in a different color to show that this is all one thing."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And of course, you keep adding that to, and I wrote AI2 times its submatrix like that, AI2, all the way to AIN times its submatrix. That's what I did in the video. That's incorrect. Let me do the incorrect in a different color to show that this is all one thing. I should have said the determinant of each of these. The determinant of this guy is equal to minus 1 to the i plus 1 times AI1 times the determinant of AI1 plus AI2 times the determinant of AI2. The determinant of the submatrix, all the way to AIN times the determinant of the submatrix, AIN."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do the incorrect in a different color to show that this is all one thing. I should have said the determinant of each of these. The determinant of this guy is equal to minus 1 to the i plus 1 times AI1 times the determinant of AI1 plus AI2 times the determinant of AI2. The determinant of the submatrix, all the way to AIN times the determinant of the submatrix, AIN. It doesn't change the logic of the proof much, but I just want to be very careful that we're not multiplying the submatrices, because that becomes a fairly complicated operation. Well, it's not that bad. It's a scalar."}, {"video_title": "(correction) scalar multiplication of row Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The determinant of the submatrix, all the way to AIN times the determinant of the submatrix, AIN. It doesn't change the logic of the proof much, but I just want to be very careful that we're not multiplying the submatrices, because that becomes a fairly complicated operation. Well, it's not that bad. It's a scalar. But we're multiplying times the determinant of the submatrix. We saw that when we first defined it using the recursive definition for the n by n determinant. But I just wanted to make that very clear."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let me just write its rows like this. Let me just write it as R1. We could call them row vectors maybe. R2. I'm not doing it too formally. This is just really to save on writing. And then it has an ith row, Ri."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "R2. I'm not doing it too formally. This is just really to save on writing. And then it has an ith row, Ri. And then you keep going. It has some, that's an i right there. Then it has a jth row, Rj."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then it has an ith row, Ri. And then you keep going. It has some, that's an i right there. Then it has a jth row, Rj. And you keep going and you get to the nth row. It has n rows and n columns. So you get to Rn just like that."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then it has a jth row, Rj. And you keep going and you get to the nth row. It has n rows and n columns. So you get to Rn just like that. That is my matrix. Where, just to make sure you get what I'm saying, R, so if I have a, let's say the kth, R sub k is equal to Ak1, Ak, maybe I'll write it as a vector, Ak2, all the way to Akn. So this is just your standard representation."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you get to Rn just like that. That is my matrix. Where, just to make sure you get what I'm saying, R, so if I have a, let's say the kth, R sub k is equal to Ak1, Ak, maybe I'll write it as a vector, Ak2, all the way to Akn. So this is just your standard representation. I wrote it this way because we're just going to be dealing with rows in this video and it makes our notation a little bit easier. Now, let's just focus on these two rows right here. And let me define another matrix B that is also an n by n matrix."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is just your standard representation. I wrote it this way because we're just going to be dealing with rows in this video and it makes our notation a little bit easier. Now, let's just focus on these two rows right here. And let me define another matrix B that is also an n by n matrix. And it's identical to matrix A except for one row. So it's identical to matrix A except for one row. You have R1 just like that."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let me define another matrix B that is also an n by n matrix. And it's identical to matrix A except for one row. So it's identical to matrix A except for one row. You have R1 just like that. It's the same as that one there. R2. Keep going."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have R1 just like that. It's the same as that one there. R2. Keep going. Go down to Ri. Even that one's identical. But Rj, I've now replaced."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Keep going. Go down to Ri. Even that one's identical. But Rj, I've now replaced. I'm replacing Rj with Rj minus a scalar multiple of Ri. Minus c times Ri. So minus a scalar multiple of that."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But Rj, I've now replaced. I'm replacing Rj with Rj minus a scalar multiple of Ri. Minus c times Ri. So minus a scalar multiple of that. I've replaced Rj with that. So this is equivalent to the row operations we do when we did our Gaussian elimination or when we put things in reduced row echelon form. And everything else in this matrix is the same as A."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So minus a scalar multiple of that. I've replaced Rj with that. So this is equivalent to the row operations we do when we did our Gaussian elimination or when we put things in reduced row echelon form. And everything else in this matrix is the same as A. So all the way down to Rn. This is our matrix B. So let's think about what the determinant of B is going to be equal to."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And everything else in this matrix is the same as A. So all the way down to Rn. This is our matrix B. So let's think about what the determinant of B is going to be equal to. We do it in blue. The determinant of B. Well, you can immediately say that B is equivalent to, well, you can imagine two vectors."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's think about what the determinant of B is going to be equal to. We do it in blue. The determinant of B. Well, you can immediately say that B is equivalent to, well, you can imagine two vectors. Let me, you can imagine two matrices. One matrix that look like this. One matrix that look like R1, R2, all the way down Ri, all the way down to Rj."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, you can immediately say that B is equivalent to, well, you can imagine two vectors. Let me, you can imagine two matrices. One matrix that look like this. One matrix that look like R1, R2, all the way down Ri, all the way down to Rj. And then you keep going down to Rn. That's one matrix. Which you may have already noticed is identical to A."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "One matrix that look like R1, R2, all the way down Ri, all the way down to Rj. And then you keep going down to Rn. That's one matrix. Which you may have already noticed is identical to A. That's one matrix. And then you could have another matrix here. And then you could have another matrix here that looks like this."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which you may have already noticed is identical to A. That's one matrix. And then you could have another matrix here. And then you could have another matrix here that looks like this. It's identical everywhere. R1, R2, Ri. Some dots there, just to show you I might have skipped some rows, skipped some more rows."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you could have another matrix here that looks like this. It's identical everywhere. R1, R2, Ri. Some dots there, just to show you I might have skipped some rows, skipped some more rows. And then you have C times R1. Sorry, C times Ri. Let me do that in a different color."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Some dots there, just to show you I might have skipped some rows, skipped some more rows. And then you have C times R1. Sorry, C times Ri. Let me do that in a different color. This is Ri right here. So Ri. And then you just keep going down to Rn."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do that in a different color. This is Ri right here. So Ri. And then you just keep going down to Rn. Now, the determinant of B, you could view as the determinant of this guy. Let me write this here. The determinant of B is equal to the determinant of this guy plus the determinant of this guy."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you just keep going down to Rn. Now, the determinant of B, you could view as the determinant of this guy. Let me write this here. The determinant of B is equal to the determinant of this guy plus the determinant of this guy. Hopefully you remember a couple of videos ago that if one matrix, let's say I have two matrices that are identical in every way except for one row. So these two matrices are completely identical except for what's going on on the j-th row. Here you have a R sub j."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The determinant of B is equal to the determinant of this guy plus the determinant of this guy. Hopefully you remember a couple of videos ago that if one matrix, let's say I have two matrices that are identical in every way except for one row. So these two matrices are completely identical except for what's going on on the j-th row. Here you have a R sub j. Here you have a C times R sub i. So it's a scalar multiple of a row that you had up here. This guy."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Here you have a R sub j. Here you have a C times R sub i. So it's a scalar multiple of a row that you had up here. This guy. So this is Ri. This is the i-th row. Here it's an Ri."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This guy. So this is Ri. This is the i-th row. Here it's an Ri. Here you have an Ri. But here you have another version of Ri, a scalar multiple of Ri. Well, here you have an Rj."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Here it's an Ri. Here you have an Ri. But here you have another version of Ri, a scalar multiple of Ri. Well, here you have an Rj. Now, if you have another matrix that is essentially identical to these two matrices except for this one row, and in that one row it looks like the addition of these two matrices. And let me put a negative here. So if you kept this matrix completely identical, but if you replace it with the sum of these two rows, so Rj minus C times Ri, you'll get this matrix right here."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, here you have an Rj. Now, if you have another matrix that is essentially identical to these two matrices except for this one row, and in that one row it looks like the addition of these two matrices. And let me put a negative here. So if you kept this matrix completely identical, but if you replace it with the sum of these two rows, so Rj minus C times Ri, you'll get this matrix right here. You'll get matrix B. And we learned that the determinant of B is equal to the determinant of this guy and that guy. Remember, B is not the sum of these two matrices."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you kept this matrix completely identical, but if you replace it with the sum of these two rows, so Rj minus C times Ri, you'll get this matrix right here. You'll get matrix B. And we learned that the determinant of B is equal to the determinant of this guy and that guy. Remember, B is not the sum of these two matrices. B is identical to these two matrices except for that one row where B's j-th row is equivalent to the j-th row of this guy plus the j-th row of that guy. And when I talk about adding rows, you're just adding their corresponding elements. So I could rewrite."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, B is not the sum of these two matrices. B is identical to these two matrices except for that one row where B's j-th row is equivalent to the j-th row of this guy plus the j-th row of that guy. And when I talk about adding rows, you're just adding their corresponding elements. So I could rewrite. Let me rewrite this just to be. So this row would look like, let me make it, it would be A, the first term would be Aj1 minus C times Ai1. That would be the first term in that row."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I could rewrite. Let me rewrite this just to be. So this row would look like, let me make it, it would be A, the first term would be Aj1 minus C times Ai1. That would be the first term in that row. The second term in that row would be Aj2 minus C times Ai2. And it would go all the way to Ajn minus Ca sub in, the n-th column. So that's all it means by that."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That would be the first term in that row. The second term in that row would be Aj2 minus C times Ai2. And it would go all the way to Ajn minus Ca sub in, the n-th column. So that's all it means by that. So the determinant of B is equal to the determinant of this plus the determinant of this. The determinant of this, well this thing is right here. This is our matrix A."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's all it means by that. So the determinant of B is equal to the determinant of this plus the determinant of this. The determinant of this, well this thing is right here. This is our matrix A. So this is going to be the determinant of A. So this is the determinant of A. And what's the determinant of this?"}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is our matrix A. So this is going to be the determinant of A. So this is the determinant of A. And what's the determinant of this? Well, let's break this down a little bit more. The determinant of this is equal to what? This is completely equivalent to A, except one of its rows, sorry, this is completely equivalent to this matrix."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what's the determinant of this? Well, let's break this down a little bit more. The determinant of this is equal to what? This is completely equivalent to A, except one of its rows, sorry, this is completely equivalent to this matrix. Not equivalent to A, be very careful. Don't listen to everything I say. It's not equivalent to A."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is completely equivalent to A, except one of its rows, sorry, this is completely equivalent to this matrix. Not equivalent to A, be very careful. Don't listen to everything I say. It's not equivalent to A. The difference is that A has an Rj here. This guy has a minus C times Ri. So this is equivalent to, this is completely equivalent to this matrix."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's not equivalent to A. The difference is that A has an Rj here. This guy has a minus C times Ri. So this is equivalent to, this is completely equivalent to this matrix. It's completely equivalent to this matrix right here. Let me do it like this. So you have an R1, R2, it has, keep going, then you have an Ri, then you have another Ri."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is equivalent to, this is completely equivalent to this matrix. It's completely equivalent to this matrix right here. Let me do it like this. So you have an R1, R2, it has, keep going, then you have an Ri, then you have another Ri. Let me clean this up a little bit. Let me clear this out. Let me clear that out just so I have some space to work with."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you have an R1, R2, it has, keep going, then you have an Ri, then you have another Ri. Let me clean this up a little bit. Let me clear this out. Let me clear that out just so I have some space to work with. There you go. You have an Ri, you have that Ri there, then you have another Ri. You have another Ri right there."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me clear that out just so I have some space to work with. There you go. You have an Ri, you have that Ri there, then you have another Ri. You have another Ri right there. You have another Ri. And then you keep, so the j-th row, this is the j-th row, has an Ri there. And then you keep going and then you have an R sub n. These two guys are completely equivalent, except for this guy has a minus C times the j-th row."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have another Ri right there. You have another Ri. And then you keep, so the j-th row, this is the j-th row, has an Ri there. And then you keep going and then you have an R sub n. These two guys are completely equivalent, except for this guy has a minus C times the j-th row. That's what this was right here. This is the j-th row. Everything we're doing is in the j-th row."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you keep going and then you have an R sub n. These two guys are completely equivalent, except for this guy has a minus C times the j-th row. That's what this was right here. This is the j-th row. Everything we're doing is in the j-th row. This has a minus C times the j-th row. So this is going to be equivalent to, so the determinant of this guy right here, let me just be clear that I'm only taking the determinant of this guy right here. It's going to be equal to minus C times the determinant of, let me write it this way."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Everything we're doing is in the j-th row. This has a minus C times the j-th row. So this is going to be equivalent to, so the determinant of this guy right here, let me just be clear that I'm only taking the determinant of this guy right here. It's going to be equal to minus C times the determinant of, let me write it this way. I'll just write minus C times the determinant of R1, R2. You have your first Ri. And then in the j-th row you have another version of the Ri and then you go down to R sub n. So times that determinant."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be equal to minus C times the determinant of, let me write it this way. I'll just write minus C times the determinant of R1, R2. You have your first Ri. And then in the j-th row you have another version of the Ri and then you go down to R sub n. So times that determinant. This is just the determinant of this. Instead of brackets I have those straight lines. And we saw this a couple of videos ago."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then in the j-th row you have another version of the Ri and then you go down to R sub n. So times that determinant. This is just the determinant of this. Instead of brackets I have those straight lines. And we saw this a couple of videos ago. If you have a matrix and you just multiply one of its rows by a scalar, in this case minus C, it's equivalent to minus C, the determinant of that new matrix is equal to minus C times the determinant of your matrix. So that's all I'm saying right here. But what is the determinant of this matrix?"}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we saw this a couple of videos ago. If you have a matrix and you just multiply one of its rows by a scalar, in this case minus C, it's equivalent to minus C, the determinant of that new matrix is equal to minus C times the determinant of your matrix. So that's all I'm saying right here. But what is the determinant of this matrix? You might have already noticed that it has duplicate rows. It has an Ri and then in the i-th row, but then it has another Ri in the j-th row. Remember we all got it by, we kind of decomposed this B matrix right here."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But what is the determinant of this matrix? You might have already noticed that it has duplicate rows. It has an Ri and then in the i-th row, but then it has another Ri in the j-th row. Remember we all got it by, we kind of decomposed this B matrix right here. It's the sum of, or it can be described as the determinant of the sum of these two things. B isn't the sum of these two things because B's, every other element, is identical to every other element in each of these guys. But this guy right here, he has duplicate Ri's and what do we know about the determinant of a matrix with duplicate entries?"}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember we all got it by, we kind of decomposed this B matrix right here. It's the sum of, or it can be described as the determinant of the sum of these two things. B isn't the sum of these two things because B's, every other element, is identical to every other element in each of these guys. But this guy right here, he has duplicate Ri's and what do we know about the determinant of a matrix with duplicate entries? The determinant is 0. So this entry right here is 0 minus C times 0, 0. So the determinant of this whole thing is 0."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this guy right here, he has duplicate Ri's and what do we know about the determinant of a matrix with duplicate entries? The determinant is 0. So this entry right here is 0 minus C times 0, 0. So the determinant of this whole thing is 0. So the big takeaway right here is that the determinant of B is equal to just the determinant of this thing, which was the determinant of A. This is a very big takeaway, it's going to make our life very easy. The determinant of B is equal to the determinant of A."}, {"video_title": "Determinant after row operations Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of this whole thing is 0. So the big takeaway right here is that the determinant of B is equal to just the determinant of this thing, which was the determinant of A. This is a very big takeaway, it's going to make our life very easy. The determinant of B is equal to the determinant of A. So if you start with some matrix and you replace the j-th row in this example, but any row, if you replace any row with that row minus some scalar multiple of another row, we picked Ri in this case, that was the Ri, the determinant will not be changed. And you have to be very particular about how you say it, because obviously if you just multiplied something by a scalar, you're going to change its determinant, or if you do other things. But if you just take a row, if you take the j-th row and you replace it with the j-th row minus C times the i-th row times some other row, which is equivalent to just a row operation that we have been doing, then it will not change your determinant, which is a very big takeaway, because now we can carefully do some row operations and know that the determinant will not change."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You've seen this before. A11, A12, all the way to A1n. When you go down the rows, you get A21. That goes all the way to A2n. And let's say that there's some row here, let's say row i. It looks like Ai1, all the way to Ain. And then you have some other row here, Aj1, all the way to Ajn, and then you keep going all the way down to An1, An2, all the way to Ann."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That goes all the way to A2n. And let's say that there's some row here, let's say row i. It looks like Ai1, all the way to Ain. And then you have some other row here, Aj1, all the way to Ajn, and then you keep going all the way down to An1, An2, all the way to Ann. This is just an n by n matrix, and you can see that I took a little trouble to write out my i-th row here and my j-th row here. And just to kind of keep things a little simple, let me just define, just for notational purposes, you can view these as row vectors if you like, but I haven't formally defined row vectors, so I won't necessarily go there. But let's just define the term ri, we'll call that row i, to be equal to Ai1, Ai2, all the way to Ain."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have some other row here, Aj1, all the way to Ajn, and then you keep going all the way down to An1, An2, all the way to Ann. This is just an n by n matrix, and you can see that I took a little trouble to write out my i-th row here and my j-th row here. And just to kind of keep things a little simple, let me just define, just for notational purposes, you can view these as row vectors if you like, but I haven't formally defined row vectors, so I won't necessarily go there. But let's just define the term ri, we'll call that row i, to be equal to Ai1, Ai2, all the way to Ain. And you can write it as a vector if you like, like a row vector, we haven't really defined operations on row vectors that well yet, but I think you get the idea. We can then replace this guy with R1, this guy with R2, all the way down. Let me do that, and I'll do that in the next couple of videos, because it'll simplify things."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But let's just define the term ri, we'll call that row i, to be equal to Ai1, Ai2, all the way to Ain. And you can write it as a vector if you like, like a row vector, we haven't really defined operations on row vectors that well yet, but I think you get the idea. We can then replace this guy with R1, this guy with R2, all the way down. Let me do that, and I'll do that in the next couple of videos, because it'll simplify things. I think it'll make things a little bit easier to understand. So I can rewrite this matrix, this n by n matrix A, I can rewrite it as just Ri. Actually, since this looks like a vector, it's just a row vector, let me write it as a vector like that."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do that, and I'll do that in the next couple of videos, because it'll simplify things. I think it'll make things a little bit easier to understand. So I can rewrite this matrix, this n by n matrix A, I can rewrite it as just Ri. Actually, since this looks like a vector, it's just a row vector, let me write it as a vector like that. And I'm being a little bit hand wavy here, because all of our vectors have been defined as column vectors, but I think you get the idea. So let's call that R1, and then we have R2 as the next row, all the way down. You keep going down, you get to Ri, that's this row right there."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Actually, since this looks like a vector, it's just a row vector, let me write it as a vector like that. And I'm being a little bit hand wavy here, because all of our vectors have been defined as column vectors, but I think you get the idea. So let's call that R1, and then we have R2 as the next row, all the way down. You keep going down, you get to Ri, that's this row right there. Ri, you keep going down, you get Rj, and then you keep going down until you get to the nth row. And each of these guys are going to have n terms, because you have n columns. So that's another way of writing the same n by n matrix."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You keep going down, you get to Ri, that's this row right there. Ri, you keep going down, you get Rj, and then you keep going down until you get to the nth row. And each of these guys are going to have n terms, because you have n columns. So that's another way of writing the same n by n matrix. Now, what I'm going to do here is I'm going to create a new matrix. Let's call that the swap matrix of i and j. So I'm going to swap i and j, those two rows."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's another way of writing the same n by n matrix. Now, what I'm going to do here is I'm going to create a new matrix. Let's call that the swap matrix of i and j. So I'm going to swap i and j, those two rows. So what's the matrix going to look like? Everything else is going to be equal. You have row 1, assuming that 1 wasn't one of the i or j's."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to swap i and j, those two rows. So what's the matrix going to look like? Everything else is going to be equal. You have row 1, assuming that 1 wasn't one of the i or j's. It could have been. Row 2, all the way down to, now instead of a row i there, you have a row j there. You have a row j, and you go down."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have row 1, assuming that 1 wasn't one of the i or j's. It could have been. Row 2, all the way down to, now instead of a row i there, you have a row j there. You have a row j, and you go down. Instead of a row j, you have a row i there. And you go down and then you get an Rn. So what did we do?"}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have a row j, and you go down. Instead of a row j, you have a row i there. And you go down and then you get an Rn. So what did we do? We just swapped these two guys. That's what the swap matrix is. Now, I think it was in the last video or a couple of videos ago."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what did we do? We just swapped these two guys. That's what the swap matrix is. Now, I think it was in the last video or a couple of videos ago. We learned that if you just swap two rows of any n by n matrix, what you get, or the determinant of the resulting matrix, will be the negative of the original determinant. So we get the determinant of S, the swap of the i-th and the j rows, is going to be equal to the minus of the determinant of A. Now, let me ask you an interesting question."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, I think it was in the last video or a couple of videos ago. We learned that if you just swap two rows of any n by n matrix, what you get, or the determinant of the resulting matrix, will be the negative of the original determinant. So we get the determinant of S, the swap of the i-th and the j rows, is going to be equal to the minus of the determinant of A. Now, let me ask you an interesting question. What happens if those two rows were actually the same? What if Ri was equal to Rj? So if we go back to all of these guys, if that row is equal to this row, that means that this guy is equal to that guy, that the second column is equal to the second column for that row, all the way to the n-th guy is equal to the n-th guy."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let me ask you an interesting question. What happens if those two rows were actually the same? What if Ri was equal to Rj? So if we go back to all of these guys, if that row is equal to this row, that means that this guy is equal to that guy, that the second column is equal to the second column for that row, all the way to the n-th guy is equal to the n-th guy. That's what I mean when I say what happens if those two rows are equal to each other. Well, if those two rows are equal to each other, then this matrix is no different than this matrix here, even though we swapped them. If you swap two identical things, you're just going to be left with the same thing again."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we go back to all of these guys, if that row is equal to this row, that means that this guy is equal to that guy, that the second column is equal to the second column for that row, all the way to the n-th guy is equal to the n-th guy. That's what I mean when I say what happens if those two rows are equal to each other. Well, if those two rows are equal to each other, then this matrix is no different than this matrix here, even though we swapped them. If you swap two identical things, you're just going to be left with the same thing again. So if, let me write this down, if Rho i is equal to Rho j, then this guy, then S, the swapped matrix, is equal to A. They'll be identical. You're swapping two rows that are the same thing."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you swap two identical things, you're just going to be left with the same thing again. So if, let me write this down, if Rho i is equal to Rho j, then this guy, then S, the swapped matrix, is equal to A. They'll be identical. You're swapping two rows that are the same thing. So that implies the determinant of the swapped matrix is equal to the determinant of A. But we just said, it's the swapped matrix, when you swap two rows, it equals a negative of the determinant of A. So this tells us it also has to equal the negative of the determinant of A."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You're swapping two rows that are the same thing. So that implies the determinant of the swapped matrix is equal to the determinant of A. But we just said, it's the swapped matrix, when you swap two rows, it equals a negative of the determinant of A. So this tells us it also has to equal the negative of the determinant of A. So what does that tell us? That tells us if A has two rows that are equal to each other. If A has two rows that are equal to each other, if we swap them, we should get the negative of the determinant."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this tells us it also has to equal the negative of the determinant of A. So what does that tell us? That tells us if A has two rows that are equal to each other. If A has two rows that are equal to each other, if we swap them, we should get the negative of the determinant. But if those two rows are equal, we're going to get the same matrix again. So if A has two rows that are equal, so if Rho i is equal to Rho j, then the determinant of A has to be equal to the negative of the determinant of A. We know that because the determinant of A, or A is the same thing as the swapped version of A, and the swapped version of A has to have the negative determinant of A."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If A has two rows that are equal to each other, if we swap them, we should get the negative of the determinant. But if those two rows are equal, we're going to get the same matrix again. So if A has two rows that are equal, so if Rho i is equal to Rho j, then the determinant of A has to be equal to the negative of the determinant of A. We know that because the determinant of A, or A is the same thing as the swapped version of A, and the swapped version of A has to have the negative determinant of A. So these two things have to be equal. Now, what number is equal to a negative version of itself? If I just told you x is equal to negative x, what number does x have to be equal to?"}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know that because the determinant of A, or A is the same thing as the swapped version of A, and the swapped version of A has to have the negative determinant of A. So these two things have to be equal. Now, what number is equal to a negative version of itself? If I just told you x is equal to negative x, what number does x have to be equal to? There's only one value that it could possibly be equal to. x would have to be equal to 0. So the takeaway here is if one row, or let's say if you have duplicate rows, you could extend this if you have three or four rows that are the same, duplicate rows lead you to the fact that the determinant of your matrix is 0."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I just told you x is equal to negative x, what number does x have to be equal to? There's only one value that it could possibly be equal to. x would have to be equal to 0. So the takeaway here is if one row, or let's say if you have duplicate rows, you could extend this if you have three or four rows that are the same, duplicate rows lead you to the fact that the determinant of your matrix is 0. And that really shouldn't be a surprise. Because if you have duplicate rows, remember what we learned a long time ago. We learned that a matrix is an invertible if and only if the reduced row echelon form is the identity matrix."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the takeaway here is if one row, or let's say if you have duplicate rows, you could extend this if you have three or four rows that are the same, duplicate rows lead you to the fact that the determinant of your matrix is 0. And that really shouldn't be a surprise. Because if you have duplicate rows, remember what we learned a long time ago. We learned that a matrix is an invertible if and only if the reduced row echelon form is the identity matrix. We learned that. But if you have two duplicate rows, let's say these two guys are equal to each other, you could perform a row operation where you replace this guy with this guy minus that guy, and you'll just get a row of 0's. And if you get a row of 0's, you're never going to be able to get the identity matrix."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We learned that a matrix is an invertible if and only if the reduced row echelon form is the identity matrix. We learned that. But if you have two duplicate rows, let's say these two guys are equal to each other, you could perform a row operation where you replace this guy with this guy minus that guy, and you'll just get a row of 0's. And if you get a row of 0's, you're never going to be able to get the identity matrix. So we know that a duplicate rows could never get reduced row echelon form to be the identity. Or duplicate rows are not invertible. And we also learned that something is not invertible if and only if its determinant is equal to 0."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if you get a row of 0's, you're never going to be able to get the identity matrix. So we know that a duplicate rows could never get reduced row echelon form to be the identity. Or duplicate rows are not invertible. And we also learned that something is not invertible if and only if its determinant is equal to 0. So we now got to the same result two different ways. One, we just used some of what we learned. When you swap rows, it should become the negative."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we also learned that something is not invertible if and only if its determinant is equal to 0. So we now got to the same result two different ways. One, we just used some of what we learned. When you swap rows, it should become the negative. But if you swap the same row, you shouldn't change the matrix. So the determinant of the matrix has to be the same as itself. So if you have duplicate rows, the determinant is 0."}, {"video_title": "Duplicate row determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When you swap rows, it should become the negative. But if you swap the same row, you shouldn't change the matrix. So the determinant of the matrix has to be the same as itself. So if you have duplicate rows, the determinant is 0. Which isn't something that we had to use using this little swapping technique. We could have gone back to our requirements for invertibility, I think it was five or six videos ago. But I just wanted to point that out."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have this matrix B here, and I want to know what the null space of B is. And we've done this multiple times, but just as a review, the null space of B is just all of the x's that are a member, it's all the vector x's that are a member of what? 1, 2, 3, 4, 5, that are members of R5, where my matrix B times any of these vector x's is equal to 0. That's the definition of the null space. I'm just trying to find the solution set to this equation right here, and we've seen before that the null set of the reduced row echelon form of B is equal to the null set of B. So what's the reduced row echelon form of B? And this is actually almost trivially easy."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's the definition of the null space. I'm just trying to find the solution set to this equation right here, and we've seen before that the null set of the reduced row echelon form of B is equal to the null set of B. So what's the reduced row echelon form of B? And this is actually almost trivially easy. So let me just take a couple of steps right here. To get a 0 here, let's just replace row 2 with row 2 minus row 1. So what do we get?"}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is actually almost trivially easy. So let me just take a couple of steps right here. To get a 0 here, let's just replace row 2 with row 2 minus row 1. So what do we get? Row 2 minus row 1. Row 1 doesn't change. It's just 1, 1, 2, 3, 2."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what do we get? Row 2 minus row 1. Row 1 doesn't change. It's just 1, 1, 2, 3, 2. And then row 2 minus row 1, 1 minus 1 is 0. 3 minus 2 is 1, 1 minus 3 is minus 2, 4 minus 2 is 2. We're almost there."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just 1, 1, 2, 3, 2. And then row 2 minus row 1, 1 minus 1 is 0. 3 minus 2 is 1, 1 minus 3 is minus 2, 4 minus 2 is 2. We're almost there. So this is a free variable right here. This is a pivot variable right here. We have a 1."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We're almost there. So this is a free variable right here. This is a pivot variable right here. We have a 1. So let me get rid of that guy right there. And I can get rid of that guy right there by replacing row 1 with row 1 minus 2 times row 2. So now row 2 is going to be the same, 0, 0, 1, minus 2, 2."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We have a 1. So let me get rid of that guy right there. And I can get rid of that guy right there by replacing row 1 with row 1 minus 2 times row 2. So now row 2 is going to be the same, 0, 0, 1, minus 2, 2. And let me replace row 1 with row 1 minus 2 times row 2. So 1 minus 2 times 0 is 1. 1 minus 2 times 0 is 1."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now row 2 is going to be the same, 0, 0, 1, minus 2, 2. And let me replace row 1 with row 1 minus 2 times row 2. So 1 minus 2 times 0 is 1. 1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. 3 minus 2 times minus 2, so that's 3 plus 4 is 7. 2 times this is minus 4, we're subtracting it."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. 3 minus 2 times minus 2, so that's 3 plus 4 is 7. 2 times this is minus 4, we're subtracting it. And then 2 minus 2 times 2, that's 2 minus 4, it's minus 2. So this is the reduced row echelon form of B. It's equal to that right there."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 times this is minus 4, we're subtracting it. And then 2 minus 2 times 2, that's 2 minus 4, it's minus 2. So this is the reduced row echelon form of B. It's equal to that right there. And then if I wanted to figure out its null space, I have x1, x2, x3, x4, and x5. Equaling, I'm going to have two 0's right here. Now I can just write this as just a system of equations."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to that right there. And then if I wanted to figure out its null space, I have x1, x2, x3, x4, and x5. Equaling, I'm going to have two 0's right here. Now I can just write this as just a system of equations. So let me do that. I get x1. I'm going to write my pivot variables in a green color."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I can just write this as just a system of equations. So let me do that. I get x1. I'm going to write my pivot variables in a green color. x1 plus 1 times x2, so plus x2, plus 0 times x3, plus 7 times x4, minus 2 times x5 is equal to that 0 right there. And then I get my x3, right? 0 times x1 plus 0 times x2 plus 1 times x3."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to write my pivot variables in a green color. x1 plus 1 times x2, so plus x2, plus 0 times x3, plus 7 times x4, minus 2 times x5 is equal to that 0 right there. And then I get my x3, right? 0 times x1 plus 0 times x2 plus 1 times x3. So I get x3 minus 2 times x4 plus 2 times x5 is equal to that 0 right there. And then if we solve for our pivot variables, right? These are our free variables."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "0 times x1 plus 0 times x2 plus 1 times x3. So I get x3 minus 2 times x4 plus 2 times x5 is equal to that 0 right there. And then if we solve for our pivot variables, right? These are our free variables. We can set them equal to anything. If we solve for our pivot variables, what do we get? We get x1 is equal to, I should do that in green."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These are our free variables. We can set them equal to anything. If we solve for our pivot variables, what do we get? We get x1 is equal to, I should do that in green. The color coding helps. I get x1 is equal to minus x2 minus 7x4 plus 2x5. Just subtracted these from both sides of the equation."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We get x1 is equal to, I should do that in green. The color coding helps. I get x1 is equal to minus x2 minus 7x4 plus 2x5. Just subtracted these from both sides of the equation. And I get x3 is equal to, I've done this multiple times, 2x4 minus 2x5. And so if I wanted to write the solution set kind of in vector form, I could write my solution set is, or my null space really is, or all the possible x's. x1, x2, x3, x4, x5."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Just subtracted these from both sides of the equation. And I get x3 is equal to, I've done this multiple times, 2x4 minus 2x5. And so if I wanted to write the solution set kind of in vector form, I could write my solution set is, or my null space really is, or all the possible x's. x1, x2, x3, x4, x5. This is my vector x that's in an R5. It is equal to a linear combination of these. So let me write it out."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x1, x2, x3, x4, x5. This is my vector x that's in an R5. It is equal to a linear combination of these. So let me write it out. It's equal to, the free variables are x2 times some vector, x2 times some vector right there, plus x4, that's my next free variable, times some vector, plus x5 times some vector, run out of space, plus x5 times some vector. And what are those vectors? Well, let's see."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me write it out. It's equal to, the free variables are x2 times some vector, x2 times some vector right there, plus x4, that's my next free variable, times some vector, plus x5 times some vector, run out of space, plus x5 times some vector. And what are those vectors? Well, let's see. Let me actually, I don't want to make this too dirty, so let me see if I can maybe move, nope, that's not what I wanted to do. Let me just rewrite this. Sometimes I haven't mastered this pen tool yet, so let me rewrite this here."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, let's see. Let me actually, I don't want to make this too dirty, so let me see if I can maybe move, nope, that's not what I wanted to do. Let me just rewrite this. Sometimes I haven't mastered this pen tool yet, so let me rewrite this here. So x3 is equal to 2x4 minus 2x5. Let me delete this right over here, so just so I get some extra space. Let's cross that out."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Sometimes I haven't mastered this pen tool yet, so let me rewrite this here. So x3 is equal to 2x4 minus 2x5. Let me delete this right over here, so just so I get some extra space. Let's cross that out. I think that's good enough. So I can go back to what I was doing before. x5 times some vector right here."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's cross that out. I think that's good enough. So I can go back to what I was doing before. x5 times some vector right here. And now what are those vectors? We just have to look at these formulas. x1 is equal to minus 1 times x2, so minus 1 times x2, minus 7 times x4, plus 2 times x5."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x5 times some vector right here. And now what are those vectors? We just have to look at these formulas. x1 is equal to minus 1 times x2, so minus 1 times x2, minus 7 times x4, plus 2 times x5. Fair enough. And what is x3 equal to? x3 is equal to 2x4, it had nothing to do with x2 right here, so it's equal to 2x4 minus 2x5."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x1 is equal to minus 1 times x2, so minus 1 times x2, minus 7 times x4, plus 2 times x5. Fair enough. And what is x3 equal to? x3 is equal to 2x4, it had nothing to do with x2 right here, so it's equal to 2x4 minus 2x5. And then what is 0 times x2? Because it had no x2 term right here. And then what is x2 equal to?"}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "x3 is equal to 2x4, it had nothing to do with x2 right here, so it's equal to 2x4 minus 2x5. And then what is 0 times x2? Because it had no x2 term right here. And then what is x2 equal to? Well, x2 is just equal to 1 times x2. And so all of these terms are 0 right there. And I want you to pay attention to that."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then what is x2 equal to? Well, x2 is just equal to 1 times x2. And so all of these terms are 0 right there. And I want you to pay attention to that. Let me write it right here. x2 is a free variable, so it's just equal to itself. Write 1 and you write a 0 and a 0. x4 is a free variable, and this is the important point of this exercise."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want you to pay attention to that. Let me write it right here. x2 is a free variable, so it's just equal to itself. Write 1 and you write a 0 and a 0. x4 is a free variable, and this is the important point of this exercise. So it's just equal to 1 times itself, and you don't have to throw in any of the other free variables. And x5 is a free variable, so it just equals 1 times itself and none of the other free variables. So right here, we now say that all of the solutions of our equation bx equals 0, or the reduced row echelon form of b times x is equal to 0, will take this form, where there are linear combinations of these vectors."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Write 1 and you write a 0 and a 0. x4 is a free variable, and this is the important point of this exercise. So it's just equal to 1 times itself, and you don't have to throw in any of the other free variables. And x5 is a free variable, so it just equals 1 times itself and none of the other free variables. So right here, we now say that all of the solutions of our equation bx equals 0, or the reduced row echelon form of b times x is equal to 0, will take this form, where there are linear combinations of these vectors. Let's call this v1, v2, and v3. These are just random real numbers. I can pick any combination here to create this solution set or to create our null space."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So right here, we now say that all of the solutions of our equation bx equals 0, or the reduced row echelon form of b times x is equal to 0, will take this form, where there are linear combinations of these vectors. Let's call this v1, v2, and v3. These are just random real numbers. I can pick any combination here to create this solution set or to create our null space. So the null space of a, which is equal to the reduced row echelon form of a, is equal to all the possible linear combinations of these three vectors, is equal to the span of my vector v1, v2, and v3, just like that. Now, the whole reason I went through this exercise, because we've done this multiple times already, is to think about whether these guys form a linear independent set. So my question is, are these guys linearly independent?"}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can pick any combination here to create this solution set or to create our null space. So the null space of a, which is equal to the reduced row echelon form of a, is equal to all the possible linear combinations of these three vectors, is equal to the span of my vector v1, v2, and v3, just like that. Now, the whole reason I went through this exercise, because we've done this multiple times already, is to think about whether these guys form a linear independent set. So my question is, are these guys linearly independent? And the reason why I care is because if they are linearly independent, then they form a basis for the null space. We know that they span the null space, but if they're linearly independent, then that's the two constraints for a basis. You have to span the subspace, and you have to be linearly independent."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So my question is, are these guys linearly independent? And the reason why I care is because if they are linearly independent, then they form a basis for the null space. We know that they span the null space, but if they're linearly independent, then that's the two constraints for a basis. You have to span the subspace, and you have to be linearly independent. So let's just inspect these guys right here. This v1, he has a 1 right here. He has a 1 in the second term, because he corresponds to x2, to the free variable x2, which is the second entry."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You have to span the subspace, and you have to be linearly independent. So let's just inspect these guys right here. This v1, he has a 1 right here. He has a 1 in the second term, because he corresponds to x2, to the free variable x2, which is the second entry. So we just threw a 1 here, and we have a 0 everywhere else in all of the other vectors in our spanning set. And that's because for the other free variables, we always wanted to multiply them times a 0, right? And this is going to be true of any kind of null space problem we do."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "He has a 1 in the second term, because he corresponds to x2, to the free variable x2, which is the second entry. So we just threw a 1 here, and we have a 0 everywhere else in all of the other vectors in our spanning set. And that's because for the other free variables, we always wanted to multiply them times a 0, right? And this is going to be true of any kind of null space problem we do. For any free variable, if this free variable represents a second entry, we're going to have a 1 in the second entry here, and then a 0 for the second entry for all of the other vectors associated with the other free variables. So can this guy ever be represented as a linear combination of this guy and that guy? Well, there's nothing that I can multiply this 0 by and add to something that I multiply this 0 by to get a 1 here."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is going to be true of any kind of null space problem we do. For any free variable, if this free variable represents a second entry, we're going to have a 1 in the second entry here, and then a 0 for the second entry for all of the other vectors associated with the other free variables. So can this guy ever be represented as a linear combination of this guy and that guy? Well, there's nothing that I can multiply this 0 by and add to something that I multiply this 0 by to get a 1 here. It's just going to get 0's. So this guy can't be represented as a linear combination of these guys. Likewise, this vector right here has a 1 in the fourth position."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, there's nothing that I can multiply this 0 by and add to something that I multiply this 0 by to get a 1 here. It's just going to get 0's. So this guy can't be represented as a linear combination of these guys. Likewise, this vector right here has a 1 in the fourth position. Why is it a fourth position? Because the fourth position corresponds to its corresponding free variable, x4. So this guy has a 1 here."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Likewise, this vector right here has a 1 in the fourth position. Why is it a fourth position? Because the fourth position corresponds to its corresponding free variable, x4. So this guy has a 1 here. These other guys will definitely always have a 0 here, so you can't take any linear combination of them to get this guy. So this guy can't be represented as a linear combination of those guys. And last, this x5 guy right here has a 1 here, and these guys have 0's here."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this guy has a 1 here. These other guys will definitely always have a 0 here, so you can't take any linear combination of them to get this guy. So this guy can't be represented as a linear combination of those guys. And last, this x5 guy right here has a 1 here, and these guys have 0's here. So no linear combination of these 0's can equal this 1. So all of these guys are linearly independent. You can't construct any of these vectors with some combination of the other."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And last, this x5 guy right here has a 1 here, and these guys have 0's here. So no linear combination of these 0's can equal this 1. So all of these guys are linearly independent. You can't construct any of these vectors with some combination of the other. So they are linearly independent. And so this is actually, so the set v1, v2, and v3 is actually a basis for the null space of B. Just for variety, I defined my initial matrix as matrix B."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can't construct any of these vectors with some combination of the other. So they are linearly independent. And so this is actually, so the set v1, v2, and v3 is actually a basis for the null space of B. Just for variety, I defined my initial matrix as matrix B. So let me be very careful here. The null space of B was equal to the null space of the reduced row echelon form of B. It's good to switch things up every now and you start thinking that every matrix is named A if you don't."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Just for variety, I defined my initial matrix as matrix B. So let me be very careful here. The null space of B was equal to the null space of the reduced row echelon form of B. It's good to switch things up every now and you start thinking that every matrix is named A if you don't. And that's equal to the span of these vectors. So these vectors, and we just said that they're linearly independent, we just showed that because there's no way to get that one from these guys, that one from these guys, or that one from these guys. These guys form a basis for the null space of B."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's good to switch things up every now and you start thinking that every matrix is named A if you don't. And that's equal to the span of these vectors. So these vectors, and we just said that they're linearly independent, we just showed that because there's no way to get that one from these guys, that one from these guys, or that one from these guys. These guys form a basis for the null space of B. Now this raises an interesting question. In the last video, I defined what dimensionality is. And maybe you missed it because that video was kind of proofy, but the dimension of a subspace, I'll redefine it here, is the number of elements in a basis for the subspace."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These guys form a basis for the null space of B. Now this raises an interesting question. In the last video, I defined what dimensionality is. And maybe you missed it because that video was kind of proofy, but the dimension of a subspace, I'll redefine it here, is the number of elements in a basis for the subspace. And in the last video, I took great pains to show that all bases for any given subspace will have the same number of elements, so this is well defined. So my question to you now is, what is the dimension of my null space of B? What is the dimension of my null space of B?"}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And maybe you missed it because that video was kind of proofy, but the dimension of a subspace, I'll redefine it here, is the number of elements in a basis for the subspace. And in the last video, I took great pains to show that all bases for any given subspace will have the same number of elements, so this is well defined. So my question to you now is, what is the dimension of my null space of B? What is the dimension of my null space of B? Well, the dimension is just the number of vectors in a basis set for B. Well, this is a basis set for B right there. And how many vectors do I have in it?"}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is the dimension of my null space of B? Well, the dimension is just the number of vectors in a basis set for B. Well, this is a basis set for B right there. And how many vectors do I have in it? I have one, two, three vectors. So the dimension of the null space of B is three. Or another way to think about it, or another name for the dimension of the null space of B is the nullity of B, and that is also equal to three."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And how many vectors do I have in it? I have one, two, three vectors. So the dimension of the null space of B is three. Or another way to think about it, or another name for the dimension of the null space of B is the nullity of B, and that is also equal to three. And let's think about it. I went through all of this exercise, but what is the nullity of any matrix going to be equal to? It's the dimension of the null space."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to think about it, or another name for the dimension of the null space of B is the nullity of B, and that is also equal to three. And let's think about it. I went through all of this exercise, but what is the nullity of any matrix going to be equal to? It's the dimension of the null space. Well, the dimension of the null space, you're always going to have as many vectors here as you have free variables. So in general, the nullity of any matrix of, let's say, matrix A is equal to the number of, I guess we could call it free variable columns, or the number of free variables in the reduced row echelon form of A. Or I guess we could say the number of non-pivot columns in the reduced row echelon form of A, because that's essentially the number of free variables."}, {"video_title": "Dimension of the null space or nullity Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's the dimension of the null space. Well, the dimension of the null space, you're always going to have as many vectors here as you have free variables. So in general, the nullity of any matrix of, let's say, matrix A is equal to the number of, I guess we could call it free variable columns, or the number of free variables in the reduced row echelon form of A. Or I guess we could say the number of non-pivot columns in the reduced row echelon form of A, because that's essentially the number of free variables. All of those free variables have an associated linearly independent vector with each of them. So the number of free variables is the number of vectors you're going to have in your basis for your null space, and the number of free variables is essentially the number of non-pivot columns in your reduced row echelon form. This was a non-pivot column, that's a non-pivot column, that's a non-pivot column, and they're associated with the free variables x2, x4, and x5."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Let's do one more Gram-Schmidt example. So let's say I have the subspace V that is spanned by the vectors. Let's say we're dealing in R4. So the first vector is 0, 0, 1, 1. The second vector is 0, 1, 1, 0. And then its third vector. So it's a three-dimensional subspace of R4."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So the first vector is 0, 0, 1, 1. The second vector is 0, 1, 1, 0. And then its third vector. So it's a three-dimensional subspace of R4. It's 1, 1, 0, 0. Just like that. Three-dimensional subspace of R4."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So it's a three-dimensional subspace of R4. It's 1, 1, 0, 0. Just like that. Three-dimensional subspace of R4. And what we want to do, we want to find an orthonormal basis for V. So we want to substitute these guys with three other vectors that are orthogonal with respect to each other and have length 1. So we do the same drill we've done before. We can say, let's call this guy V1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Three-dimensional subspace of R4. And what we want to do, we want to find an orthonormal basis for V. So we want to substitute these guys with three other vectors that are orthogonal with respect to each other and have length 1. So we do the same drill we've done before. We can say, let's call this guy V1. This guy is V2. Let's call this guy V3. So the first thing we want to do is replace V1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We can say, let's call this guy V1. This guy is V2. Let's call this guy V3. So the first thing we want to do is replace V1. And I'm just picking this guy at random, just because he was the first guy on the left-hand side. I want to replace V1 with an orthogonal version of V1. So let me call U1 is equal to, well, let me just find out the length of V1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So the first thing we want to do is replace V1. And I'm just picking this guy at random, just because he was the first guy on the left-hand side. I want to replace V1 with an orthogonal version of V1. So let me call U1 is equal to, well, let me just find out the length of V1. I don't think I have to explain too much of the theory at this point. I just want to show another example. So the length of V1 is equal to the square root of 0 squared plus 0 squared plus 1 squared plus 1 squared, which equals the square root of 2."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So let me call U1 is equal to, well, let me just find out the length of V1. I don't think I have to explain too much of the theory at this point. I just want to show another example. So the length of V1 is equal to the square root of 0 squared plus 0 squared plus 1 squared plus 1 squared, which equals the square root of 2. So let me define my new vector, U1, to be equal to 1 over the length of V1, 1 over the square root of 2, times V1, times 0, 0, 1, 1. And just like that, the span of V1, V2, V3 is the same thing as the span of U1, V2, and V3. So this is my first thing that I've normalized."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So the length of V1 is equal to the square root of 0 squared plus 0 squared plus 1 squared plus 1 squared, which equals the square root of 2. So let me define my new vector, U1, to be equal to 1 over the length of V1, 1 over the square root of 2, times V1, times 0, 0, 1, 1. And just like that, the span of V1, V2, V3 is the same thing as the span of U1, V2, and V3. So this is my first thing that I've normalized. So I can say that V is now equal to the span of the vectors U1, V2, and V3. Because I can replace V1 with this guy, because this guy is just a scaled-up version of this guy. So I can definitely represent him with him."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So this is my first thing that I've normalized. So I can say that V is now equal to the span of the vectors U1, V2, and V3. Because I can replace V1 with this guy, because this guy is just a scaled-up version of this guy. So I can definitely represent him with him. So I can represent any linear combination of these guys with any linear combination of those guys right there. Now, we just did our first vector. We've just normalized this one."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So I can definitely represent him with him. So I can represent any linear combination of these guys with any linear combination of those guys right there. Now, we just did our first vector. We've just normalized this one. We need to replace these other vectors with vectors that are orthogonal to this guy right here. So let's do V2 first. So let's replace, we could say, let's call Y2 is equal to V2 minus the projection of V2 onto the space spanned by U1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We've just normalized this one. We need to replace these other vectors with vectors that are orthogonal to this guy right here. So let's do V2 first. So let's replace, we could say, let's call Y2 is equal to V2 minus the projection of V2 onto the space spanned by U1. Or onto, I could call it C times U1, or in the past videos we called that subspace V1. The space spanned by U1. And that's just going to be equal to, Y2 is equal to V2, which is 0, 1, 1, 0 minus V2 projected onto that space is just the dot product of V2."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So let's replace, we could say, let's call Y2 is equal to V2 minus the projection of V2 onto the space spanned by U1. Or onto, I could call it C times U1, or in the past videos we called that subspace V1. The space spanned by U1. And that's just going to be equal to, Y2 is equal to V2, which is 0, 1, 1, 0 minus V2 projected onto that space is just the dot product of V2. It's just the dot product of V2, 0, 1, 1, 0. With the spanning vector of that space, and there's only one of them, so we're only going to have one term like this. With U1, so dotted with 1 over the square root of 2 times 0, 0, 1, 1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And that's just going to be equal to, Y2 is equal to V2, which is 0, 1, 1, 0 minus V2 projected onto that space is just the dot product of V2. It's just the dot product of V2, 0, 1, 1, 0. With the spanning vector of that space, and there's only one of them, so we're only going to have one term like this. With U1, so dotted with 1 over the square root of 2 times 0, 0, 1, 1. And then all of that times U1. So 1 over the square root of 2 times the vector 0, 0, 1, 1. And so this is going to be equal to V2, which is 0, 1, 1, 0, the square root of 2's, let's factor them out."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "With U1, so dotted with 1 over the square root of 2 times 0, 0, 1, 1. And then all of that times U1. So 1 over the square root of 2 times the vector 0, 0, 1, 1. And so this is going to be equal to V2, which is 0, 1, 1, 0, the square root of 2's, let's factor them out. So then you just get, or just take them, kind of reassociate them out. So then you get this is 1 over the square root of 2 times 1 over the square root of 2 is minus 1 half U times, what's the dot product of these two guys? You get 0 times 0 plus 1 times 0, which is still 0, plus 1 times 1 plus 0 times 0."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And so this is going to be equal to V2, which is 0, 1, 1, 0, the square root of 2's, let's factor them out. So then you just get, or just take them, kind of reassociate them out. So then you get this is 1 over the square root of 2 times 1 over the square root of 2 is minus 1 half U times, what's the dot product of these two guys? You get 0 times 0 plus 1 times 0, which is still 0, plus 1 times 1 plus 0 times 0. So you're just going to have times 1 times this out here. 0, 0, 1, 1. Write that a little bit neater."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "You get 0 times 0 plus 1 times 0, which is still 0, plus 1 times 1 plus 0 times 0. So you're just going to have times 1 times this out here. 0, 0, 1, 1. Write that a little bit neater. Getting careless. 1, 1. So this is just going to be equal to 0, 1, 1, 0 minus 1 half times 0 is 0."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Write that a little bit neater. Getting careless. 1, 1. So this is just going to be equal to 0, 1, 1, 0 minus 1 half times 0 is 0. 1 half times 0 is 0. Then I have 2 halves here. So Y2 is equal to, let's see, 0 minus 0 is 0."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So this is just going to be equal to 0, 1, 1, 0 minus 1 half times 0 is 0. 1 half times 0 is 0. Then I have 2 halves here. So Y2 is equal to, let's see, 0 minus 0 is 0. 1 minus 0 is 1. 1 minus 1 half is 1 half. And then 0 minus 1 half is minus 1 half."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So Y2 is equal to, let's see, 0 minus 0 is 0. 1 minus 0 is 1. 1 minus 1 half is 1 half. And then 0 minus 1 half is minus 1 half. So now the span of V, we can now write as the span of U1, Y2, and V3. And this is progress. U1 is orthogonal."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And then 0 minus 1 half is minus 1 half. So now the span of V, we can now write as the span of U1, Y2, and V3. And this is progress. U1 is orthogonal. Y2, sorry, U1 is normalized. It has length 1. Y2 is orthogonal to it, or they're orthogonal with respect to each other."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "U1 is orthogonal. Y2, sorry, U1 is normalized. It has length 1. Y2 is orthogonal to it, or they're orthogonal with respect to each other. But Y2 still has not been normalized. So let me replace Y2 with a normalized version of it. The length of Y2 is equal to the square root of 0 plus 1 squared, which is 1, plus 1 half squared, which is 1 fourth, plus minus 1 half squared, which is also 1 fourth, so plus 1 fourth."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Y2 is orthogonal to it, or they're orthogonal with respect to each other. But Y2 still has not been normalized. So let me replace Y2 with a normalized version of it. The length of Y2 is equal to the square root of 0 plus 1 squared, which is 1, plus 1 half squared, which is 1 fourth, plus minus 1 half squared, which is also 1 fourth, so plus 1 fourth. So this is 1 and 1 half. So it's equal to the square root of 3 halves. So let me define another vector here."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "The length of Y2 is equal to the square root of 0 plus 1 squared, which is 1, plus 1 half squared, which is 1 fourth, plus minus 1 half squared, which is also 1 fourth, so plus 1 fourth. So this is 1 and 1 half. So it's equal to the square root of 3 halves. So let me define another vector here. U2, which is equal to 1 over the square root of 3 halves, or we could say it's the square root of 2 thirds. I'm just inverting it. It's 1 over the length of Y2."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So let me define another vector here. U2, which is equal to 1 over the square root of 3 halves, or we could say it's the square root of 2 thirds. I'm just inverting it. It's 1 over the length of Y2. So I'll just find the reciprocal. So it's the square root of 2 over 3 times Y2, times this guy right here, times 0, 1, 1 half, and minus 1 half. And so this span is going to be the same thing as the span of U1, U2, and V3."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "It's 1 over the length of Y2. So I'll just find the reciprocal. So it's the square root of 2 over 3 times Y2, times this guy right here, times 0, 1, 1 half, and minus 1 half. And so this span is going to be the same thing as the span of U1, U2, and V3. And there's our second basis vector. And we're making a lot of progress. These guys are orthogonal with respect to each other."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And so this span is going to be the same thing as the span of U1, U2, and V3. And there's our second basis vector. And we're making a lot of progress. These guys are orthogonal with respect to each other. They both have length 1. We just have to do something about V3. And we do it the same way."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "These guys are orthogonal with respect to each other. They both have length 1. We just have to do something about V3. And we do it the same way. Let's find a vector that is orthogonal to these guys. And if I sum that vector to some linear combination of these guys, I'm going to get V3. And I'm going to call that vector Y3."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And we do it the same way. Let's find a vector that is orthogonal to these guys. And if I sum that vector to some linear combination of these guys, I'm going to get V3. And I'm going to call that vector Y3. Y3 is going to be equal to V3 minus the projection of V3 onto the subspace spanned by U1 and U2. So I could call that subspace, let me just write it here, the span of U1 and U2. Just for notation, I'm going to call it V2."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And I'm going to call that vector Y3. Y3 is going to be equal to V3 minus the projection of V3 onto the subspace spanned by U1 and U2. So I could call that subspace, let me just write it here, the span of U1 and U2. Just for notation, I'm going to call it V2. So it's V3, and actually I don't even have to write that, minus the projection of V3 onto that. What's that going to be? That's going to be V3 dot U1 times U1 times the vector U1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Just for notation, I'm going to call it V2. So it's V3, and actually I don't even have to write that, minus the projection of V3 onto that. What's that going to be? That's going to be V3 dot U1 times U1 times the vector U1. And actually, let me just, well, plus V3 dot U2 times the vector U2. Since this is an orthonormal basis, the projection onto it, you just take the dot product of V3 with each of the orthonormal basis vectors and multiply them times the orthonormal basis vectors. We saw that several videos ago."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be V3 dot U1 times U1 times the vector U1. And actually, let me just, well, plus V3 dot U2 times the vector U2. Since this is an orthonormal basis, the projection onto it, you just take the dot product of V3 with each of the orthonormal basis vectors and multiply them times the orthonormal basis vectors. We saw that several videos ago. That's one of the neat things about orthonormal bases. So what is this going to be equal to? A little bit more computation here."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We saw that several videos ago. That's one of the neat things about orthonormal bases. So what is this going to be equal to? A little bit more computation here. Y3 is equal to V3, which was up here. So that's V3. V3 looks like this."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "A little bit more computation here. Y3 is equal to V3, which was up here. So that's V3. V3 looks like this. It's 1, 1, 0, 0 minus V3 dot U1. So this is minus V3, 1, 1, 0, 0, dot U1. So it's dot 1 over the square root of 2 times 0, 0, 1, 1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "V3 looks like this. It's 1, 1, 0, 0 minus V3 dot U1. So this is minus V3, 1, 1, 0, 0, dot U1. So it's dot 1 over the square root of 2 times 0, 0, 1, 1. That's U1. So that's this part right here. Times U1, so times 1 over the square root of 2 times 0, 0, 1, 1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So it's dot 1 over the square root of 2 times 0, 0, 1, 1. That's U1. So that's this part right here. Times U1, so times 1 over the square root of 2 times 0, 0, 1, 1. This piece right there is this piece right there. And then we can distribute this minus sign. So it's going to be plus."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Times U1, so times 1 over the square root of 2 times 0, 0, 1, 1. This piece right there is this piece right there. And then we can distribute this minus sign. So it's going to be plus. You have a plus, but then there's this minus over here. So let's put a minus. V3, let me switch colors, minus V3, which is 1, 1, 0, 0 dotted with U2, dotted with the square root of 2 thirds times 0, 1, 1 half minus 1 half times U2, times the vector U2, times the square root of 2 over 3 times the vector 0, 1, 1 half minus 1 half."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be plus. You have a plus, but then there's this minus over here. So let's put a minus. V3, let me switch colors, minus V3, which is 1, 1, 0, 0 dotted with U2, dotted with the square root of 2 thirds times 0, 1, 1 half minus 1 half times U2, times the vector U2, times the square root of 2 over 3 times the vector 0, 1, 1 half minus 1 half. And what do we get? Let's calculate this. So this is going to be equal to the vector 1, 1, 0, 0 minus."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "V3, let me switch colors, minus V3, which is 1, 1, 0, 0 dotted with U2, dotted with the square root of 2 thirds times 0, 1, 1 half minus 1 half times U2, times the vector U2, times the square root of 2 over 3 times the vector 0, 1, 1 half minus 1 half. And what do we get? Let's calculate this. So this is going to be equal to the vector 1, 1, 0, 0 minus. So the 1 over square root of 2, 1 over square root of 2, multiply them, you're going to get a 1 half. And then when you take the dot product of these two, 1 times 0, let's see, this is actually all going to be, if you take the dot product of all of these, it gets a 0. So this guy, V3, was actually already orthogonal to U1."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to the vector 1, 1, 0, 0 minus. So the 1 over square root of 2, 1 over square root of 2, multiply them, you're going to get a 1 half. And then when you take the dot product of these two, 1 times 0, let's see, this is actually all going to be, if you take the dot product of all of these, it gets a 0. So this guy, V3, was actually already orthogonal to U1. They're already orthogonal to U1. So this will just go straight to 0, which is nice. We don't have to have a term right there."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So this guy, V3, was actually already orthogonal to U1. They're already orthogonal to U1. So this will just go straight to 0, which is nice. We don't have to have a term right there. And I took the dot product, 1 times 0 plus 1 times 0 plus 0 times 1 plus 0 times 1, all gets 0. So this whole term drops out. We can ignore it, which makes our computation simpler."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We don't have to have a term right there. And I took the dot product, 1 times 0 plus 1 times 0 plus 0 times 1 plus 0 times 1, all gets 0. So this whole term drops out. We can ignore it, which makes our computation simpler. And then over here, we have minus square root of 2 thirds times square root of 2 thirds is just 2 thirds times the dot product of these two guys. So that's 1 times 0, which is 0, plus 1 times 1, which is 1, plus 0 times 1 half, which is 0, plus 0 times minus 1 half, which is 0. So we just get a 1 there."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We can ignore it, which makes our computation simpler. And then over here, we have minus square root of 2 thirds times square root of 2 thirds is just 2 thirds times the dot product of these two guys. So that's 1 times 0, which is 0, plus 1 times 1, which is 1, plus 0 times 1 half, which is 0, plus 0 times minus 1 half, which is 0. So we just get a 1 there. Times this vector, times the vector 0, 1, 1 half, minus 1 half, and then what do we get? We get, this is the home stretch, 1, 1, 0, 0 minus 2 thirds times all of these guys. So 2 thirds times 0 is 0."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So we just get a 1 there. Times this vector, times the vector 0, 1, 1 half, minus 1 half, and then what do we get? We get, this is the home stretch, 1, 1, 0, 0 minus 2 thirds times all of these guys. So 2 thirds times 0 is 0. 2 thirds times 1 is 2 thirds. 2 thirds times 1 half is 1 third. And then 2 thirds times minus 1 half is minus 1 third."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So 2 thirds times 0 is 0. 2 thirds times 1 is 2 thirds. 2 thirds times 1 half is 1 third. And then 2 thirds times minus 1 half is minus 1 third. So then this is going to be equal to 1 minus 0 is 1. 1 minus 2 thirds is 1 third. 0 minus 1 third is minus 1 third."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And then 2 thirds times minus 1 half is minus 1 third. So then this is going to be equal to 1 minus 0 is 1. 1 minus 2 thirds is 1 third. 0 minus 1 third is minus 1 third. And then 0 minus minus 1 third is positive 1 third. So this vector y3 is orthogonal to these two other vectors, which is nice. But it still hasn't been normalized."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 1 third is minus 1 third. And then 0 minus minus 1 third is positive 1 third. So this vector y3 is orthogonal to these two other vectors, which is nice. But it still hasn't been normalized. So we finally have to normalize this guy, and then we're done. Then we have an orthonormal basis. We'll have u1, u2, and now we'll find u3."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "But it still hasn't been normalized. So we finally have to normalize this guy, and then we're done. Then we have an orthonormal basis. We'll have u1, u2, and now we'll find u3. So the length of my vector y, actually, let's do something even better. It'll simplify things a little bit. Instead of writing y this way, I could scale up y, right?"}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "We'll have u1, u2, and now we'll find u3. So the length of my vector y, actually, let's do something even better. It'll simplify things a little bit. Instead of writing y this way, I could scale up y, right? All I want is a vector that's orthogonal to the other two that still spans the same space. So I can scale this guy up. So I could say, I don't know, let me just call it y3 prime."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "Instead of writing y this way, I could scale up y, right? All I want is a vector that's orthogonal to the other two that still spans the same space. So I can scale this guy up. So I could say, I don't know, let me just call it y3 prime. I'm just doing this to ease the computation. I could just scale this guy up, multiply him by 3. So what do I get?"}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So I could say, I don't know, let me just call it y3 prime. I'm just doing this to ease the computation. I could just scale this guy up, multiply him by 3. So what do I get? I probably should have done it with some of the other ones. 3, 1, minus 1, and 1. And so I could replace y3 with this guy, and then I could just normalize this guy."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So what do I get? I probably should have done it with some of the other ones. 3, 1, minus 1, and 1. And so I could replace y3 with this guy, and then I could just normalize this guy. It'll be a little bit easier. So the length of y3 prime that I just defined is equal to the square root of 3 squared, which is 9 plus 1 squared, plus minus 1 squared, plus 1 squared, which is equal to the square root of 12, which is what? That's 2 square roots of 3."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "And so I could replace y3 with this guy, and then I could just normalize this guy. It'll be a little bit easier. So the length of y3 prime that I just defined is equal to the square root of 3 squared, which is 9 plus 1 squared, plus minus 1 squared, plus 1 squared, which is equal to the square root of 12, which is what? That's 2 square roots of 3. That is equal to 2 square roots of 3, right? Square root of 4 times square root of 3, which is 2 square roots of 3. So now I can define u3 is equal to y3 times 1 over the length of y3."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "That's 2 square roots of 3. That is equal to 2 square roots of 3, right? Square root of 4 times square root of 3, which is 2 square roots of 3. So now I can define u3 is equal to y3 times 1 over the length of y3. So it's equal to 1 over 2 square roots of 3 times the vector 3, 1, minus 1, and 1. And then we're done. If we have a basis, an orthonormal basis would be this guy."}, {"video_title": "Gram-Schmidt example with 3 basis vectors Linear Algebra Khan Academy.mp3", "Sentence": "So now I can define u3 is equal to y3 times 1 over the length of y3. So it's equal to 1 over 2 square roots of 3 times the vector 3, 1, minus 1, and 1. And then we're done. If we have a basis, an orthonormal basis would be this guy. Let me take the other ones down here. And these guys, all of these form, let me bring it all the way down, if I have a collection of these three vectors, I now have an orthonormal basis for v. These three right here. They form, that set is an orthonormal basis for my original subspace v that I started off with."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say the coordinates of that point are x0, x0, y0, and z0. Or it could be specified as a position vector. I could draw the position vector like this. So the position vector, let me draw better dotted lines. The position vector for this could be x0i plus y0j plus z0k. It specifies this coordinate right over here. What I want to do is find the distance between this point and the plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the position vector, let me draw better dotted lines. The position vector for this could be x0i plus y0j plus z0k. It specifies this coordinate right over here. What I want to do is find the distance between this point and the plane. Obviously, there could be a lot of distance. I could find the distance between this point and that point, and this point and this point, and this point and this point. When I say I want to find the distance, I want to find the minimum distance."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What I want to do is find the distance between this point and the plane. Obviously, there could be a lot of distance. I could find the distance between this point and that point, and this point and this point, and this point and this point. When I say I want to find the distance, I want to find the minimum distance. You're actually going to get the minimum distance when you go the perpendicular distance to the plane, or the normal distance to the plane. We'll hopefully see that visually as we try to figure out how to calculate the distance. The first thing we can do is let's just construct a vector between this point that's off the plane and some point that's on the plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "When I say I want to find the distance, I want to find the minimum distance. You're actually going to get the minimum distance when you go the perpendicular distance to the plane, or the normal distance to the plane. We'll hopefully see that visually as we try to figure out how to calculate the distance. The first thing we can do is let's just construct a vector between this point that's off the plane and some point that's on the plane. We already have a point from the last video that's on the plane, this xp, yp, zp. Let's construct a vector here. Let's construct this orange vector that starts on the plane, its tail is on the plane, and it goes off the plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The first thing we can do is let's just construct a vector between this point that's off the plane and some point that's on the plane. We already have a point from the last video that's on the plane, this xp, yp, zp. Let's construct a vector here. Let's construct this orange vector that starts on the plane, its tail is on the plane, and it goes off the plane. I want to do that in orange. It goes off the plane to this vector that's to this position, x0, y0, z0. What would be, how could we specify this vector right over here?"}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's construct this orange vector that starts on the plane, its tail is on the plane, and it goes off the plane. I want to do that in orange. It goes off the plane to this vector that's to this position, x0, y0, z0. What would be, how could we specify this vector right over here? That vector, let me call that vector, what letters have I not used yet, let me call that vector f. Vector f is just going to be this yellow position vector minus this green position vector. It's going to be, this x component is going to be the difference of the x coordinates, its y coordinates is going to be the difference of the y coordinates. Its y coordinates is going to be x0 minus x sub p, I subtracted the x coordinates, i, plus y0 minus ypj, plus z0 minus zp minus zpk."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What would be, how could we specify this vector right over here? That vector, let me call that vector, what letters have I not used yet, let me call that vector f. Vector f is just going to be this yellow position vector minus this green position vector. It's going to be, this x component is going to be the difference of the x coordinates, its y coordinates is going to be the difference of the y coordinates. Its y coordinates is going to be x0 minus x sub p, I subtracted the x coordinates, i, plus y0 minus ypj, plus z0 minus zp minus zpk. Fair enough, that's just some vector that comes off of the plane and onto this point. What we want to find out is this distance. We want to find out this distance in yellow, the distance that if I were to take a normal off of the plane and go straight to the point, that's going to be the shortest distance."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Its y coordinates is going to be x0 minus x sub p, I subtracted the x coordinates, i, plus y0 minus ypj, plus z0 minus zp minus zpk. Fair enough, that's just some vector that comes off of the plane and onto this point. What we want to find out is this distance. We want to find out this distance in yellow, the distance that if I were to take a normal off of the plane and go straight to the point, that's going to be the shortest distance. Actually, you can see it visually now, because if you look at, we can actually form a right triangle here. This base of the right triangle is along the plane, this side is normal to the plane, so this is a right angle. You can see, if I take any point, any other point on the plane, it will form a hypotenuse on a right triangle."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We want to find out this distance in yellow, the distance that if I were to take a normal off of the plane and go straight to the point, that's going to be the shortest distance. Actually, you can see it visually now, because if you look at, we can actually form a right triangle here. This base of the right triangle is along the plane, this side is normal to the plane, so this is a right angle. You can see, if I take any point, any other point on the plane, it will form a hypotenuse on a right triangle. Obviously, the shortest side here, or the shortest way to get to the plane, is going to be this distance right here, as opposed to the hypotenuse. This side will always be shorter than that side. Given that we know this vector here, how can we figure out this length here in blue?"}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can see, if I take any point, any other point on the plane, it will form a hypotenuse on a right triangle. Obviously, the shortest side here, or the shortest way to get to the plane, is going to be this distance right here, as opposed to the hypotenuse. This side will always be shorter than that side. Given that we know this vector here, how can we figure out this length here in blue? We could figure out the magnitude of this vector. The length of this side right here is going to be the magnitude of the vector, so it's going to be the magnitude of the vector f. That will just give us this length, but we want this blue length. We could think about it, if this was some angle, I know the writing is getting small, if this was some angle theta, we could use some pretty straight-up, pretty straightforward trigonometry."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Given that we know this vector here, how can we figure out this length here in blue? We could figure out the magnitude of this vector. The length of this side right here is going to be the magnitude of the vector, so it's going to be the magnitude of the vector f. That will just give us this length, but we want this blue length. We could think about it, if this was some angle, I know the writing is getting small, if this was some angle theta, we could use some pretty straight-up, pretty straightforward trigonometry. If the distance under question is d, you could say cosine of theta is equal to the adjacent side over the hypotenuse. d is the adjacent side, is equal to d over the hypotenuse. The hypotenuse is the magnitude of this vector."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We could think about it, if this was some angle, I know the writing is getting small, if this was some angle theta, we could use some pretty straight-up, pretty straightforward trigonometry. If the distance under question is d, you could say cosine of theta is equal to the adjacent side over the hypotenuse. d is the adjacent side, is equal to d over the hypotenuse. The hypotenuse is the magnitude of this vector. It's the magnitude of the vector f. Or we could say the magnitude of the vector f times the cosine of theta, I'm just multiplying both sides times the magnitude of the vector x. f is equal to d. But that's still, you might say, well Sal, we know what f is, we can figure that out, we can figure out its magnitude, but we don't know what theta is. How do we figure out what theta? To do that, let's just think about it a little bit."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The hypotenuse is the magnitude of this vector. It's the magnitude of the vector f. Or we could say the magnitude of the vector f times the cosine of theta, I'm just multiplying both sides times the magnitude of the vector x. f is equal to d. But that's still, you might say, well Sal, we know what f is, we can figure that out, we can figure out its magnitude, but we don't know what theta is. How do we figure out what theta? To do that, let's just think about it a little bit. This angle, this angle theta, is the same angle, so this distance here isn't necessarily the same as the length of the normal vector, but it's definitely going in the same direction. This angle here is really the same thing as the angle between this vector and the normal vector. You might remember from earlier linear algebra when we talked about the dot products of two vectors, it involves something with the cosine of the angle between them."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "To do that, let's just think about it a little bit. This angle, this angle theta, is the same angle, so this distance here isn't necessarily the same as the length of the normal vector, but it's definitely going in the same direction. This angle here is really the same thing as the angle between this vector and the normal vector. You might remember from earlier linear algebra when we talked about the dot products of two vectors, it involves something with the cosine of the angle between them. To make that fresh in your mind, let's multiply and divide both sides, let me multiply and divide the left side of this equation by the magnitude of the normal vector. I'm obviously not changing its value, I'm multiplying and dividing by the same number. I'm going to multiply by the magnitude of the normal vector and I'm going to divide by the magnitude of the normal vector."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You might remember from earlier linear algebra when we talked about the dot products of two vectors, it involves something with the cosine of the angle between them. To make that fresh in your mind, let's multiply and divide both sides, let me multiply and divide the left side of this equation by the magnitude of the normal vector. I'm obviously not changing its value, I'm multiplying and dividing by the same number. I'm going to multiply by the magnitude of the normal vector and I'm going to divide by the magnitude of the normal vector. I'm essentially multiplying by one, so I have not changed this. But when you do it in this, it might ring a bell. This expression up here, this expression right here, is the dot product of the normal vector and this vector right here, f. This right here is the dot product, this is n dot f up there."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to multiply by the magnitude of the normal vector and I'm going to divide by the magnitude of the normal vector. I'm essentially multiplying by one, so I have not changed this. But when you do it in this, it might ring a bell. This expression up here, this expression right here, is the dot product of the normal vector and this vector right here, f. This right here is the dot product, this is n dot f up there. It's equal to the product of their magnitudes times the cosine of the angle between them. The distance, that shortest distance we care about, is the dot product between this vector, the normal vector, divided by the magnitude of the normal vector. Let's do that, let's take the dot product between the normal and this."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This expression up here, this expression right here, is the dot product of the normal vector and this vector right here, f. This right here is the dot product, this is n dot f up there. It's equal to the product of their magnitudes times the cosine of the angle between them. The distance, that shortest distance we care about, is the dot product between this vector, the normal vector, divided by the magnitude of the normal vector. Let's do that, let's take the dot product between the normal and this. We already figured out in the last video, the normal vector, if you have the equation of a plane, the normal vector is literally, its components are just the coefficients on the x, y, and z terms. This is the normal vector right over here. Let's literally take the dot product."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's do that, let's take the dot product between the normal and this. We already figured out in the last video, the normal vector, if you have the equation of a plane, the normal vector is literally, its components are just the coefficients on the x, y, and z terms. This is the normal vector right over here. Let's literally take the dot product. n dot f is going to be equal to a times x naught minus xp. I'll do that in pink. It'll be a x naught minus a xp, and then plus b times the y component here."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's literally take the dot product. n dot f is going to be equal to a times x naught minus xp. I'll do that in pink. It'll be a x naught minus a xp, and then plus b times the y component here. Plus b y naught, I'm just distributing the b, minus b yp, and then plus c times the z component. Plus c z naught minus c zp. All of that over the magnitude of the normal vector."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It'll be a x naught minus a xp, and then plus b times the y component here. Plus b y naught, I'm just distributing the b, minus b yp, and then plus c times the z component. Plus c z naught minus c zp. All of that over the magnitude of the normal vector. What's the magnitude of the normal vector going to be? It's just the square root of the normal vector dotted with itself. It's just each of these guys squared, added to themselves, and you're taking the square root."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All of that over the magnitude of the normal vector. What's the magnitude of the normal vector going to be? It's just the square root of the normal vector dotted with itself. It's just each of these guys squared, added to themselves, and you're taking the square root. It's the square root of a squared plus b squared plus c squared. What does this up here simplify to? Let me just rewrite this."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's just each of these guys squared, added to themselves, and you're taking the square root. It's the square root of a squared plus b squared plus c squared. What does this up here simplify to? Let me just rewrite this. This is the distance in question. This right here is equal to the distance. Let's see if we can simplify it."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me just rewrite this. This is the distance in question. This right here is equal to the distance. Let's see if we can simplify it. First, we can take all of the terms with the x naught. These involve the point that sits off the plane. Remember, x naught, y naught, z naught sat off the plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if we can simplify it. First, we can take all of the terms with the x naught. These involve the point that sits off the plane. Remember, x naught, y naught, z naught sat off the plane. This is a x naught plus b y naught plus c z naught. Then what are these terms equal to? What are these terms?"}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Remember, x naught, y naught, z naught sat off the plane. This is a x naught plus b y naught plus c z naught. Then what are these terms equal to? What are these terms? Negative a xp minus b yp minus c zp. If you remember here, d in the equation of a plane, d, when we started in the last video, when we tried to figure out what the normal to a plane is, d is, if this point xp sits on the plane, d is a xp plus b yp plus c zp. Or another way you could say it is negative d would be negative a, and it's just the difference between lower case and upper case here."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What are these terms? Negative a xp minus b yp minus c zp. If you remember here, d in the equation of a plane, d, when we started in the last video, when we tried to figure out what the normal to a plane is, d is, if this point xp sits on the plane, d is a xp plus b yp plus c zp. Or another way you could say it is negative d would be negative a, and it's just the difference between lower case and upper case here. We say that lower case a is the same as this upper case a. It's negative a xp minus b yp minus c zp. I'm just using what we got from the last video."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or another way you could say it is negative d would be negative a, and it's just the difference between lower case and upper case here. We say that lower case a is the same as this upper case a. It's negative a xp minus b yp minus c zp. I'm just using what we got from the last video. This is what d is, so negative d will be this business. That's exactly what we have over here. Negative a xp, negative b yp, negative c zp."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just using what we got from the last video. This is what d is, so negative d will be this business. That's exactly what we have over here. Negative a xp, negative b yp, negative c zp. All of this term, this term, and this term simplifies to a minus d. Remember, this negative capital d, this is the d from the equation of the plane, not the d, not the distance d. This is the numerator of our distance, and then the denominator of our distance is just the square root of a squared plus b squared plus c squared. We're done. This tells us the distance between any point and a plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Negative a xp, negative b yp, negative c zp. All of this term, this term, and this term simplifies to a minus d. Remember, this negative capital d, this is the d from the equation of the plane, not the d, not the distance d. This is the numerator of our distance, and then the denominator of our distance is just the square root of a squared plus b squared plus c squared. We're done. This tells us the distance between any point and a plane. This is a pretty intuitive formula here because all we're doing, if I'm giving you, let me give you an example. Let's say I have the plane 1x minus 2y plus 3z is equal to 5. That's some plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This tells us the distance between any point and a plane. This is a pretty intuitive formula here because all we're doing, if I'm giving you, let me give you an example. Let's say I have the plane 1x minus 2y plus 3z is equal to 5. That's some plane. Let me pick some point that's not on the plane. Let's say I have the point 2, 2, 3. Let me make sure it's not on the plane."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's some plane. Let me pick some point that's not on the plane. Let's say I have the point 2, 2, 3. Let me make sure it's not on the plane. 2 minus 6 is negative, let me just pick a random 1. This definitely is not on the plane because we have 2 minus 6 plus 3. That gives us negative 1, which is not 5."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me make sure it's not on the plane. 2 minus 6 is negative, let me just pick a random 1. This definitely is not on the plane because we have 2 minus 6 plus 3. That gives us negative 1, which is not 5. This is definitely not on the plane. We can find the distance between this point and the plane using the formula we just derived. We literally just evaluate it."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That gives us negative 1, which is not 5. This is definitely not on the plane. We can find the distance between this point and the plane using the formula we just derived. We literally just evaluate it. This will just be 1 times 2 minus 2 times, I'm going to fill it in, plus 3 times something minus 5. All of that over, and I haven't put these guys in. I'm going to do that right now."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We literally just evaluate it. This will just be 1 times 2 minus 2 times, I'm going to fill it in, plus 3 times something minus 5. All of that over, and I haven't put these guys in. I'm going to do that right now. 1 times 2 minus 2 times 3 plus 3 times 1, this one minus 5, kind of bringing it over to the left-hand side. All of that over the square root of 1 squared, which is 1, plus negative 2 squared, which is 4, plus 3 squared, which is 9. It's going to be equal to 2 minus 6, or negative 6, plus 3, and then minus 5."}, {"video_title": "Point distance to plane Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to do that right now. 1 times 2 minus 2 times 3 plus 3 times 1, this one minus 5, kind of bringing it over to the left-hand side. All of that over the square root of 1 squared, which is 1, plus negative 2 squared, which is 4, plus 3 squared, which is 9. It's going to be equal to 2 minus 6, or negative 6, plus 3, and then minus 5. This is 5, 2 plus 3 is 5, minus 5, so those cancel out. This is negative 6, so it's equal to negative 6 over the square root of 5 plus 9 is 14. This is negative 6 over the square root of 14, and you're done."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "We learned several videos ago that its row space is the same thing as the column space of its transpose, so that right there is the row space of A. That this thing's orthogonal complement, so the set of all of the vectors that are orthogonal to this, so its orthogonal complement is equal to the null space of A. And essentially the same result if you switch A and A transpose, we also learned that the orthogonal complement of the column space of A is equal to the left null space of A, which is the same thing as the null space of A transpose. So we could write this just to understand the terminology. That's the left null space, which is the same thing as the null space of A transpose. Now, what is the orthogonal complement of the null space of A? What is the orthogonal complement of the null space of A?"}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "So we could write this just to understand the terminology. That's the left null space, which is the same thing as the null space of A transpose. Now, what is the orthogonal complement of the null space of A? What is the orthogonal complement of the null space of A? Well, you might guess that it's the row space of A, but we didn't have the tools until the last video to figure that out. In the last video, we saw that if we take the orthogonal complement of the orthogonal complement, it equals the original subspace. So now what are we doing?"}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "What is the orthogonal complement of the null space of A? Well, you might guess that it's the row space of A, but we didn't have the tools until the last video to figure that out. In the last video, we saw that if we take the orthogonal complement of the orthogonal complement, it equals the original subspace. So now what are we doing? We're taking the orthogonal complement of the null space of A. Well, the null space of A is just this thing right here. The null space of A is that thing right there."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "So now what are we doing? We're taking the orthogonal complement of the null space of A. Well, the null space of A is just this thing right here. The null space of A is that thing right there. So this is equal to taking the orthogonal complement of the null space of A, but the null space of A is this thing. It's the row space's orthogonal complement. Now, we're essentially taking the orthogonal complement of the orthogonal complement."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "The null space of A is that thing right there. So this is equal to taking the orthogonal complement of the null space of A, but the null space of A is this thing. It's the row space's orthogonal complement. Now, we're essentially taking the orthogonal complement of the orthogonal complement. So we can use this property, which we just proved in the last video, to say that this is equal to just the row space of A, which is the same thing as the column space of A transpose. So the orthogonal complement of the row space is the null space, and the orthogonal complement of the null space is the row space. We can apply that same property on this side right here."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "Now, we're essentially taking the orthogonal complement of the orthogonal complement. So we can use this property, which we just proved in the last video, to say that this is equal to just the row space of A, which is the same thing as the column space of A transpose. So the orthogonal complement of the row space is the null space, and the orthogonal complement of the null space is the row space. We can apply that same property on this side right here. What is the orthogonal complement of the left null space of A? What is this? Well, this is going to be equal to the orthogonal complement of this thing, because that's what the left null space of A is equal to."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "We can apply that same property on this side right here. What is the orthogonal complement of the left null space of A? What is this? Well, this is going to be equal to the orthogonal complement of this thing, because that's what the left null space of A is equal to. So it's equal to the orthogonal complement of the column space. And we just learned in the last video, if you take the orthogonal complement of the orthogonal complement, it equals the original subspace. So this is just equal to the column space of A."}, {"video_title": "Orthogonal complement of the nullspace Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is going to be equal to the orthogonal complement of this thing, because that's what the left null space of A is equal to. So it's equal to the orthogonal complement of the column space. And we just learned in the last video, if you take the orthogonal complement of the orthogonal complement, it equals the original subspace. So this is just equal to the column space of A. So we now see some nice symmetry. The null space is the orthogonal complement of the row space, and then we see that the row space is the orthogonal complement of the null space. Similarly, the left null space is the orthogonal complement of the column space, and the column space is the orthogonal complement of the left null space."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let me call my set B. And let's say I have the vectors v1, v2, all the way through vk. Let's say this isn't just any set of vectors. There's some interesting things about these vectors. The first thing is that all of these guys have length of 1. So we could say the length of vector vi is equal to 1 for i is equal to, well, we could say between 1 and k, or i is equal to 1, 2, all the way to k. All of these guys have length equal 1. Or another way to say it is that the square of their lengths are 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "There's some interesting things about these vectors. The first thing is that all of these guys have length of 1. So we could say the length of vector vi is equal to 1 for i is equal to, well, we could say between 1 and k, or i is equal to 1, 2, all the way to k. All of these guys have length equal 1. Or another way to say it is that the square of their lengths are 1. The square of vi's length is equal to 1. Or vi dot vi is equal to 1 for i is, any of these guys. Any i can be 1, 2, 3, all the way to k. Or i is equal to 1, 2, all the way to k. So that's the first interesting thing about it."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is that the square of their lengths are 1. The square of vi's length is equal to 1. Or vi dot vi is equal to 1 for i is, any of these guys. Any i can be 1, 2, 3, all the way to k. Or i is equal to 1, 2, all the way to k. So that's the first interesting thing about it. Let me write it in regular words. All the vectors in B have length 1. Or another way to say it is that they've all been normalized."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Any i can be 1, 2, 3, all the way to k. Or i is equal to 1, 2, all the way to k. So that's the first interesting thing about it. Let me write it in regular words. All the vectors in B have length 1. Or another way to say it is that they've all been normalized. Or they're all unit vectors. Normalized vectors are vectors that you've made their lengths 1. You've turned them into unit vectors."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is that they've all been normalized. Or they're all unit vectors. Normalized vectors are vectors that you've made their lengths 1. You've turned them into unit vectors. They have all been normalized. So that's the first interesting thing about my set B, and then the next interesting thing about my set B, is that all of the vectors are orthogonal to each other. So if you dot it with itself, you get length 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "You've turned them into unit vectors. They have all been normalized. So that's the first interesting thing about my set B, and then the next interesting thing about my set B, is that all of the vectors are orthogonal to each other. So if you dot it with itself, you get length 1. But if you take a vector and you dot it with any other vector, if you take vi and you were to dot it with vj, so if you took v2 and dotted it with v1, it's going to be equal to 0 for i does not equal j. It's going to be 0. All of these guys are orthogonal."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So if you dot it with itself, you get length 1. But if you take a vector and you dot it with any other vector, if you take vi and you were to dot it with vj, so if you took v2 and dotted it with v1, it's going to be equal to 0 for i does not equal j. It's going to be 0. All of these guys are orthogonal. So let me write that down. All of the vectors are orthogonal to each other. And of course, they're not orthogonal to themselves because they all have length 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "All of these guys are orthogonal. So let me write that down. All of the vectors are orthogonal to each other. And of course, they're not orthogonal to themselves because they all have length 1. So if you take the dot product with itself, you get 1. If you take a dot product with some other guy in your set, you're going to get 0. Maybe I can write it this way."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And of course, they're not orthogonal to themselves because they all have length 1. So if you take the dot product with itself, you get 1. If you take a dot product with some other guy in your set, you're going to get 0. Maybe I can write it this way. vi dot vj for all the members of the set is going to be equal to 0 for i does not equal j. And then if these guys are the same vector, I'm dotting with myself, I'm going to have length 1. So it would equal length 1 for i is equal to j."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I can write it this way. vi dot vj for all the members of the set is going to be equal to 0 for i does not equal j. And then if these guys are the same vector, I'm dotting with myself, I'm going to have length 1. So it would equal length 1 for i is equal to j. So I've got a special set. All of these guys have length 1. And they are all orthogonal with each other."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So it would equal length 1 for i is equal to j. So I've got a special set. All of these guys have length 1. And they are all orthogonal with each other. They're normalized and they're all orthogonal. And we have a special word for this. This is called an orthonormal set."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And they are all orthogonal with each other. They're normalized and they're all orthogonal. And we have a special word for this. This is called an orthonormal set. So B is an ortho for orthogonal, orthonormal set. Normal for normalized. Everything is orthogonal."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This is called an orthonormal set. So B is an ortho for orthogonal, orthonormal set. Normal for normalized. Everything is orthogonal. They're all orthogonal relative to each other. And everything has been normalized. Everything has length 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Everything is orthogonal. They're all orthogonal relative to each other. And everything has been normalized. Everything has length 1. Now, the first interesting thing about an orthonormal set is that it's also going to be a linearly independent set. So if B is orthonormal, B is also going to be linearly independent. And how can I show that to you?"}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Everything has length 1. Now, the first interesting thing about an orthonormal set is that it's also going to be a linearly independent set. So if B is orthonormal, B is also going to be linearly independent. And how can I show that to you? Well, let's assume that it isn't linearly independent. Let's assume, so they're clearly all of, let me take vi and let me take vj that are members of my set. And let's assume that i does not equal j."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And how can I show that to you? Well, let's assume that it isn't linearly independent. Let's assume, so they're clearly all of, let me take vi and let me take vj that are members of my set. And let's assume that i does not equal j. Now, we already know that it's an orthonormal set. So vi dot vj is going to be equal to 0. They're orthogonal."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And let's assume that i does not equal j. Now, we already know that it's an orthonormal set. So vi dot vj is going to be equal to 0. They're orthogonal. These are two vectors in my set. Now, let's assume that they are linearly dependent. I want to prove that they're linearly independent."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "They're orthogonal. These are two vectors in my set. Now, let's assume that they are linearly dependent. I want to prove that they're linearly independent. And the way I'm going to prove that is by assuming they're linearly dependent and then arriving at a contradiction. So let's assume that vi and vj are linearly dependent. Linearly dependent."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "I want to prove that they're linearly independent. And the way I'm going to prove that is by assuming they're linearly dependent and then arriving at a contradiction. So let's assume that vi and vj are linearly dependent. Linearly dependent. Well, then that means that I can represent one of these guys as a scalar multiple of the other. And I can pick either way. So let's just say, for the sake of argument, that I can represent vi."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Linearly dependent. Well, then that means that I can represent one of these guys as a scalar multiple of the other. And I can pick either way. So let's just say, for the sake of argument, that I can represent vi. Let's say that vi is equal to some scalar c times vj. That's what linear dependency means, that one of them can be represented as a scalar multiple of the other. Well, if this is true, then I can just substitute this back in for vi, and what do I get?"}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let's just say, for the sake of argument, that I can represent vi. Let's say that vi is equal to some scalar c times vj. That's what linear dependency means, that one of them can be represented as a scalar multiple of the other. Well, if this is true, then I can just substitute this back in for vi, and what do I get? I get c times vj, which is just another way of writing vi, because I assumed linear dependence. That dot vj has got to be equal to 0. This guy was vi, this is vj."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Well, if this is true, then I can just substitute this back in for vi, and what do I get? I get c times vj, which is just another way of writing vi, because I assumed linear dependence. That dot vj has got to be equal to 0. This guy was vi, this is vj. They are orthogonal to each other. But this right here is just equal to c times vj dot vj, which is just equal to c times the length of vj squared. And that has to equal 0."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "This guy was vi, this is vj. They are orthogonal to each other. But this right here is just equal to c times vj dot vj, which is just equal to c times the length of vj squared. And that has to equal 0. They're orthogonal, so it has to be equal to 0, which implies that the length of vj has to be equal to 0. If we assume that this is some non-zero multiple, and this has to be some non-zero multiple. I should have written it there."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And that has to equal 0. They're orthogonal, so it has to be equal to 0, which implies that the length of vj has to be equal to 0. If we assume that this is some non-zero multiple, and this has to be some non-zero multiple. I should have written it there. c does not equal 0. Why does this have to be a non-zero multiple? Because these were both non-zero vectors."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "I should have written it there. c does not equal 0. Why does this have to be a non-zero multiple? Because these were both non-zero vectors. This is a non-zero vector, so this guy can't be 0. This guy has length 1. So if this is a non-zero vector, there's no way that I could just put a 0 here, because if I put a 0, then I would get a 0 vector."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Because these were both non-zero vectors. This is a non-zero vector, so this guy can't be 0. This guy has length 1. So if this is a non-zero vector, there's no way that I could just put a 0 here, because if I put a 0, then I would get a 0 vector. So c can't be 0. So if c isn't 0, then this guy right here has to be 0. And so we get that the length of vj is 0, which we know is false."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So if this is a non-zero vector, there's no way that I could just put a 0 here, because if I put a 0, then I would get a 0 vector. So c can't be 0. So if c isn't 0, then this guy right here has to be 0. And so we get that the length of vj is 0, which we know is false. The length of vj is 1. This is an orthonormal set. The length of all of the members of B are 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And so we get that the length of vj is 0, which we know is false. The length of vj is 1. This is an orthonormal set. The length of all of the members of B are 1. So we reach a contradiction. This is our contradiction. vj is not the 0 vector."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "The length of all of the members of B are 1. So we reach a contradiction. This is our contradiction. vj is not the 0 vector. It has length 1. Contradiction. So if you have a bunch of vectors that are orthogonal and they're non-zero, they have to be linearly independent, which is pretty interesting."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "vj is not the 0 vector. It has length 1. Contradiction. So if you have a bunch of vectors that are orthogonal and they're non-zero, they have to be linearly independent, which is pretty interesting. So if I have this set, this orthonormal set right here, it's also a set of linearly independent vectors. So it can be a basis for a subspace. So let's say that B is the basis for some subspace v, or we could say that v is equal to the span of v1, v2, all the way to vk."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So if you have a bunch of vectors that are orthogonal and they're non-zero, they have to be linearly independent, which is pretty interesting. So if I have this set, this orthonormal set right here, it's also a set of linearly independent vectors. So it can be a basis for a subspace. So let's say that B is the basis for some subspace v, or we could say that v is equal to the span of v1, v2, all the way to vk. Then we call B, if it was just a set, we call it an orthonormal set. But it can be an orthonormal basis. When it spans some subspace."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that B is the basis for some subspace v, or we could say that v is equal to the span of v1, v2, all the way to vk. Then we call B, if it was just a set, we call it an orthonormal set. But it can be an orthonormal basis. When it spans some subspace. So we can write, we can say that B is an orthonormal basis for v. Now everything I've done is very abstract, but let me do some quick examples for you, just so you understand what an orthonormal basis looks like with real numbers. So let's say I have 2 vectors. Let's say I have the vector v1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "When it spans some subspace. So we can write, we can say that B is an orthonormal basis for v. Now everything I've done is very abstract, but let me do some quick examples for you, just so you understand what an orthonormal basis looks like with real numbers. So let's say I have 2 vectors. Let's say I have the vector v1. That is, let's say we're dealing in R3. So it's 1 third, 2 thirds, 2 thirds, and 2 thirds. And let's say I have another vector v2 that is equal to 2 thirds, 1 third, and minus 2 thirds."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have the vector v1. That is, let's say we're dealing in R3. So it's 1 third, 2 thirds, 2 thirds, and 2 thirds. And let's say I have another vector v2 that is equal to 2 thirds, 1 third, and minus 2 thirds. And let's say that B is the set of v1 and v2. So the first question is, what are the lengths of these guys? So let's take the length of v1 squared is just v1 dot v1, which is just 1 third squared, which is just 1 over 9, plus 2 thirds squared, which is 4 over 9, plus 2 thirds squared, which is 4 over 9, which is equal to 1."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And let's say I have another vector v2 that is equal to 2 thirds, 1 third, and minus 2 thirds. And let's say that B is the set of v1 and v2. So the first question is, what are the lengths of these guys? So let's take the length of v1 squared is just v1 dot v1, which is just 1 third squared, which is just 1 over 9, plus 2 thirds squared, which is 4 over 9, plus 2 thirds squared, which is 4 over 9, which is equal to 1. So the length squared is 1, then that tells us that the length of our first vector is equal to 1. If the square of the length is 1, you take the square root, so the length is 1. What about vector 2?"}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So let's take the length of v1 squared is just v1 dot v1, which is just 1 third squared, which is just 1 over 9, plus 2 thirds squared, which is 4 over 9, plus 2 thirds squared, which is 4 over 9, which is equal to 1. So the length squared is 1, then that tells us that the length of our first vector is equal to 1. If the square of the length is 1, you take the square root, so the length is 1. What about vector 2? Well, the length of vector 2 squared is equal to v2 dot v2, which is equal to 2 thirds squared is 4 ninths, plus 1 third squared is 1 ninth, plus 2 thirds squared is 4 ninths. So that is 9 ninths, which is equal to 1, which tells us that the length of v2 is equal to 1. So we know that these guys are definitely normalized."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "What about vector 2? Well, the length of vector 2 squared is equal to v2 dot v2, which is equal to 2 thirds squared is 4 ninths, plus 1 third squared is 1 ninth, plus 2 thirds squared is 4 ninths. So that is 9 ninths, which is equal to 1, which tells us that the length of v2 is equal to 1. So we know that these guys are definitely normalized. We could call this a normalized set, but is it an orthonormal set? Are these guys orthogonal to each other? And to test that out, we just take their dot product."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "So we know that these guys are definitely normalized. We could call this a normalized set, but is it an orthonormal set? Are these guys orthogonal to each other? And to test that out, we just take their dot product. So v1 dot v2 is equal to 1 third times 2 thirds, which is 2 ninths, plus 2 thirds times 1 third, which is 2 ninths, plus 2 thirds times minus 2 thirds. That's minus 4 ninths. 2 plus 2 minus 4 is 0, so it equals 0."}, {"video_title": "Introduction to orthonormal bases Linear Algebra Khan Academy.mp3", "Sentence": "And to test that out, we just take their dot product. So v1 dot v2 is equal to 1 third times 2 thirds, which is 2 ninths, plus 2 thirds times 1 third, which is 2 ninths, plus 2 thirds times minus 2 thirds. That's minus 4 ninths. 2 plus 2 minus 4 is 0, so it equals 0. So these guys indeed are orthogonal. So B is an orthonormal set. And if I have some subspace, let's say that B is equal to the span of v1 and v2, then we can say that the basis for B, or we could say that B is an orthonormal basis for v."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So if you dot it with yourself, you get 1. If you dot it with any of the other columns, you get 0. We've seen this multiple times. It's orthogonal to everything else. If you have a matrix like this, and I actually forgot to tell you the name of this, this is called an orthogonal matrix, we've already seen that the transpose of this matrix is the same thing as the inverse of this matrix, which makes it super, super duper useful to deal with. The transpose of this matrix is equal to the inverse. Now this statement leads to some other interesting things about this."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "It's orthogonal to everything else. If you have a matrix like this, and I actually forgot to tell you the name of this, this is called an orthogonal matrix, we've already seen that the transpose of this matrix is the same thing as the inverse of this matrix, which makes it super, super duper useful to deal with. The transpose of this matrix is equal to the inverse. Now this statement leads to some other interesting things about this. So so far, we've been dealing this mainly with the change of basis. I can kind of draw the diagram that you're probably tired of by now. If I have some, let's say that's the standard basis."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Now this statement leads to some other interesting things about this. So so far, we've been dealing this mainly with the change of basis. I can kind of draw the diagram that you're probably tired of by now. If I have some, let's say that's the standard basis. Let's say that I have x in coordinates with another basis. We've seen I can multiply this guy times C to get that up there. I could multiply that guy by C inverse to get this guy right here."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "If I have some, let's say that's the standard basis. Let's say that I have x in coordinates with another basis. We've seen I can multiply this guy times C to get that up there. I could multiply that guy by C inverse to get this guy right here. And in that world, we viewed C as just a change of a basis. We're representing the same matrix, we're representing the same vector, we're just changing the coordinates of how we represent it. But we also know that any matrix product, any matrix vector product, is also a linear transformation."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "I could multiply that guy by C inverse to get this guy right here. And in that world, we viewed C as just a change of a basis. We're representing the same matrix, we're representing the same vector, we're just changing the coordinates of how we represent it. But we also know that any matrix product, any matrix vector product, is also a linear transformation. So this change of basis is really just a linear transformation. What I want to show you in this video, and you can view it either as a change of basis or as a linear transformation, is that when you multiply this orthogonal matrix times some vector, it preserves lengths and angles. So let's have a little touchy-feely discussion of what that means."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "But we also know that any matrix product, any matrix vector product, is also a linear transformation. So this change of basis is really just a linear transformation. What I want to show you in this video, and you can view it either as a change of basis or as a linear transformation, is that when you multiply this orthogonal matrix times some vector, it preserves lengths and angles. So let's have a little touchy-feely discussion of what that means. So let's view it as a transformation. Let's say I have some set of vectors in my domain. Let's say they look like this."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So let's have a little touchy-feely discussion of what that means. So let's view it as a transformation. Let's say I have some set of vectors in my domain. Let's say they look like this. Let's say that it looks like this. If I draw that one like that guy, and then this guy like that, there's some angle between them. And angles are easy to visualize in R2, R3, maybe a little harder once we get to higher dimensions."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Let's say they look like this. Let's say that it looks like this. If I draw that one like that guy, and then this guy like that, there's some angle between them. And angles are easy to visualize in R2, R3, maybe a little harder once we get to higher dimensions. But that's the angle between them. Now, if we're saying that we're preserving the angles and the lengths, that means if I were to multiply these vectors times c, then we can view it as a transformation. Maybe I rotate them or do something like that."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "And angles are easy to visualize in R2, R3, maybe a little harder once we get to higher dimensions. But that's the angle between them. Now, if we're saying that we're preserving the angles and the lengths, that means if I were to multiply these vectors times c, then we can view it as a transformation. Maybe I rotate them or do something like that. So maybe that pink vector will now look like this. But it's going to have the same length. This length is going to be the same thing as that length."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Maybe I rotate them or do something like that. So maybe that pink vector will now look like this. But it's going to have the same length. This length is going to be the same thing as that length. And even more, when I said it preserves lengths and angles, this yellow vector is going to look something like this, where the angle is going to be the same, where this theta is going to be that theta. That's what I mean by preserves angles. If we didn't have this case, we could imagine a transformation that doesn't preserve angles."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "This length is going to be the same thing as that length. And even more, when I said it preserves lengths and angles, this yellow vector is going to look something like this, where the angle is going to be the same, where this theta is going to be that theta. That's what I mean by preserves angles. If we didn't have this case, we could imagine a transformation that doesn't preserve angles. Let me draw one that doesn't. If this got transformed to, I don't know, let's say this guy got a lot longer, and then let's say this guy also got longer. And actually, I want to show that the angle also doesn't get preserved."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "If we didn't have this case, we could imagine a transformation that doesn't preserve angles. Let me draw one that doesn't. If this got transformed to, I don't know, let's say this guy got a lot longer, and then let's say this guy also got longer. And actually, I want to show that the angle also doesn't get preserved. Not only did it get longer, but it got distorted a little bit. So the angle also changed. This transformation right there is not preserving angles."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "And actually, I want to show that the angle also doesn't get preserved. Not only did it get longer, but it got distorted a little bit. So the angle also changed. This transformation right there is not preserving angles. So when you have a change of basis matrix that's orthogonal, or when you have a transformation matrix that's orthogonal, all it's essentially doing to your vectors is it can kind of rotate them around, but it's not going to really distort them. So, and I'll write that in quotes, because I've never thought of mathematically rigorous terms. So no distortion of vectors."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "This transformation right there is not preserving angles. So when you have a change of basis matrix that's orthogonal, or when you have a transformation matrix that's orthogonal, all it's essentially doing to your vectors is it can kind of rotate them around, but it's not going to really distort them. So, and I'll write that in quotes, because I've never thought of mathematically rigorous terms. So no distortion of vectors. So I've kind of shown you the intuition of what that means. Let's actually prove it to ourselves that this is the case. So let's say I'm saying that if I have, let's say this pink vector here is x, and that this pink vector here is c times x, I'm claiming that the length of x is equal to the length of c times x."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So no distortion of vectors. So I've kind of shown you the intuition of what that means. Let's actually prove it to ourselves that this is the case. So let's say I'm saying that if I have, let's say this pink vector here is x, and that this pink vector here is c times x, I'm claiming that the length of x is equal to the length of c times x. Let's see if that's actually the case. So the length of x squared, or let me do it, the length of, let me write it this way, the length of cx squared is the same thing as cx dot cx. And here it's always useful for me to kind of remind myself that if I take two vectors, let me do it over here, let's say I have y dot y, this is the same thing as y transpose, if you view them as matrices, y transpose times y. y transpose y is just y1, y2, all the way to yn, times y1, y2, all the way to yn."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I'm saying that if I have, let's say this pink vector here is x, and that this pink vector here is c times x, I'm claiming that the length of x is equal to the length of c times x. Let's see if that's actually the case. So the length of x squared, or let me do it, the length of, let me write it this way, the length of cx squared is the same thing as cx dot cx. And here it's always useful for me to kind of remind myself that if I take two vectors, let me do it over here, let's say I have y dot y, this is the same thing as y transpose, if you view them as matrices, y transpose times y. y transpose y is just y1, y2, all the way to yn, times y1, y2, all the way to yn. And if you were to do this 1 by n times n by 1 matrix product, you're going to get a 1 by 1 matrix, or just a number, that's going to be y1 times y1 plus y2 times y2, all the way to yn times yn. This is the same thing as y dot y. So let's, I think I did this 10 or 20 videos ago, but it's always a good refresher."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "And here it's always useful for me to kind of remind myself that if I take two vectors, let me do it over here, let's say I have y dot y, this is the same thing as y transpose, if you view them as matrices, y transpose times y. y transpose y is just y1, y2, all the way to yn, times y1, y2, all the way to yn. And if you were to do this 1 by n times n by 1 matrix product, you're going to get a 1 by 1 matrix, or just a number, that's going to be y1 times y1 plus y2 times y2, all the way to yn times yn. This is the same thing as y dot y. So let's, I think I did this 10 or 20 videos ago, but it's always a good refresher. So let's use this property right here. So these two dotted with each other, this is the same thing as taking one of their transpose times the other one. So turn this from a vector dot product to a matrix matrix product."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So let's, I think I did this 10 or 20 videos ago, but it's always a good refresher. So let's use this property right here. So these two dotted with each other, this is the same thing as taking one of their transpose times the other one. So turn this from a vector dot product to a matrix matrix product. So this is the same thing as cx transpose, so you can do this as a 1 by n matrix now, times the n by 1 matrix, which is just the column vector cx. These are the same thing. Now we also know that a times b transpose is the same thing as b transpose a transpose."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So turn this from a vector dot product to a matrix matrix product. So this is the same thing as cx transpose, so you can do this as a 1 by n matrix now, times the n by 1 matrix, which is just the column vector cx. These are the same thing. Now we also know that a times b transpose is the same thing as b transpose a transpose. We saw that a long time ago. So this thing right here is going to be equal to x transpose c transpose. I just switched their order and take the transpose of each."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Now we also know that a times b transpose is the same thing as b transpose a transpose. We saw that a long time ago. So this thing right here is going to be equal to x transpose c transpose. I just switched their order and take the transpose of each. x transpose times c transpose, and then you have that times cx, and now we know that c transpose is the same thing as c inverse. This is what we needed, the orthogonality, this is where we need the orthogonality of the matrix c. This is where we need it to be a square matrix where all of its columns are mutually orthogonal and they're all normal, and so this thing is just going to become the identity matrix. That's just going to be the identity matrix."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "I just switched their order and take the transpose of each. x transpose times c transpose, and then you have that times cx, and now we know that c transpose is the same thing as c inverse. This is what we needed, the orthogonality, this is where we need the orthogonality of the matrix c. This is where we need it to be a square matrix where all of its columns are mutually orthogonal and they're all normal, and so this thing is just going to become the identity matrix. That's just going to be the identity matrix. So I could write the identity matrix there, but that's just going to disappear, so this is going to be equal to x transpose x. x transpose x is the same thing as x dot x, which is the same thing as the length of x squared. So the length of cx squared is the same thing as the length of x squared. Well, that tells us that the length of x, or the length of cx is the length of x, right?"}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "That's just going to be the identity matrix. So I could write the identity matrix there, but that's just going to disappear, so this is going to be equal to x transpose x. x transpose x is the same thing as x dot x, which is the same thing as the length of x squared. So the length of cx squared is the same thing as the length of x squared. Well, that tells us that the length of x, or the length of cx is the length of x, right? Because both of these are going to be positive quantities. So I've shown you that orthogonal matrices definitely preserve length. Let's see if they preserve angles."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Well, that tells us that the length of x, or the length of cx is the length of x, right? Because both of these are going to be positive quantities. So I've shown you that orthogonal matrices definitely preserve length. Let's see if they preserve angles. So we actually had to define angles, because throughout our mathematical careers we understood what angles mean in kind of R2 or R3, but in linear algebra we like to be general, and we defined an angle using the dot product. We use the law of cosines, and we took an analogy to kind of a triangle in R2, but we defined an angle, or we said that the dot product v dot w, it's equal to the lengths of those two vectors, the products of the lengths of those two vectors, times the cosine of the angle between them. Or you could say that the cosine of the angle between two vectors we defined as the dot product of those two vectors divided by the lengths of those two vectors."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if they preserve angles. So we actually had to define angles, because throughout our mathematical careers we understood what angles mean in kind of R2 or R3, but in linear algebra we like to be general, and we defined an angle using the dot product. We use the law of cosines, and we took an analogy to kind of a triangle in R2, but we defined an angle, or we said that the dot product v dot w, it's equal to the lengths of those two vectors, the products of the lengths of those two vectors, times the cosine of the angle between them. Or you could say that the cosine of the angle between two vectors we defined as the dot product of those two vectors divided by the lengths of those two vectors. This was a definition so that we could extend the idea of an angle to an arbitrarily high dimension, to R Google if we had to. So let's see if it preserves. So let's see what the angle is if we multiply these guys by c. So if we want to, let's say our new angle, so cosine of theta c, once we perform our transformation."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Or you could say that the cosine of the angle between two vectors we defined as the dot product of those two vectors divided by the lengths of those two vectors. This was a definition so that we could extend the idea of an angle to an arbitrarily high dimension, to R Google if we had to. So let's see if it preserves. So let's see what the angle is if we multiply these guys by c. So if we want to, let's say our new angle, so cosine of theta c, once we perform our transformation. So our new angle, well, we're going to perform the transformation on all of these characters. It's going to be cv dot cw over the lengths of cv times the length of cw. Now, we already know that lengths are preserved."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So let's see what the angle is if we multiply these guys by c. So if we want to, let's say our new angle, so cosine of theta c, once we perform our transformation. So our new angle, well, we're going to perform the transformation on all of these characters. It's going to be cv dot cw over the lengths of cv times the length of cw. Now, we already know that lengths are preserved. We already know that the length of cw and cv are just going to be w and v. We just proved that, so let me write that. So the cosine of theta c is equal to cv dot cw over the lengths of v times w. Because we know, we've already shown that it preserves length. So let's see what this top part equals."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "Now, we already know that lengths are preserved. We already know that the length of cw and cv are just going to be w and v. We just proved that, so let me write that. So the cosine of theta c is equal to cv dot cw over the lengths of v times w. Because we know, we've already shown that it preserves length. So let's see what this top part equals. So we could just use the general property, the dot product is equal to the transpose of one guy as kind of a matrix times the second guy. So this is equal to cw transpose times cv and all of that over these lengths, the length of w. And then this is going to be equal to, I'm going to write it down here to have some space. We can switch these guys and take their transpose."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So let's see what this top part equals. So we could just use the general property, the dot product is equal to the transpose of one guy as kind of a matrix times the second guy. So this is equal to cw transpose times cv and all of that over these lengths, the length of w. And then this is going to be equal to, I'm going to write it down here to have some space. We can switch these guys and take their transpose. So it's w transpose times c transpose times cv, all of that over their lengths, the product of their lengths, v and w. And this is the identity matrix. That's the identity matrix. So this is going to be equal to w transpose times v over the products of their lengths."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "We can switch these guys and take their transpose. So it's w transpose times c transpose times cv, all of that over their lengths, the product of their lengths, v and w. And this is the identity matrix. That's the identity matrix. So this is going to be equal to w transpose times v over the products of their lengths. And this is the same thing as v dot w over their lengths, which is cosine of theta. So notice, by our definition of an angle is the dot product divided by the vector lengths. When you perform a transformation, or you could imagine a change of basis either way, with an orthogonal matrix C, the angle between the transformed vectors does not change."}, {"video_title": "Orthogonal matrices preserve angles and lengths Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to w transpose times v over the products of their lengths. And this is the same thing as v dot w over their lengths, which is cosine of theta. So notice, by our definition of an angle is the dot product divided by the vector lengths. When you perform a transformation, or you could imagine a change of basis either way, with an orthogonal matrix C, the angle between the transformed vectors does not change. It's the same as the angle between the vectors before they were transformed. Which is a really neat thing to know, that change of bases or transformations with orthogonal matrices don't distort the vectors. They might just kind of rotate them around or shift them a little bit, but it doesn't change the angles between them."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what I'm going to do is I'm going to solve it using an augmented matrix. And I'm going to put it in reduced row echelon form. So what's the augmented matrix for this system of equations? Three unknowns with three equations. So it'll be, I'll just have to do the coefficients. So the coefficients of the x terms are just 1, 1, 1. Coefficients of the y terms are 1, 2, and 3."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Three unknowns with three equations. So it'll be, I'll just have to do the coefficients. So the coefficients of the x terms are just 1, 1, 1. Coefficients of the y terms are 1, 2, and 3. Coefficients of the z terms are 1, 3, and 4. And let me show that it's augmented. And then they equal 3, 0, and minus 2."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Coefficients of the y terms are 1, 2, and 3. Coefficients of the z terms are 1, 3, and 4. And let me show that it's augmented. And then they equal 3, 0, and minus 2. Now I want to get this augmented matrix into reduced row echelon form. So the first thing, I have a 1 leading 1 here. That's a pivot entry."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then they equal 3, 0, and minus 2. Now I want to get this augmented matrix into reduced row echelon form. So the first thing, I have a 1 leading 1 here. That's a pivot entry. Let me make everything else in that column equal to a 0. So I'm not going to change my first row. So it'll just be a 1, a 1, a 1, my dividing line, and then I have a 3."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's a pivot entry. Let me make everything else in that column equal to a 0. So I'm not going to change my first row. So it'll just be a 1, a 1, a 1, my dividing line, and then I have a 3. Now to zero this out, let me just replace the second row with the first row minus the second row. So 1 minus 1 is 0. 1 minus 2 is 1 minus 2."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it'll just be a 1, a 1, a 1, my dividing line, and then I have a 3. Now to zero this out, let me just replace the second row with the first row minus the second row. So 1 minus 1 is 0. 1 minus 2 is 1 minus 2. Actually, a better thing to do, because I eventually want this to be a 1 anyway. Let me replace this row with the second row minus the first row, instead of the first row minus the second row. I can do it either way."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 is 1 minus 2. Actually, a better thing to do, because I eventually want this to be a 1 anyway. Let me replace this row with the second row minus the first row, instead of the first row minus the second row. I can do it either way. So the second row minus the first row. So 1 minus 1 is 0. 2 minus 1 is 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can do it either way. So the second row minus the first row. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. And then 0 minus 3 is minus 3. Now I want to also zero this out."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 1 is 1. 3 minus 1 is 2. And then 0 minus 3 is minus 3. Now I want to also zero this out. So let me replace this guy with this equation minus that equation, so 1 minus 1 is 0. 3 minus 1 is 2. 4 minus 1 is 3."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to also zero this out. So let me replace this guy with this equation minus that equation, so 1 minus 1 is 0. 3 minus 1 is 2. 4 minus 1 is 3. Minus 2 minus 3 is minus 5. Fair enough. So I got my pivot entry here."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4 minus 1 is 3. Minus 2 minus 3 is minus 5. Fair enough. So I got my pivot entry here. I have another pivot entry here. It's to the right of this one, which is what I want for reduced row echelon form. Now I need to target this entry and that entry."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I got my pivot entry here. I have another pivot entry here. It's to the right of this one, which is what I want for reduced row echelon form. Now I need to target this entry and that entry. I need to zero them out. So let's do it. So we get, so I'm going to keep my second row the same."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I need to target this entry and that entry. I need to zero them out. So let's do it. So we get, so I'm going to keep my second row the same. My second row is 0, 1, 2, and then I have a minus 3, the augmented part of it. And to zero this guy out, what I can do is I can replace the first row with the first row minus the second row. So I get 1 minus 0 is 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we get, so I'm going to keep my second row the same. My second row is 0, 1, 2, and then I have a minus 3, the augmented part of it. And to zero this guy out, what I can do is I can replace the first row with the first row minus the second row. So I get 1 minus 0 is 1. 1 minus 1, there's a bird outside. Let me close my window. So where was I?"}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I get 1 minus 0 is 1. 1 minus 1, there's a bird outside. Let me close my window. So where was I? I'm replacing the first row with the first row minus the second row. So 1 minus 0 is 1. 1 minus 1 is 0."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So where was I? I'm replacing the first row with the first row minus the second row. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1, and then 3 minus minus 3. So that's equal to 3 plus 3. So that's equal to 6."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 1 is 0. 1 minus 2 is minus 1, and then 3 minus minus 3. So that's equal to 3 plus 3. So that's equal to 6. 1 minus 0 is 1. 1 minus 1 is 0, negative 1, and then 3 minus negative 3, that's 6. I always want to make sure I don't make a careless mistake."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's equal to 6. 1 minus 0 is 1. 1 minus 1 is 0, negative 1, and then 3 minus negative 3, that's 6. I always want to make sure I don't make a careless mistake. Now let me get rid of this entry right here. Let me zero that out. So let me replace the third row with the third row minus 2 times the second row."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I always want to make sure I don't make a careless mistake. Now let me get rid of this entry right here. Let me zero that out. So let me replace the third row with the third row minus 2 times the second row. So we have 0 minus, well, 2 times 0, that's just going to be 0. 2 minus 2 times 1, that's 2 minus 2, that's 0. 3 minus 2 times 2, that's 3 minus 4, or minus 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me replace the third row with the third row minus 2 times the second row. So we have 0 minus, well, 2 times 0, that's just going to be 0. 2 minus 2 times 1, that's 2 minus 2, that's 0. 3 minus 2 times 2, that's 3 minus 4, or minus 1. And then finally, minus 5 minus 2 times minus 3. Let me write that down. Minus 5 minus 2 times minus 3."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "3 minus 2 times 2, that's 3 minus 4, or minus 1. And then finally, minus 5 minus 2 times minus 3. Let me write that down. Minus 5 minus 2 times minus 3. That's minus 5 minus minus 6. That's minus 5 plus 6 is equal to 1. I really wanted to make sure I didn't make a careless mistake there."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Minus 5 minus 2 times minus 3. That's minus 5 minus minus 6. That's minus 5 plus 6 is equal to 1. I really wanted to make sure I didn't make a careless mistake there. So that is equal to 1. So I'm almost done, but I'm still not in reduced row echelon form. This has to be a positive 1 in order to get there."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I really wanted to make sure I didn't make a careless mistake there. So that is equal to 1. So I'm almost done, but I'm still not in reduced row echelon form. This has to be a positive 1 in order to get there. It can't be anything other than a 1. That's just the style of reduced row echelon form. And then these guys up here have to be zeroed out."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This has to be a positive 1 in order to get there. It can't be anything other than a 1. That's just the style of reduced row echelon form. And then these guys up here have to be zeroed out. Well, the easy thing to do, let me just multiply this equation by minus 1. So then this becomes a plus 1, and then that becomes a minus 1. And then I just need to zero out these two guys up here."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then these guys up here have to be zeroed out. Well, the easy thing to do, let me just multiply this equation by minus 1. So then this becomes a plus 1, and then that becomes a minus 1. And then I just need to zero out these two guys up here. So let's do it. So my equation, I'm going to keep my third row the same. My third row is now 0, 0, 1, minus 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I just need to zero out these two guys up here. So let's do it. So my equation, I'm going to keep my third row the same. My third row is now 0, 0, 1, minus 1. And now I want to zero this guy out. So what I can do is I can set my first row equal to my first row plus my last row. Because if these two add up, they're going to be equal to 0."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "My third row is now 0, 0, 1, minus 1. And now I want to zero this guy out. So what I can do is I can set my first row equal to my first row plus my last row. Because if these two add up, they're going to be equal to 0. So what do I get? 1 plus 0 is 1. 0 plus 0 is 0."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because if these two add up, they're going to be equal to 0. So what do I get? 1 plus 0 is 1. 0 plus 0 is 0. Minus 1 plus 1 is 0. 6 plus minus 1 is 5. Now I want to zero this row out."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "0 plus 0 is 0. Minus 1 plus 1 is 0. 6 plus minus 1 is 5. Now I want to zero this row out. And to zero this row out, what I can do is I'll replace it with the second row minus 2 times the first row. So 0 minus 2 times 0 is just 0. 1 minus 2 times 0 is just 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to zero this row out. And to zero this row out, what I can do is I'll replace it with the second row minus 2 times the first row. So 0 minus 2 times 0 is just 0. 1 minus 2 times 0 is just 1. 2 minus 2 times 1 is 0. Minus 3 minus 2 times negative 1. Let me write that down."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 times 0 is just 1. 2 minus 2 times 1 is 0. Minus 3 minus 2 times negative 1. Let me write that down. Minus 3 minus 2 times minus 1. Don't want to make a careless mistake. So what is that equal to?"}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. Minus 3 minus 2 times minus 1. Don't want to make a careless mistake. So what is that equal to? This is equal to minus 3 minus minus 2. Or minus 3 plus 2, which is equal to minus 1. So that's equal to minus 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what is that equal to? This is equal to minus 3 minus minus 2. Or minus 3 plus 2, which is equal to minus 1. So that's equal to minus 1. And now I have my augmented matrix in reduced row echelon form. My pivot entries are the only entries in their columns. Each pivot entry in each successive row is to the right of the pivot entry before it."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's equal to minus 1. And now I have my augmented matrix in reduced row echelon form. My pivot entries are the only entries in their columns. Each pivot entry in each successive row is to the right of the pivot entry before it. And actually I have no free variables. Every column has a pivot entry. So let's go back from the augmented matrix world and kind of put back our variables there."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Each pivot entry in each successive row is to the right of the pivot entry before it. And actually I have no free variables. Every column has a pivot entry. So let's go back from the augmented matrix world and kind of put back our variables there. So what do we get? We get x plus 0y plus 0z is equal to 5. That's that row right there."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's go back from the augmented matrix world and kind of put back our variables there. So what do we get? We get x plus 0y plus 0z is equal to 5. That's that row right there. We get 0x plus 1y plus 0z is equal to minus 1. That's that row right there. And then finally, we have 0x plus 0y plus 1z is equal to minus 1."}, {"video_title": "Matrices Reduced row echelon form 2 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's that row right there. We get 0x plus 1y plus 0z is equal to minus 1. That's that row right there. And then finally, we have 0x plus 0y plus 1z is equal to minus 1. That's that row right there. And just like that, we've actually solved our system of three equations with three unknowns. That's the solution right there."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We've already seen that you can visually represent a vector as an arrow, where the length of the arrow is the magnitude of the vector, and the direction of the arrow is the direction of the vector. And if we want to represent this mathematically, we could just think about, well, starting from the tail of the vector, how far away is the head of the vector in the horizontal direction, and how far away is it in the vertical direction? So for example, in the horizontal direction, you would have to go this distance. And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector, and let's call this vector v. We could represent vector v as an ordered list, or a 2-tuple of, so we could say we move 2 in the horizontal direction and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2, 3, like that."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector, and let's call this vector v. We could represent vector v as an ordered list, or a 2-tuple of, so we could say we move 2 in the horizontal direction and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2, 3, like that. What I now want to introduce you to, and we could come up with other ways of representing this 2-tuple, is another notation. This really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could represent vector v like this, where it is 2, 3, like that. What I now want to introduce you to, and we could come up with other ways of representing this 2-tuple, is another notation. This really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that. So let's define a unit vector i. And the way that we denote that it is a unit vector, the way that we denote it's a unit vector, is instead of putting an arrow on top, we put this kind of hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes one unit in the horizontal direction, and it doesn't go at all in the vertical direction."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so let's do that. So let's define a unit vector i. And the way that we denote that it is a unit vector, the way that we denote it's a unit vector, is instead of putting an arrow on top, we put this kind of hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes one unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like that. That is the unit vector i. And then we can define another unit vector."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes one unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like that. That is the unit vector i. And then we can define another unit vector. Let's call that unit vector, or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And it goes one unit in the vertical direction. So this went one unit in the horizontal."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then we can define another unit vector. Let's call that unit vector, or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And it goes one unit in the vertical direction. So this went one unit in the horizontal. Now j is going to go one unit in the vertical. So j, just like that. Now, any vector, any two-dimensional vector, we can now represent as a sum of scaled-up versions of i and j."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this went one unit in the horizontal. Now j is going to go one unit in the vertical. So j, just like that. Now, any vector, any two-dimensional vector, we can now represent as a sum of scaled-up versions of i and j. And you say, well, how do we do that? Well, you could imagine vector v right here is the sum of a vector that moves purely in the horizontal direction that has length 2 and a vector that moves purely in the vertical direction that has length 3. So we could say that vector v is equal to."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, any vector, any two-dimensional vector, we can now represent as a sum of scaled-up versions of i and j. And you say, well, how do we do that? Well, you could imagine vector v right here is the sum of a vector that moves purely in the horizontal direction that has length 2 and a vector that moves purely in the vertical direction that has length 3. So we could say that vector v is equal to. So if we want a vector that has length 2 and it moves purely in the horizontal direction, well, we could just scale up the unit vector i. We could just multiply 2 times i. So let's do that."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that vector v is equal to. So if we want a vector that has length 2 and it moves purely in the horizontal direction, well, we could just scale up the unit vector i. We could just multiply 2 times i. So let's do that. It's equal to 2 times our unit vector i. So 2i is going to be this whole thing right over here, or this whole vector. Let me do it in this yellow color."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. It's equal to 2 times our unit vector i. So 2i is going to be this whole thing right over here, or this whole vector. Let me do it in this yellow color. This vector right over here you could view as 2i. And then to that, we're going to add 3 times j. So plus 3 times j."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in this yellow color. This vector right over here you could view as 2i. And then to that, we're going to add 3 times j. So plus 3 times j. Let me write it like this. Let me get the color. 3 times j."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So plus 3 times j. Let me write it like this. Let me get the color. 3 times j. Once again, 3 times j is going to be this vector right over here. And if you add this yellow vector right over here to the magenta vector, you're going to get. Notice, we're putting the tail of the magenta vector at the head of the yellow vector."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "3 times j. Once again, 3 times j is going to be this vector right over here. And if you add this yellow vector right over here to the magenta vector, you're going to get. Notice, we're putting the tail of the magenta vector at the head of the yellow vector. And if you start at the tail of the yellow vector and you go all the way to the head of the magenta vector, you have now constructed vector v. So vector v, you could represent it as a column vector like this, 2, 3. You could represent it as 2, 3. Or you could represent it as 2 times i with this little hat over it, plus 3 times j with this little hat over it."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Notice, we're putting the tail of the magenta vector at the head of the yellow vector. And if you start at the tail of the yellow vector and you go all the way to the head of the magenta vector, you have now constructed vector v. So vector v, you could represent it as a column vector like this, 2, 3. You could represent it as 2, 3. Or you could represent it as 2 times i with this little hat over it, plus 3 times j with this little hat over it. i is the unit vector in the horizontal direction, in the positive horizontal direction. If you want to go the other way, you would multiply it by a negative. And j is the unit vector in the vertical direction."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or you could represent it as 2 times i with this little hat over it, plus 3 times j with this little hat over it. i is the unit vector in the horizontal direction, in the positive horizontal direction. If you want to go the other way, you would multiply it by a negative. And j is the unit vector in the vertical direction. As we'll see in future videos, once you go to three dimensions, you'll introduce a k. But it's very natural to translate between these two things. Notice, 2, 3. And so with that, let's actually do some vector operations using this notation."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And j is the unit vector in the vertical direction. As we'll see in future videos, once you go to three dimensions, you'll introduce a k. But it's very natural to translate between these two things. Notice, 2, 3. And so with that, let's actually do some vector operations using this notation. So let's say that I define another vector. Let's say it's vector b. And vector b is equal to negative 1 times i, times the unit vector i, plus 4 times the unit vector in the horizontal direction."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so with that, let's actually do some vector operations using this notation. So let's say that I define another vector. Let's say it's vector b. And vector b is equal to negative 1 times i, times the unit vector i, plus 4 times the unit vector in the horizontal direction. So given these two vector definitions, what would the vector v plus b be equal to? I encourage you to pause the video and think about it. Well, once again, we just literally have to add corresponding components."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And vector b is equal to negative 1 times i, times the unit vector i, plus 4 times the unit vector in the horizontal direction. So given these two vector definitions, what would the vector v plus b be equal to? I encourage you to pause the video and think about it. Well, once again, we just literally have to add corresponding components. We could say, OK, well, let's think about what we're doing in the horizontal direction. We're going 2 in the horizontal direction here. Now we're going negative 1."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, once again, we just literally have to add corresponding components. We could say, OK, well, let's think about what we're doing in the horizontal direction. We're going 2 in the horizontal direction here. Now we're going negative 1. So our horizontal component is going to be 2 plus negative 1 in the horizontal direction. And we're going to multiply that times the unit vector i. This, once again, just goes back to adding the corresponding components of the vector."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now we're going negative 1. So our horizontal component is going to be 2 plus negative 1 in the horizontal direction. And we're going to multiply that times the unit vector i. This, once again, just goes back to adding the corresponding components of the vector. And then we're going to have plus 4, or plus 3 plus 4. Let me write it that way. Plus 3 plus 4 times the unit vector j in the vertical direction."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This, once again, just goes back to adding the corresponding components of the vector. And then we're going to have plus 4, or plus 3 plus 4. Let me write it that way. Plus 3 plus 4 times the unit vector j in the vertical direction. And so that's going to give us 2. I'll do this all in this one color. 2 plus negative 1 is 1i."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Plus 3 plus 4 times the unit vector j in the vertical direction. And so that's going to give us 2. I'll do this all in this one color. 2 plus negative 1 is 1i. And we could literally write that just as i. Actually, let's do that. Let's just write that as i."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 plus negative 1 is 1i. And we could literally write that just as i. Actually, let's do that. Let's just write that as i. But we got that from 2 plus negative 1 is 1. So 1 times a vector is just going to be that vector, plus 3 plus 4 is 7j. And you see, this is exactly how we saw vector addition in the past, is that we could also represent vector b like this."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's just write that as i. But we got that from 2 plus negative 1 is 1. So 1 times a vector is just going to be that vector, plus 3 plus 4 is 7j. And you see, this is exactly how we saw vector addition in the past, is that we could also represent vector b like this. We could represent it like this. Negative 1, negative 1, 4. And so if you were to add v to b, you add the corresponding terms."}, {"video_title": "Unit vector notation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you see, this is exactly how we saw vector addition in the past, is that we could also represent vector b like this. We could represent it like this. Negative 1, negative 1, 4. And so if you were to add v to b, you add the corresponding terms. So this should be equal to, if we were to add corresponding terms looking at them as column vectors, that is going to be equal to 2 plus negative 1, which is 1. 3 plus 4 is 7. So this is the exact same representation as this."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The whole premise of the last series of videos was really just trying to get at, trying to figure out whether some transformation, let's have some transformation that is a mapping from, let's say it's a mapping from Rn to Rm. The whole question is, is T invertible? And we showed several videos ago that a function in the transformation is really just a function, that a function is invertible if it meets two conditions. So invertible, I don't have to keep writing the word over then, you have to have two conditions. It has to be onto, or essentially, it has to map to every member of your codomain. And it also has to be one-to-one. Another way of saying one-to-one is that every member of your codomain is mapped to at most one member of your domain."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So invertible, I don't have to keep writing the word over then, you have to have two conditions. It has to be onto, or essentially, it has to map to every member of your codomain. And it also has to be one-to-one. Another way of saying one-to-one is that every member of your codomain is mapped to at most one member of your domain. And we did several videos where we thought, well, if we had a transformation, a linear transformation that's defined by a matrix A, where this would be an m by n matrix, we said that this is going to be met if the rank, this is met if the rank of, and this is only met if the rank of A is equal to the number of rows in your transformation matrix is equal to m. And in the last video, I just showed that this is only true if every one of your column vectors are linearly independent, or that they all are basis vectors for your column space, or that the rank of your matrix has to be equal to n. Now, in order for something to be invertible, in order for the transformation to be invertible, both of these things have to be true. Your rank of A has to be equal to m, and your rank of A has to be equal to n. So in order to be invertible, a couple of things have to happen. In order to be invertible, your rank of your transformation matrix has to be equal to m, which has to be equal to n. So m has to be equal to n. So we have an interesting condition."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Another way of saying one-to-one is that every member of your codomain is mapped to at most one member of your domain. And we did several videos where we thought, well, if we had a transformation, a linear transformation that's defined by a matrix A, where this would be an m by n matrix, we said that this is going to be met if the rank, this is met if the rank of, and this is only met if the rank of A is equal to the number of rows in your transformation matrix is equal to m. And in the last video, I just showed that this is only true if every one of your column vectors are linearly independent, or that they all are basis vectors for your column space, or that the rank of your matrix has to be equal to n. Now, in order for something to be invertible, in order for the transformation to be invertible, both of these things have to be true. Your rank of A has to be equal to m, and your rank of A has to be equal to n. So in order to be invertible, a couple of things have to happen. In order to be invertible, your rank of your transformation matrix has to be equal to m, which has to be equal to n. So m has to be equal to n. So we have an interesting condition. You have to have a square matrix. You have to have a square matrix. Your matrix has to be n by n. That's what this implies."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In order to be invertible, your rank of your transformation matrix has to be equal to m, which has to be equal to n. So m has to be equal to n. So we have an interesting condition. You have to have a square matrix. You have to have a square matrix. Your matrix has to be n by n. That's what this implies. If both of these are true, then m has to be equal to n, and you're dealing with a square matrix. Even more, you're dealing with a square matrix where every one of the columns are linearly independent. So this is going to be, so this is our A."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Your matrix has to be n by n. That's what this implies. If both of these are true, then m has to be equal to n, and you're dealing with a square matrix. Even more, you're dealing with a square matrix where every one of the columns are linearly independent. So this is going to be, so this is our A. A looks like this, A1, A2, all the way to An. Since the rank of A is equal to n, and this is, of course, an n by n matrix, we just said that this has to be the case, because both its rank, its rank has to be equal to m, which is the number of rows, and its rank has to be equal to n, which is the number of columns. So the rows and columns have to be the same."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be, so this is our A. A looks like this, A1, A2, all the way to An. Since the rank of A is equal to n, and this is, of course, an n by n matrix, we just said that this has to be the case, because both its rank, its rank has to be equal to m, which is the number of rows, and its rank has to be equal to n, which is the number of columns. So the rows and columns have to be the same. But the fact that your rank is equal to the number of columns, that means that all of your column vectors are bases for your column space, or that if you put them into reduced row echelon form, so if you put this into reduced row echelon form, what are you going to get? Well, all of these guys are basis vectors. So they're all going to be associated with pivot vectors, or they're all going to be associated with pivot columns."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the rows and columns have to be the same. But the fact that your rank is equal to the number of columns, that means that all of your column vectors are bases for your column space, or that if you put them into reduced row echelon form, so if you put this into reduced row echelon form, what are you going to get? Well, all of these guys are basis vectors. So they're all going to be associated with pivot vectors, or they're all going to be associated with pivot columns. So this is going to be 1, 0, a bunch of 0's, and then you're going to have 0, 1, a bunch of 0's like this. They're all going to be associated with pivot columns. So all associated with pivot columns when you go into reduced row echelon form."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So they're all going to be associated with pivot vectors, or they're all going to be associated with pivot columns. So this is going to be 1, 0, a bunch of 0's, and then you're going to have 0, 1, a bunch of 0's like this. They're all going to be associated with pivot columns. So all associated with pivot columns when you go into reduced row echelon form. So all of them are pivot columns. It's an n by n matrix. So what is an n by n matrix where every column is a pivot column?"}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So all associated with pivot columns when you go into reduced row echelon form. So all of them are pivot columns. It's an n by n matrix. So what is an n by n matrix where every column is a pivot column? What is an n by n matrix? Let me write this. So you have an n, so the reduced row echelon form of a has to be equal to an n by n matrix, because a is n by n, where every column is a linearly independent pivot column."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is an n by n matrix where every column is a pivot column? What is an n by n matrix? Let me write this. So you have an n, so the reduced row echelon form of a has to be equal to an n by n matrix, because a is n by n, where every column is a linearly independent pivot column. And I mean, by definition of reduced row echelon form, you can't have the same pivot column twice, where every column is a linearly independent pivot column. It's a little bit redundant, but I think you get the idea. So what is an n by n matrix where every column is a linearly independent pivot column?"}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you have an n, so the reduced row echelon form of a has to be equal to an n by n matrix, because a is n by n, where every column is a linearly independent pivot column. And I mean, by definition of reduced row echelon form, you can't have the same pivot column twice, where every column is a linearly independent pivot column. It's a little bit redundant, but I think you get the idea. So what is an n by n matrix where every column is a linearly independent pivot column? Well, that is just a matrix that has 1's down the diagonal, and everything else is a 0. Or, we've seen this matrix before, this is just an n by n identity matrix, or the identity matrix on n, or on rn. So if you multiply this matrix times any member of rn, you're just going to get that matrix again."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is an n by n matrix where every column is a linearly independent pivot column? Well, that is just a matrix that has 1's down the diagonal, and everything else is a 0. Or, we've seen this matrix before, this is just an n by n identity matrix, or the identity matrix on n, or on rn. So if you multiply this matrix times any member of rn, you're just going to get that matrix again. But this is interesting. We now have a pretty usable condition for invertibility. We can say that the transformation T that is a mapping from rn to, well, it has to map to the same dimension space, so from rn to rn, it's equal to some square matrix, n by n, has to be an n by n matrix, times our vectors in our domain."}, {"video_title": "Simplifying conditions for invertibility Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you multiply this matrix times any member of rn, you're just going to get that matrix again. But this is interesting. We now have a pretty usable condition for invertibility. We can say that the transformation T that is a mapping from rn to, well, it has to map to the same dimension space, so from rn to rn, it's equal to some square matrix, n by n, has to be an n by n matrix, times our vectors in our domain. And it's only going to be only invertible if the reduced row echelon form of our transformation matrix is equal to the identity matrix on n. I mean, I could have written an m here, and I could have said this is an m by n matrix, but the only way that this is going to be true is if this is also an n, and this is also an n. But maybe I could leave them there. Let me leave those m's there, because that's the big takeaway, is that if in order for the transformation matrix to be invertible, the only way it's invertible is if the reduced row echelon form of our transformation matrix is equal to an n by n identity matrix. The identity matrix is always going to be n by n. So that's a big takeaway."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I have the transformation s, which is a function or a transformation from Rn to Rm. And I also have the transformation t, which is also a transformation from Rn to Rm. I'm going to define right now what it means to add the two transformations. So this is a definition. Let me write it as a definition. I'm going to define the addition of our two transformations. So if I add our two transformations, the addition of two transformations operating on some vector x, this is a definition."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is a definition. Let me write it as a definition. I'm going to define the addition of our two transformations. So if I add our two transformations, the addition of two transformations operating on some vector x, this is a definition. I'm going to say that this is the same thing as the first transformation operating on the vector x plus the second transformation operating on the vector x. And obviously, this is going to end up being a vector in Rm. So this whole thing is going to be a vector in Rm."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I add our two transformations, the addition of two transformations operating on some vector x, this is a definition. I'm going to say that this is the same thing as the first transformation operating on the vector x plus the second transformation operating on the vector x. And obviously, this is going to end up being a vector in Rm. So this whole thing is going to be a vector in Rm. So we still are, by definition, this s plus t transformation is still a transformation because it takes an input from Rn. It's still a transformation from Rn to Rm. Now, let me make another definition."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this whole thing is going to be a vector in Rm. So we still are, by definition, this s plus t transformation is still a transformation because it takes an input from Rn. It's still a transformation from Rn to Rm. Now, let me make another definition. Let me define a scalar multiple of a transformation. So I'm going to define, let's say, c, where c is just any real number, c times the transformation s of some vector x. I'm going to say that this is equal to c times the transformation of x. This is equal to c times the transformation of x right there."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let me make another definition. Let me define a scalar multiple of a transformation. So I'm going to define, let's say, c, where c is just any real number, c times the transformation s of some vector x. I'm going to say that this is equal to c times the transformation of x. This is equal to c times the transformation of x right there. And so similarly, this is also the transformation of x. Obviously, it's going to be in Rm. So if you multiply any vector in Rm times some scalar, you're still going to have another vector in Rm."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to c times the transformation of x right there. And so similarly, this is also the transformation of x. Obviously, it's going to be in Rm. So if you multiply any vector in Rm times some scalar, you're still going to have another vector in Rm. So luckily for us, this definition of a scalar multiple, so if I have this new transformation called c times s, this is still a mapping from Rn to Rm. This is still a vector in Rm, and this is still a vector in Rn. Fair enough."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you multiply any vector in Rm times some scalar, you're still going to have another vector in Rm. So luckily for us, this definition of a scalar multiple, so if I have this new transformation called c times s, this is still a mapping from Rn to Rm. This is still a vector in Rm, and this is still a vector in Rn. Fair enough. Now, let's see what happens if we look at their corresponding matrices for these transformations. We've seen in a previous video that any linear transformation can be represented as a matrix vector product. So let's say that s of a vector x is equivalent to the matrix A times that vector x."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now, let's see what happens if we look at their corresponding matrices for these transformations. We've seen in a previous video that any linear transformation can be represented as a matrix vector product. So let's say that s of a vector x is equivalent to the matrix A times that vector x. And let's say that T of x is equal to the matrix B times a vector x. And of course, since both of these guys are mappings from Rn to Rm, both of these matrices are going to be m by n. Both of these are m by n matrices. Now, let's just go back to these definitions that I just constructed."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that s of a vector x is equivalent to the matrix A times that vector x. And let's say that T of x is equal to the matrix B times a vector x. And of course, since both of these guys are mappings from Rn to Rm, both of these matrices are going to be m by n. Both of these are m by n matrices. Now, let's just go back to these definitions that I just constructed. What is s of T of x can then be written as? So let me write it this way. I'll do it in that same color."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's just go back to these definitions that I just constructed. What is s of T of x can then be written as? So let me write it this way. I'll do it in that same color. So you have s. No, I was going to do it in red. So you have, maybe I'll do it right here. You have s plus T of x. I'm just rewriting this up here."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it in that same color. So you have s. No, I was going to do it in red. So you have, maybe I'll do it right here. You have s plus T of x. I'm just rewriting this up here. Is equal to s of x plus T of x, or the transformation T of x, which we now know is equal to these two things. So this is equal to this term right there. The transformation s of x is equal to Ax."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have s plus T of x. I'm just rewriting this up here. Is equal to s of x plus T of x, or the transformation T of x, which we now know is equal to these two things. So this is equal to this term right there. The transformation s of x is equal to Ax. That's that one right there. And then the transformation T of x is equal to b, the matrix b times x. Now, what are these things?"}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation s of x is equal to Ax. That's that one right there. And then the transformation T of x is equal to b, the matrix b times x. Now, what are these things? Well, I can write, let me write our two matrices in a form that you're probably familiar with right now. Let's say the matrix A is just a bunch of column vectors. A1, A2, all the way to An."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what are these things? Well, I can write, let me write our two matrices in a form that you're probably familiar with right now. Let's say the matrix A is just a bunch of column vectors. A1, A2, all the way to An. And similarly, the matrix B is just a bunch of column vectors. The matrix B is B1, B2, all the way to Bn. These are each column vectors with m components, one for each of the rows."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "A1, A2, all the way to An. And similarly, the matrix B is just a bunch of column vectors. The matrix B is B1, B2, all the way to Bn. These are each column vectors with m components, one for each of the rows. And there's n of these, because there are n columns in each of these vectors. So when you multiply this guy times, let me make it very clear. If I multiply an x, the vector x is going to look like this."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are each column vectors with m components, one for each of the rows. And there's n of these, because there are n columns in each of these vectors. So when you multiply this guy times, let me make it very clear. If I multiply an x, the vector x is going to look like this. The vector x is going to be x1, x2, all the way down to xn. And we've shown this multiple, multiple times. It's a very handy way of thinking about matrix vector products."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply an x, the vector x is going to look like this. The vector x is going to be x1, x2, all the way down to xn. And we've shown this multiple, multiple times. It's a very handy way of thinking about matrix vector products. But we know that this product right here can be written as each of these scalar terms in x times the corresponding column vector in A. I've done this in probably the fifth video that I'm doing this. So this can be written as x1 times A1 plus x2 times A times A2, all the way to xn times An. It's equal to this."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's a very handy way of thinking about matrix vector products. But we know that this product right here can be written as each of these scalar terms in x times the corresponding column vector in A. I've done this in probably the fifth video that I'm doing this. So this can be written as x1 times A1 plus x2 times A times A2, all the way to xn times An. It's equal to this. That's what Ax can be rewritten as, as just as kind of a weighted combination of these column vectors where the weights are each of the values of our vector x. And I have to add this guy to Bx. So Bx, by the same argument, so plus is just going to be, let me do it in the blue, is going to be x1 times B1, x1 times B1 plus x2 times B2, all the way to xn times Bn."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to this. That's what Ax can be rewritten as, as just as kind of a weighted combination of these column vectors where the weights are each of the values of our vector x. And I have to add this guy to Bx. So Bx, by the same argument, so plus is just going to be, let me do it in the blue, is going to be x1 times B1, x1 times B1 plus x2 times B2, all the way to xn times Bn. Now, what is this equal to? Well, we know that scalar multiplication times vectors exhibits the distributive property. So we can just add these two guys right here and factor out the x1."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So Bx, by the same argument, so plus is just going to be, let me do it in the blue, is going to be x1 times B1, x1 times B1 plus x2 times B2, all the way to xn times Bn. Now, what is this equal to? Well, we know that scalar multiplication times vectors exhibits the distributive property. So we can just add these two guys right here and factor out the x1. And what do we get? We get this is equal to this whole expression right here. Let me draw a line here because I'm not saying that this matrix is equal to that."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can just add these two guys right here and factor out the x1. And what do we get? We get this is equal to this whole expression right here. Let me draw a line here because I'm not saying that this matrix is equal to that. I'm saying that this is equal to this term plus this term, which is equal to x1 times A1 plus B1 plus x2 times A2. Right, I'm just adding these two terms up. x2 times A2 plus B2, all the way to plus xn times An plus Bn."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw a line here because I'm not saying that this matrix is equal to that. I'm saying that this is equal to this term plus this term, which is equal to x1 times A1 plus B1 plus x2 times A2. Right, I'm just adding these two terms up. x2 times A2 plus B2, all the way to plus xn times An plus Bn. So what is this thing equal to? Well, this is equal to some matrix, some new matrix. And let's define this new matrix."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "x2 times A2 plus B2, all the way to plus xn times An plus Bn. So what is this thing equal to? Well, this is equal to some matrix, some new matrix. And let's define this new matrix. This is equivalent to some new matrix. I'll make it pretty big right here, times our vector x. I'll do the vector x in green. Vector x we know is x1, x2, all the way down to xn."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's define this new matrix. This is equivalent to some new matrix. I'll make it pretty big right here, times our vector x. I'll do the vector x in green. Vector x we know is x1, x2, all the way down to xn. But what is the new matrix going to be? Well, this product is going to be each of these scalar terms times the column vectors of this matrix. So these guys right here are the columns of my matrix."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Vector x we know is x1, x2, all the way down to xn. But what is the new matrix going to be? Well, this product is going to be each of these scalar terms times the column vectors of this matrix. So these guys right here are the columns of my matrix. This thing is equivalent to a matrix where the first column right here is A1 plus B1. We're essentially adding the column vectors of those two guys. The second column right here, let me draw a little line right there to show you that these are different expressions."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these guys right here are the columns of my matrix. This thing is equivalent to a matrix where the first column right here is A1 plus B1. We're essentially adding the column vectors of those two guys. The second column right here, let me draw a little line right there to show you that these are different expressions. The second one would be A2 plus B2. And then we'll just have a bunch of them. And then the last one will just be An plus Bn."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "The second column right here, let me draw a little line right there to show you that these are different expressions. The second one would be A2 plus B2. And then we'll just have a bunch of them. And then the last one will just be An plus Bn. So what happens is that by definition, when I added these two transformations, I just used their corresponding matrices. And I said, you know what? The addition of these two transformations created a new transformation that is essentially some matrix times my vector."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the last one will just be An plus Bn. So what happens is that by definition, when I added these two transformations, I just used their corresponding matrices. And I said, you know what? The addition of these two transformations created a new transformation that is essentially some matrix times my vector. And that matrix ended up being the sum of the corresponding column vectors of our two original transformation matrices. This new matrix that I got, and I haven't defined matrix addition yet, but we got here just by thinking about vector addition. This matrix is constructed by adding the corresponding vectors of the matrices A and B."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "The addition of these two transformations created a new transformation that is essentially some matrix times my vector. And that matrix ended up being the sum of the corresponding column vectors of our two original transformation matrices. This new matrix that I got, and I haven't defined matrix addition yet, but we got here just by thinking about vector addition. This matrix is constructed by adding the corresponding vectors of the matrices A and B. Now, why did I go through all of this trouble? Well, I can make a new definition here that'll make everything fit together well. I'm going to define this matrix right here as A plus B."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This matrix is constructed by adding the corresponding vectors of the matrices A and B. Now, why did I go through all of this trouble? Well, I can make a new definition here that'll make everything fit together well. I'm going to define this matrix right here as A plus B. So my new matrix definition, if I have two matrices that have the same dimensions, and they have to have the same dimensions, I'm defining A plus B to be equal to a new matrix, some new matrix where you add up their corresponding columns. So A1 plus B1, just like what I did here. I don't have to rewrite it all the way up to An plus Bn is the last column."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to define this matrix right here as A plus B. So my new matrix definition, if I have two matrices that have the same dimensions, and they have to have the same dimensions, I'm defining A plus B to be equal to a new matrix, some new matrix where you add up their corresponding columns. So A1 plus B1, just like what I did here. I don't have to rewrite it all the way up to An plus Bn is the last column. And you've seen this before in your Algebra 2 class, but I wanted to here to do it because this shows you the motivation for it. Because now we can say that the sum of two transformations, so S plus T of x, which is equal to S of x, this is a vector, S of x plus T of x, which we know is equal to A times x plus B times x. We can now say is equal to, because it's equal to some new matrix, which we can now call A plus B times x. I just showed, this part is from the definition of our transformations and the sum of our transformations that I defined earlier in this video."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "I don't have to rewrite it all the way up to An plus Bn is the last column. And you've seen this before in your Algebra 2 class, but I wanted to here to do it because this shows you the motivation for it. Because now we can say that the sum of two transformations, so S plus T of x, which is equal to S of x, this is a vector, S of x plus T of x, which we know is equal to A times x plus B times x. We can now say is equal to, because it's equal to some new matrix, which we can now call A plus B times x. I just showed, this part is from the definition of our transformations and the sum of our transformations that I defined earlier in this video. And then when we just worked this out and kind of expressed these products as products of, or as weighted combinations of the column vectors of these guys, we got to this new matrix. And I defined this new matrix as A plus B. And I did that because it has this neat property now."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can now say is equal to, because it's equal to some new matrix, which we can now call A plus B times x. I just showed, this part is from the definition of our transformations and the sum of our transformations that I defined earlier in this video. And then when we just worked this out and kind of expressed these products as products of, or as weighted combinations of the column vectors of these guys, we got to this new matrix. And I defined this new matrix as A plus B. And I did that because it has this neat property now. Because now the sum of two linear transformations operating on x is equivalent to, when you think of it as a matrix vector product, as the sum of their two matrices. Now let's do the same thing with scalar multiplication. We know that c times our transformation of x, by definition I'm saying is c times the transformation of x."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I did that because it has this neat property now. Because now the sum of two linear transformations operating on x is equivalent to, when you think of it as a matrix vector product, as the sum of their two matrices. Now let's do the same thing with scalar multiplication. We know that c times our transformation of x, by definition I'm saying is c times the transformation of x. So c times whatever vector this is in Rm. And so we know that S of x can be, this right here can be rewritten as Ax. So this is c times A times x."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know that c times our transformation of x, by definition I'm saying is c times the transformation of x. So c times whatever vector this is in Rm. And so we know that S of x can be, this right here can be rewritten as Ax. So this is c times A times x. And we know that Ax can be rewritten as, this is equal to c times x1 times the first column vector in A. So A1 plus x2 times A2. All the way to plus xn times An."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is c times A times x. And we know that Ax can be rewritten as, this is equal to c times x1 times the first column vector in A. So A1 plus x2 times A2. All the way to plus xn times An. And what is this? This is scalar multiple. This is just going to be, we can just distribute this c. And then what do we get?"}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "All the way to plus xn times An. And what is this? This is scalar multiple. This is just going to be, we can just distribute this c. And then what do we get? We get x. And I'll just, and multiplication is associative. c is a scalar, x1 is a scalar."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is just going to be, we can just distribute this c. And then what do we get? We get x. And I'll just, and multiplication is associative. c is a scalar, x1 is a scalar. So we can switch them around if we want. We know that scalar multiplication is distributive. So we can write this as x1 times ca1 plus x2 times ca2."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "c is a scalar, x1 is a scalar. So we can switch them around if we want. We know that scalar multiplication is distributive. So we can write this as x1 times ca1 plus x2 times ca2. All the way to xn times can. Now what is this equal to? This is equal to some new matrix times x."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can write this as x1 times ca1 plus x2 times ca2. All the way to xn times can. Now what is this equal to? This is equal to some new matrix times x. This is equal to some new matrix. Let me make that here. Times x1, x2, all the way to xn."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to some new matrix times x. This is equal to some new matrix. Let me make that here. Times x1, x2, all the way to xn. And what is that new matrix? What are the columns of the new matrix? Well the columns are now that, that, all the way to that."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times x1, x2, all the way to xn. And what is that new matrix? What are the columns of the new matrix? Well the columns are now that, that, all the way to that. So the columns of this new matrix are ca1, ca2, all the way to can. Now why did I go through this exercise? Well, wouldn't it be nice?"}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well the columns are now that, that, all the way to that. So the columns of this new matrix are ca1, ca2, all the way to can. Now why did I go through this exercise? Well, wouldn't it be nice? I already said that by definition, a scalar multiple of a transformation is equal to the scalar times the transformation of any vector that you kind of input into it. And of course, that is equal to c times ax. Now, wouldn't it be nice if I could define this thing as some new matrix times a vector x?"}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, wouldn't it be nice? I already said that by definition, a scalar multiple of a transformation is equal to the scalar times the transformation of any vector that you kind of input into it. And of course, that is equal to c times ax. Now, wouldn't it be nice if I could define this thing as some new matrix times a vector x? Because this should also be a linear transformation. And this new matrix I'm going to define. This is a definition again."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, wouldn't it be nice if I could define this thing as some new matrix times a vector x? Because this should also be a linear transformation. And this new matrix I'm going to define. This is a definition again. I'm going to define this new matrix as being c times a. So now we have this definition that c times a, if I take any scalar times any matrix a, it's just equal to c times each of the column vectors. And we know what happens when you take a scalar times each of the, let me write this, is equal to c times a1, c times a2."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a definition again. I'm going to define this new matrix as being c times a. So now we have this definition that c times a, if I take any scalar times any matrix a, it's just equal to c times each of the column vectors. And we know what happens when you take a scalar times each of the, let me write this, is equal to c times a1, c times a2. I'm just rewriting what I just wrote there. All the way to c times an. But what is this in effect?"}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we know what happens when you take a scalar times each of the, let me write this, is equal to c times a1, c times a2. I'm just rewriting what I just wrote there. All the way to c times an. But what is this in effect? We know that when you multiply c times a vector, you multiply the scalar times each of the vector's elements. So this is the equivalent of multiplying c times every entry up in this matrix right here. And while this video, you're probably saying, hey Sal, I already knew how to, in algebra 2 in 10th grade or 9th grade, I already was exposed to multiplying a scalar times a matrix or adding two matrices with the same dimensions."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "But what is this in effect? We know that when you multiply c times a vector, you multiply the scalar times each of the vector's elements. So this is the equivalent of multiplying c times every entry up in this matrix right here. And while this video, you're probably saying, hey Sal, I already knew how to, in algebra 2 in 10th grade or 9th grade, I already was exposed to multiplying a scalar times a matrix or adding two matrices with the same dimensions. Why did you go through all of this trouble of defining the sum of transformations and the sum of matrices? And I went through the trouble because I wanted you to kind of understand that there's nothing, I mean it is natural, but there's nothing about the universe that said matrices had to be defined this way. Matrix addition or matrix scalar multiplication or the addition of two transformations."}, {"video_title": "Sums and scalar multiples of linear transformations Linear Algebra Khan Academy.mp3", "Sentence": "And while this video, you're probably saying, hey Sal, I already knew how to, in algebra 2 in 10th grade or 9th grade, I already was exposed to multiplying a scalar times a matrix or adding two matrices with the same dimensions. Why did you go through all of this trouble of defining the sum of transformations and the sum of matrices? And I went through the trouble because I wanted you to kind of understand that there's nothing, I mean it is natural, but there's nothing about the universe that said matrices had to be defined this way. Matrix addition or matrix scalar multiplication or the addition of two transformations. I wanted you to see that it's all kind of, the mathematical world has constructed it in this way because it seems to have nice properties that are useful. And that's what I've done in this video. In the next video, I'll do a couple of scalar multiplications and matrix additions just to make sure that you remember what you had learned in your 9th or 10th grade algebra class."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's an n by k matrix. Let's say it's not just any n by k matrix. This matrix A has a bunch of columns that are all linearly independent. So A1, A2, all the way through Ak are linearly independent. Let me write that down here. So A1, A2, all the column vectors of A, all the way through Ak are linearly independent. Now what does that mean?"}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A1, A2, all the way through Ak are linearly independent. Let me write that down here. So A1, A2, all the column vectors of A, all the way through Ak are linearly independent. Now what does that mean? That means that the only solution to x1 times A1 plus x2 times A2, that's a 1, plus all the way to xk times Ak, that the only solution to this is all of these x's have to be 0, so all xi's must be equal to 0. That's what linear independence implies. Or another way to write it is all the solutions to this equation, x1, x2, all the way down, xk, equaling the 0 vector, that all of the solutions to this are all of these entries have to be equal to 0."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what does that mean? That means that the only solution to x1 times A1 plus x2 times A2, that's a 1, plus all the way to xk times Ak, that the only solution to this is all of these x's have to be 0, so all xi's must be equal to 0. That's what linear independence implies. Or another way to write it is all the solutions to this equation, x1, x2, all the way down, xk, equaling the 0 vector, that all of the solutions to this are all of these entries have to be equal to 0. This is just another way of writing this right there. We've seen it multiple times. That's the 0 vector right there."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to write it is all the solutions to this equation, x1, x2, all the way down, xk, equaling the 0 vector, that all of the solutions to this are all of these entries have to be equal to 0. This is just another way of writing this right there. We've seen it multiple times. That's the 0 vector right there. So if all of these have to be 0, that's like saying that the only solution to Ax is equal to 0 is x is equal to the 0 vector. Or another way to say it, and this is all coming out of the fact that this guy's columns are linearly independent. So linear independence of columns, based on that, we can say since the only solution to Ax is equal to 0 is x is equal to the 0 vector, we know that the null space of A must be equal to the 0 vector."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the 0 vector right there. So if all of these have to be 0, that's like saying that the only solution to Ax is equal to 0 is x is equal to the 0 vector. Or another way to say it, and this is all coming out of the fact that this guy's columns are linearly independent. So linear independence of columns, based on that, we can say since the only solution to Ax is equal to 0 is x is equal to the 0 vector, we know that the null space of A must be equal to the 0 vector. Or it's a set with just the 0 vector in it. And that is all a bit of review. Now, n by k, we don't know its dimensions."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So linear independence of columns, based on that, we can say since the only solution to Ax is equal to 0 is x is equal to the 0 vector, we know that the null space of A must be equal to the 0 vector. Or it's a set with just the 0 vector in it. And that is all a bit of review. Now, n by k, we don't know its dimensions. It may or may not be a square matrix, so we don't know necessarily whether it's invertible and all of that. But maybe we can construct an invertible matrix with it. So if we take A, let's study A transpose times A."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, n by k, we don't know its dimensions. It may or may not be a square matrix, so we don't know necessarily whether it's invertible and all of that. But maybe we can construct an invertible matrix with it. So if we take A, let's study A transpose times A. A is an n by k matrix. A transpose will be a k by n matrix. So A transpose A is going to be a k by k matrix."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we take A, let's study A transpose times A. A is an n by k matrix. A transpose will be a k by n matrix. So A transpose A is going to be a k by k matrix. So it's a square matrix. So that's a nice place to start for an invertible matrix. So let's see if it is actually invertible."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A transpose A is going to be a k by k matrix. So it's a square matrix. So that's a nice place to start for an invertible matrix. So let's see if it is actually invertible. We don't know anything about A. All we know is its columns are linearly independent. Let's see if A transpose A is invertible."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see if it is actually invertible. We don't know anything about A. All we know is its columns are linearly independent. Let's see if A transpose A is invertible. And essentially, to show that it's invertible, if we could show that all of its columns are linearly independent, then we'll know it's invertible. I'll get back to this at the end of the video. But if you have a square matrix with linearly independent columns, remember, the linearly independent columns all are associated with pivot columns when you put them in reduced row echelon form."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if A transpose A is invertible. And essentially, to show that it's invertible, if we could show that all of its columns are linearly independent, then we'll know it's invertible. I'll get back to this at the end of the video. But if you have a square matrix with linearly independent columns, remember, the linearly independent columns all are associated with pivot columns when you put them in reduced row echelon form. So if you have a square matrix, then you're going to have exactly, so if it's a k by k matrix, that means you're going to have k, that means that the reduced row echelon form of a matrix will have k pivot columns and be a square k by k matrix. And there's only one k by k matrix with k pivot columns, and that's the identity matrix. And that is the identity matrix."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But if you have a square matrix with linearly independent columns, remember, the linearly independent columns all are associated with pivot columns when you put them in reduced row echelon form. So if you have a square matrix, then you're going to have exactly, so if it's a k by k matrix, that means you're going to have k, that means that the reduced row echelon form of a matrix will have k pivot columns and be a square k by k matrix. And there's only one k by k matrix with k pivot columns, and that's the identity matrix. And that is the identity matrix. The k by k identity matrix. And if when you reduce something to reduced row echelon form and you get the identity matrix, that means that your matrix is invertible. That means it's invertible."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that is the identity matrix. The k by k identity matrix. And if when you reduce something to reduced row echelon form and you get the identity matrix, that means that your matrix is invertible. That means it's invertible. I could have probably left that to the end of the video, but I just want to show you, if we can show that we already know that this guy is square, that A transpose A is a square matrix, if we can show that given that A has linearly independent columns, that A transpose times A also has linearly independent columns, then given that the columns are linearly independent and it's a square matrix, that tells us that when we put it into reduced row echelon form, we'll get the identity matrix, and that tells us that this thing would be invertible. So let's see if we can prove that all of this guy's columns are linearly independent. So let's say I have some vector v. Let's say my vector v is a member of the null space of A transpose A."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That means it's invertible. I could have probably left that to the end of the video, but I just want to show you, if we can show that we already know that this guy is square, that A transpose A is a square matrix, if we can show that given that A has linearly independent columns, that A transpose times A also has linearly independent columns, then given that the columns are linearly independent and it's a square matrix, that tells us that when we put it into reduced row echelon form, we'll get the identity matrix, and that tells us that this thing would be invertible. So let's see if we can prove that all of this guy's columns are linearly independent. So let's say I have some vector v. Let's say my vector v is a member of the null space of A transpose A. That means that if I take A transpose A times my vector v, I'm going to get the 0 vector. Fair enough. Now, what happens if I multiply both sides of this equation times the transpose of this guy?"}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have some vector v. Let's say my vector v is a member of the null space of A transpose A. That means that if I take A transpose A times my vector v, I'm going to get the 0 vector. Fair enough. Now, what happens if I multiply both sides of this equation times the transpose of this guy? So I'll get v transpose. Actually, let me just do it right here. If I multiply v transpose on this side and v transpose on this side."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what happens if I multiply both sides of this equation times the transpose of this guy? So I'll get v transpose. Actually, let me just do it right here. If I multiply v transpose on this side and v transpose on this side. You could view this as a matrix vector product, or in general, if you take a row vector times a column vector, it's essentially their dot product. So this right-hand side of the equation, you dot anything with the 0 vector. This is just going to be the 0 vector."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I multiply v transpose on this side and v transpose on this side. You could view this as a matrix vector product, or in general, if you take a row vector times a column vector, it's essentially their dot product. So this right-hand side of the equation, you dot anything with the 0 vector. This is just going to be the 0 vector. Now, what is the left-hand side of this going to be? We've seen this before. If you have the transpose of, we could view this as, even though it's a transpose of a vector, you could view it as a row vector, but you could also view it as a matrix."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is just going to be the 0 vector. Now, what is the left-hand side of this going to be? We've seen this before. If you have the transpose of, we could view this as, even though it's a transpose of a vector, you could view it as a row vector, but you could also view it as a matrix. A column vector is a, let's say v is a member, v is a k by 1 matrix, v transpose would be a 1 by k matrix. Anyway, we've seen this before, that that is equal to the reverse product, the transpose of the reverse product, or if we take the product of two things and transpose it, that's the same thing as taking the reverse product of the transposes of either of those two matrices. So given that, we can replace this right here with a times a vector v transpose, and we're multiplying this vector times av times this vector right here, and that that is going to be equal to the 0 vector."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you have the transpose of, we could view this as, even though it's a transpose of a vector, you could view it as a row vector, but you could also view it as a matrix. A column vector is a, let's say v is a member, v is a k by 1 matrix, v transpose would be a 1 by k matrix. Anyway, we've seen this before, that that is equal to the reverse product, the transpose of the reverse product, or if we take the product of two things and transpose it, that's the same thing as taking the reverse product of the transposes of either of those two matrices. So given that, we can replace this right here with a times a vector v transpose, and we're multiplying this vector times av times this vector right here, and that that is going to be equal to the 0 vector. Now, what is this? If I'm taking some vectors transpose, and let's say, this is a vector, remember, even though I have a matrix vector product right here, when I multiply a matrix times this vector, it will result in another vector. So this is a vector, and this is a vector right here."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So given that, we can replace this right here with a times a vector v transpose, and we're multiplying this vector times av times this vector right here, and that that is going to be equal to the 0 vector. Now, what is this? If I'm taking some vectors transpose, and let's say, this is a vector, remember, even though I have a matrix vector product right here, when I multiply a matrix times this vector, it will result in another vector. So this is a vector, and this is a vector right here. And if I take some vector and I multiply its transpose times that vector, we've seen this before, that is the same thing as y dot y. These two statements are identical. So this thing right here is the same thing as av dot av."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is a vector, and this is a vector right here. And if I take some vector and I multiply its transpose times that vector, we've seen this before, that is the same thing as y dot y. These two statements are identical. So this thing right here is the same thing as av dot av. This is the same thing as av dot av. And so what is the right-hand side equal? The right-hand side is going to be equal to 0."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here is the same thing as av dot av. This is the same thing as av dot av. And so what is the right-hand side equal? The right-hand side is going to be equal to 0. Actually, let me just make a correction up here. When I take v transpose times the 0 vector, v transpose is going to have k elements, and then the 0 vector is also going to have k elements. And when I take this product, that's like dotting it."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The right-hand side is going to be equal to 0. Actually, let me just make a correction up here. When I take v transpose times the 0 vector, v transpose is going to have k elements, and then the 0 vector is also going to have k elements. And when I take this product, that's like dotting it. You're taking the dot product of v and 0. So this is the dot product of v with the 0 vector, which is equal to 0, the scalar 0. So this right here is the scalar 0."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And when I take this product, that's like dotting it. You're taking the dot product of v and 0. So this is the dot product of v with the 0 vector, which is equal to 0, the scalar 0. So this right here is the scalar 0. I want to make sure I clarify that. It wouldn't have made sense otherwise. So the right-hand side, when I multiply the 0 vector times the transpose of v, it gets just the number 0, no vector 0 there."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is the scalar 0. I want to make sure I clarify that. It wouldn't have made sense otherwise. So the right-hand side, when I multiply the 0 vector times the transpose of v, it gets just the number 0, no vector 0 there. So this av dot av is going to be equal to 0. Or we could say that the magnitude or the length of av squared is equal to 0. Or that tells us that av has to be equal to 0."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the right-hand side, when I multiply the 0 vector times the transpose of v, it gets just the number 0, no vector 0 there. So this av dot av is going to be equal to 0. Or we could say that the magnitude or the length of av squared is equal to 0. Or that tells us that av has to be equal to 0. The only vector whose length is 0 is the 0 vector. So av, let me switch colors. I'm using that a little bit too much."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or that tells us that av has to be equal to 0. The only vector whose length is 0 is the 0 vector. So av, let me switch colors. I'm using that a little bit too much. So we know that av must be equal to 0, to the 0 vector. This must be equal to the 0 vector, since its length is 0. Now, we started off with saying v is a member of the null space, is a member of the null space of A transpose A. v could be any member of the null space of A transpose A."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm using that a little bit too much. So we know that av must be equal to 0, to the 0 vector. This must be equal to the 0 vector, since its length is 0. Now, we started off with saying v is a member of the null space, is a member of the null space of A transpose A. v could be any member of the null space of A transpose A. But then from that assumption, it turns out that v also has to be a member of the null space of A, that av is equal to 0. Let's write that down. If v is a member of the null space of A transpose A, then v is a member of the null space of A."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we started off with saying v is a member of the null space, is a member of the null space of A transpose A. v could be any member of the null space of A transpose A. But then from that assumption, it turns out that v also has to be a member of the null space of A, that av is equal to 0. Let's write that down. If v is a member of the null space of A transpose A, then v is a member of the null space of A. Now, our null space of A, because A's columns are linearly independent, it only contains 1 vector. It only contains the 0 vector. So if this guy is a member of the null space of A transpose A, and he has to be a member of the null space of A, there's only one thing he can be."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If v is a member of the null space of A transpose A, then v is a member of the null space of A. Now, our null space of A, because A's columns are linearly independent, it only contains 1 vector. It only contains the 0 vector. So if this guy is a member of the null space of A transpose A, and he has to be a member of the null space of A, there's only one thing he can be. There's only one entry there. So then v has to be equal to the 0 vector. Or another way to say that is any v that's in the null space of A transpose A has to be the 0 vector."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this guy is a member of the null space of A transpose A, and he has to be a member of the null space of A, there's only one thing he can be. There's only one entry there. So then v has to be equal to the 0 vector. Or another way to say that is any v that's in the null space of A transpose A has to be the 0 vector. Or the null space of A transpose A is equal to the null space of A, which is equal to just the 0 vector sitting there. Now, what does that do for us? That tells us that the only solution to A transpose A times some vector x equal to 0, this says that the only solution is the 0 vector."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say that is any v that's in the null space of A transpose A has to be the 0 vector. Or the null space of A transpose A is equal to the null space of A, which is equal to just the 0 vector sitting there. Now, what does that do for us? That tells us that the only solution to A transpose A times some vector x equal to 0, this says that the only solution is the 0 vector. Is, let me write, is x is equal to the 0 vector, right? Because the null space of A transpose A is the same as the null space of A, and that just has the 0 vector in it. The null space is just the solutions to this."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That tells us that the only solution to A transpose A times some vector x equal to 0, this says that the only solution is the 0 vector. Is, let me write, is x is equal to the 0 vector, right? Because the null space of A transpose A is the same as the null space of A, and that just has the 0 vector in it. The null space is just the solutions to this. So if the only solutions to the null space is this, that means that the columns of A are linearly independent. You could essentially write all of the linear combinations of the columns by the weights of the entries of x. We actually did that at the beginning."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The null space is just the solutions to this. So if the only solutions to the null space is this, that means that the columns of A are linearly independent. You could essentially write all of the linear combinations of the columns by the weights of the entries of x. We actually did that at the beginning. It's the same argument we used up here. So if all of their columns are linearly independent, and I said it over here, A transpose A has linearly independent columns, and it's a square matrix. That was kind of from the definition of it."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We actually did that at the beginning. It's the same argument we used up here. So if all of their columns are linearly independent, and I said it over here, A transpose A has linearly independent columns, and it's a square matrix. That was kind of from the definition of it. So we now know that A transpose A, if I were to put it, let me do it this way. That tells me that the reduced row echelon form of A transpose A is going to be equal to the identity, the k by k identity matrix, which tells me that A transpose A is invertible, which is a pretty neat result. I started with a matrix that has linearly independent columns, so it wasn't just any matrix."}, {"video_title": "Showing that A-transpose x A is invertible Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That was kind of from the definition of it. So we now know that A transpose A, if I were to put it, let me do it this way. That tells me that the reduced row echelon form of A transpose A is going to be equal to the identity, the k by k identity matrix, which tells me that A transpose A is invertible, which is a pretty neat result. I started with a matrix that has linearly independent columns, so it wasn't just any matrix. It wasn't just any run-of-the-mill matrix. It did have linearly independent columns, but it might have weird dimensions. It's not necessarily a square matrix."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I thought I would at least show you that what we've been covering so far is very consistent with a way of determining or finding determinants that you might have been exposed to in your Algebra 2 class. It's called the Rule of Saris. And essentially, well, let me just prove it for you. So let's say we want to find this determinant. So our matrix is A, B, C, D, E, F, G, H, I. We know how to do this. This is equal to, let's just go down that first row, A times the determinant of E, F, H, I minus B times the determinant of D, G, F, I plus C times the determinant of D, E, G, H. And what are these equal to?"}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we want to find this determinant. So our matrix is A, B, C, D, E, F, G, H, I. We know how to do this. This is equal to, let's just go down that first row, A times the determinant of E, F, H, I minus B times the determinant of D, G, F, I plus C times the determinant of D, E, G, H. And what are these equal to? This is going to be equal to A times EI minus FH. And this is going to be minus B times DI minus FG. And this is going to be plus C times DH minus EG."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, let's just go down that first row, A times the determinant of E, F, H, I minus B times the determinant of D, G, F, I plus C times the determinant of D, E, G, H. And what are these equal to? This is going to be equal to A times EI minus FH. And this is going to be minus B times DI minus FG. And this is going to be plus C times DH minus EG. And if we multiply this out, we get this as being equal to AEI minus AFH minus BDI plus, right, minus times a minus, so plus BFG plus CDH minus CEG. Now let me group the positive and the negative terms. So I have this term is positive, this term is positive, and that term is positive."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is going to be plus C times DH minus EG. And if we multiply this out, we get this as being equal to AEI minus AFH minus BDI plus, right, minus times a minus, so plus BFG plus CDH minus CEG. Now let me group the positive and the negative terms. So I have this term is positive, this term is positive, and that term is positive. So we have this being equal to AEI plus BFG plus CDH. Those are our positive terms. And then our negative terms are here."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I have this term is positive, this term is positive, and that term is positive. So we have this being equal to AEI plus BFG plus CDH. Those are our positive terms. And then our negative terms are here. We have that term, that term, and that term. So we have minus AFH minus BDI minus CEG. So this is a formula for the determinant of this matrix right here."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then our negative terms are here. We have that term, that term, and that term. So we have minus AFH minus BDI minus CEG. So this is a formula for the determinant of this matrix right here. Let's see what it actually looks like. So let me rewrite it. Let me rewrite our matrix."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is a formula for the determinant of this matrix right here. Let's see what it actually looks like. So let me rewrite it. Let me rewrite our matrix. So if we do it in green. So we have A, B, C, D, E, F, G, H, I. And we wanted to find its determinant."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me rewrite our matrix. So if we do it in green. So we have A, B, C, D, E, F, G, H, I. And we wanted to find its determinant. So let me show you something interesting here. AEI is what? AEI is a product of this guy, this guy, and that guy."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we wanted to find its determinant. So let me show you something interesting here. AEI is what? AEI is a product of this guy, this guy, and that guy. So you're essentially going along that diagonal right there. Now what is BFG? BFG, you're going this guy, this guy, and then you're going all the way down to this guy."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "AEI is a product of this guy, this guy, and that guy. So you're essentially going along that diagonal right there. Now what is BFG? BFG, you're going this guy, this guy, and then you're going all the way down to this guy. So it's like if you imagine that when you come out of this side, you come out of this side, and there's some video games where you know one end, you end up showing up on the other end like that. It would also be a diagonal. Or even a better way to visualize it, let me redraw these two columns."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "BFG, you're going this guy, this guy, and then you're going all the way down to this guy. So it's like if you imagine that when you come out of this side, you come out of this side, and there's some video games where you know one end, you end up showing up on the other end like that. It would also be a diagonal. Or even a better way to visualize it, let me redraw these two columns. Let me kind of augment this determinant. That's not an official terminology, but I think you'll get what I'm trying to do. So if I write these first two columns again, ADG and BEH, this guy right here, BFG, it's this one right here."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or even a better way to visualize it, let me redraw these two columns. Let me kind of augment this determinant. That's not an official terminology, but I think you'll get what I'm trying to do. So if I write these first two columns again, ADG and BEH, this guy right here, BFG, it's this one right here. It's this diagonal right there. And then you might guess what's about to happen. Where is CDH?"}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I write these first two columns again, ADG and BEH, this guy right here, BFG, it's this one right here. It's this diagonal right there. And then you might guess what's about to happen. Where is CDH? It's this diagonal. It's that diagonal right there. So you take this product, add it to this product, add it to this product, and then you subtract these guys."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Where is CDH? It's this diagonal. It's that diagonal right there. So you take this product, add it to this product, add it to this product, and then you subtract these guys. Now where are these guys? Where is the AFH? AFH, it's that one right there."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you take this product, add it to this product, add it to this product, and then you subtract these guys. Now where are these guys? Where is the AFH? AFH, it's that one right there. So you subtract out AFH, and then you subtract out BDI. It's that one right there. And then you have CEG, which is this one right there."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "AFH, it's that one right there. So you subtract out AFH, and then you subtract out BDI. It's that one right there. And then you have CEG, which is this one right there. So the rule of Saurus sounds like something in Lord of the Rings, the rule of Saurus. The rule of Saurus is essentially a quick way of memorizing this little technique, where you write the two columns again, and you say, OK, this product plus this product plus this product minus this product minus this product minus that product. Let's actually do it with a 3 by 3 matrix to make it clear that the rule of Saurus can be useful."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have CEG, which is this one right there. So the rule of Saurus sounds like something in Lord of the Rings, the rule of Saurus. The rule of Saurus is essentially a quick way of memorizing this little technique, where you write the two columns again, and you say, OK, this product plus this product plus this product minus this product minus this product minus that product. Let's actually do it with a 3 by 3 matrix to make it clear that the rule of Saurus can be useful. So let's say we have the matrix, we want the determinant of the matrix 1, 2, 4, 2, minus 1, 2, minus 1, 3, and then we have 4, 0, minus 1. We want to find that determinant. So by the rule of Saurus, we can rewrite these first two columns."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's actually do it with a 3 by 3 matrix to make it clear that the rule of Saurus can be useful. So let's say we have the matrix, we want the determinant of the matrix 1, 2, 4, 2, minus 1, 2, minus 1, 3, and then we have 4, 0, minus 1. We want to find that determinant. So by the rule of Saurus, we can rewrite these first two columns. So 1, 2, 2, minus 1, 4, 0. We rewrote those first two columns. And to figure out this determinant, we take this guy."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So by the rule of Saurus, we can rewrite these first two columns. So 1, 2, 2, minus 1, 4, 0. We rewrote those first two columns. And to figure out this determinant, we take this guy. So what is this going to be? 1 times minus 1 times minus 1. That is just a 1, right?"}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And to figure out this determinant, we take this guy. So what is this going to be? 1 times minus 1 times minus 1. That is just a 1, right? The minuses cancel out. Plus this guy, plus this product right here. I should draw it a little bit neater."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is just a 1, right? The minuses cancel out. Plus this guy, plus this product right here. I should draw it a little bit neater. So what is this? 2 times 3 times 4. 2 times 3 is 6."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I should draw it a little bit neater. So what is this? 2 times 3 times 4. 2 times 3 is 6. 6 times 4 is 24. Plus 24. And then we take this guy right here."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "2 times 3 is 6. 6 times 4 is 24. Plus 24. And then we take this guy right here. 4 times 2 times 0. Anything times 0 is a 0, so it's going to be plus 0. And then we subtract out these guys."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we take this guy right here. 4 times 2 times 0. Anything times 0 is a 0, so it's going to be plus 0. And then we subtract out these guys. So you have 4 times 4 times minus 1. That's minus 16. But this is on the minus side of things."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we subtract out these guys. So you have 4 times 4 times minus 1. That's minus 16. But this is on the minus side of things. So 4 times minus 1 times 4 is minus 16. But since we're going to do a minus on it, it's going to be plus 16. So it's 16."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this is on the minus side of things. So 4 times minus 1 times 4 is minus 16. But since we're going to do a minus on it, it's going to be plus 16. So it's 16. Then you have a 0 times 3 times 1. That, of course, is going to be a 0. It would be a minus 0, but we can ignore it."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's 16. Then you have a 0 times 3 times 1. That, of course, is going to be a 0. It would be a minus 0, but we can ignore it. So we could say plus 0 or minus 0. Same thing. And then you have a minus 1 times 2 times 2."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It would be a minus 0, but we can ignore it. So we could say plus 0 or minus 0. Same thing. And then you have a minus 1 times 2 times 2. So that's 4 times minus 1, which is minus 4. When you go in this direction, from the top right to the bottom left, you're subtracting. So this would be a minus 4."}, {"video_title": "Rule of Sarrus of determinants Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you have a minus 1 times 2 times 2. So that's 4 times minus 1, which is minus 4. When you go in this direction, from the top right to the bottom left, you're subtracting. So this would be a minus 4. But since we're subtracting, this becomes a plus 4. So the value of our determinant is equal to, by the rule of Soris, so these guys, we're going to have 16 plus 4 is a 20, plus 1 plus 20 is 25, which is equal to 45. So that actually is, I'd have to say, a faster way of computing this 3 by 3 derivative."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "It can be represented by the matrix A. So the transformation of x is equal to A times x. We saw in the last video, it's interesting to find the vectors that only get scaled up or down by the transformation. So we're interested in the vectors where, if I take the transformation of some special vector v, it equals, of course, A times v. And we say it only gets scaled up by some factor, lambda times v. And these are interesting because they make for interesting basis vectors. The transformation matrix and the alternate basis, this is one of the basis vectors, might be easier to compute, might make for good coordinate systems. But they're, in general, interesting. And we call vectors v that satisfy this, we call them eigenvectors."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "So we're interested in the vectors where, if I take the transformation of some special vector v, it equals, of course, A times v. And we say it only gets scaled up by some factor, lambda times v. And these are interesting because they make for interesting basis vectors. The transformation matrix and the alternate basis, this is one of the basis vectors, might be easier to compute, might make for good coordinate systems. But they're, in general, interesting. And we call vectors v that satisfy this, we call them eigenvectors. And we call their scaling factors the eigenvalues associated with this transformation and that eigenvector. And hopefully from that last video, we have a little bit of appreciation of why they're useful. But now in this video, let's at least try to determine what some of them are."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "And we call vectors v that satisfy this, we call them eigenvectors. And we call their scaling factors the eigenvalues associated with this transformation and that eigenvector. And hopefully from that last video, we have a little bit of appreciation of why they're useful. But now in this video, let's at least try to determine what some of them are. Based on what we know so far, if you show me an eigenvector, I can verify that it definitely is the case, or an eigenvalue. I could verify the case. But I don't know a systematic way of solving for either of them."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "But now in this video, let's at least try to determine what some of them are. Based on what we know so far, if you show me an eigenvector, I can verify that it definitely is the case, or an eigenvalue. I could verify the case. But I don't know a systematic way of solving for either of them. So let's see if we can come up with something. So in general, we're looking for solutions to the equation A times v is equal to lambda v. It's equal to lambda times the vector. Now one solution might immediately pop out at you."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "But I don't know a systematic way of solving for either of them. So let's see if we can come up with something. So in general, we're looking for solutions to the equation A times v is equal to lambda v. It's equal to lambda times the vector. Now one solution might immediately pop out at you. And that's just v is equal to the 0 vector. And that definitely is a solution. Although it's not normally considered to be an eigenvector."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Now one solution might immediately pop out at you. And that's just v is equal to the 0 vector. And that definitely is a solution. Although it's not normally considered to be an eigenvector. Just because one, it's not a useful basis vector. It doesn't add anything to a basis. It doesn't add really the amount of vectors that you can span when you throw the basis vector in there."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Although it's not normally considered to be an eigenvector. Just because one, it's not a useful basis vector. It doesn't add anything to a basis. It doesn't add really the amount of vectors that you can span when you throw the basis vector in there. And also, it's not clear what is your eigenvalue that's associated with it. Because if v is equal to 0, any eigenvalue will work for that. So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "It doesn't add really the amount of vectors that you can span when you throw the basis vector in there. And also, it's not clear what is your eigenvalue that's associated with it. Because if v is equal to 0, any eigenvalue will work for that. So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors. So we're looking for vectors that are not equal to the 0 vector. So given that, let's see if we can play around with this equation a little bit. And see if we can at least come up with eigenvalues maybe in this video."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors. So we're looking for vectors that are not equal to the 0 vector. So given that, let's see if we can play around with this equation a little bit. And see if we can at least come up with eigenvalues maybe in this video. So we subtract Av from both sides. We get the 0 vector is equal to lambda v minus A times v. Now we can rewrite v as v is just the same thing as the identity matrix times v. v is a member of Rn. The identity matrix n by n, you just multiply."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "And see if we can at least come up with eigenvalues maybe in this video. So we subtract Av from both sides. We get the 0 vector is equal to lambda v minus A times v. Now we can rewrite v as v is just the same thing as the identity matrix times v. v is a member of Rn. The identity matrix n by n, you just multiply. We're just going to get v again. So if I rewrite v this way, at least on this part of the expression, let me swap sides. So then I'll get lambda times, instead of v, I'll write the identity matrix, the n by n identity matrix, times v minus A times v is equal to the 0 vector."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "The identity matrix n by n, you just multiply. We're just going to get v again. So if I rewrite v this way, at least on this part of the expression, let me swap sides. So then I'll get lambda times, instead of v, I'll write the identity matrix, the n by n identity matrix, times v minus A times v is equal to the 0 vector. Now I have one matrix times v minus another matrix times v. Matrix vector products, they have the distributive property. So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v. And that's going to be equal to 0. This is just some matrix right here."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "So then I'll get lambda times, instead of v, I'll write the identity matrix, the n by n identity matrix, times v minus A times v is equal to the 0 vector. Now I have one matrix times v minus another matrix times v. Matrix vector products, they have the distributive property. So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v. And that's going to be equal to 0. This is just some matrix right here. And the whole reason why I made this substitution is so that I could write this as a matrix vector product instead of just a scalar vector product. And that way I was able to essentially factor out the v and just write this whole equation as essentially some matrix vector product is equal to 0. Now, if we assume that this is the case, and we're assuming, remember, we're assuming that v does not equal 0."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "This is just some matrix right here. And the whole reason why I made this substitution is so that I could write this as a matrix vector product instead of just a scalar vector product. And that way I was able to essentially factor out the v and just write this whole equation as essentially some matrix vector product is equal to 0. Now, if we assume that this is the case, and we're assuming, remember, we're assuming that v does not equal 0. So what does this mean? So we know that v is a member of the null space of this matrix right here. Let me write this down."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Now, if we assume that this is the case, and we're assuming, remember, we're assuming that v does not equal 0. So what does this mean? So we know that v is a member of the null space of this matrix right here. Let me write this down. v is a member of the null space of lambda i sub n minus A. I know that might look a little convoluted to you right now, but just imagine this is just some matrix B. It might make it simpler. This is just some matrix here, right?"}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this down. v is a member of the null space of lambda i sub n minus A. I know that might look a little convoluted to you right now, but just imagine this is just some matrix B. It might make it simpler. This is just some matrix here, right? That's B. Let's make that substitution. Then this equation just becomes Bv is equal to 0."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "This is just some matrix here, right? That's B. Let's make that substitution. Then this equation just becomes Bv is equal to 0. Now, if we want to look at the null space of this, the null space of B is all of the vectors x that are a member of Rn such that B times x is equal to 0. Well, v is clearly one of those guys, right? Because B times v is equal to 0."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Then this equation just becomes Bv is equal to 0. Now, if we want to look at the null space of this, the null space of B is all of the vectors x that are a member of Rn such that B times x is equal to 0. Well, v is clearly one of those guys, right? Because B times v is equal to 0. We're assuming v solves this equation. That gets us all the way to the assumption that v must solve this equation. And v is not equal to 0."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Because B times v is equal to 0. We're assuming v solves this equation. That gets us all the way to the assumption that v must solve this equation. And v is not equal to 0. So v is a member of the null space, and this is a non-trivial member of the null space. We already said that the 0 vector is always going to be a member of the null space, and it would make this true. But we're assuming v is non-zero."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "And v is not equal to 0. So v is a member of the null space, and this is a non-trivial member of the null space. We already said that the 0 vector is always going to be a member of the null space, and it would make this true. But we're assuming v is non-zero. We're only interested in non-zero eigenvectors, and that means that this guy's null space has to be non-trivial. So this means that the null space of lambda in minus a is non-trivial. The 0 vector is not the only member."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "But we're assuming v is non-zero. We're only interested in non-zero eigenvectors, and that means that this guy's null space has to be non-trivial. So this means that the null space of lambda in minus a is non-trivial. The 0 vector is not the only member. And you might remember before that the only time we write this in general, if I have some matrix, I don't know if I've used A and B, let's say I have some matrix D. D is D's D's columns are linearly independent if and only if the null space of D only contains the 0 vector. So if we have some matrix here whose null space does not only contain the 0 vector, then it has linearly dependent columns. So we know that, and I just wrote that there to kind of show you what we do know."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "The 0 vector is not the only member. And you might remember before that the only time we write this in general, if I have some matrix, I don't know if I've used A and B, let's say I have some matrix D. D is D's D's columns are linearly independent if and only if the null space of D only contains the 0 vector. So if we have some matrix here whose null space does not only contain the 0 vector, then it has linearly dependent columns. So we know that, and I just wrote that there to kind of show you what we do know. And the fact that this one doesn't have a trivial null space tells us that we're dealing with linearly dependent columns. So lambda in minus a, it looks all fancy, but this is just a matrix, must have linearly dependent columns. Or another way to say that is if you have linearly dependent columns, you're not invertible."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "So we know that, and I just wrote that there to kind of show you what we do know. And the fact that this one doesn't have a trivial null space tells us that we're dealing with linearly dependent columns. So lambda in minus a, it looks all fancy, but this is just a matrix, must have linearly dependent columns. Or another way to say that is if you have linearly dependent columns, you're not invertible. Which also means that your determinant must be equal to 0. All of these are true. If your determinant is equal to 0, you're not going to be invertible, you're going to have linearly dependent columns, if your determinant is equal to 0, then that also means that you have non-trivial members in your null space."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say that is if you have linearly dependent columns, you're not invertible. Which also means that your determinant must be equal to 0. All of these are true. If your determinant is equal to 0, you're not going to be invertible, you're going to have linearly dependent columns, if your determinant is equal to 0, then that also means that you have non-trivial members in your null space. And so if your determinant is equal to 0, that means there's some lambdas for which this is true, for non-zero vectors v. So if there are some solutions, if there are some non-zero vector v's that satisfy this equation, then this matrix right here must have a determinant of 0. And it goes the other way. If this guy has a determinant of 0, then there must be, or there's some lambdas that make this guy have a determinant of 0, then those lambdas are going to satisfy this equation."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "If your determinant is equal to 0, you're not going to be invertible, you're going to have linearly dependent columns, if your determinant is equal to 0, then that also means that you have non-trivial members in your null space. And so if your determinant is equal to 0, that means there's some lambdas for which this is true, for non-zero vectors v. So if there are some solutions, if there are some non-zero vector v's that satisfy this equation, then this matrix right here must have a determinant of 0. And it goes the other way. If this guy has a determinant of 0, then there must be, or there's some lambdas that make this guy have a determinant of 0, then those lambdas are going to satisfy this equation. And you can go the other way. If there's some lambdas that satisfy this, then those lambdas are going to make this matrix have a 0 determinant. So the determinant, let me write this."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "If this guy has a determinant of 0, then there must be, or there's some lambdas that make this guy have a determinant of 0, then those lambdas are going to satisfy this equation. And you can go the other way. If there's some lambdas that satisfy this, then those lambdas are going to make this matrix have a 0 determinant. So the determinant, let me write this. So if and only if, so let me write this. A v is equal to lambda v for non-zero v's if and only if the determinant of lambda i n minus a is equal to the 0 vector, is equal to, no, not the 0 vector, sorry, it's just equal to 0. The determinant is just a scalar factor."}, {"video_title": "Proof of formula for determining eigenvalues Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant, let me write this. So if and only if, so let me write this. A v is equal to lambda v for non-zero v's if and only if the determinant of lambda i n minus a is equal to the 0 vector, is equal to, no, not the 0 vector, sorry, it's just equal to 0. The determinant is just a scalar factor. So that's our big takeaway. I know what you're saying, now how is that useful for me, Sal, you know, we did all of this manipulation, I talked a little bit about the null spaces, and my big takeaway is that in order for this to be true for some non-zero vectors v, then lambda has to be some value such that if I take the determinant of lambda times the identity matrix minus a, it has got to be equal to 0. And the reason why this is useful is that you can actually set this equation up for your matrices and then solve for your lambdas."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I'm trying to determine the null space of A, I'm essentially just asking, look, if we set up the equation Ax is equal to the 0 vector, the null space of A is all the x's that satisfy this equation. So it's all the x's that satisfy that equation. Ax is equal to the 0 vector, or you could call it the system. And the way you would solve it, and we've done this many times, this was many videos ago, you would make an augmented matrix with this. So the augmented matrix would look like that. You'd have the 0 vector on the right-hand side. And then you'd perform a bunch of row operations to put the left-hand side into reduced row echelon form."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the way you would solve it, and we've done this many times, this was many videos ago, you would make an augmented matrix with this. So the augmented matrix would look like that. You'd have the 0 vector on the right-hand side. And then you'd perform a bunch of row operations to put the left-hand side into reduced row echelon form. So you'd do a bunch of operations. The left-hand side would go into reduced row echelon form. Let's call that reduced row echelon form of A."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then you'd perform a bunch of row operations to put the left-hand side into reduced row echelon form. So you'd do a bunch of operations. The left-hand side would go into reduced row echelon form. Let's call that reduced row echelon form of A. And then the right-hand side is just going to stay 0, because you perform the same row operations. But when you perform those row operations on 0, you just get the 0 vector right here. And then when you create the system back from this right here, because these two systems are equivalent, you're essentially going to have your solution set look something like this."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that reduced row echelon form of A. And then the right-hand side is just going to stay 0, because you perform the same row operations. But when you perform those row operations on 0, you just get the 0 vector right here. And then when you create the system back from this right here, because these two systems are equivalent, you're essentially going to have your solution set look something like this. You're going to have your, let me write it like this, your solution set is going to be equal to some scalar multiple, you know, let's say that of your free variables, your free variables are going to be the scalar multiples. And you've seen this multiple times, so I'll state fairly general. But it's going to be some multiple times, let's say, vector 1 plus some other scalar times vector 2."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then when you create the system back from this right here, because these two systems are equivalent, you're essentially going to have your solution set look something like this. You're going to have your, let me write it like this, your solution set is going to be equal to some scalar multiple, you know, let's say that of your free variables, your free variables are going to be the scalar multiples. And you've seen this multiple times, so I'll state fairly general. But it's going to be some multiple times, let's say, vector 1 plus some other scalar times vector 2. These scalars tend to be your free variables. Times vector 2, all the way to, I don't know, whatever, c times your nth vector. I'm just trying to say general."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But it's going to be some multiple times, let's say, vector 1 plus some other scalar times vector 2. These scalars tend to be your free variables. Times vector 2, all the way to, I don't know, whatever, c times your nth vector. I'm just trying to say general. We haven't seen any examples that had more than 2 or 3 vectors here, but this is what, essentially, your null space is spanned by these vectors right there. You get an equation, you get a solution set that looks something like that, and you call that your null space. We've done that multiple times."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just trying to say general. We haven't seen any examples that had more than 2 or 3 vectors here, but this is what, essentially, your null space is spanned by these vectors right there. You get an equation, you get a solution set that looks something like that, and you call that your null space. We've done that multiple times. Your null space is that, so it's all the linear combinations, or it's the span of these little vectors that you get here, n1, n2, all the way to nn. This is nothing new. I'm just restating something that we've seen multiple, multiple times."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've done that multiple times. Your null space is that, so it's all the linear combinations, or it's the span of these little vectors that you get here, n1, n2, all the way to nn. This is nothing new. I'm just restating something that we've seen multiple, multiple times. We've actually did this in the previous video. I just maybe never wrote it exactly like this. But what about the case when you're solving the inhomogeneous equation?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm just restating something that we've seen multiple, multiple times. We've actually did this in the previous video. I just maybe never wrote it exactly like this. But what about the case when you're solving the inhomogeneous equation? So the inhomogeneous equation looks like this. So if I want to solve Ax is equal to b, I would do something very similar to this. I will create an augmented matrix."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But what about the case when you're solving the inhomogeneous equation? So the inhomogeneous equation looks like this. So if I want to solve Ax is equal to b, I would do something very similar to this. I will create an augmented matrix. I have a on the left-hand side, and I'd put b on the right-hand side. And then I'll perform a bunch of row operations to put a into reduced row echelon form. So let me do that."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I will create an augmented matrix. I have a on the left-hand side, and I'd put b on the right-hand side. And then I'll perform a bunch of row operations to put a into reduced row echelon form. So let me do that. So this left-hand side will be the reduced row echelon form of a. And then the right-hand side, whatever operations I did on a I have to do on the entire row, so I'll also be doing them to b. So I'll have some new vector here."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me do that. So this left-hand side will be the reduced row echelon form of a. And then the right-hand side, whatever operations I did on a I have to do on the entire row, so I'll also be doing them to b. So I'll have some new vector here. Maybe I'll call it the vector b prime. It's going to be different than b, but let's just call it b prime. And so when you go back to your, I guess, when you go out of the augmented matrix world and rewrite it as a system and you solve for it, and we did this in the last video, you'll get your solution set."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'll have some new vector here. Maybe I'll call it the vector b prime. It's going to be different than b, but let's just call it b prime. And so when you go back to your, I guess, when you go out of the augmented matrix world and rewrite it as a system and you solve for it, and we did this in the last video, you'll get your solution set. Your solution set that satisfies this is going to be x is going to be equal to this b prime, whatever this new vector is, this b prime plus something that looks exactly like this. That looks exactly like that. In fact, I'll copy and paste it."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so when you go back to your, I guess, when you go out of the augmented matrix world and rewrite it as a system and you solve for it, and we did this in the last video, you'll get your solution set. Your solution set that satisfies this is going to be x is going to be equal to this b prime, whatever this new vector is, this b prime plus something that looks exactly like this. That looks exactly like that. In fact, I'll copy and paste it. It'll look exactly like this. Let me see if I did copy. Let me copy and paste it."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In fact, I'll copy and paste it. It'll look exactly like this. Let me see if I did copy. Let me copy and paste it. Edit, copy, and then let me paste it. So it'll look something that looks exactly like that. And we said in the last video that that, given this, you can kind of think of the solution set to the inhomogeneous equation is equivalent to some particular solution is equal to some particular solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me copy and paste it. Edit, copy, and then let me paste it. So it'll look something that looks exactly like that. And we said in the last video that that, given this, you can kind of think of the solution set to the inhomogeneous equation is equivalent to some particular solution is equal to some particular solution. Let's call that x particular. Some particular solution plus some member of your null space, so you could say plus some homogeneous solution. So if you just pick particular values for a, b, and c, all of the different multiples of the vectors that span your null space, you'll get some particular homogeneous solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we said in the last video that that, given this, you can kind of think of the solution set to the inhomogeneous equation is equivalent to some particular solution is equal to some particular solution. Let's call that x particular. Some particular solution plus some member of your null space, so you could say plus some homogeneous solution. So if you just pick particular values for a, b, and c, all of the different multiples of the vectors that span your null space, you'll get some particular homogeneous solution. So what I implied in the last video, and I didn't show it to you rigorously, is that any solution to the inhomogeneous system. Let me write it this way. Any solution, and do it in white."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you just pick particular values for a, b, and c, all of the different multiples of the vectors that span your null space, you'll get some particular homogeneous solution. So what I implied in the last video, and I didn't show it to you rigorously, is that any solution to the inhomogeneous system. Let me write it this way. Any solution, and do it in white. Any solution to the inhomogeneous system, ax is equal to b. This is a claim I made. We'll take the form some particular solution, that was this right here, maybe I should do it in green."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Any solution, and do it in white. Any solution to the inhomogeneous system, ax is equal to b. This is a claim I made. We'll take the form some particular solution, that was this right here, maybe I should do it in green. This is this right here. When you do the reduced row echelon form, it becomes that b prime vector, plus some homogeneous solution. So some member of the null space."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We'll take the form some particular solution, that was this right here, maybe I should do it in green. This is this right here. When you do the reduced row echelon form, it becomes that b prime vector, plus some homogeneous solution. So some member of the null space. Now I didn't prove it to you, but I implied that this is the case. And what I want to do in this video is actually do a little bit of a more rigorous proof, but it's actually fairly straightforward. So let's first of all verify that this is a solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So some member of the null space. Now I didn't prove it to you, but I implied that this is the case. And what I want to do in this video is actually do a little bit of a more rigorous proof, but it's actually fairly straightforward. So let's first of all verify that this is a solution. So let's verify that this is a solution. So let's just put this into our original equation. So if we, let's remember our original equation was ax is equal to b."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's first of all verify that this is a solution. So let's verify that this is a solution. So let's just put this into our original equation. So if we, let's remember our original equation was ax is equal to b. So let's verify. So is, let me write it as a question, is that particular solution plus some homogeneous solution a solution to ax is equal to b? Well to do that, you just put that in the place of x."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we, let's remember our original equation was ax is equal to b. So let's verify. So is, let me write it as a question, is that particular solution plus some homogeneous solution a solution to ax is equal to b? Well to do that, you just put that in the place of x. So let's try it out. So a times this guy right here, times some particular solution, plus some homogeneous solution is going to be equal to a times the particular solution, plus a times some member of my null space. And what is this equal to?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well to do that, you just put that in the place of x. So let's try it out. So a times this guy right here, times some particular solution, plus some homogeneous solution is going to be equal to a times the particular solution, plus a times some member of my null space. And what is this equal to? That is going to be equal to b. We're saying that this is a particular solution to this equation. That is going to be equal to b."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? That is going to be equal to b. We're saying that this is a particular solution to this equation. That is going to be equal to b. And that this is going to be equal to the 0 vector, because this is a solution to our homogeneous equation. So this is going to be equal to b plus 0, or it's equal to b. So a times this vector right here is indeed equal to b."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is going to be equal to b. And that this is going to be equal to the 0 vector, because this is a solution to our homogeneous equation. So this is going to be equal to b plus 0, or it's equal to b. So a times this vector right here is indeed equal to b. So this is a solution. This is a solution. Yes."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So a times this vector right here is indeed equal to b. So this is a solution. This is a solution. Yes. Now the next question is, does every solution to the inhomogeneous system, or does any solution to the inhomogeneous system, take this form? So does any solution x to Ax equal to b take the form x is equal to some particular solution, plus a member of our null space, or plus a homogeneous solution? So to do that, let's test out what happens when we multiply the vector a times x."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Yes. Now the next question is, does every solution to the inhomogeneous system, or does any solution to the inhomogeneous system, take this form? So does any solution x to Ax equal to b take the form x is equal to some particular solution, plus a member of our null space, or plus a homogeneous solution? So to do that, let's test out what happens when we multiply the vector a times x. Let's just write it this way. Let's say that x is any solution to Ax is equal to b. Let's start off with that."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So to do that, let's test out what happens when we multiply the vector a times x. Let's just write it this way. Let's say that x is any solution to Ax is equal to b. Let's start off with that. And let's see what happens when we take a times x minus some particular solution to this. So when we distribute the matrix vector product, you get a times our any solution minus a times our particular solution. Now what is this going to be equal to?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's start off with that. And let's see what happens when we take a times x minus some particular solution to this. So when we distribute the matrix vector product, you get a times our any solution minus a times our particular solution. Now what is this going to be equal to? We're saying that this is a solution to Ax equal to b. So this is going to be equal to b. And of course, any particular solution to this, when you multiply it by a, is also going to be equal to b."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what is this going to be equal to? We're saying that this is a solution to Ax equal to b. So this is going to be equal to b. And of course, any particular solution to this, when you multiply it by a, is also going to be equal to b. So it's going to be b minus b. So that's going to be equal to the 0 vector. Or another way to think about it is, the vector x minus our particular solution is a solution to a times x is equal to 0."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And of course, any particular solution to this, when you multiply it by a, is also going to be equal to b. So it's going to be b minus b. So that's going to be equal to the 0 vector. Or another way to think about it is, the vector x minus our particular solution is a solution to a times x is equal to 0. Think about this. If you take this in parentheses right here, and you put it right there, when you multiply it times a, you get the 0 vector. We just did that."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to think about it is, the vector x minus our particular solution is a solution to a times x is equal to 0. Think about this. If you take this in parentheses right here, and you put it right there, when you multiply it times a, you get the 0 vector. We just did that. You get the 0 vector, because when you multiply each of these guys by a, you get b, and then you get b minus b. And so then you get 0. So you can say that x minus, so our any solution x minus the particular solution of x is a member of our null space."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We just did that. You get the 0 vector, because when you multiply each of these guys by a, you get b, and then you get b minus b. And so then you get 0. So you can say that x minus, so our any solution x minus the particular solution of x is a member of our null space. By definition, your null space is all of the x's that satisfy this equation. And let's say, so since it's a member of our null space, we can say that it is equal to, so our any solution minus our particular solution is equal to some member of our null space. We could say that it's equal to a homogeneous solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you can say that x minus, so our any solution x minus the particular solution of x is a member of our null space. By definition, your null space is all of the x's that satisfy this equation. And let's say, so since it's a member of our null space, we can say that it is equal to, so our any solution minus our particular solution is equal to some member of our null space. We could say that it's equal to a homogeneous solution. There might be more than one. A homogeneous solution. Now, if we just add our particular solution to both sides of this, we get that any solution, remember we assume that x is any solution to this, that any solution is equal to our homogeneous solution, is equal to a homogeneous solution plus our particular solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could say that it's equal to a homogeneous solution. There might be more than one. A homogeneous solution. Now, if we just add our particular solution to both sides of this, we get that any solution, remember we assume that x is any solution to this, that any solution is equal to our homogeneous solution, is equal to a homogeneous solution plus our particular solution. So we've proven it both ways. That this is a solution to our inhomogeneous equation, and that any solution to our inhomogeneous equation takes this form right here. Now, why am I so concerned with this?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, if we just add our particular solution to both sides of this, we get that any solution, remember we assume that x is any solution to this, that any solution is equal to our homogeneous solution, is equal to a homogeneous solution plus our particular solution. So we've proven it both ways. That this is a solution to our inhomogeneous equation, and that any solution to our inhomogeneous equation takes this form right here. Now, why am I so concerned with this? I've been kind of fixated on this inhomogeneous equation for some time. But we've been talking about the notion of a transformation being one-to-one. That was one of the two conditions for a transformation to be invertible."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, why am I so concerned with this? I've been kind of fixated on this inhomogeneous equation for some time. But we've been talking about the notion of a transformation being one-to-one. That was one of the two conditions for a transformation to be invertible. Now, to be one-to-one, so let me draw a transformation here. So let's say this is my domain, x, and this is my co-domain right here, y. And I have a transformation that maps from x to y."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That was one of the two conditions for a transformation to be invertible. Now, to be one-to-one, so let me draw a transformation here. So let's say this is my domain, x, and this is my co-domain right here, y. And I have a transformation that maps from x to y. In order for t to be one-to-one, so I'll write like this, one-to-one. In order for t to be one-to-one, that means for any b that you pick here, for any b that is a member of our co-domain, there's at most one solution to A times x. And I'm assuming that A is our transformation matrix, so we can write our transformation t as being equal to some matrix times our vector in our domain."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I have a transformation that maps from x to y. In order for t to be one-to-one, so I'll write like this, one-to-one. In order for t to be one-to-one, that means for any b that you pick here, for any b that is a member of our co-domain, there's at most one solution to A times x. And I'm assuming that A is our transformation matrix, so we can write our transformation t as being equal to some matrix times our vector in our domain. So this would be Ax if this is x right here. So t would map from that to that right there. So in order for our transformation to be one-to-one, that means you pick any b here, there has to be at most one solution to Ax is equal to b."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I'm assuming that A is our transformation matrix, so we can write our transformation t as being equal to some matrix times our vector in our domain. So this would be Ax if this is x right here. So t would map from that to that right there. So in order for our transformation to be one-to-one, that means you pick any b here, there has to be at most one solution to Ax is equal to b. Or another way to say that is that there is at most one guy that maps into that element of our co-domain. There might be none, so there could be no solution to this, but there has to be at most one solution. Now, we just said that any solution to an inhomogeneous so we just said that any, let me write it in blue, any solution takes the form, if there is a solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So in order for our transformation to be one-to-one, that means you pick any b here, there has to be at most one solution to Ax is equal to b. Or another way to say that is that there is at most one guy that maps into that element of our co-domain. There might be none, so there could be no solution to this, but there has to be at most one solution. Now, we just said that any solution to an inhomogeneous so we just said that any, let me write it in blue, any solution takes the form, if there is a solution. So if there isn't a solution, that's fine. That'll still satisfy one-to-one. But if there is a solution, any solution is going to take the form x particular plus a member of your null space."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we just said that any solution to an inhomogeneous so we just said that any, let me write it in blue, any solution takes the form, if there is a solution. So if there isn't a solution, that's fine. That'll still satisfy one-to-one. But if there is a solution, any solution is going to take the form x particular plus a member of your null space. With this guy right here as a member of the null space. This thing right here just applies to that guy right here. Any solution, if they exist."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But if there is a solution, any solution is going to take the form x particular plus a member of your null space. With this guy right here as a member of the null space. This thing right here just applies to that guy right here. Any solution, if they exist. If there are no solutions, that's fine. You can still be one-to-one. But if you do have a solution, you can have at most one person that maps to it, and any solution will take this form."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Any solution, if they exist. If there are no solutions, that's fine. You can still be one-to-one. But if you do have a solution, you can have at most one person that maps to it, and any solution will take this form. I just showed you that. Now, in order to be one-to-one, this can only be one solution. The solution set can only be one solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But if you do have a solution, you can have at most one person that maps to it, and any solution will take this form. I just showed you that. Now, in order to be one-to-one, this can only be one solution. The solution set can only be one solution. This can only be one solution. We can only have one solution here, right? Which means, what does that mean?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The solution set can only be one solution. This can only be one solution. We can only have one solution here, right? Which means, what does that mean? That means that this guy right here cannot be more than one vector. It just has to be one vector. There's only one particular solution right there, but this guy right here has to be, well, for any solution set, depending on how you define it, there's only one particular vector there."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which means, what does that mean? That means that this guy right here cannot be more than one vector. It just has to be one vector. There's only one particular solution right there, but this guy right here has to be, well, for any solution set, depending on how you define it, there's only one particular vector there. But this guy, the only way that you're only going to have one solution is if your null space is trivial, if it only contains the zero vector. Your null space will always, at minimum, contain the zero vector, and the last video, I think I just off the cuff said, oh, your null space has to be empty. But no, your null space will always, by definition, by the fact that it is a subspace, it will always contain a zero vector."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "There's only one particular solution right there, but this guy right here has to be, well, for any solution set, depending on how you define it, there's only one particular vector there. But this guy, the only way that you're only going to have one solution is if your null space is trivial, if it only contains the zero vector. Your null space will always, at minimum, contain the zero vector, and the last video, I think I just off the cuff said, oh, your null space has to be empty. But no, your null space will always, by definition, by the fact that it is a subspace, it will always contain a zero vector. You can always multiply a times zero to get zero. So your null space will always contain that. But in order to have only one solution, your null space can only have the zero vector, so that this can only be zero."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But no, your null space will always, by definition, by the fact that it is a subspace, it will always contain a zero vector. You can always multiply a times zero to get zero. So your null space will always contain that. But in order to have only one solution, your null space can only have the zero vector, so that this can only be zero. And so that your only solution is going to be the particular solution that you found, depending on how you got there. But it's only going to be your particular solution. So let me put this this way."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But in order to have only one solution, your null space can only have the zero vector, so that this can only be zero. And so that your only solution is going to be the particular solution that you found, depending on how you got there. But it's only going to be your particular solution. So let me put this this way. So one to one, in order to be one to one, your null space of your transformation matrix has to be trivial. It has to contain only the zero vector. Now, we've covered this many, many videos ago."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me put this this way. So one to one, in order to be one to one, your null space of your transformation matrix has to be trivial. It has to contain only the zero vector. Now, we've covered this many, many videos ago. What does it mean if your null space only contains the trivial vector? Let me make this clear. So if your transformation vector looks like this, a1, a2, a2, all the way to an."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we've covered this many, many videos ago. What does it mean if your null space only contains the trivial vector? Let me make this clear. So if your transformation vector looks like this, a1, a2, a2, all the way to an. And you're multiplying it times x1, x2, all the way to xn. And the null space is all of the x's that satisfy this equation. Zero, and you're going to have m zeros right there."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if your transformation vector looks like this, a1, a2, a2, all the way to an. And you're multiplying it times x1, x2, all the way to xn. And the null space is all of the x's that satisfy this equation. Zero, and you're going to have m zeros right there. So if your null space is trivial, and we're saying that that is a condition for you to be one to one, for your transformation to be one to one, the transformation that's specified by this matrix. If your null space is trivial, what does that mean? That means that the only solution to another way of writing this is x1 times a1 plus x2 times a2, all the way to xn times an is equal to the zero vector."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Zero, and you're going to have m zeros right there. So if your null space is trivial, and we're saying that that is a condition for you to be one to one, for your transformation to be one to one, the transformation that's specified by this matrix. If your null space is trivial, what does that mean? That means that the only solution to another way of writing this is x1 times a1 plus x2 times a2, all the way to xn times an is equal to the zero vector. These are equivalent statements right here. I just multiplied each of these terms times these respective column vectors. These are the same thing."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That means that the only solution to another way of writing this is x1 times a1 plus x2 times a2, all the way to xn times an is equal to the zero vector. These are equivalent statements right here. I just multiplied each of these terms times these respective column vectors. These are the same thing. Now, if you say that your null space has to be equal to zero, you're saying that the only solution to this equation right here, the only scalars that satisfy this equation, oh sorry, these aren't, let me actually, because I wrote the scalars as vectors. So this guy right here, this statement right here is equivalent to x1 times a1 plus x2 times a2 plus all the way to xn times an is equal to the zero vector, where the x1's through xn's are scalars. Now, if we say the null space is zero, we're saying the only way that this is satisfied is if your x1 all the way to xn is equal to zero."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are the same thing. Now, if you say that your null space has to be equal to zero, you're saying that the only solution to this equation right here, the only scalars that satisfy this equation, oh sorry, these aren't, let me actually, because I wrote the scalars as vectors. So this guy right here, this statement right here is equivalent to x1 times a1 plus x2 times a2 plus all the way to xn times an is equal to the zero vector, where the x1's through xn's are scalars. Now, if we say the null space is zero, we're saying the only way that this is satisfied is if your x1 all the way to xn is equal to zero. And this means, this is our definition actually, of linear independence. That means that a1, so the null space being zero also means that your column vectors of a, let me write it this way, it also means that a1, a2, all the way through an are linearly independent. Now what does that mean?"}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, if we say the null space is zero, we're saying the only way that this is satisfied is if your x1 all the way to xn is equal to zero. And this means, this is our definition actually, of linear independence. That means that a1, so the null space being zero also means that your column vectors of a, let me write it this way, it also means that a1, a2, all the way through an are linearly independent. Now what does that mean? If all of these guys are linearly independent, what is going to be the basis for your column space? Remember, the column space is the span. The column space of a is equal to the span of a1, a2, all the way to an."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what does that mean? If all of these guys are linearly independent, what is going to be the basis for your column space? Remember, the column space is the span. The column space of a is equal to the span of a1, a2, all the way to an. Well, we just said if we're dealing with a one-to-one, or one of the conditions, or the condition to be one-to-one is that your null space has to be zero, or only contain the zero vector. If your null space contains a zero vector, then all of your columns are linearly independent. If all of these columns span your column space and they're linearly independent, then they form a basis."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The column space of a is equal to the span of a1, a2, all the way to an. Well, we just said if we're dealing with a one-to-one, or one of the conditions, or the condition to be one-to-one is that your null space has to be zero, or only contain the zero vector. If your null space contains a zero vector, then all of your columns are linearly independent. If all of these columns span your column space and they're linearly independent, then they form a basis. So that means that a1, a2, all the way to an are a basis for our column space. And then that means if all of our column vectors here are linearly independent, they obviously span our column space by definition, and they all are linearly independent, they form the basis. So the dimension of our basis, so the dimension of our column space, that's essentially the number of vectors you need to form the basis, is going to be equal to n. We have n columns."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If all of these columns span your column space and they're linearly independent, then they form a basis. So that means that a1, a2, all the way to an are a basis for our column space. And then that means if all of our column vectors here are linearly independent, they obviously span our column space by definition, and they all are linearly independent, they form the basis. So the dimension of our basis, so the dimension of our column space, that's essentially the number of vectors you need to form the basis, is going to be equal to n. We have n columns. So it's going to be equal to n. Or another way to say it is that the rank of your matrix is going to be equal to n. So now we have a condition for something to be one-to-one. Something is going to be one-to-one if and only if the rank of your matrix is equal to n. And you could go both ways. If you assume something is one-to-one, then that means that its null space here has to only have the zero vector, so it only has one solution."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the dimension of our basis, so the dimension of our column space, that's essentially the number of vectors you need to form the basis, is going to be equal to n. We have n columns. So it's going to be equal to n. Or another way to say it is that the rank of your matrix is going to be equal to n. So now we have a condition for something to be one-to-one. Something is going to be one-to-one if and only if the rank of your matrix is equal to n. And you could go both ways. If you assume something is one-to-one, then that means that its null space here has to only have the zero vector, so it only has one solution. If its null space only has the zero vector, then that means its columns are linearly independent, which means that they all are part of the basis, which means that you have n basis vectors or you have a rank of n. Go the other way. If you have a rank of n, that means that all of these guys are linearly independent. If all of these guys are linearly independent, then the null space is just the zero vector."}, {"video_title": "Matrix condition for one-to-one trans Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you assume something is one-to-one, then that means that its null space here has to only have the zero vector, so it only has one solution. If its null space only has the zero vector, then that means its columns are linearly independent, which means that they all are part of the basis, which means that you have n basis vectors or you have a rank of n. Go the other way. If you have a rank of n, that means that all of these guys are linearly independent. If all of these guys are linearly independent, then the null space is just the zero vector. And then if the null space is just the zero vector, this part of your solution disappears, and then you're only left with one solution. So you're one-to-one. So you're one-to-one if and only if the rank of your transformation matrix is equal to n."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw some plane in R3. Maybe it looks something like that. That is my subspace. I think that's good enough. Let me see if I can draw it even a little bit better than that. There you go. So that is my plane in R3."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "I think that's good enough. Let me see if I can draw it even a little bit better than that. There you go. So that is my plane in R3. This is V. This is a subspace. And let's say that I have some other vector x, any vector in R3. So my vector x looks like this."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So that is my plane in R3. This is V. This is a subspace. And let's say that I have some other vector x, any vector in R3. So my vector x looks like this. That is my vector x. Now, what I want to show you in this video is that the projection of x onto our subspace, and let's say that this is our 0 vector right there, I want to show that the projection of x onto our subspace is the closest vector in our subspace 2x. So let me draw that out and maybe it'll make a little more sense."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So my vector x looks like this. That is my vector x. Now, what I want to show you in this video is that the projection of x onto our subspace, and let's say that this is our 0 vector right there, I want to show that the projection of x onto our subspace is the closest vector in our subspace 2x. So let me draw that out and maybe it'll make a little more sense. So the projection of x onto the subspace will look something like this. It will look something like that. That right there, that green vector right there, is the projection of the vector x onto our subspace V. That's our vector x."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw that out and maybe it'll make a little more sense. So the projection of x onto the subspace will look something like this. It will look something like that. That right there, that green vector right there, is the projection of the vector x onto our subspace V. That's our vector x. Now, let's take some arbitrary other vector in our subspace. Let's just take this one. This is just some other arbitrary vector in our subspace."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "That right there, that green vector right there, is the projection of the vector x onto our subspace V. That's our vector x. Now, let's take some arbitrary other vector in our subspace. Let's just take this one. This is just some other arbitrary vector in our subspace. Let me draw it a little bit differently. Let me draw it like that. Let's call that vector, let's say that's vector V. That's clearly another vector in our subspace."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "This is just some other arbitrary vector in our subspace. Let me draw it a little bit differently. Let me draw it like that. Let's call that vector, let's say that's vector V. That's clearly another vector in our subspace. It lies on that plane. What I want to show you is that the distance from x to our projection of x onto V is shorter than the distance from x to any other vector. And obviously the way I've drawn it, it looks pretty clear that this line is shorter than that line."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that vector, let's say that's vector V. That's clearly another vector in our subspace. It lies on that plane. What I want to show you is that the distance from x to our projection of x onto V is shorter than the distance from x to any other vector. And obviously the way I've drawn it, it looks pretty clear that this line is shorter than that line. But that was just a particular choice that I picked. Let's prove it that's general. So what I want to prove is that the distance between x and its projection onto the subspace, and the way we can get that is essentially just take the length of the vector of x minus the projection of x onto my subspace."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "And obviously the way I've drawn it, it looks pretty clear that this line is shorter than that line. But that was just a particular choice that I picked. Let's prove it that's general. So what I want to prove is that the distance between x and its projection onto the subspace, and the way we can get that is essentially just take the length of the vector of x minus the projection of x onto my subspace. This length right here is this length right here. And actually, let me call this vector. So x minus the projection of x onto V, that's going to be this vector right there."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So what I want to prove is that the distance between x and its projection onto the subspace, and the way we can get that is essentially just take the length of the vector of x minus the projection of x onto my subspace. This length right here is this length right here. And actually, let me call this vector. So x minus the projection of x onto V, that's going to be this vector right there. Let me do it in a different color. Don't want to reuse colors too often. That's going to be that vector right there."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So x minus the projection of x onto V, that's going to be this vector right there. Let me do it in a different color. Don't want to reuse colors too often. That's going to be that vector right there. We could call that vector A. It's clearly in the orthogonal complement of V, because it's orthogonal to this guy. I mean, that's the definition of a projection actually."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be that vector right there. We could call that vector A. It's clearly in the orthogonal complement of V, because it's orthogonal to this guy. I mean, that's the definition of a projection actually. So this is equal to A. My claim, what I want to show you is that this distance A is shorter than any distance here. Is less than or equal to the distance between x and V, where V is any member."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "I mean, that's the definition of a projection actually. So this is equal to A. My claim, what I want to show you is that this distance A is shorter than any distance here. Is less than or equal to the distance between x and V, where V is any member. So that's this distance right here. This vector right here. That distance right, let me draw this vector."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Is less than or equal to the distance between x and V, where V is any member. So that's this distance right here. This vector right here. That distance right, let me draw this vector. The vector x minus V looks like this. It looks like that. That is the vector x minus V, right?"}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "That distance right, let me draw this vector. The vector x minus V looks like this. It looks like that. That is the vector x minus V, right? If you take V plus x minus V, you're going to go to x. So what I want to show is that this distance, the length of A of the difference between x and its projection, is always going to be less than the distance between x and any other vector in the subspace. So that is x minus V. So let's see if we can prove that."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "That is the vector x minus V, right? If you take V plus x minus V, you're going to go to x. So what I want to show is that this distance, the length of A of the difference between x and its projection, is always going to be less than the distance between x and any other vector in the subspace. So that is x minus V. So let's see if we can prove that. So let's figure out, let's take the square of this distance. So the square of x, actually even before we, let me do that, let me write it that way. So let's say x, so we want to concern ourselves with the square of the distance of x minus V. Where x is some vector in R3, and V is some vector in R3 that's also a member of our subspace."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So that is x minus V. So let's see if we can prove that. So let's figure out, let's take the square of this distance. So the square of x, actually even before we, let me do that, let me write it that way. So let's say x, so we want to concern ourselves with the square of the distance of x minus V. Where x is some vector in R3, and V is some vector in R3 that's also a member of our subspace. It sits on this plane. So what's the square of this going to be? Well, x minus V is equal to this vector."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So let's say x, so we want to concern ourselves with the square of the distance of x minus V. Where x is some vector in R3, and V is some vector in R3 that's also a member of our subspace. It sits on this plane. So what's the square of this going to be? Well, x minus V is equal to this vector. Let me draw a new vector here. It is equal to this. Let me draw it in this yellow."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Well, x minus V is equal to this vector. Let me draw a new vector here. It is equal to this. Let me draw it in this yellow. It's equal to this vector. It's equal to this yellow vector plus a. x minus V is this magenta vector that starts here and goes there. Clearly equal to this yellow vector plus this orange vector, so let me call that yellow vector b."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw it in this yellow. It's equal to this vector. It's equal to this yellow vector plus a. x minus V is this magenta vector that starts here and goes there. Clearly equal to this yellow vector plus this orange vector, so let me call that yellow vector b. Now, what is b equal to? Well, b is going to be equal to the vector b is equal to this vector, this green vector, which is the projection of x onto V minus this purple vector, minus this mauve vector, I guess. Minus V. That's what b is."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Clearly equal to this yellow vector plus this orange vector, so let me call that yellow vector b. Now, what is b equal to? Well, b is going to be equal to the vector b is equal to this vector, this green vector, which is the projection of x onto V minus this purple vector, minus this mauve vector, I guess. Minus V. That's what b is. So we could write x minus V as being equal to the sum of the vector b plus the vector a. So x minus V is equal to b plus a. And if we're taking the length of x minus V squared, that's the same thing as the length of b plus a squared."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Minus V. That's what b is. So we could write x minus V as being equal to the sum of the vector b plus the vector a. So x minus V is equal to b plus a. And if we're taking the length of x minus V squared, that's the same thing as the length of b plus a squared. And that's just equal to b plus a dotted with b plus a, which is the same thing as b dot b. Let me write that a little bit neater. Let me write it down here."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "And if we're taking the length of x minus V squared, that's the same thing as the length of b plus a squared. And that's just equal to b plus a dotted with b plus a, which is the same thing as b dot b. Let me write that a little bit neater. Let me write it down here. This is going to be equal to, I'll switch colors, b dot b plus b dot a plus a dot b. So plus 2 times a dot b plus a dot a. Now, a and b are clearly orthogonal."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it down here. This is going to be equal to, I'll switch colors, b dot b plus b dot a plus a dot b. So plus 2 times a dot b plus a dot a. Now, a and b are clearly orthogonal. b is the difference between two vectors in our subspace. The subspace is closed under addition and subtraction. So b is a member of our subspace."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Now, a and b are clearly orthogonal. b is the difference between two vectors in our subspace. The subspace is closed under addition and subtraction. So b is a member of our subspace. a is orthogonal to everything in our subspace. Die definition. So since a is clearly orthogonal to b, a by definition is going to be in the orthogonal complement of the subspace, this is going to be 0."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So b is a member of our subspace. a is orthogonal to everything in our subspace. Die definition. So since a is clearly orthogonal to b, a by definition is going to be in the orthogonal complement of the subspace, this is going to be 0. And then this right here will simplify to the length of b squared. And then this right here is going to be plus the length of a squared. So we get the distance between x and some arbitrary member of our subspace squared is equal to the length of b right here plus the length of a squared."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So since a is clearly orthogonal to b, a by definition is going to be in the orthogonal complement of the subspace, this is going to be 0. And then this right here will simplify to the length of b squared. And then this right here is going to be plus the length of a squared. So we get the distance between x and some arbitrary member of our subspace squared is equal to the length of b right here plus the length of a squared. Now, a was the distance between our vector x and our projection. That's what the definition of a was. a was the distance between our vector x and our projection."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So we get the distance between x and some arbitrary member of our subspace squared is equal to the length of b right here plus the length of a squared. Now, a was the distance between our vector x and our projection. That's what the definition of a was. a was the distance between our vector x and our projection. Now, this number right here is going to be at least 0 or positive. So this right here is definitely going to be greater than or equal to a squared. Or another way to say it is that the distance between x minus v squared is definitely going to be greater than or equal to the distance of a squared."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "a was the distance between our vector x and our projection. Now, this number right here is going to be at least 0 or positive. So this right here is definitely going to be greater than or equal to a squared. Or another way to say it is that the distance between x minus v squared is definitely going to be greater than or equal to the distance of a squared. Or the distance between x minus v, this is still going to be a positive quantity. Length is always going to be positive. Is greater than or equal to the length of vector a."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it is that the distance between x minus v squared is definitely going to be greater than or equal to the distance of a squared. Or the distance between x minus v, this is still going to be a positive quantity. Length is always going to be positive. Is greater than or equal to the length of vector a. Or what's that length of vector a? a is just this thing right here. So let's write our result."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "Is greater than or equal to the length of vector a. Or what's that length of vector a? a is just this thing right here. So let's write our result. The length of the vector x minus v, or the distance between x and some arbitrary member of our subspace, is always going to be greater than or equal to the length of a. Which is just the distance between x and the projection of x onto our subspace. So there you have it."}, {"video_title": "Projection is closest vector in subspace Linear Algebra Khan Academy.mp3", "Sentence": "So let's write our result. The length of the vector x minus v, or the distance between x and some arbitrary member of our subspace, is always going to be greater than or equal to the length of a. Which is just the distance between x and the projection of x onto our subspace. So there you have it. We've shown, and the original graph kind of hinted at it, that the projection of x onto v is the closest vector in our subspace to x. It's closer than any other vector in v to our arbitrary vector in R3x. And we've pruned it right there."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's just start off, so this is a plane, I'm drawing part of it, obviously it keeps going in every direction. So let's say that that is our plane, and let's say that this is a normal vector to the plane. So that is our normal vector to the plane, and it's given by ai plus bj plus ck. So that is our normal vector to the plane, and let's say that we have, so it's perpendicular to every other vector that's on the plane. And let's say we have some point on the plane. We have some point, it's the point x sub p, I'll say p for plane, so it's a point on the plane, xp, yp, zp. If we pick the origin, so let's say that our axes are here."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that is our normal vector to the plane, and let's say that we have, so it's perpendicular to every other vector that's on the plane. And let's say we have some point on the plane. We have some point, it's the point x sub p, I'll say p for plane, so it's a point on the plane, xp, yp, zp. If we pick the origin, so let's say that our axes are here. So let me draw our coordinate axes. So let's say our coordinate axes look like that. This is our z-axis, this is, let's say that's our y-axis, and let's say that this is our x-axis."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we pick the origin, so let's say that our axes are here. So let me draw our coordinate axes. So let's say our coordinate axes look like that. This is our z-axis, this is, let's say that's our y-axis, and let's say that this is our x-axis. Let's say this is our x-axis coming out like this. This is our x-axis. You could specify this as a position vector."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is our z-axis, this is, let's say that's our y-axis, and let's say that this is our x-axis. Let's say this is our x-axis coming out like this. This is our x-axis. You could specify this as a position vector. There is a position vector, let me draw it like this. Let me draw it like this, then it would be behind the plane right over there. You have a position vector."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could specify this as a position vector. There is a position vector, let me draw it like this. Let me draw it like this, then it would be behind the plane right over there. You have a position vector. That position vector would be xpi plus ypj plus zpk. It specifies this coordinate right here that sits on the plane. Let me just call that something."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You have a position vector. That position vector would be xpi plus ypj plus zpk. It specifies this coordinate right here that sits on the plane. Let me just call that something. Let me call that position vector, I don't know, let me call that p1. So this is a point on the plane. So it's p, it is p1, and it is equal to this."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me just call that something. Let me call that position vector, I don't know, let me call that p1. So this is a point on the plane. So it's p, it is p1, and it is equal to this. Now, we could take another point on the plane. Let's say we're taking, this is a particular point on the plane. Let's just say any other point on the plane, xyz, but we're saying that xyz sits on the plane."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's p, it is p1, and it is equal to this. Now, we could take another point on the plane. Let's say we're taking, this is a particular point on the plane. Let's just say any other point on the plane, xyz, but we're saying that xyz sits on the plane. So let's say we take this point right over here, xyz. That clearly, same logic, can be specified by another position vector. We could have a position vector that looks like this, and dotted line, it's going under the plane right over here."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's just say any other point on the plane, xyz, but we're saying that xyz sits on the plane. So let's say we take this point right over here, xyz. That clearly, same logic, can be specified by another position vector. We could have a position vector that looks like this, and dotted line, it's going under the plane right over here. And this position vector, let me just call it p, instead of that particular, that p1. This would just be xi plus yj plus zk. Now, the whole reason why I did this setup is because I want to find, given some particular point that I know is on the plane, and any other xyz that is on the plane, I can find, I can construct a vector that is definitely on the plane."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We could have a position vector that looks like this, and dotted line, it's going under the plane right over here. And this position vector, let me just call it p, instead of that particular, that p1. This would just be xi plus yj plus zk. Now, the whole reason why I did this setup is because I want to find, given some particular point that I know is on the plane, and any other xyz that is on the plane, I can find, I can construct a vector that is definitely on the plane. And we've done this before when we try to figure out what the equations of a plane are. A vector that's definitely on the plane is going to be the difference of these two vectors, and I'll do that in blue. So if you take the yellow vector minus the green vector, or you take this position, or you take, you will get the vector, if you view it that way, that connects this point and that point, although you can shift the vector."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, the whole reason why I did this setup is because I want to find, given some particular point that I know is on the plane, and any other xyz that is on the plane, I can find, I can construct a vector that is definitely on the plane. And we've done this before when we try to figure out what the equations of a plane are. A vector that's definitely on the plane is going to be the difference of these two vectors, and I'll do that in blue. So if you take the yellow vector minus the green vector, or you take this position, or you take, you will get the vector, if you view it that way, that connects this point and that point, although you can shift the vector. But you'll get a vector that definitely lies along the plane, even if you, so if you start one of these points, it'll definitely lie along the plane. So the vector will look like this, and it would be lying along our plane. So this vector lies along our plane."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you take the yellow vector minus the green vector, or you take this position, or you take, you will get the vector, if you view it that way, that connects this point and that point, although you can shift the vector. But you'll get a vector that definitely lies along the plane, even if you, so if you start one of these points, it'll definitely lie along the plane. So the vector will look like this, and it would be lying along our plane. So this vector lies along our plane. That vector is P minus P1. This is the vector P minus P1. It's this position vector minus that position vector gives you this one."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this vector lies along our plane. That vector is P minus P1. This is the vector P minus P1. It's this position vector minus that position vector gives you this one. Or another way to view it is this green position vector plus this blue vector that sits on the plane will clearly equal this yellow vector, right? Heads to tails, it clearly equals it. And the whole reason why I did that is we can now take the dot product between this blue thing and this magenta thing, and we've done it before, and they have to be equal to zero because this lies on the plane, this is perpendicular to everything that sits on the plane, and it equals zero, and so we will get the equation for the plane."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's this position vector minus that position vector gives you this one. Or another way to view it is this green position vector plus this blue vector that sits on the plane will clearly equal this yellow vector, right? Heads to tails, it clearly equals it. And the whole reason why I did that is we can now take the dot product between this blue thing and this magenta thing, and we've done it before, and they have to be equal to zero because this lies on the plane, this is perpendicular to everything that sits on the plane, and it equals zero, and so we will get the equation for the plane. But before I do that, let me make sure we know what the components of this blue vector are. So P minus P1, that's the blue vector. You're just going to subtract each of the components."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And the whole reason why I did that is we can now take the dot product between this blue thing and this magenta thing, and we've done it before, and they have to be equal to zero because this lies on the plane, this is perpendicular to everything that sits on the plane, and it equals zero, and so we will get the equation for the plane. But before I do that, let me make sure we know what the components of this blue vector are. So P minus P1, that's the blue vector. You're just going to subtract each of the components. So it's going to be X minus XPI plus Y minus YPJ plus Z minus ZPK. And we just said this is in the plane, and the normal vector is normal to the plane. You take their dot product, it's going to be equal to zero."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're just going to subtract each of the components. So it's going to be X minus XPI plus Y minus YPJ plus Z minus ZPK. And we just said this is in the plane, and the normal vector is normal to the plane. You take their dot product, it's going to be equal to zero. So N dot, this vector, is going to be equal to zero, but it's also equal to this A times this expression. I'll do it right over here so these find some good colors. So A times that, which is AX minus AXP plus B times that, so that is plus BY minus BYP, and then it's going to be plus that times that."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You take their dot product, it's going to be equal to zero. So N dot, this vector, is going to be equal to zero, but it's also equal to this A times this expression. I'll do it right over here so these find some good colors. So A times that, which is AX minus AXP plus B times that, so that is plus BY minus BYP, and then it's going to be plus that times that. So that's plus CZ minus CZP, and all of this is equal to zero. Now what I'm going to do is I'm going to rewrite this, so we have all of these terms. I'm looking for the right color."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So A times that, which is AX minus AXP plus B times that, so that is plus BY minus BYP, and then it's going to be plus that times that. So that's plus CZ minus CZP, and all of this is equal to zero. Now what I'm going to do is I'm going to rewrite this, so we have all of these terms. I'm looking for the right color. We have all of the X terms, AX. Remember, any X that's on the plane will satisfy this. So AX, BY, and CZ."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm looking for the right color. We have all of the X terms, AX. Remember, any X that's on the plane will satisfy this. So AX, BY, and CZ. Let me leave that on the right-hand side. So we have AX plus BY plus CZ is equal to, and what I want to do is I'm going to subtract each of these from both sides. Another way is I'm going to move them all over."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So AX, BY, and CZ. Let me leave that on the right-hand side. So we have AX plus BY plus CZ is equal to, and what I want to do is I'm going to subtract each of these from both sides. Another way is I'm going to move them all over. Let me not do too many things. I'm going to move them over to the left-hand side. So I'm going to add positive AXP to both sides."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Another way is I'm going to move them all over. Let me not do too many things. I'm going to move them over to the left-hand side. So I'm going to add positive AXP to both sides. That's the equivalent of subtracting negative AXP. So this is going to be positive AXP, and then we're going to have positive BYP plus, do that same green, plus BYP, and then finally plus CZP. Plus CZP is going to be equal to that."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to add positive AXP to both sides. That's the equivalent of subtracting negative AXP. So this is going to be positive AXP, and then we're going to have positive BYP plus, do that same green, plus BYP, and then finally plus CZP. Plus CZP is going to be equal to that. Now the whole reason why I did this, and I've done this in previous videos where we're trying to find the formula or we're trying to find the equation of a plane, is now we said, hey, if you have a normal vector and if you're given a point on the plane, where in this case it's XP, YP, ZP, we now have a very quick way of figuring out the equation. But I want to go the other way. I want you to be able to, if I were to give you a equation for a plane, where I were to say AX plus BY plus CZ is equal to D, so this is the general equation for a plane, if I were to give you this, I want you to be able to figure out the normal vector very quickly."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Plus CZP is going to be equal to that. Now the whole reason why I did this, and I've done this in previous videos where we're trying to find the formula or we're trying to find the equation of a plane, is now we said, hey, if you have a normal vector and if you're given a point on the plane, where in this case it's XP, YP, ZP, we now have a very quick way of figuring out the equation. But I want to go the other way. I want you to be able to, if I were to give you a equation for a plane, where I were to say AX plus BY plus CZ is equal to D, so this is the general equation for a plane, if I were to give you this, I want you to be able to figure out the normal vector very quickly. So how could you do that? Well, this AX plus BY plus CZ is completely analogous to this part right up over here. Let me rewrite this over here so it becomes clear."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I want you to be able to, if I were to give you a equation for a plane, where I were to say AX plus BY plus CZ is equal to D, so this is the general equation for a plane, if I were to give you this, I want you to be able to figure out the normal vector very quickly. So how could you do that? Well, this AX plus BY plus CZ is completely analogous to this part right up over here. Let me rewrite this over here so it becomes clear. This part is AX plus BY plus CZ is equal to all of this stuff on the left-hand side. So let me copy and paste it. So I just essentially flipped this expression."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me rewrite this over here so it becomes clear. This part is AX plus BY plus CZ is equal to all of this stuff on the left-hand side. So let me copy and paste it. So I just essentially flipped this expression. But now you see, all of this, this A has to be this A, this B has to be this B, this C has to be this thing, and then the D is all of this, and this is just going to be a number. This is just going to be a number, assuming that you knew what the normal vector is, what your A, Bs, and Cs are, and you know a particular value. So this is what D is."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I just essentially flipped this expression. But now you see, all of this, this A has to be this A, this B has to be this B, this C has to be this thing, and then the D is all of this, and this is just going to be a number. This is just going to be a number, assuming that you knew what the normal vector is, what your A, Bs, and Cs are, and you know a particular value. So this is what D is. So this is how you can get the equation for a plane. Now, if I were to give you an equation for a plane, what is the normal vector? Well, we just saw the normal vector, this A corresponds to that A, this B corresponds to that B, that C corresponds to that C. The normal vector to this plane we started off with has the components A, B, and C. So if you're given an equation for a plane here, the normal vector to this plane right over here is going to be Ai plus Bj plus Ck."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is what D is. So this is how you can get the equation for a plane. Now, if I were to give you an equation for a plane, what is the normal vector? Well, we just saw the normal vector, this A corresponds to that A, this B corresponds to that B, that C corresponds to that C. The normal vector to this plane we started off with has the components A, B, and C. So if you're given an equation for a plane here, the normal vector to this plane right over here is going to be Ai plus Bj plus Ck. So it's a very easy thing to do. If I were to give you the equation of a plane, let me give you a particular example. If I were to tell you that I have some plane in three dimensions, let's say it's negative 3, although it'll work for more dimensions, let's say I have negative 3x plus square root of 2y, let me put it this way, minus, or let's say plus 7z is equal to pi."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, we just saw the normal vector, this A corresponds to that A, this B corresponds to that B, that C corresponds to that C. The normal vector to this plane we started off with has the components A, B, and C. So if you're given an equation for a plane here, the normal vector to this plane right over here is going to be Ai plus Bj plus Ck. So it's a very easy thing to do. If I were to give you the equation of a plane, let me give you a particular example. If I were to tell you that I have some plane in three dimensions, let's say it's negative 3, although it'll work for more dimensions, let's say I have negative 3x plus square root of 2y, let me put it this way, minus, or let's say plus 7z is equal to pi. So you have this crazy, I mean it's not crazy, it's just a plane in three dimensions. And I say, what is a normal vector to this plane? You literally can just pick out these coefficients and you say a normal vector to this plane is negative 3i plus square root of 2j plus 7k."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If I were to tell you that I have some plane in three dimensions, let's say it's negative 3, although it'll work for more dimensions, let's say I have negative 3x plus square root of 2y, let me put it this way, minus, or let's say plus 7z is equal to pi. So you have this crazy, I mean it's not crazy, it's just a plane in three dimensions. And I say, what is a normal vector to this plane? You literally can just pick out these coefficients and you say a normal vector to this plane is negative 3i plus square root of 2j plus 7k. And you could ignore the d part there. And the reason why you could ignore that is that'll just shift the plane, but it won't fundamentally change how the plane is tilted. So a normal vector to this normal vector will also be normal if this was e or if this was 100."}, {"video_title": "Normal vector from plane equation Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You literally can just pick out these coefficients and you say a normal vector to this plane is negative 3i plus square root of 2j plus 7k. And you could ignore the d part there. And the reason why you could ignore that is that'll just shift the plane, but it won't fundamentally change how the plane is tilted. So a normal vector to this normal vector will also be normal if this was e or if this was 100. It would be normal to all of those planes because all of those planes are just shifted, they all have the same inclination, so they would all kind of point in the same direction, and so their normal vectors would point in the same direction. So hopefully you found that vaguely useful. We'll now build on this to find the distance between any point in three dimensions and some plane, and what is the distance that we can get to that plane?"}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this will span a subspace of dimension k. Let's assume that each of these guys are members of Rn. So v1, v2, all the way to vk. They're all members of Rn. Now in the last video, we saw that we can define a change of basis matrix. And it's a fancy word, but all it means is a matrix that has these basis vectors as its columns. So v1, v2, all the way to vk as its columns. So we're going to have k columns."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now in the last video, we saw that we can define a change of basis matrix. And it's a fancy word, but all it means is a matrix that has these basis vectors as its columns. So v1, v2, all the way to vk as its columns. So we're going to have k columns. And we're going to have n rows, because each of these guys are members of Rn. So they're going to have n entries. So we're going to have n rows."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to have k columns. And we're going to have n rows, because each of these guys are members of Rn. So they're going to have n entries. So we're going to have n rows. So it's going to be an n by k matrix. And we saw in the last video that if I have some matrix A, if we have, sorry, if I have some vector A that is a member of Rn, then in assuming that A is in the span of B, I can represent A, I could say that A is equal to the change of basis matrix times the coordinates of A with respect to our basis. This is what we saw in the last video."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to have n rows. So it's going to be an n by k matrix. And we saw in the last video that if I have some matrix A, if we have, sorry, if I have some vector A that is a member of Rn, then in assuming that A is in the span of B, I can represent A, I could say that A is equal to the change of basis matrix times the coordinates of A with respect to our basis. This is what we saw in the last video. If I have the coordinates of A with respect to B, I can multiply it by the change of basis matrix, and I'll get my vector A in standard coordinates. Or if I have my vector A in standard coordinates, then I can solve for my vector A in coordinates with respect to B. We saw that in the last video."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is what we saw in the last video. If I have the coordinates of A with respect to B, I can multiply it by the change of basis matrix, and I'll get my vector A in standard coordinates. Or if I have my vector A in standard coordinates, then I can solve for my vector A in coordinates with respect to B. We saw that in the last video. Now let's take the special case. Let's assume that C is invertible. What does that mean?"}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "We saw that in the last video. Now let's take the special case. Let's assume that C is invertible. What does that mean? Or what does that tell us about C? Well, if C is invertible, it's two things. It means that C is a square matrix, or it has the same number of rows and columns."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "What does that mean? Or what does that tell us about C? Well, if C is invertible, it's two things. It means that C is a square matrix, or it has the same number of rows and columns. And that its rows or columns, you can pick either of them, have to be linearly independent. So linearly independent, let's just pick columns. Now the second statement is a bit redundant."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "It means that C is a square matrix, or it has the same number of rows and columns. And that its rows or columns, you can pick either of them, have to be linearly independent. So linearly independent, let's just pick columns. Now the second statement is a bit redundant. We know that C has linearly independent columns, because its columns are bases for a subspace. So basis by definition, all of the vectors have to be linearly independent. So we know this is a bit redundant."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now the second statement is a bit redundant. We know that C has linearly independent columns, because its columns are bases for a subspace. So basis by definition, all of the vectors have to be linearly independent. So we know this is a bit redundant. But what's interesting is if we know that C is invertible, C has to be square. And if all of these vectors are members of Rn, then k has to be equal to n. So C is square means that k is equal to n, or that we have n basis vectors. Now if that's the case, what is the span of B?"}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we know this is a bit redundant. But what's interesting is if we know that C is invertible, C has to be square. And if all of these vectors are members of Rn, then k has to be equal to n. So C is square means that k is equal to n, or that we have n basis vectors. Now if that's the case, what is the span of B? Think about it. We have n linearly independent vectors in Rn. So any time you have n linearly independent vectors in Rn, those guys are a basis for Rn."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now if that's the case, what is the span of B? Think about it. We have n linearly independent vectors in Rn. So any time you have n linearly independent vectors in Rn, those guys are a basis for Rn. Because any basis that has n entries, and they're all linearly independent, is going to be a basis for Rn. So then B is a basis for Rn. So if we know that C is invertible, we also know that you can get to any vector in Rn by some linear combination of your basis vectors right there."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So any time you have n linearly independent vectors in Rn, those guys are a basis for Rn. Because any basis that has n entries, and they're all linearly independent, is going to be a basis for Rn. So then B is a basis for Rn. So if we know that C is invertible, we also know that you can get to any vector in Rn by some linear combination of your basis vectors right there. So in the last video, we had to make sure that this guy was in the span of these vectors. But now we don't have to make sure. Because if C is invertible, then the span of B is going to be equal to Rn."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So if we know that C is invertible, we also know that you can get to any vector in Rn by some linear combination of your basis vectors right there. So in the last video, we had to make sure that this guy was in the span of these vectors. But now we don't have to make sure. Because if C is invertible, then the span of B is going to be equal to Rn. Or another way you could say it is if the span of B is equal to Rn, if we have n vectors here, if k was equal to n, then we know that the span of B would be equal to Rn. And so we'd have n vectors here, n linearly independent columns here, and it would be an n by n matrix with all of the columns linearly independent. So then C would be invertible."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Because if C is invertible, then the span of B is going to be equal to Rn. Or another way you could say it is if the span of B is equal to Rn, if we have n vectors here, if k was equal to n, then we know that the span of B would be equal to Rn. And so we'd have n vectors here, n linearly independent columns here, and it would be an n by n matrix with all of the columns linearly independent. So then C would be invertible. So we could write if and only if. So we could write it the other way. If the span of B is Rn, then C is invertible."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So then C would be invertible. So we could write if and only if. So we could write it the other way. If the span of B is Rn, then C is invertible. And that's useful. Because if either of these things are true, then we can rewrite the same equation. So let's say if we know this and we're looking for that, we can just multiply C times that."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "If the span of B is Rn, then C is invertible. And that's useful. Because if either of these things are true, then we can rewrite the same equation. So let's say if we know this and we're looking for that, we can just multiply C times that. Let's say we know this and we're looking for that. Before we had to do that augmented matrix and solve for it and whatnot. But if we know C is invertible, then one, we know that any vector here can be represented in the span of our basis."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's say if we know this and we're looking for that, we can just multiply C times that. Let's say we know this and we're looking for that. Before we had to do that augmented matrix and solve for it and whatnot. But if we know C is invertible, then one, we know that any vector here can be represented in the span of our basis. So any vector here can be represented as linear combinations of these guys. So you know that any vector can be represented in these coordinates or with coordinates with respect to our basis. And so we can just take, we can multiply both sides of this equation times C inverse."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "But if we know C is invertible, then one, we know that any vector here can be represented in the span of our basis. So any vector here can be represented as linear combinations of these guys. So you know that any vector can be represented in these coordinates or with coordinates with respect to our basis. And so we can just take, we can multiply both sides of this equation times C inverse. And what do you get? If you multiply, so it becomes C inverse C times our coordinates of A with respect to B is equal to C inverse times A. This is just the identity matrix right there."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And so we can just take, we can multiply both sides of this equation times C inverse. And what do you get? If you multiply, so it becomes C inverse C times our coordinates of A with respect to B is equal to C inverse times A. This is just the identity matrix right there. So we just get, another way of writing this is that the coordinates of A with respect to our basis B, which spans all of our n, is equal to C inverse times our vector A. Let's apply this a little bit. Let's apply this."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is just the identity matrix right there. So we just get, another way of writing this is that the coordinates of A with respect to our basis B, which spans all of our n, is equal to C inverse times our vector A. Let's apply this a little bit. Let's apply this. Let's use this information, what we've done in this video. Let's do some concrete examples. So let's say I have some basis."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's apply this. Let's use this information, what we've done in this video. Let's do some concrete examples. So let's say I have some basis. Let's say my basis is equal to, let me define two vectors. I'll do it this way. So let's say I have v1 is equal to the vector 1, 3."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have some basis. Let's say my basis is equal to, let me define two vectors. I'll do it this way. So let's say I have v1 is equal to the vector 1, 3. And let's say v2 is equal to the vector 2, 1. And I have a basis that is equal to the set of v1 and v2. Now I'll leave it for you to verify that these guys are linearly independent."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have v1 is equal to the vector 1, 3. And let's say v2 is equal to the vector 2, 1. And I have a basis that is equal to the set of v1 and v2. Now I'll leave it for you to verify that these guys are linearly independent. But if I have two linearly independent vectors in R2, then v is a basis for R2. And if we write the change of basis matrix, if we say C is equal to 1, 3, 2, 1, we know that C is invertible. And actually to show that C is invertible, we can actually just calculate its inverse."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now I'll leave it for you to verify that these guys are linearly independent. But if I have two linearly independent vectors in R2, then v is a basis for R2. And if we write the change of basis matrix, if we say C is equal to 1, 3, 2, 1, we know that C is invertible. And actually to show that C is invertible, we can actually just calculate its inverse. So what's the determinant of C? The determinant of C is equal to 1 times 1 minus 2 times 3. So it's equal to minus 5."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And actually to show that C is invertible, we can actually just calculate its inverse. So what's the determinant of C? The determinant of C is equal to 1 times 1 minus 2 times 3. So it's equal to minus 5. That's the determinant of C. And so C inverse, we figured out a general formula for doing this for 2 by 2 matrices, is equal to 1 over the determinant of C. So 1 over 5, or it's 1 over minus 5, times you switch these two guys. So you switch the 1's. And you make these two guys negative."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to minus 5. That's the determinant of C. And so C inverse, we figured out a general formula for doing this for 2 by 2 matrices, is equal to 1 over the determinant of C. So 1 over 5, or it's 1 over minus 5, times you switch these two guys. So you switch the 1's. And you make these two guys negative. So minus 2 and then minus 3. And the very fact that this guy was the determinant of C was non-zero told us that this was invertible. But anyway, this is C inverse."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And you make these two guys negative. So minus 2 and then minus 3. And the very fact that this guy was the determinant of C was non-zero told us that this was invertible. But anyway, this is C inverse. So let's say that I have some vector A that is a member of R2, and let's say that A, I'm just going to pick some random numbers, A is equal to 7, 2. And I want to find out what the coordinates of A are with respect to my basis B. Well, we go to this situation."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "But anyway, this is C inverse. So let's say that I have some vector A that is a member of R2, and let's say that A, I'm just going to pick some random numbers, A is equal to 7, 2. And I want to find out what the coordinates of A are with respect to my basis B. Well, we go to this situation. We know what A is, so we just multiply A times C inverse to get this guy right here. To get the coordinates of A with respect to B. So let me write that down."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well, we go to this situation. We know what A is, so we just multiply A times C inverse to get this guy right here. To get the coordinates of A with respect to B. So let me write that down. So what is C? So C is that, C inverse is that. So we could write the coordinates of A with respect to B is equal to C inverse times the standard coordinates of A."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that down. So what is C? So C is that, C inverse is that. So we could write the coordinates of A with respect to B is equal to C inverse times the standard coordinates of A. Or this is the same thing. Let me put the actual numbers here. The coordinates of A with respect to B are going to be equal to C inverse, which is minus 1 5th, times 1 minus 3 minus 2 1, times A."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we could write the coordinates of A with respect to B is equal to C inverse times the standard coordinates of A. Or this is the same thing. Let me put the actual numbers here. The coordinates of A with respect to B are going to be equal to C inverse, which is minus 1 5th, times 1 minus 3 minus 2 1, times A. Times A. Times 7, 2. And what is this equal to?"}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "The coordinates of A with respect to B are going to be equal to C inverse, which is minus 1 5th, times 1 minus 3 minus 2 1, times A. Times A. Times 7, 2. And what is this equal to? This is equal to minus 1 5th, and then we're going to get 1 times 7 minus 2 times 2, so it's minus 4. So 7 minus 4 is 3. And then we're going to get minus 3 times 7, which is minus 21, plus 1 times 2."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? This is equal to minus 1 5th, and then we're going to get 1 times 7 minus 2 times 2, so it's minus 4. So 7 minus 4 is 3. And then we're going to get minus 3 times 7, which is minus 21, plus 1 times 2. So minus 21 plus 2 is minus 19. So the coordinates of A with respect to the basis B, so the coordinates of A with respect to B are going to be equal to, let me just multiply the negative 1 5th, you get minus 3 5ths, and then you get plus 19 5ths. So 19 over 5."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then we're going to get minus 3 times 7, which is minus 21, plus 1 times 2. So minus 21 plus 2 is minus 19. So the coordinates of A with respect to the basis B, so the coordinates of A with respect to B are going to be equal to, let me just multiply the negative 1 5th, you get minus 3 5ths, and then you get plus 19 5ths. So 19 over 5. Just like that. And let's verify that. Let's take, this means that A is equal to minus 3 5ths times our first basis vector, plus 19 5ths times our second basis vector."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "So 19 over 5. Just like that. And let's verify that. Let's take, this means that A is equal to minus 3 5ths times our first basis vector, plus 19 5ths times our second basis vector. Let's verify that that's the case. So let's see, minus 3 5ths times our 1 3, plus 19 5ths times 2 1. Let's see what this is going to be equal to."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's take, this means that A is equal to minus 3 5ths times our first basis vector, plus 19 5ths times our second basis vector. Let's verify that that's the case. So let's see, minus 3 5ths times our 1 3, plus 19 5ths times 2 1. Let's see what this is going to be equal to. This is equal to, let me write the two vectors. This is minus 3 5ths. This is minus 3 5ths times 3 is minus 9 5ths."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Let's see what this is going to be equal to. This is equal to, let me write the two vectors. This is minus 3 5ths. This is minus 3 5ths times 3 is minus 9 5ths. And then we're going to add it to this guy. So this guy is 2 times 19 is 38 5ths. And then 19 5ths times 1 is 19 5ths."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is minus 3 5ths times 3 is minus 9 5ths. And then we're going to add it to this guy. So this guy is 2 times 19 is 38 5ths. And then 19 5ths times 1 is 19 5ths. And then if you add these two vectors together, what do we get, we get minus 3 5ths plus 38 5ths. That's 35 5ths. 35 5ths is 7."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then 19 5ths times 1 is 19 5ths. And then if you add these two vectors together, what do we get, we get minus 3 5ths plus 38 5ths. That's 35 5ths. 35 5ths is 7. Minus 9 5ths plus 19 5ths, that's 10 5ths, or 2. And there you have it. That was our original A."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "35 5ths is 7. Minus 9 5ths plus 19 5ths, that's 10 5ths, or 2. And there you have it. That was our original A. So we see that A can definitely be represented as minus 3 5ths times our first basis vector, plus 19 5ths times our second basis vector. Now, that was the case where we had some vector A, and we wanted to represent it in coordinates with respect to B. What if we had the other way?"}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "That was our original A. So we see that A can definitely be represented as minus 3 5ths times our first basis vector, plus 19 5ths times our second basis vector. Now, that was the case where we had some vector A, and we wanted to represent it in coordinates with respect to B. What if we had the other way? What if we said that some vector, let's say some vector W's coordinates with respect to B are, I'll do something simple, are 1 1. Then what is W in standard coordinates? Well there, we can just multiply the change of, remember, W is just equal to the change of basis matrix times W's coordinates with respect to the basis B."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "What if we had the other way? What if we said that some vector, let's say some vector W's coordinates with respect to B are, I'll do something simple, are 1 1. Then what is W in standard coordinates? Well there, we can just multiply the change of, remember, W is just equal to the change of basis matrix times W's coordinates with respect to the basis B. So W is going to be equal to the change of basis matrix, which is just 1 3 2 1, times the coordinates of W with respect to B, times 1 1, which is equal to 1 times 1 plus 2 times 1 is 3, and then 3 times 1 plus 1 plus 1, so 3 times 1 is 3 plus 1 is 4. So W is just equal to the vector 3 4. So there you see if our change of basis matrix is invertible, which is really just another way of saying that our basis spans Rn, in this example it was R2, then you can easily go back and forth between coordinate representations in our standard coordinates and coordinate representations with respect to our basis."}, {"video_title": "Invertible change of basis matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well there, we can just multiply the change of, remember, W is just equal to the change of basis matrix times W's coordinates with respect to the basis B. So W is going to be equal to the change of basis matrix, which is just 1 3 2 1, times the coordinates of W with respect to B, times 1 1, which is equal to 1 times 1 plus 2 times 1 is 3, and then 3 times 1 plus 1 plus 1, so 3 times 1 is 3 plus 1 is 4. So W is just equal to the vector 3 4. So there you see if our change of basis matrix is invertible, which is really just another way of saying that our basis spans Rn, in this example it was R2, then you can easily go back and forth between coordinate representations in our standard coordinates and coordinate representations with respect to our basis. This is with respect to the basis, this is in standard coordinates, this was with respect to the basis, this is in standard coordinates. And you can do that just simply by either using this information or just saying, oh, the coordinates with respect to the basis equal to C inverse times A are the inverse of our change of basis matrix times A. Or saying, our coordinates with respect to the standard basis is just equal to the change of basis matrix times the coordinates with respect to the basis."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we know that all linear transformations can be expressed as a multiplication of a matrix. But this one is equal to the matrix 1, 3, 2, 6 times whatever vector you give me in my domain, times x1, x2. Now, let's say I have some subset in my codomain. So let me draw this right here. So my domain looks like that. It's R2. And of course, my function or my transformation maps elements of R2 into elements of its codomain, which also happens to be R2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw this right here. So my domain looks like that. It's R2. And of course, my function or my transformation maps elements of R2 into elements of its codomain, which also happens to be R2. I could show it mapping into itself, but for the sake of simplicity, let's draw my codomain here. And our transformation, of course, maps for any element here. The transformation of that will be an association or a mapping into R2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And of course, my function or my transformation maps elements of R2 into elements of its codomain, which also happens to be R2. I could show it mapping into itself, but for the sake of simplicity, let's draw my codomain here. And our transformation, of course, maps for any element here. The transformation of that will be an association or a mapping into R2. Now, what if we take some subset of R2? And let's just say it's a set of two vectors, the 0 vector in R2 and the vector 1, 2. So it's literally, let's say it's this point."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation of that will be an association or a mapping into R2. Now, what if we take some subset of R2? And let's just say it's a set of two vectors, the 0 vector in R2 and the vector 1, 2. So it's literally, let's say it's this point. Let me do it in a different color. Let's say this is my 0 vector in R2. I'm not plotting them."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's literally, let's say it's this point. Let me do it in a different color. Let's say this is my 0 vector in R2. I'm not plotting them. I'm just showing that they're in R2. That's my 0 vector. And let's say the vector 1, 2 is here."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm not plotting them. I'm just showing that they're in R2. That's my 0 vector. And let's say the vector 1, 2 is here. What I want to know is what are all of the vectors in my domain whose transformations map to this subset, map to this point? So I want to know the preimage of S. So the preimage of S under T. And I said be careful, because when you just say preimage of something without saying under something else, it implies that you're taking the image of an entire transformation, like I showed you, I think, two videos ago. But when you're taking the image or preimage of a set, yeah, make sure you say under what transformation."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say the vector 1, 2 is here. What I want to know is what are all of the vectors in my domain whose transformations map to this subset, map to this point? So I want to know the preimage of S. So the preimage of S under T. And I said be careful, because when you just say preimage of something without saying under something else, it implies that you're taking the image of an entire transformation, like I showed you, I think, two videos ago. But when you're taking the image or preimage of a set, yeah, make sure you say under what transformation. So we want to know the preimage of this subset of our codomain under the transformation T. And we write this as T to the, or T inverse of S. And we saw in the last video, this is all of the elements in our domain where the transformation of those X's is a member of the subset of our codomain that we're trying to find the preimage of. Now what is another way of writing this? Well, you could write this as we're trying to look for all of the X's in our domain such that this vector, let's call this A, that matrix A, such that A times X is a member of S. So that means A times X has to be equal to this or has to be equal to that."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But when you're taking the image or preimage of a set, yeah, make sure you say under what transformation. So we want to know the preimage of this subset of our codomain under the transformation T. And we write this as T to the, or T inverse of S. And we saw in the last video, this is all of the elements in our domain where the transformation of those X's is a member of the subset of our codomain that we're trying to find the preimage of. Now what is another way of writing this? Well, you could write this as we're trying to look for all of the X's in our domain such that this vector, let's call this A, that matrix A, such that A times X is a member of S. So that means A times X has to be equal to this or has to be equal to that. So that means A times our vector X has to be equal to the 0 vector. Or A times our vector X has to be equal to this vector 1, 2. This is the exact same statement as this one right here."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, you could write this as we're trying to look for all of the X's in our domain such that this vector, let's call this A, that matrix A, such that A times X is a member of S. So that means A times X has to be equal to this or has to be equal to that. So that means A times our vector X has to be equal to the 0 vector. Or A times our vector X has to be equal to this vector 1, 2. This is the exact same statement as this one right here. I just made it a little bit more explicit in terms of our actual transformation, A times X, and in terms of what our actual set is. Our set is just two vectors. So if we wanted to determine the preimage, if we wanted to determine the preimage of S, so we write that the preimage of S under T, that set there, we essentially have to just find all of the X's that satisfy these two equations right there."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the exact same statement as this one right here. I just made it a little bit more explicit in terms of our actual transformation, A times X, and in terms of what our actual set is. Our set is just two vectors. So if we wanted to determine the preimage, if we wanted to determine the preimage of S, so we write that the preimage of S under T, that set there, we essentially have to just find all of the X's that satisfy these two equations right there. So this equation, so we have to find all of the X's that satisfy the first one right here is the matrix 1, 3, 2, 6 times X1, X2 is equal to the 0 vector. That's this equation right there. We need to find all of the solutions to that."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we wanted to determine the preimage, if we wanted to determine the preimage of S, so we write that the preimage of S under T, that set there, we essentially have to just find all of the X's that satisfy these two equations right there. So this equation, so we have to find all of the X's that satisfy the first one right here is the matrix 1, 3, 2, 6 times X1, X2 is equal to the 0 vector. That's this equation right there. We need to find all of the solutions to that. And you might already recognize that all of the solutions to this, all of the X's that satisfy this is the null space of this matrix. I just thought I would point that out on the side. Now, that's not the only one."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We need to find all of the solutions to that. And you might already recognize that all of the solutions to this, all of the X's that satisfy this is the null space of this matrix. I just thought I would point that out on the side. Now, that's not the only one. We also have to solve this guy over here. I'll do that in blue. So the preimage of S under T is going to be all of the solutions to this plus all of the solutions to 1, 3, 2, 6 times X1, X2 is equal to 1, 2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, that's not the only one. We also have to solve this guy over here. I'll do that in blue. So the preimage of S under T is going to be all of the solutions to this plus all of the solutions to 1, 3, 2, 6 times X1, X2 is equal to 1, 2. Now, we could just solve this with an augmented matrix. So my augmented matrix would look like 1, 3, 2, 6, 0, 0. And here my augmented matrix would be 1, 3, 2, 6, 1, 2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the preimage of S under T is going to be all of the solutions to this plus all of the solutions to 1, 3, 2, 6 times X1, X2 is equal to 1, 2. Now, we could just solve this with an augmented matrix. So my augmented matrix would look like 1, 3, 2, 6, 0, 0. And here my augmented matrix would be 1, 3, 2, 6, 1, 2. Let's put this in reduced row echelon form. So let's replace our second row with our second row minus 2 times the first row. So what do we get?"}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And here my augmented matrix would be 1, 3, 2, 6, 1, 2. Let's put this in reduced row echelon form. So let's replace our second row with our second row minus 2 times the first row. So what do we get? Our first row stays the same, 1, 3, 0. Let me do them simultaneously. Let me solve these systems in parallel."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what do we get? Our first row stays the same, 1, 3, 0. Let me do them simultaneously. Let me solve these systems in parallel. So my first system stays, first row stays the same, 1, 3, 1. And in both cases, because I just want to get the left-hand side of my augmented matrix into reduced row echelon form, I can apply the same row operation. So I'm replacing my second row with 2 times my first row."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me solve these systems in parallel. So my first system stays, first row stays the same, 1, 3, 1. And in both cases, because I just want to get the left-hand side of my augmented matrix into reduced row echelon form, I can apply the same row operation. So I'm replacing my second row with 2 times my first row. So 2 minus 2 times 1 is 0. 6 minus 2 times 3 is 0. And of course, 0 minus 2 times 0 is 0."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm replacing my second row with 2 times my first row. So 2 minus 2 times 1 is 0. 6 minus 2 times 3 is 0. And of course, 0 minus 2 times 0 is 0. Here, 2 minus 2 times 1 is 0. 6 minus 2 times 3 is 0. And 2 minus 2 times 1 is 0."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And of course, 0 minus 2 times 0 is 0. Here, 2 minus 2 times 1 is 0. 6 minus 2 times 3 is 0. And 2 minus 2 times 1 is 0. So we get all these 0's, and we're actually done. We have both of these augmented matrices in reduced row echelon form. And how do we go back to solve all of the x1's and x2's that satisfy these?"}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And 2 minus 2 times 1 is 0. So we get all these 0's, and we're actually done. We have both of these augmented matrices in reduced row echelon form. And how do we go back to solve all of the x1's and x2's that satisfy these? Well, you recognize that our first columns right here are pivot columns. And these are associated with our variable x1. So we know that x1 is a pivot variable."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And how do we go back to solve all of the x1's and x2's that satisfy these? Well, you recognize that our first columns right here are pivot columns. And these are associated with our variable x1. So we know that x1 is a pivot variable. And we know that the second column is a non-pivot column because it has no 1's in it. And so it's associated with x2. And since x2 isn't a pivot column, we know that x2 is a free variable, which essentially means we can set x2 to be anything."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we know that x1 is a pivot variable. And we know that the second column is a non-pivot column because it has no 1's in it. And so it's associated with x2. And since x2 isn't a pivot column, we know that x2 is a free variable, which essentially means we can set x2 to be anything. So let's just set x2 is equal to t, where t is a member of the reals. In this case, what is x1 going to be equal to? So here, this top equation says, let me write it."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And since x2 isn't a pivot column, we know that x2 is a free variable, which essentially means we can set x2 to be anything. So let's just set x2 is equal to t, where t is a member of the reals. In this case, what is x1 going to be equal to? So here, this top equation says, let me write it. If we just kind of go back to this world right here, this means that x1 plus 3x2 is equal to this 0 here. This top line right here says x1 plus 3x2 is equal to 1. And so if we say x2 is equal to t, this equation becomes x1 plus 3t is equal to 0."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So here, this top equation says, let me write it. If we just kind of go back to this world right here, this means that x1 plus 3x2 is equal to this 0 here. This top line right here says x1 plus 3x2 is equal to 1. And so if we say x2 is equal to t, this equation becomes x1 plus 3t is equal to 0. Subtract 3t from both sides. You get x1 is equal to minus 3t. And then this equation, you get x1 is equal to, if we substitute x2 with t, is x1 is equal to 1 minus 3t."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so if we say x2 is equal to t, this equation becomes x1 plus 3t is equal to 0. Subtract 3t from both sides. You get x1 is equal to minus 3t. And then this equation, you get x1 is equal to, if we substitute x2 with t, is x1 is equal to 1 minus 3t. Now, if we wanted to write the solution sets in kind of vector notations, the solution set for this guy right here, for this first equation right there, is going to be x1 x2 is equal to what? It's going to be equal to, well, x2 is just going to be equal to t times, it's just going to be equal to t. So let me just write it there. So it's going to be equal to t times, x2 is just t times 1."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then this equation, you get x1 is equal to, if we substitute x2 with t, is x1 is equal to 1 minus 3t. Now, if we wanted to write the solution sets in kind of vector notations, the solution set for this guy right here, for this first equation right there, is going to be x1 x2 is equal to what? It's going to be equal to, well, x2 is just going to be equal to t times, it's just going to be equal to t. So let me just write it there. So it's going to be equal to t times, x2 is just t times 1. It's just equal to t. I made that definition up here. And then what's x1 equal to? It's equal to minus 3 times t. So if I put a t out here as a scalar, it's just minus 3 times t. So this is a solution for this first equation."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be equal to t times, x2 is just t times 1. It's just equal to t. I made that definition up here. And then what's x1 equal to? It's equal to minus 3 times t. So if I put a t out here as a scalar, it's just minus 3 times t. So this is a solution for this first equation. Sum where t is a member of the reals. So it's just scalar multiples of the vector minus 3, 1. And if we think of this as a position vector, this will be a line in R2, and I'll draw that in a second."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to minus 3 times t. So if I put a t out here as a scalar, it's just minus 3 times t. So this is a solution for this first equation. Sum where t is a member of the reals. So it's just scalar multiples of the vector minus 3, 1. And if we think of this as a position vector, this will be a line in R2, and I'll draw that in a second. So that's the solution for this first equation. And then the solution for the second equation, how can we set this up? It's going to be, let me make sure you can see it, it is x1, x2, and let's see, x2, once again, is just t times 1."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if we think of this as a position vector, this will be a line in R2, and I'll draw that in a second. So that's the solution for this first equation. And then the solution for the second equation, how can we set this up? It's going to be, let me make sure you can see it, it is x1, x2, and let's see, x2, once again, is just t times 1. So let me just write it like this. So it's t times 1. That's x2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be, let me make sure you can see it, it is x1, x2, and let's see, x2, once again, is just t times 1. So let me just write it like this. So it's t times 1. That's x2. Now what's x1? x1 is equal to 1 minus 3 times t. So if we do a minus 3, that gives us our minus 3 times t. But we need to do a 1 minus that, or 1 plus minus 3 times t, so what we can do is we can say that all the solution set here is equal to the vector 1 here. So now we have 1 plus minus 3 times t for x1."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's x2. Now what's x1? x1 is equal to 1 minus 3 times t. So if we do a minus 3, that gives us our minus 3 times t. But we need to do a 1 minus that, or 1 plus minus 3 times t, so what we can do is we can say that all the solution set here is equal to the vector 1 here. So now we have 1 plus minus 3 times t for x1. And then here we could say 0. x2 is equal to 0 plus t, or x2 is equal to t. So this is the solution set for the second equation. So our preimage of s, and remember, s was just these two points in our codomain. The preimage of s under t is essentially all of the x's that satisfy these two equations."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So now we have 1 plus minus 3 times t for x1. And then here we could say 0. x2 is equal to 0 plus t, or x2 is equal to t. So this is the solution set for the second equation. So our preimage of s, and remember, s was just these two points in our codomain. The preimage of s under t is essentially all of the x's that satisfy these two equations. And let's actually graph those. Let me turn on my graphs. That makes it look a little bit messy, but let me graph it down here."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The preimage of s under t is essentially all of the x's that satisfy these two equations. And let's actually graph those. Let me turn on my graphs. That makes it look a little bit messy, but let me graph it down here. Let me copy and paste my two results. Then I want to paste it. So those are my two results."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That makes it look a little bit messy, but let me graph it down here. Let me copy and paste my two results. Then I want to paste it. So those are my two results. Let me put my pen back on. And now what I can do is I can graph it right here. So let's see."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So those are my two results. Let me put my pen back on. And now what I can do is I can graph it right here. So let's see. This is the solution set for that first equation is all the multiples of the vector minus 3, 1. So the vector minus 3, 1 looks like this. This solution, minus 3, 1 looks like that."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's see. This is the solution set for that first equation is all the multiples of the vector minus 3, 1. So the vector minus 3, 1 looks like this. This solution, minus 3, 1 looks like that. That's the vector minus 3, 1. But my solution set is all the scalar multiples of minus 3, 1. This is a comma right here."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This solution, minus 3, 1 looks like that. That's the vector minus 3, 1. But my solution set is all the scalar multiples of minus 3, 1. This is a comma right here. So if I just take all the scalar multiples of minus 3, 1, it's going to look like this. So if you take 2 times it, you're going to have minus 6, 2. So you're going to get like that."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a comma right here. So if I just take all the scalar multiples of minus 3, 1, it's going to look like this. So if you take 2 times it, you're going to have minus 6, 2. So you're going to get like that. So it's going to be all of these points right here. I wish I could draw it a little bit neater, but I think you get the idea. It's going to be a line like that."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to get like that. So it's going to be all of these points right here. I wish I could draw it a little bit neater, but I think you get the idea. It's going to be a line like that. That is that solution set right there. And then what is this solution set right here? It's the vector 1, 0."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be a line like that. That is that solution set right there. And then what is this solution set right here? It's the vector 1, 0. So we go out 1, 0, so that's there, plus scalar multiples of minus 3, 1. So plus scalar multiples of this. So plus scalar multiples."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's the vector 1, 0. So we go out 1, 0, so that's there, plus scalar multiples of minus 3, 1. So plus scalar multiples of this. So plus scalar multiples. So if we just add one scalar multiple of minus 3, 1, we'll end up right there. But we want to put all the scalar multiples of it, because we have this t right here. So we're going to end up with another line with the same slope, essentially, that's just shifted a little bit."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So plus scalar multiples. So if we just add one scalar multiple of minus 3, 1, we'll end up right there. But we want to put all the scalar multiples of it, because we have this t right here. So we're going to end up with another line with the same slope, essentially, that's just shifted a little bit. It shifted 1 to the right. Now why did we do all of this? Remember, what we wanted to find out is what were all of the vectors, let me turn off the graph paper."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to end up with another line with the same slope, essentially, that's just shifted a little bit. It shifted 1 to the right. Now why did we do all of this? Remember, what we wanted to find out is what were all of the vectors, let me turn off the graph paper. What were all of the vectors in our domain that when we apply the transformation, map to vectors within our subset of our codomain, map to either 0, 0 or the vector 1, 2. And we figured all of those vectors out by solving these two equations. And we were able to see that these two lines, when I turn my graph paper on, they map to the points."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, what we wanted to find out is what were all of the vectors, let me turn off the graph paper. What were all of the vectors in our domain that when we apply the transformation, map to vectors within our subset of our codomain, map to either 0, 0 or the vector 1, 2. And we figured all of those vectors out by solving these two equations. And we were able to see that these two lines, when I turn my graph paper on, they map to the points. So when these guys, when you apply the transformation, I'll draw it all on the same graph, they map, when you apply the transformation, to the points 0, 0 and to the point 1, 2, which was right here. So all of these points, when you apply the transformation, and actually all of the ones in the blue, they map to 0, 0, because they solved this top equation. All the ones in orange, when you apply the transformation, map to the point 1, 2."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we were able to see that these two lines, when I turn my graph paper on, they map to the points. So when these guys, when you apply the transformation, I'll draw it all on the same graph, they map, when you apply the transformation, to the points 0, 0 and to the point 1, 2, which was right here. So all of these points, when you apply the transformation, and actually all of the ones in the blue, they map to 0, 0, because they solved this top equation. All the ones in orange, when you apply the transformation, map to the point 1, 2. Now, this blue line right here, this has a special name for it, and I actually touched on it a little bit before. Everything in this blue line, if I call this set right here, I don't know, let's call it b for blue, this is the blue line, that's this set of vectors right here. Everything there, when I apply my transformation of those blue vectors, or if I take the image of my blue set under t, it all maps to the 0 vector."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "All the ones in orange, when you apply the transformation, map to the point 1, 2. Now, this blue line right here, this has a special name for it, and I actually touched on it a little bit before. Everything in this blue line, if I call this set right here, I don't know, let's call it b for blue, this is the blue line, that's this set of vectors right here. Everything there, when I apply my transformation of those blue vectors, or if I take the image of my blue set under t, it all maps to the 0 vector. It equals the set of the 0 vector right there. And we saw that right there. And I remember, earlier in the video, I pointed out that, look, this set right here, this is equivalent to the null space, right?"}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Everything there, when I apply my transformation of those blue vectors, or if I take the image of my blue set under t, it all maps to the 0 vector. It equals the set of the 0 vector right there. And we saw that right there. And I remember, earlier in the video, I pointed out that, look, this set right here, this is equivalent to the null space, right? The null space of a matrix is all of the vectors that, if you multiply it by that matrix, you get 0. So the similar idea here is this transformation is defined by a matrix, and we're saying, what are all of the x's that when you transform them, you get the 0 vector? And so this idea, this blue thing right here, is called the kernel of t. And sometimes it's written as just shorthand, k-e-r of t. And this literally is all of the vectors that, if you apply the transformation, let me write it this way, it's all of the vectors in our domain, which was R2, such that the transformation of those vectors is equal to the 0 vector."}, {"video_title": "Preimage and kernel example Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I remember, earlier in the video, I pointed out that, look, this set right here, this is equivalent to the null space, right? The null space of a matrix is all of the vectors that, if you multiply it by that matrix, you get 0. So the similar idea here is this transformation is defined by a matrix, and we're saying, what are all of the x's that when you transform them, you get the 0 vector? And so this idea, this blue thing right here, is called the kernel of t. And sometimes it's written as just shorthand, k-e-r of t. And this literally is all of the vectors that, if you apply the transformation, let me write it this way, it's all of the vectors in our domain, which was R2, such that the transformation of those vectors is equal to the 0 vector. This is the definition of the kernel. And if the transformation, if our transformation is equal to some matrix times some vector, and we know that any linear transformation can be written as a matrix vector product, then the kernel of t is the same thing as the null space of a, and we saw that earlier in the video. Anyway, hopefully you found that reasonably useful."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "I have some subspace of it that we'll call v right here. So that is my subspace v. We know that the orthogonal complement of v is equal to the set of all of the members of Rn, so x is a member of Rn, such that x dot v is equal to 0 for every v that is a member of our subspace. So our orthogonal complement of our subspace is only all of the vectors that are orthogonal to all of these vectors. And we've seen before that they only overlap, there's only one vector that's a member of both, and that's the 0 vector, let's say it's right there. Let's say the orthogonal complement, let's say it's this set right here in pink. So that's the orthogonal complement. Fair enough."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "And we've seen before that they only overlap, there's only one vector that's a member of both, and that's the 0 vector, let's say it's right there. Let's say the orthogonal complement, let's say it's this set right here in pink. So that's the orthogonal complement. Fair enough. Now, what if we were to think about the orthogonal complement of the orthogonal complement? So we want the orthogonal complement, so that's the orthogonal complement, that's the thing in pink. But we want the orthogonal complement of that."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now, what if we were to think about the orthogonal complement of the orthogonal complement? So we want the orthogonal complement, so that's the orthogonal complement, that's the thing in pink. But we want the orthogonal complement of that. So this is going to be all of the x's, let's just write it like this, all of the x's that are members of Rn, such that x dot w is equal to 0 for every w that is a member of the orthogonal complement of v. That's what that thing is saying. So it's all of the vectors in Rn that are orthogonal to everything here. Now, obviously all of the things in v are going to be a member of that, because these guys are orthogonal to everything in these guys."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "But we want the orthogonal complement of that. So this is going to be all of the x's, let's just write it like this, all of the x's that are members of Rn, such that x dot w is equal to 0 for every w that is a member of the orthogonal complement of v. That's what that thing is saying. So it's all of the vectors in Rn that are orthogonal to everything here. Now, obviously all of the things in v are going to be a member of that, because these guys are orthogonal to everything in these guys. But maybe this is just a subset of the orthogonal complement of the orthogonal complement. So maybe this thing in blue right here looks like this. Maybe it's a slightly larger set than v. Maybe there are some things, these things that I'm shading in blue, maybe there are some vectors that are orthogonal to the orthogonal complement of v, but that are outside of v. We don't know that yet."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Now, obviously all of the things in v are going to be a member of that, because these guys are orthogonal to everything in these guys. But maybe this is just a subset of the orthogonal complement of the orthogonal complement. So maybe this thing in blue right here looks like this. Maybe it's a slightly larger set than v. Maybe there are some things, these things that I'm shading in blue, maybe there are some vectors that are orthogonal to the orthogonal complement of v, but that are outside of v. We don't know that yet. We don't know whether this area right here exists, or maybe the orthogonal complement of the orthogonal complement, maybe that takes us back to v. Maybe it's like the transpose or an inverse function, where it just goes back to our original subspace. So let's see if we can think about that a little bit better. So let's say that I have some member of the orthogonal complement of the orthogonal complement."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Maybe it's a slightly larger set than v. Maybe there are some things, these things that I'm shading in blue, maybe there are some vectors that are orthogonal to the orthogonal complement of v, but that are outside of v. We don't know that yet. We don't know whether this area right here exists, or maybe the orthogonal complement of the orthogonal complement, maybe that takes us back to v. Maybe it's like the transpose or an inverse function, where it just goes back to our original subspace. So let's see if we can think about that a little bit better. So let's say that I have some member of the orthogonal complement of the orthogonal complement. So let's say I have some vector x that is a member of the orthogonal complement of the orthogonal complement. Now, we saw in the last video that any vector in Rn can be represented by a sum of some vector in a subspace and the subspace's complement. So we know that x can be represented."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that I have some member of the orthogonal complement of the orthogonal complement. So let's say I have some vector x that is a member of the orthogonal complement of the orthogonal complement. Now, we saw in the last video that any vector in Rn can be represented by a sum of some vector in a subspace and the subspace's complement. So we know that x can be represented. We can say that x can be represented as a sum of two vectors, one that's in v and one that's in the orthogonal complement of v. So one, let's call that the vector that's in v, and let's call w the vector that's in the orthogonal complement of v. Let me write it like this. Where v is a member of the subspace V, and the vector w is a member of the orthogonal complement of v. So this is some member. It could be some guy out here."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So we know that x can be represented. We can say that x can be represented as a sum of two vectors, one that's in v and one that's in the orthogonal complement of v. So one, let's call that the vector that's in v, and let's call w the vector that's in the orthogonal complement of v. Let me write it like this. Where v is a member of the subspace V, and the vector w is a member of the orthogonal complement of v. So this is some member. It could be some guy out here. It could be some guy over here. He's a member of the orthogonal complement of the orthogonal complement, which is this whole area here, which v is a subset of, but we're not sure whether v equals that thing. But we say, look, anything that's in the orthogonal complement of the orthogonal complement is going to be a member of Rn, and anything in Rn can be represented as a sum of a vector in v and a vector in the orthogonal complement of v. So that's all I wrote right there."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "It could be some guy out here. It could be some guy over here. He's a member of the orthogonal complement of the orthogonal complement, which is this whole area here, which v is a subset of, but we're not sure whether v equals that thing. But we say, look, anything that's in the orthogonal complement of the orthogonal complement is going to be a member of Rn, and anything in Rn can be represented as a sum of a vector in v and a vector in the orthogonal complement of v. So that's all I wrote right there. Now, what happens if I dot, if I take the dot product of x with w? What is this going to be equal to? Well, this is the orthogonal complement of the orthogonal complement."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "But we say, look, anything that's in the orthogonal complement of the orthogonal complement is going to be a member of Rn, and anything in Rn can be represented as a sum of a vector in v and a vector in the orthogonal complement of v. So that's all I wrote right there. Now, what happens if I dot, if I take the dot product of x with w? What is this going to be equal to? Well, this is the orthogonal complement of the orthogonal complement. So if you take the dot product of any vector in this with any vector in the orthogonal complement, which this vector is, it's a member of the orthogonal complement, you're going to get 0 by definition. These are all of the vectors. This vector is definitely orthogonal to anything in just v perp."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Well, this is the orthogonal complement of the orthogonal complement. So if you take the dot product of any vector in this with any vector in the orthogonal complement, which this vector is, it's a member of the orthogonal complement, you're going to get 0 by definition. These are all of the vectors. This vector is definitely orthogonal to anything in just v perp. Anything in v perp perp is orthogonal to anything in v perp. So this thing is going to be equal to 0. But what's another way of writing x dot w?"}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "This vector is definitely orthogonal to anything in just v perp. Anything in v perp perp is orthogonal to anything in v perp. So this thing is going to be equal to 0. But what's another way of writing x dot w? We could write it like this. This is the same thing as v plus w dot w, which is the same thing as v dot w plus w dot w. Now, what is v dot w? v is a member of our original subspace."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "But what's another way of writing x dot w? We could write it like this. This is the same thing as v plus w dot w, which is the same thing as v dot w plus w dot w. Now, what is v dot w? v is a member of our original subspace. And if you take the dot product of anything in our original subspace with anything in its orthogonal complement, you're going to get 0. So this term right here is going to be 0. And you're just going to get this term, which is the same thing as the length of our vector w squared."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "v is a member of our original subspace. And if you take the dot product of anything in our original subspace with anything in its orthogonal complement, you're going to get 0. So this term right here is going to be 0. And you're just going to get this term, which is the same thing as the length of our vector w squared. Now, that has to equal 0. Remember, we just wrote x dot w. x is a member of the orthogonal complement of the orthogonal complement. So you dot that with anything in the orthogonal complement."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "And you're just going to get this term, which is the same thing as the length of our vector w squared. Now, that has to equal 0. Remember, we just wrote x dot w. x is a member of the orthogonal complement of the orthogonal complement. So you dot that with anything in the orthogonal complement. That's got to be equal to 0. But if we write it the other way, if we write it as the sum of v plus w and distribute this w, we say that's the same thing as the magnitude of w squared. So the magnitude of w squared has got to be equal to 0."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So you dot that with anything in the orthogonal complement. That's got to be equal to 0. But if we write it the other way, if we write it as the sum of v plus w and distribute this w, we say that's the same thing as the magnitude of w squared. So the magnitude of w squared has got to be equal to 0. The magnitude of w squared, or the length of w squared, has got to be equal to 0, which tells us that w is the 0 vector. That's the only vector in Rn that when you take its length, and especially when you square it, you get 0. But you could just take its length."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So the magnitude of w squared has got to be equal to 0. The magnitude of w squared, or the length of w squared, has got to be equal to 0, which tells us that w is the 0 vector. That's the only vector in Rn that when you take its length, and especially when you square it, you get 0. But you could just take its length. So what does that mean? That means that our original vector x is equal to v plus w, but w is just equal to 0. So that implies that our original vector x is equal to v, and v is a member of our subspace V. So that tells us that x is a member of our subspace V. So what we just showed, that if something is a member of the orthogonal complement of the orthogonal complement, then that same vector has to be a member of the original subspace."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "But you could just take its length. So what does that mean? That means that our original vector x is equal to v plus w, but w is just equal to 0. So that implies that our original vector x is equal to v, and v is a member of our subspace V. So that tells us that x is a member of our subspace V. So what we just showed, that if something is a member of the orthogonal complement of the orthogonal complement, then that same vector has to be a member of the original subspace. So there is no such thing as something being in the orthogonal complement of the orthogonal complement and not being a member of our original subspace. All of this has to be inside of this right there. So there is no outside blue space like that."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So that implies that our original vector x is equal to v, and v is a member of our subspace V. So that tells us that x is a member of our subspace V. So what we just showed, that if something is a member of the orthogonal complement of the orthogonal complement, then that same vector has to be a member of the original subspace. So there is no such thing as something being in the orthogonal complement of the orthogonal complement and not being a member of our original subspace. All of this has to be inside of this right there. So there is no outside blue space like that. All of that is our original subspace if you want to view it that way. Now I just, at the beginning of this video, said, hey, anything in our subspace is going to be a member of our orthogonal complement. And you can kind of reason that in your head."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So there is no outside blue space like that. All of that is our original subspace if you want to view it that way. Now I just, at the beginning of this video, said, hey, anything in our subspace is going to be a member of our orthogonal complement. And you can kind of reason that in your head. But let's use the same argument to just be a little bit more rigorous about it. So let's say, right now we say, look, if anything is in the orthogonal complement of the orthogonal complement, then it's going to be in the original subspace. Now let's go the other way."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "And you can kind of reason that in your head. But let's use the same argument to just be a little bit more rigorous about it. So let's say, right now we say, look, if anything is in the orthogonal complement of the orthogonal complement, then it's going to be in the original subspace. Now let's go the other way. Let's say that something is in the original subspace. Just like that. If anything, let me draw another graph right here, because this might be useful."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Now let's go the other way. Let's say that something is in the original subspace. Just like that. If anything, let me draw another graph right here, because this might be useful. Let me draw Rn again. Let me draw all of this. Let me draw all of Rn like that."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "If anything, let me draw another graph right here, because this might be useful. Let me draw Rn again. Let me draw all of this. Let me draw all of Rn like that. Now we start, we have the orthogonal complement. Let me just draw that first. So V perp, and then you have the orthogonal complement of the orthogonal complement, which could be this set right here."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw all of Rn like that. Now we start, we have the orthogonal complement. Let me just draw that first. So V perp, and then you have the orthogonal complement of the orthogonal complement, which could be this set right here. This is V perp. I haven't even drawn the subspace V. All I've shown is I have some subspace here, which I happen to call V perp, and then I have the orthogonal complement of that subspace. So this means that anything in Rn can be represented as a sum of a vector that's here and a vector that's here."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So V perp, and then you have the orthogonal complement of the orthogonal complement, which could be this set right here. This is V perp. I haven't even drawn the subspace V. All I've shown is I have some subspace here, which I happen to call V perp, and then I have the orthogonal complement of that subspace. So this means that anything in Rn can be represented as a sum of a vector that's here and a vector that's here. So if I say that W, let me do it in purple. If I say that the vector W, or let me write it this way, the vector V can be represented as a sum of the vector W and the vector X, where W is a member of the orthogonal complement of V, or V perp, and X is a member of its orthogonal complement. Notice all I'm saying, I could have called this set S, and then this would have been S and its orthogonal complement."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So this means that anything in Rn can be represented as a sum of a vector that's here and a vector that's here. So if I say that W, let me do it in purple. If I say that the vector W, or let me write it this way, the vector V can be represented as a sum of the vector W and the vector X, where W is a member of the orthogonal complement of V, or V perp, and X is a member of its orthogonal complement. Notice all I'm saying, I could have called this set S, and then this would have been S and its orthogonal complement. And we learned that anything in Rn can be represented as a sum of something in a subspace and the subspace's orthogonal complement. So it doesn't matter that V is somehow related to this. It can be represented as a sum of a vector here plus a vector there."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Notice all I'm saying, I could have called this set S, and then this would have been S and its orthogonal complement. And we learned that anything in Rn can be represented as a sum of something in a subspace and the subspace's orthogonal complement. So it doesn't matter that V is somehow related to this. It can be represented as a sum of a vector here plus a vector there. Fair enough. Now what happens if I dot V with W? I'm doing the exact same argument that I did before."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "It can be represented as a sum of a vector here plus a vector there. Fair enough. Now what happens if I dot V with W? I'm doing the exact same argument that I did before. Well, if you take anything that's a member of our original subspace and you dot it with anything in its orthogonal complement, that's going to give us 0. What else is that going to be equal to? Well, if we write V in this way, V dot W is the same thing as this thing dot W. So W plus X dot W. And this is going to be equal to W dot W plus X dot W. And then what's X dot W?"}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "I'm doing the exact same argument that I did before. Well, if you take anything that's a member of our original subspace and you dot it with anything in its orthogonal complement, that's going to give us 0. What else is that going to be equal to? Well, if we write V in this way, V dot W is the same thing as this thing dot W. So W plus X dot W. And this is going to be equal to W dot W plus X dot W. And then what's X dot W? X is in the orthogonal complement of the orthogonal complement, and W is in the orthogonal complement. So if you take their dot product, you're going to get 0. They're orthogonal to each other."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "Well, if we write V in this way, V dot W is the same thing as this thing dot W. So W plus X dot W. And this is going to be equal to W dot W plus X dot W. And then what's X dot W? X is in the orthogonal complement of the orthogonal complement, and W is in the orthogonal complement. So if you take their dot product, you're going to get 0. They're orthogonal to each other. So this is just equal to W dot W, or the length of W squared. And since that has to equal 0, we just have a bunch of equals here. That tells us that once again, the vector W has to be equal to 0, and that V, so that tells us V is equal to W plus X."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "They're orthogonal to each other. So this is just equal to W dot W, or the length of W squared. And since that has to equal 0, we just have a bunch of equals here. That tells us that once again, the vector W has to be equal to 0, and that V, so that tells us V is equal to W plus X. But if W is equal to 0, then V is going to be equal to X. So we've just shown that if V is a member of the subspace V, then V is a member of the orthogonal complement of the orthogonal complement. V is equal to X, which is a member of the orthogonal complement of the orthogonal complement."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "That tells us that once again, the vector W has to be equal to 0, and that V, so that tells us V is equal to W plus X. But if W is equal to 0, then V is going to be equal to X. So we've just shown that if V is a member of the subspace V, then V is a member of the orthogonal complement of the orthogonal complement. V is equal to X, which is a member of the orthogonal complement of the orthogonal complement. So we've proven it both ways. If you look at the original statement, we wrote here that if you're a member of the orthogonal complement of the orthogonal complement, you're a member of the original subspace. So we proved this, and earlier in the video, we proved that if X is a member of the orthogonal complement of the orthogonal complement, then X is a member of our subspace."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "V is equal to X, which is a member of the orthogonal complement of the orthogonal complement. So we've proven it both ways. If you look at the original statement, we wrote here that if you're a member of the orthogonal complement of the orthogonal complement, you're a member of the original subspace. So we proved this, and earlier in the video, we proved that if X is a member of the orthogonal complement of the orthogonal complement, then X is a member of our subspace. So these two things are equivalent. Anything that's in the subspace is a member of V perp perp. Anything in V perp perp is a member of our subspace."}, {"video_title": "Orthogonal complement of the orthogonal complement Linear Algebra Khan Academy.mp3", "Sentence": "So we proved this, and earlier in the video, we proved that if X is a member of the orthogonal complement of the orthogonal complement, then X is a member of our subspace. So these two things are equivalent. Anything that's in the subspace is a member of V perp perp. Anything in V perp perp is a member of our subspace. So our subspace and V perp perp are the same set. And of course, it overlaps. This equals this."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So we're going to start with these two things, this definition of a cross product in R3, the only place it really is defined, and then this result, and we want to get to the result that the length of the cross product of two vectors, and so obviously when you take a cross product you get a vector, but if you take its length you get a number again, you just get a scalar value, is equal to the product of each of the vectors' lengths, so the product of the length of a times the product of the length of b, times the sine of the angle between them. Which is a pretty neat outcome, because it kind of shows that they're two sides of the same coin, dot product is cosine, cross product is sine, I'm sure you've seen this before, well you definitely have seen it if you've watched my physics playlist, and I even do a whole video where I talk about the intuition behind what this really means, and I encourage you to re-watch that, and I'll probably do that again in the linear algebra context, but the point of this video is to prove this to you, is to prove that with this and this I can get to this, now if you just believe me, and you just say oh I've seen that before and I just think it definitely is the case, then you don't have to watch the rest of this video, because I'll tell you right now, it's going to get dirty, it's going to be a hairy, hairy proof. But if you're willing to watch, and bear with me, let's start on this, start proving this result. So the place I'm going to start is with the idea of taking the length of a cross b squared, that's a cross b right there, so I'm essentially taking the length of this vector squared. And we saw in many videos, and I've used this idea multiple times, that if I just have some arbitrary vector, let me just say some arbitrary vector, and I take its length squared, that's just equal to that vector dotted with itself, or the square of each of its terms summed up, all the way to xn squared. So what will this be equal to? Well this is just equal to that vector, and we only have three components, so it's equal to the sum of the squares of each of these components."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So the place I'm going to start is with the idea of taking the length of a cross b squared, that's a cross b right there, so I'm essentially taking the length of this vector squared. And we saw in many videos, and I've used this idea multiple times, that if I just have some arbitrary vector, let me just say some arbitrary vector, and I take its length squared, that's just equal to that vector dotted with itself, or the square of each of its terms summed up, all the way to xn squared. So what will this be equal to? Well this is just equal to that vector, and we only have three components, so it's equal to the sum of the squares of each of these components. So it's equal to, let me write this down, it's equal to this term squared, so let me write that down, a2b3 minus a3b2 squared, plus this term squared, so plus a3b1 minus a1b3 squared, and then finally plus that term squared. So plus a1b2 minus a2b1 squared. And what is this equal to?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Well this is just equal to that vector, and we only have three components, so it's equal to the sum of the squares of each of these components. So it's equal to, let me write this down, it's equal to this term squared, so let me write that down, a2b3 minus a3b2 squared, plus this term squared, so plus a3b1 minus a1b3 squared, and then finally plus that term squared. So plus a1b2 minus a2b1 squared. And what is this equal to? Well let's just expand it out. Let's expand that out. So this term right here, we're just going to have to do our expansion of the square of a binomial, and we've done this multiple times, so this is going to be equal to a2 squared b3 squared, and then we're going to have these two multiplied by each other twice, so minus 2, I'm just multiplying this out, minus 2 times a2 a3 b2 b3, I'm just rearranging them to get the order right, plus a3 squared b2 squared, that term squared, and then I'll have, then I have to add this term, so plus a3 squared b1 squared minus 2 times both of these terms multiplied, minus 2 times a1 a3 b1 b3, plus that term squared, a1 squared b3 squared, and then finally this term squared."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And what is this equal to? Well let's just expand it out. Let's expand that out. So this term right here, we're just going to have to do our expansion of the square of a binomial, and we've done this multiple times, so this is going to be equal to a2 squared b3 squared, and then we're going to have these two multiplied by each other twice, so minus 2, I'm just multiplying this out, minus 2 times a2 a3 b2 b3, I'm just rearranging them to get the order right, plus a3 squared b2 squared, that term squared, and then I'll have, then I have to add this term, so plus a3 squared b1 squared minus 2 times both of these terms multiplied, minus 2 times a1 a3 b1 b3, plus that term squared, a1 squared b3 squared, and then finally this term squared. So plus a1 squared b2 squared minus 2 times a1 a2 b1 b2 plus a2 squared b1 squared. So there you go. Let's see if we can write this in a form, well I'm going to write this in a form that I know will be useful later."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So this term right here, we're just going to have to do our expansion of the square of a binomial, and we've done this multiple times, so this is going to be equal to a2 squared b3 squared, and then we're going to have these two multiplied by each other twice, so minus 2, I'm just multiplying this out, minus 2 times a2 a3 b2 b3, I'm just rearranging them to get the order right, plus a3 squared b2 squared, that term squared, and then I'll have, then I have to add this term, so plus a3 squared b1 squared minus 2 times both of these terms multiplied, minus 2 times a1 a3 b1 b3, plus that term squared, a1 squared b3 squared, and then finally this term squared. So plus a1 squared b2 squared minus 2 times a1 a2 b1 b2 plus a2 squared b1 squared. So there you go. Let's see if we can write this in a form, well I'm going to write this in a form that I know will be useful later. So what I'm going to do is I'm going to factor out the a2 a1 a3 squared terms, so I could write this as, let me pick a new neutral color, so this is equal to, if I just write a1 squared, where's my a1 squared terms? I've got that one right there, and I have that one right there. So a1 squared times b2 squared plus b3 squared."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if we can write this in a form, well I'm going to write this in a form that I know will be useful later. So what I'm going to do is I'm going to factor out the a2 a1 a3 squared terms, so I could write this as, let me pick a new neutral color, so this is equal to, if I just write a1 squared, where's my a1 squared terms? I've got that one right there, and I have that one right there. So a1 squared times b2 squared plus b3 squared. Good enough. Now where are my a2 squared terms? a plus a2 squared times, I have that one and that one, so times b1 squared, that's that, plus b3 squared, and then finally, let me pick another new color, I'll go back to yellow, plus a3 squared times, well that's that term and that term, so b1 and b2, so b1 squared plus b2 squared, and obviously I can't forget about all of that mess that I have in the middle, all of this stuff right here, all of that stuff right there."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So a1 squared times b2 squared plus b3 squared. Good enough. Now where are my a2 squared terms? a plus a2 squared times, I have that one and that one, so times b1 squared, that's that, plus b3 squared, and then finally, let me pick another new color, I'll go back to yellow, plus a3 squared times, well that's that term and that term, so b1 and b2, so b1 squared plus b2 squared, and obviously I can't forget about all of that mess that I have in the middle, all of this stuff right here, all of that stuff right there. So plus, or maybe I should write minus 2, minus 2 times all of this stuff, let me just write it real fast. So it's a2 a3 b2 b3 plus a1 a3 b1 b3 plus a1 a2 b1 b2. There you go."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "a plus a2 squared times, I have that one and that one, so times b1 squared, that's that, plus b3 squared, and then finally, let me pick another new color, I'll go back to yellow, plus a3 squared times, well that's that term and that term, so b1 and b2, so b1 squared plus b2 squared, and obviously I can't forget about all of that mess that I have in the middle, all of this stuff right here, all of that stuff right there. So plus, or maybe I should write minus 2, minus 2 times all of this stuff, let me just write it real fast. So it's a2 a3 b2 b3 plus a1 a3 b1 b3 plus a1 a2 b1 b2. There you go. Now let's put this aside for a little bit. Let me put this on the side for a little bit. We'll let that equation rest for a little while."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "There you go. Now let's put this aside for a little bit. Let me put this on the side for a little bit. We'll let that equation rest for a little while. And remember, this is just an expansion of the length of a cross b squared. That's all this is. So just remember that."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "We'll let that equation rest for a little while. And remember, this is just an expansion of the length of a cross b squared. That's all this is. So just remember that. And now let's do another equally hairy and cumbersome computation. Let's take this result up here. We know that the magnitude or the length of a times the length of b times the angle between them is equal to a dot b, which is the same thing as if we actually do the dot product, a1 times b1 plus a2 times b2 plus a3 times b3."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So just remember that. And now let's do another equally hairy and cumbersome computation. Let's take this result up here. We know that the magnitude or the length of a times the length of b times the angle between them is equal to a dot b, which is the same thing as if we actually do the dot product, a1 times b1 plus a2 times b2 plus a3 times b3. Now, just to make sure that I get to do the hairiest problem possible, let's take the square of both sides. So if we square this side, you get a squared, b squared, cosine squared, then you get a dot b squared, or you get this whole thing squared. So what's this whole thing squared?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "We know that the magnitude or the length of a times the length of b times the angle between them is equal to a dot b, which is the same thing as if we actually do the dot product, a1 times b1 plus a2 times b2 plus a3 times b3. Now, just to make sure that I get to do the hairiest problem possible, let's take the square of both sides. So if we square this side, you get a squared, b squared, cosine squared, then you get a dot b squared, or you get this whole thing squared. So what's this whole thing squared? For me, it's easier to just write out the thing again. Instead of writing a square, just multiply that times a1 b1 plus a2 b2 plus a3 b3. And let's do some polynomial multiplication."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So what's this whole thing squared? For me, it's easier to just write out the thing again. Instead of writing a square, just multiply that times a1 b1 plus a2 b2 plus a3 b3. And let's do some polynomial multiplication. So first, let's multiply this guy times each of those guys. So you have a1 b1 times, well, they're a1 b1, so I'm going to do it right here. You get a1 squared, b1 squared, plus this guy times this guy, plus a1 a2 times b1 b2, plus this guy times that guy, plus a1 a3 times b1 b3."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And let's do some polynomial multiplication. So first, let's multiply this guy times each of those guys. So you have a1 b1 times, well, they're a1 b1, so I'm going to do it right here. You get a1 squared, b1 squared, plus this guy times this guy, plus a1 a2 times b1 b2, plus this guy times that guy, plus a1 a3 times b1 b3. Fair enough. Now let's do the second term. We have to multiply this guy times each of those guys."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "You get a1 squared, b1 squared, plus this guy times this guy, plus a1 a2 times b1 b2, plus this guy times that guy, plus a1 a3 times b1 b3. Fair enough. Now let's do the second term. We have to multiply this guy times each of those guys. So a2 b2 times a1 b1, well, that's this one right here. a2 b2 times a1 b1. I wrote it right here because this is really the same term, and eventually we want to simplify that."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "We have to multiply this guy times each of those guys. So a2 b2 times a1 b1, well, that's this one right here. a2 b2 times a1 b1. I wrote it right here because this is really the same term, and eventually we want to simplify that. So that's that times that guy. Then we have this guy times that over there. So let me write it over here."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "I wrote it right here because this is really the same term, and eventually we want to simplify that. So that's that times that guy. Then we have this guy times that over there. So let me write it over here. So that's a2 squared b2 squared. Put a plus right there. And then finally, this middle guy times this third guy."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So let me write it over here. So that's a2 squared b2 squared. Put a plus right there. And then finally, this middle guy times this third guy. So let me write it over here. Plus, plus, so a2 a2 a3 b2 b3. Now we only have one left, and I'll do it in this blue color."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, this middle guy times this third guy. So let me write it over here. Plus, plus, so a2 a2 a3 b2 b3. Now we only have one left, and I'll do it in this blue color. I have to multiply this guy times each of those guys. So a3 b3 times a1 b1. That's the same thing as this term right here, right?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Now we only have one left, and I'll do it in this blue color. I have to multiply this guy times each of those guys. So a3 b3 times a1 b1. That's the same thing as this term right here, right? Because you have a3 b3 times a1 b1. Then you have this guy times that guy, which is this, because it's a3 b3 times a2 b2. Let me put a little plus sign there."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "That's the same thing as this term right here, right? Because you have a3 b3 times a1 b1. Then you have this guy times that guy, which is this, because it's a3 b3 times a2 b2. Let me put a little plus sign there. And then finally, you have this guy times himself. So you have a3 squared b3 squared. And so if you add up all of this business here, what do you get?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Let me put a little plus sign there. And then finally, you have this guy times himself. So you have a3 squared b3 squared. And so if you add up all of this business here, what do you get? I'll switch to another color. You have a1 squared b1 squared plus, and I'm doing these colors a certain way on purpose, plus a2 squared b2 squared plus a3 squared b3 squared plus, and let me do it in this, I'll do it in white, plus, what do you have here? You have this term times 2."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And so if you add up all of this business here, what do you get? I'll switch to another color. You have a1 squared b1 squared plus, and I'm doing these colors a certain way on purpose, plus a2 squared b2 squared plus a3 squared b3 squared plus, and let me do it in this, I'll do it in white, plus, what do you have here? You have this term times 2. You have this term times 2. And then you have this term times 2. So plus 2 times a1, let me write that down, plus 2 times a1 a2 b1 b2, that's that term, plus this one right here, plus a1 a3 b1 b3, finally plus this one, a2 a3 plus b2 b3."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "You have this term times 2. You have this term times 2. And then you have this term times 2. So plus 2 times a1, let me write that down, plus 2 times a1 a2 b1 b2, that's that term, plus this one right here, plus a1 a3 b1 b3, finally plus this one, a2 a3 plus b2 b3. And you might have noticed something interesting already. If you compare this term right here, if you compare that guy right there to this guy right there, they're the same thing. a1 a2 b1 b2, this term and that term are the same."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So plus 2 times a1, let me write that down, plus 2 times a1 a2 b1 b2, that's that term, plus this one right here, plus a1 a3 b1 b3, finally plus this one, a2 a3 plus b2 b3. And you might have noticed something interesting already. If you compare this term right here, if you compare that guy right there to this guy right there, they're the same thing. a1 a2 b1 b2, this term and that term are the same. Let's look at the other terms. Let me pick a nice color. a1 a3 b1 b3, that term and that term is the same."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "a1 a2 b1 b2, this term and that term are the same. Let's look at the other terms. Let me pick a nice color. a1 a3 b1 b3, that term and that term is the same. And then finally, if you compare a2 a3 b2 b3, this shouldn't be a plus, this is just this one. a2 a3 b2 b3, this term and this term is the same. And this expression, when we expanded it out, we have 2 times this, positive 2 times this."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "a1 a3 b1 b3, that term and that term is the same. And then finally, if you compare a2 a3 b2 b3, this shouldn't be a plus, this is just this one. a2 a3 b2 b3, this term and this term is the same. And this expression, when we expanded it out, we have 2 times this, positive 2 times this. And this term, right here when we expanded it out, we have minus 2 times this. So you might see, so let's see if we can simplify things a little bit. So what happens if we add this guy to this guy?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And this expression, when we expanded it out, we have 2 times this, positive 2 times this. And this term, right here when we expanded it out, we have minus 2 times this. So you might see, so let's see if we can simplify things a little bit. So what happens if we add this guy to this guy? Let's do it. Let's do it. So it's a little exciting."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So what happens if we add this guy to this guy? Let's do it. Let's do it. So it's a little exciting. So we get a cross b, the length of that squared, and we're going to add to that this expression right here. So plus the length of a squared times the length of b squared times the cosine of the angle between them squared. What's that going to be equal?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So it's a little exciting. So we get a cross b, the length of that squared, and we're going to add to that this expression right here. So plus the length of a squared times the length of b squared times the cosine of the angle between them squared. What's that going to be equal? It's going to be equal to this thing plus this thing. Let's do a simplification. What's this thing plus this thing?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "What's that going to be equal? It's going to be equal to this thing plus this thing. Let's do a simplification. What's this thing plus this thing? Well, we already said that this is the minus 2 times this, this is the plus 2 times this. So this guy, let me be very clear, this right here is going to cancel out when we add the two terms, it's going to cancel out with this guy. These guys are going to cancel out, thank God."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "What's this thing plus this thing? Well, we already said that this is the minus 2 times this, this is the plus 2 times this. So this guy, let me be very clear, this right here is going to cancel out when we add the two terms, it's going to cancel out with this guy. These guys are going to cancel out, thank God. Cancel, cancel out. Makes our life a little bit easier. And what are we left with?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "These guys are going to cancel out, thank God. Cancel, cancel out. Makes our life a little bit easier. And what are we left with? We're left with this right here plus that right there. So then we see we have an a1 squared term, so we just add the coefficients on the a1 squared, we add the coefficients on the a2 squared, we add the coefficients on the a3 squared, and what do we get? We get a1 squared times this coefficient plus this coefficient."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And what are we left with? We're left with this right here plus that right there. So then we see we have an a1 squared term, so we just add the coefficients on the a1 squared, we add the coefficients on the a2 squared, we add the coefficients on the a3 squared, and what do we get? We get a1 squared times this coefficient plus this coefficient. So you get b1 squared plus b2 squared plus b3 squared. Things are starting to look a little bit orderly all of a sudden. And then you have plus a2 squared times all of their coefficients added up."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "We get a1 squared times this coefficient plus this coefficient. So you get b1 squared plus b2 squared plus b3 squared. Things are starting to look a little bit orderly all of a sudden. And then you have plus a2 squared times all of their coefficients added up. So b1 squared plus b2 squared plus b3 squared. And then finally, in yellow, you have plus a3 squared. Sorry, I'm not trying to do that in yellow."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And then you have plus a2 squared times all of their coefficients added up. So b1 squared plus b2 squared plus b3 squared. And then finally, in yellow, you have plus a3 squared. Sorry, I'm not trying to do that in yellow. You have a3 squared, and you have that, you have b1 squared, b2 squared, and b3 squared. So b1 squared plus b2 squared plus b3 squared. And if you see, we're multiplying all of these things by this b1 squared plus b2 squared plus b3 squared, so we can actually factor that out, and we get something very interesting."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Sorry, I'm not trying to do that in yellow. You have a3 squared, and you have that, you have b1 squared, b2 squared, and b3 squared. So b1 squared plus b2 squared plus b3 squared. And if you see, we're multiplying all of these things by this b1 squared plus b2 squared plus b3 squared, so we can actually factor that out, and we get something very interesting. So this is equal to, if we factor this thing out, of all of the terms, we get b1 squared plus b2 squared plus b3 squared times my a squared terms, times a1 squared plus a2 squared plus a3 squared. So these two things are equal to each other. But what's this thing?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And if you see, we're multiplying all of these things by this b1 squared plus b2 squared plus b3 squared, so we can actually factor that out, and we get something very interesting. So this is equal to, if we factor this thing out, of all of the terms, we get b1 squared plus b2 squared plus b3 squared times my a squared terms, times a1 squared plus a2 squared plus a3 squared. So these two things are equal to each other. But what's this thing? What's another way I could write this? This is the same thing as b dot b or the length of my vector b squared. And what's that?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "But what's this thing? What's another way I could write this? This is the same thing as b dot b or the length of my vector b squared. And what's that? That's the length of my vector a squared. This is my length of my vector a squared. That's just a dot a."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "And what's that? That's the length of my vector a squared. This is my length of my vector a squared. That's just a dot a. So we have, let me rewrite everything. So we have the length of a, that's a darker green, a cross b squared, plus this thing, plus the length of, let me actually just copy and paste it. It's monotonous."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "That's just a dot a. So we have, let me rewrite everything. So we have the length of a, that's a darker green, a cross b squared, plus this thing, plus the length of, let me actually just copy and paste it. It's monotonous. Plus that thing right there. Why isn't it? If I control, copy, and paste, it's not working."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "It's monotonous. Plus that thing right there. Why isn't it? If I control, copy, and paste, it's not working. So plus that thing, the length of a squared times the length of b squared times the cosine of the angle squared between them is equal to that. Now what if we subtract this from both sides? What do we get?"}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "If I control, copy, and paste, it's not working. So plus that thing, the length of a squared times the length of b squared times the cosine of the angle squared between them is equal to that. Now what if we subtract this from both sides? What do we get? We get the length of a cross b squared is equal to this minus this. So we can factor, let me write that. Actually let me just subtract this on this line."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "What do we get? We get the length of a cross b squared is equal to this minus this. So we can factor, let me write that. Actually let me just subtract this on this line. So if I subtract it from both sides, I can get that out there and I'll put the minus, the length of a squared times the length of b squared times the cosine squared of the angle between them. And we can factor this a squared b squared, the length of the two vectors out. I'm just switching the order."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Actually let me just subtract this on this line. So if I subtract it from both sides, I can get that out there and I'll put the minus, the length of a squared times the length of b squared times the cosine squared of the angle between them. And we can factor this a squared b squared, the length of the two vectors out. I'm just switching the order. This is equal to the length of a squared times the length of b squared times, and this is exciting, times, when you factor this out of this, you just get a one, minus cosine squared of theta. And what is one minus cosine squared of theta? Well, sine squared of theta plus cosine, this is the most basic trig identity."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "I'm just switching the order. This is equal to the length of a squared times the length of b squared times, and this is exciting, times, when you factor this out of this, you just get a one, minus cosine squared of theta. And what is one minus cosine squared of theta? Well, sine squared of theta plus cosine, this is the most basic trig identity. Sine squared of theta plus cosine squared of theta is equal to one. So if you subtract cosine squared from both sides, you get sine squared of theta is equal to one minus cosine squared of theta. So this is sine squared of theta."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "Well, sine squared of theta plus cosine, this is the most basic trig identity. Sine squared of theta plus cosine squared of theta is equal to one. So if you subtract cosine squared from both sides, you get sine squared of theta is equal to one minus cosine squared of theta. So this is sine squared of theta. And then what happens if you take the square root of both sides? And this is really exciting. You get the length of vector a crossed with vector b is equal to the length of vector a times the length of vector b times the sine of the angle between them."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "So this is sine squared of theta. And then what happens if you take the square root of both sides? And this is really exciting. You get the length of vector a crossed with vector b is equal to the length of vector a times the length of vector b times the sine of the angle between them. I just took the square root of both sides of this, and we finally get our result. I never thought I would get here. And so hopefully you're satisfied."}, {"video_title": "Proof Relationship between cross product and sin of angle Linear Algebra Khan Academy.mp3", "Sentence": "You get the length of vector a crossed with vector b is equal to the length of vector a times the length of vector b times the sine of the angle between them. I just took the square root of both sides of this, and we finally get our result. I never thought I would get here. And so hopefully you're satisfied. You don't ever have to take this as kind of a leap of faith anymore. And hopefully you're satisfied with this. I'm going to stop recording this video before I make a careless mistake or the power goes out that would ruin everything."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The orthogonal complement of V is the set. And this is shorthand notation right here, would be the orthogonal complement of V. So we write this little orthogonal notation as a superscript on V. And you can pronounce this as Vperp. Not for perpetrator, but for perpendicular. So Vperp is equal to the set of all the vectors x that are a member of our Rn such that x dot V is equal to 0 for every vector V that is a member of our subspace. So we're essentially saying, look, you have some subspace. It's got a bunch of vectors in it. Now, if I can find some other set of vectors where every member of that set is orthogonal to every member of the subspace in question, then the set of those vectors is called the orthogonal complement of V. And you write it this way, Vperp right there."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So Vperp is equal to the set of all the vectors x that are a member of our Rn such that x dot V is equal to 0 for every vector V that is a member of our subspace. So we're essentially saying, look, you have some subspace. It's got a bunch of vectors in it. Now, if I can find some other set of vectors where every member of that set is orthogonal to every member of the subspace in question, then the set of those vectors is called the orthogonal complement of V. And you write it this way, Vperp right there. So the first thing that we just tend to do when we're defining a space or defining some set is to see, hey, is this a subspace? Is Vperp, or is the orthogonal complement of V, is this a subspace? Well, you might remember from many, many videos ago that we had just a couple of conditions for a subspace."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, if I can find some other set of vectors where every member of that set is orthogonal to every member of the subspace in question, then the set of those vectors is called the orthogonal complement of V. And you write it this way, Vperp right there. So the first thing that we just tend to do when we're defining a space or defining some set is to see, hey, is this a subspace? Is Vperp, or is the orthogonal complement of V, is this a subspace? Well, you might remember from many, many videos ago that we had just a couple of conditions for a subspace. That if, let's say, that A and B are both a member of Vperp, then we have to wonder whether A plus B is a member of Vperp. That's our first condition. It needs to be closed under addition in order for this to be a subspace."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, you might remember from many, many videos ago that we had just a couple of conditions for a subspace. That if, let's say, that A and B are both a member of Vperp, then we have to wonder whether A plus B is a member of Vperp. That's our first condition. It needs to be closed under addition in order for this to be a subspace. And then the next condition is, well, if A is a member of Vperp, is some scalar multiple of A also a member of Vperp? And then the last one is it has to contain the 0 vector, which is a little bit redundant with this. Because if any scalar multiple of A is a member of our orthogonal complement of V, then you can just multiply it by 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It needs to be closed under addition in order for this to be a subspace. And then the next condition is, well, if A is a member of Vperp, is some scalar multiple of A also a member of Vperp? And then the last one is it has to contain the 0 vector, which is a little bit redundant with this. Because if any scalar multiple of A is a member of our orthogonal complement of V, then you can just multiply it by 0. So it would imply that the 0 vector is a member of V. So what does this imply? What is the fact that A and B are members of Vperp? That means that A dot V, where this V is any member of our original subspace V, is equal to 0 for any V that is a member of our subspace V. And it also means that B, since B is also a member of Vperp, that V dot any member of our subspace is also going to be 0 for any V that is a member of V. So what happens if we take A plus B dot V?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Because if any scalar multiple of A is a member of our orthogonal complement of V, then you can just multiply it by 0. So it would imply that the 0 vector is a member of V. So what does this imply? What is the fact that A and B are members of Vperp? That means that A dot V, where this V is any member of our original subspace V, is equal to 0 for any V that is a member of our subspace V. And it also means that B, since B is also a member of Vperp, that V dot any member of our subspace is also going to be 0 for any V that is a member of V. So what happens if we take A plus B dot V? Let's do that. So if I do A plus B dot V, what is this going to be equal to? This is going to be equal to A dot V plus B dot V. And we just said the fact that both A and B are members of our orthogonal complement means that both of these quantities are going to be equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That means that A dot V, where this V is any member of our original subspace V, is equal to 0 for any V that is a member of our subspace V. And it also means that B, since B is also a member of Vperp, that V dot any member of our subspace is also going to be 0 for any V that is a member of V. So what happens if we take A plus B dot V? Let's do that. So if I do A plus B dot V, what is this going to be equal to? This is going to be equal to A dot V plus B dot V. And we just said the fact that both A and B are members of our orthogonal complement means that both of these quantities are going to be equal to 0. So this is going to be equal to 0 plus 0, which is equal to 0. So A plus B is definitely a member of our orthogonal complement. So we got our checkbox right there."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be equal to A dot V plus B dot V. And we just said the fact that both A and B are members of our orthogonal complement means that both of these quantities are going to be equal to 0. So this is going to be equal to 0 plus 0, which is equal to 0. So A plus B is definitely a member of our orthogonal complement. So we got our checkbox right there. Do it in a different color than the question mark. Check for the first condition for being a subspace. Now, is CA a member of Vperp?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we got our checkbox right there. Do it in a different color than the question mark. Check for the first condition for being a subspace. Now, is CA a member of Vperp? Well, let's just take C. If we take CA and dot it with any member of our original subspace, this is the same thing as C times A dot V. And what is this equal to? By definition, A was a member of our orthogonal complement. So this is going to be equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, is CA a member of Vperp? Well, let's just take C. If we take CA and dot it with any member of our original subspace, this is the same thing as C times A dot V. And what is this equal to? By definition, A was a member of our orthogonal complement. So this is going to be equal to 0. So it's going to be C times 0, which is equal to 0. So this is also a member of our orthogonal complement to V. And of course, I could multiply C times 0, and I would get to 0. Or you could just say, look, 0 is orthogonal to everything."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to 0. So it's going to be C times 0, which is equal to 0. So this is also a member of our orthogonal complement to V. And of course, I could multiply C times 0, and I would get to 0. Or you could just say, look, 0 is orthogonal to everything. You take the 0 vector and dot it with anything, you're going to get 0. So the 0 vector is always going to be a member of any orthogonal complement, because it obviously is always going to be true for this condition right here. So we know that Vperp, or the orthogonal complement of V, is a subspace, which is nice, because now we can apply to it all of the properties that we know of subspaces."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or you could just say, look, 0 is orthogonal to everything. You take the 0 vector and dot it with anything, you're going to get 0. So the 0 vector is always going to be a member of any orthogonal complement, because it obviously is always going to be true for this condition right here. So we know that Vperp, or the orthogonal complement of V, is a subspace, which is nice, because now we can apply to it all of the properties that we know of subspaces. Now, the next question, and I touched on this in the last video. I said, I kind of just touched on the idea, I said that if I have some matrix A, and let's just say it's an m by n matrix, in the last video, I said that the row space of A is, well, let me write it this way. I wrote that the null space of A, let me switch the order, is equal to the orthogonal complement of the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we know that Vperp, or the orthogonal complement of V, is a subspace, which is nice, because now we can apply to it all of the properties that we know of subspaces. Now, the next question, and I touched on this in the last video. I said, I kind of just touched on the idea, I said that if I have some matrix A, and let's just say it's an m by n matrix, in the last video, I said that the row space of A is, well, let me write it this way. I wrote that the null space of A, let me switch the order, is equal to the orthogonal complement of the row space. And in a way that we can write the row space of A, this thing right here, the row space of A, is the same thing as the column space of A transpose. So one way you could rewrite this sentence right here is that the null space of A is the orthogonal complement of the row space. The row space is the column space of the transpose matrix."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I wrote that the null space of A, let me switch the order, is equal to the orthogonal complement of the row space. And in a way that we can write the row space of A, this thing right here, the row space of A, is the same thing as the column space of A transpose. So one way you could rewrite this sentence right here is that the null space of A is the orthogonal complement of the row space. The row space is the column space of the transpose matrix. And the claim, which I have not proven to you, is that this is the orthogonal complement of this. So this is equal to that, the little perpendicular superscript. That's the claim."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The row space is the column space of the transpose matrix. And the claim, which I have not proven to you, is that this is the orthogonal complement of this. So this is equal to that, the little perpendicular superscript. That's the claim. And at least in the particular example that I did in the last two videos, this was the case where I actually showed you that 2 by 3 matrix. But let's see if this applies generally. So let me write my matrix A like this."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That's the claim. And at least in the particular example that I did in the last two videos, this was the case where I actually showed you that 2 by 3 matrix. But let's see if this applies generally. So let me write my matrix A like this. So my matrix A, I can write it as just a bunch of row vectors. But just to be consistent with our notation, with vectors we tend to associate as column vectors, so to represent the row vectors here, I'm just going to write them as transpose vectors. Because in our reality, vectors will always be column vectors, and row vectors are just transposes of those."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let me write my matrix A like this. So my matrix A, I can write it as just a bunch of row vectors. But just to be consistent with our notation, with vectors we tend to associate as column vectors, so to represent the row vectors here, I'm just going to write them as transpose vectors. Because in our reality, vectors will always be column vectors, and row vectors are just transposes of those. R1 transpose, R2 transpose, and you go all the way down. We have m rows. So you're going to get Rm transpose."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Because in our reality, vectors will always be column vectors, and row vectors are just transposes of those. R1 transpose, R2 transpose, and you go all the way down. We have m rows. So you're going to get Rm transpose. Don't let the transpose part confuse you. I'm just saying that these are row vectors. I'm writing transposes there just to say that look, these are the transposes of column vectors that represent these rows."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to get Rm transpose. Don't let the transpose part confuse you. I'm just saying that these are row vectors. I'm writing transposes there just to say that look, these are the transposes of column vectors that represent these rows. But if it's helpful for you to imagine them, just imagine this is the first row of the matrix. This is the second row of that matrix. So on and so forth."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I'm writing transposes there just to say that look, these are the transposes of column vectors that represent these rows. But if it's helpful for you to imagine them, just imagine this is the first row of the matrix. This is the second row of that matrix. So on and so forth. Now, what is the null space of A? Well, that's all of the vectors here. Let me do it like this."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So on and so forth. Now, what is the null space of A? Well, that's all of the vectors here. Let me do it like this. The null space of A is, let me write it like this. It's all of the vectors x that satisfy the equation that this is going to be equal to the 0 vector. Now, to solve this equation, what can we do?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it like this. The null space of A is, let me write it like this. It's all of the vectors x that satisfy the equation that this is going to be equal to the 0 vector. Now, to solve this equation, what can we do? We've seen this multiple times. This matrix vector product is essentially the same thing as saying, let me actually write it like this a little bit. Let me write it like this."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, to solve this equation, what can we do? We've seen this multiple times. This matrix vector product is essentially the same thing as saying, let me actually write it like this a little bit. Let me write it like this. It's going to be equal to the 0 vector in Rm. So you're going to have m0's all the way down to the mth 0. So another way to write this equation, you've seen it before, is when you take the matrix vector product, you're essentially taking the dot product."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it like this. It's going to be equal to the 0 vector in Rm. So you're going to have m0's all the way down to the mth 0. So another way to write this equation, you've seen it before, is when you take the matrix vector product, you're essentially taking the dot product. So to get to this entry right here, it's going to be this row dotted with my vector x. So this is R1. We're calling this row vector R1 transpose."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So another way to write this equation, you've seen it before, is when you take the matrix vector product, you're essentially taking the dot product. So to get to this entry right here, it's going to be this row dotted with my vector x. So this is R1. We're calling this row vector R1 transpose. This is a transpose of some column vector that can represent that row. But that dot, dot my vector x, this vector x is going to be equal to that 0. Now, if I take this guy, let me do it in a different color."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We're calling this row vector R1 transpose. This is a transpose of some column vector that can represent that row. But that dot, dot my vector x, this vector x is going to be equal to that 0. Now, if I take this guy, let me do it in a different color. If I take this guy and I dot him with vector x, it's going to be equal to that 0. So R2 transpose dot x is going to be equal to that 0 right there. So by definition, or I guess let me write it this way."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, if I take this guy, let me do it in a different color. If I take this guy and I dot him with vector x, it's going to be equal to that 0. So R2 transpose dot x is going to be equal to that 0 right there. So by definition, or I guess let me write it this way. Another way to write this equation is that R1 transpose dot x is equal to 0. R2 transpose dot x is equal to 0. All the way down to Rn transpose dot x is equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So by definition, or I guess let me write it this way. Another way to write this equation is that R1 transpose dot x is equal to 0. R2 transpose dot x is equal to 0. All the way down to Rn transpose dot x is equal to 0. And by definition, the null space of A is equal to all of the x's that are members of, well in this case it's an m by n matrix, so you're going to have n columns. So it's all the x's that are members of Rn such that Ax is equal to 0. Or you could alternately write it this way."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "All the way down to Rn transpose dot x is equal to 0. And by definition, the null space of A is equal to all of the x's that are members of, well in this case it's an m by n matrix, so you're going to have n columns. So it's all the x's that are members of Rn such that Ax is equal to 0. Or you could alternately write it this way. That if you were to dot each of the rows with x, you're going to be equal to 0. So you could write it this way. Well, let me just write such that Ax is equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or you could alternately write it this way. That if you were to dot each of the rows with x, you're going to be equal to 0. So you could write it this way. Well, let me just write such that Ax is equal to 0. That's an easier way to write it. So let's think about it. If someone is a member, if by definition I give you some vector v, if I say that, let me switch colors, if I were to tell you that v is a member of the null space of A. I just tell you, let's call it v1."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, let me just write such that Ax is equal to 0. That's an easier way to write it. So let's think about it. If someone is a member, if by definition I give you some vector v, if I say that, let me switch colors, if I were to tell you that v is a member of the null space of A. I just tell you, let's call it v1. v1 is a member of null space of A. That means it satisfies this equation right here. That means A times v is equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "If someone is a member, if by definition I give you some vector v, if I say that, let me switch colors, if I were to tell you that v is a member of the null space of A. I just tell you, let's call it v1. v1 is a member of null space of A. That means it satisfies this equation right here. That means A times v is equal to 0. A times v is equal to 0 means that when you dot each of these rows with v, you get equal to 0. Or another way of saying that is that v1 is orthogonal to all of these rows. To R1 transpose, that's just the first row, R2 transpose, all the way to Rm transpose."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That means A times v is equal to 0. A times v is equal to 0 means that when you dot each of these rows with v, you get equal to 0. Or another way of saying that is that v1 is orthogonal to all of these rows. To R1 transpose, that's just the first row, R2 transpose, all the way to Rm transpose. So this is orthogonal to all of these guys, by definition, any member of the null space. Well, if you're orthogonal to all of these rows in your matrix, you're also orthogonal to any linear combination of them. You're also orthogonal to any linear combination."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "To R1 transpose, that's just the first row, R2 transpose, all the way to Rm transpose. So this is orthogonal to all of these guys, by definition, any member of the null space. Well, if you're orthogonal to all of these rows in your matrix, you're also orthogonal to any linear combination of them. You're also orthogonal to any linear combination. Orthogonal to any linear combination. You can imagine, let's say that we have some vector that is a linear combination of these guys right here. So let's say vector, let me just write it, well, it's going to be a member of the R space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "You're also orthogonal to any linear combination. Orthogonal to any linear combination. You can imagine, let's say that we have some vector that is a linear combination of these guys right here. So let's say vector, let me just write it, well, it's going to be a member of the R space. Let me think of a good, I want to do it different. Let's say vector w is equal to some linear combination of these vectors right here. I wrote them as transposes just because they're row vectors, but I could just write them as regular column vectors, just to show that w could be just a regular column vector."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's say vector, let me just write it, well, it's going to be a member of the R space. Let me think of a good, I want to do it different. Let's say vector w is equal to some linear combination of these vectors right here. I wrote them as transposes just because they're row vectors, but I could just write them as regular column vectors, just to show that w could be just a regular column vector. So let's say w is equal to c1 times R1 plus c2 times R2, all the way to cm times Rm. That's what w is equal to. So what happens when you take v, which is a member of our null space, and you dot it with w?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I wrote them as transposes just because they're row vectors, but I could just write them as regular column vectors, just to show that w could be just a regular column vector. So let's say w is equal to c1 times R1 plus c2 times R2, all the way to cm times Rm. That's what w is equal to. So what happens when you take v, which is a member of our null space, and you dot it with w? So if we were to take v and dot it with w, it's going to be v dotted with each of these guys, because our dot product has the distributive property. So if you dot v with each of these guys, it's going to be equal to c1, I'm just going to take the scalar out, c1 times v dot R1 plus c2 times v dot R2. This is an R right here, not a v. Plus all the way to plus cm times v dot Rm."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So what happens when you take v, which is a member of our null space, and you dot it with w? So if we were to take v and dot it with w, it's going to be v dotted with each of these guys, because our dot product has the distributive property. So if you dot v with each of these guys, it's going to be equal to c1, I'm just going to take the scalar out, c1 times v dot R1 plus c2 times v dot R2. This is an R right here, not a v. Plus all the way to plus cm times v dot Rm. And we know, we already just said that each of these Rs are going to be equal to 0. So all of these are going to be equal to 0. So this whole expression is going to be equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is an R right here, not a v. Plus all the way to plus cm times v dot Rm. And we know, we already just said that each of these Rs are going to be equal to 0. So all of these are going to be equal to 0. So this whole expression is going to be equal to 0. So if you have any vector that's a linear combination of these row vectors, if you dot it with any member of your null space, you're going to get 0. So let me write it this way. What is any vector that's any linear combination of these guys?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this whole expression is going to be equal to 0. So if you have any vector that's a linear combination of these row vectors, if you dot it with any member of your null space, you're going to get 0. So let me write it this way. What is any vector that's any linear combination of these guys? Well, that's the span of these guys, or you could say that the row space. So if w is a member of the row space, which you can just represent as a column space of A transpose, then we know, and let's say we know that v is a member of our null space, we know that v dot w is going to be equal to 0. I just showed that to you right there."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "What is any vector that's any linear combination of these guys? Well, that's the span of these guys, or you could say that the row space. So if w is a member of the row space, which you can just represent as a column space of A transpose, then we know, and let's say we know that v is a member of our null space, we know that v dot w is going to be equal to 0. I just showed that to you right there. So every member of our null space is definitely orthogonal to every member of our row space. So that's what we know so far. Every member of null space of A is orthogonal to every member of the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I just showed that to you right there. So every member of our null space is definitely orthogonal to every member of our row space. So that's what we know so far. Every member of null space of A is orthogonal to every member of the row space. Now, that only gives us halfway. That still doesn't tell us that this is equivalent to the orthogonal complement of the null space. For example, there might be members of our orthogonal complement of the row space that aren't a member of our null space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Every member of null space of A is orthogonal to every member of the row space. Now, that only gives us halfway. That still doesn't tell us that this is equivalent to the orthogonal complement of the null space. For example, there might be members of our orthogonal complement of the row space that aren't a member of our null space. So let's say that I have some other vector. Let me say I have some vector u. Let's say that u is a member of the orthogonal complement of our row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "For example, there might be members of our orthogonal complement of the row space that aren't a member of our null space. So let's say that I have some other vector. Let me say I have some vector u. Let's say that u is a member of the orthogonal complement of our row space. I know the notation is a little convoluted. Maybe I should write an r there. But I want to really get set in your mind that the row space is just the column space of the transpose."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that u is a member of the orthogonal complement of our row space. I know the notation is a little convoluted. Maybe I should write an r there. But I want to really get set in your mind that the row space is just the column space of the transpose. So let's say that u is some member of our orthogonal complement. What I want to do is show you that u has to be in your null space, and when I show you that, then we know. So far we just said that everything in the null space is orthogonal to the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But I want to really get set in your mind that the row space is just the column space of the transpose. So let's say that u is some member of our orthogonal complement. What I want to do is show you that u has to be in your null space, and when I show you that, then we know. So far we just said that everything in the null space is orthogonal to the row space. But we don't know that everything that's orthogonal to the row space, which is represented by this set, is also going to be in your null space. That's what we have to show in order for those two sets to be equivalent, in order for the null space to be equal to this. So if we know this is true, then this means that u dot w, where w is a member of our row space, is going to be equal to 0."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So far we just said that everything in the null space is orthogonal to the row space. But we don't know that everything that's orthogonal to the row space, which is represented by this set, is also going to be in your null space. That's what we have to show in order for those two sets to be equivalent, in order for the null space to be equal to this. So if we know this is true, then this means that u dot w, where w is a member of our row space, is going to be equal to 0. Let me write this down right here. That is going to be equal to 0. And what does that mean?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if we know this is true, then this means that u dot w, where w is a member of our row space, is going to be equal to 0. Let me write this down right here. That is going to be equal to 0. And what does that mean? That means that u is also orthogonal. So this implies that u dot rj, any of the row vectors, is also equal to 0, where j is equal to 1 all the way through m. How do I know that? Well, I'm saying that, look, you take u as a member of the orthogonal complement of the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And what does that mean? That means that u is also orthogonal. So this implies that u dot rj, any of the row vectors, is also equal to 0, where j is equal to 1 all the way through m. How do I know that? Well, I'm saying that, look, you take u as a member of the orthogonal complement of the row space. So that means u is orthogonal to any member of your row space. So in particular, the basis vectors of your row space, we don't know whether all of these guys are basis vectors, but that means we can take, these guys are definitely all members of the row space. Some of them are actually the basis for the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, I'm saying that, look, you take u as a member of the orthogonal complement of the row space. So that means u is orthogonal to any member of your row space. So in particular, the basis vectors of your row space, we don't know whether all of these guys are basis vectors, but that means we can take, these guys are definitely all members of the row space. Some of them are actually the basis for the row space. So that means if you take u dot in any of these guys, it's going to be equal to 0. So if u dot any of these guys is equal to 0, that means that u dot r1 is 0. Let me write this down."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Some of them are actually the basis for the row space. So that means if you take u dot in any of these guys, it's going to be equal to 0. So if u dot any of these guys is equal to 0, that means that u dot r1 is 0. Let me write this down. u dot r1 is equal to 0, u dot r2 is equal to 0, all the way to u dot rm is equal to 0. Well, if all of this is true, that means that a times the vector u is equal to 0. That implies this, right?"}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this down. u dot r1 is equal to 0, u dot r2 is equal to 0, all the way to u dot rm is equal to 0. Well, if all of this is true, that means that a times the vector u is equal to 0. That implies this, right? You stick u there, you take all the dot products, it's going to satisfy this equation, which implies that u is a member of our null space. So we've just shown you that every member of your null space is definitely a member of the orthogonal complement. And now we said that every member of our orthogonal complement is a member of our null space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That implies this, right? You stick u there, you take all the dot products, it's going to satisfy this equation, which implies that u is a member of our null space. So we've just shown you that every member of your null space is definitely a member of the orthogonal complement. And now we said that every member of our orthogonal complement is a member of our null space. Actually, I just noticed that I made a slight error here. This dot product, I don't have to write the transpose here, because we've defined our dot product as the dot product of column vectors. So this is the transpose of some column vector, so you can un-transpose it here and just take the dot product."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And now we said that every member of our orthogonal complement is a member of our null space. Actually, I just noticed that I made a slight error here. This dot product, I don't have to write the transpose here, because we've defined our dot product as the dot product of column vectors. So this is the transpose of some column vector, so you can un-transpose it here and just take the dot product. But anyway, minor error there. But that diverts me from my main takeaway, my punch line, the big picture. We now showed you any member of our null space is a member of the orthogonal complement."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is the transpose of some column vector, so you can un-transpose it here and just take the dot product. But anyway, minor error there. But that diverts me from my main takeaway, my punch line, the big picture. We now showed you any member of our null space is a member of the orthogonal complement. So we just showed you this first statement here is another way of saying any member of the null space, or that the null space is a subset of the orthogonal complement of the row space. So that's our row space, and that's the orthogonal complement of our row space. And here we just showed that any member of the orthogonal complement of our row space is also a member of your null space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We now showed you any member of our null space is a member of the orthogonal complement. So we just showed you this first statement here is another way of saying any member of the null space, or that the null space is a subset of the orthogonal complement of the row space. So that's our row space, and that's the orthogonal complement of our row space. And here we just showed that any member of the orthogonal complement of our row space is also a member of your null space. Well, if these two guys are subsets of each other, they must be equal to each other. So we now know that the null space of A is equal to the orthogonal complement of the row space of A, or the column space of A transpose. Now, this is I related the null space with the row space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And here we just showed that any member of the orthogonal complement of our row space is also a member of your null space. Well, if these two guys are subsets of each other, they must be equal to each other. So we now know that the null space of A is equal to the orthogonal complement of the row space of A, or the column space of A transpose. Now, this is I related the null space with the row space. Now I could just as easily make a bit of a substitution here. Let's say that A is equal to some other matrix B transpose. It's going to be the transpose of some matrix."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Now, this is I related the null space with the row space. Now I could just as easily make a bit of a substitution here. Let's say that A is equal to some other matrix B transpose. It's going to be the transpose of some matrix. You could transpose either way. So if I just make that substitution here, what do we get? We get the null space of B transpose is equal to the column space of, let me write it out, B transpose transpose."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be the transpose of some matrix. You could transpose either way. So if I just make that substitution here, what do we get? We get the null space of B transpose is equal to the column space of, let me write it out, B transpose transpose. A transpose is B transpose transposed. Let me get my parentheses right. And then that thing's orthogonal complement."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We get the null space of B transpose is equal to the column space of, let me write it out, B transpose transpose. A transpose is B transpose transposed. Let me get my parentheses right. And then that thing's orthogonal complement. So what is this equal to? The transpose of the transpose is just equal to B. So I could write it as the null space of B transpose is equal to the orthogonal complement of the column space of B."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then that thing's orthogonal complement. So what is this equal to? The transpose of the transpose is just equal to B. So I could write it as the null space of B transpose is equal to the orthogonal complement of the column space of B. So just like this, we just showed that the left, B and A, they're just arbitrary matrices. So this showed us that the null space, so this is null space, sometimes it's nice to write in words, is orthogonal complement of row space. And this right here is showing us that the left null space, which is just the same thing as the null space of a transpose matrix, is equal to is orthogonal, I'll just shorthand it, complement of the column space."}, {"video_title": "Orthogonal complements Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So I could write it as the null space of B transpose is equal to the orthogonal complement of the column space of B. So just like this, we just showed that the left, B and A, they're just arbitrary matrices. So this showed us that the null space, so this is null space, sometimes it's nice to write in words, is orthogonal complement of row space. And this right here is showing us that the left null space, which is just the same thing as the null space of a transpose matrix, is equal to is orthogonal, I'll just shorthand it, complement of the column space. Which are two pretty neat takeaways. We saw a particular example of it a couple of videos ago. And now you see that it's true for all matrices."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And in this case, we'll calculate the null space of matrix A. So null space is literally just a set of all the vectors that when I multiply A times any of those vectors. So let me say that the vector x1, x2, x3, x4 is a member of our null space. When I multiply this matrix times this vector, I should get the zero vector. I should get the vector. And just to make a few points here, this has exactly four columns. This is a 3 by 4 matrix."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "When I multiply this matrix times this vector, I should get the zero vector. I should get the vector. And just to make a few points here, this has exactly four columns. This is a 3 by 4 matrix. So I've only legitimately defined multiplication of this times a four-component vector, or a member of Rn. Let me call this x. If this is our vector x, this is a member of R4."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is a 3 by 4 matrix. So I've only legitimately defined multiplication of this times a four-component vector, or a member of Rn. Let me call this x. If this is our vector x, this is a member of R4. It has four components. And then when you multiply these, we need to produce the zero vector. The null space is the set of all the vectors, and when I multiply it times A, I produce the zero vector."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "If this is our vector x, this is a member of R4. It has four components. And then when you multiply these, we need to produce the zero vector. The null space is the set of all the vectors, and when I multiply it times A, I produce the zero vector. And what am I going to get? I'm going to have one row times this, and that's going to be the first entry. Then this row times that's the second entry, then the third row."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "The null space is the set of all the vectors, and when I multiply it times A, I produce the zero vector. And what am I going to get? I'm going to have one row times this, and that's going to be the first entry. Then this row times that's the second entry, then the third row. So I should have three zeros. So my zero vector is going to be kind of the zero vector in R3. So how do we figure out the set of all of these x's that satisfy this?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Then this row times that's the second entry, then the third row. So I should have three zeros. So my zero vector is going to be kind of the zero vector in R3. So how do we figure out the set of all of these x's that satisfy this? Let me just write kind of our formal notation. The null space of A is the set of all vectors that are a member of, generally you say Rn, but this is a 3 by 4 matrix, so these are all the vectors that are going to be members of R4, because I'm using this particular A, such that my matrix A times any of these vectors is equal to the zero vector, and in this case it's going to be the zero vector in R3. So how do we do this?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So how do we figure out the set of all of these x's that satisfy this? Let me just write kind of our formal notation. The null space of A is the set of all vectors that are a member of, generally you say Rn, but this is a 3 by 4 matrix, so these are all the vectors that are going to be members of R4, because I'm using this particular A, such that my matrix A times any of these vectors is equal to the zero vector, and in this case it's going to be the zero vector in R3. So how do we do this? Well, this is just a straight up linear equation. We can write it that way. If we were to actually do perform the matrix multiplication, we get 1 times x1."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So how do we do this? Well, this is just a straight up linear equation. We can write it that way. If we were to actually do perform the matrix multiplication, we get 1 times x1. Let me write it here. Let me do it in a different color. 1 times x1 plus 1 times x2 plus 1 times x3 plus 1 times x4 is equal to this zero there."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "If we were to actually do perform the matrix multiplication, we get 1 times x1. Let me write it here. Let me do it in a different color. 1 times x1 plus 1 times x2 plus 1 times x3 plus 1 times x4 is equal to this zero there. It's equal to that zero. So that times that is equal to that zero. And then this times this should be equal to that zero."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 times x1 plus 1 times x2 plus 1 times x3 plus 1 times x4 is equal to this zero there. It's equal to that zero. So that times that is equal to that zero. And then this times this should be equal to that zero. So 1 times x1, so you get x1 plus 2 times x2 plus 3 times x3 plus 4 times x4 is going to be equal to that zero. And then finally, we have that times this vector, should be equal to that zero. So you can have the dot product of this row vector with this convector should be equal to that zero."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then this times this should be equal to that zero. So 1 times x1, so you get x1 plus 2 times x2 plus 3 times x3 plus 4 times x4 is going to be equal to that zero. And then finally, we have that times this vector, should be equal to that zero. So you can have the dot product of this row vector with this convector should be equal to that zero. So you get 4x1 plus 3x2 plus 2x1 plus x4. Oh, sorry. What am I doing?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So you can have the dot product of this row vector with this convector should be equal to that zero. So you get 4x1 plus 3x2 plus 2x1 plus x4. Oh, sorry. What am I doing? Plus 2x3 plus x4 is equal to zero. 4x1 plus 3x2 plus 2x3 plus x4 is equal to zero. And we just have to find the solution set to this, and we'll essentially have figured out our null space."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "What am I doing? Plus 2x3 plus x4 is equal to zero. 4x1 plus 3x2 plus 2x3 plus x4 is equal to zero. And we just have to find the solution set to this, and we'll essentially have figured out our null space. Now, we've figured out the solution set to a system of equations like this. We have three equations with four unknowns. And we can do that."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And we just have to find the solution set to this, and we'll essentially have figured out our null space. Now, we've figured out the solution set to a system of equations like this. We have three equations with four unknowns. And we can do that. We can represent this by an augmented matrix and then put that in reduced row echelon form. So let's do that. So I can represent this problem as the augmented matrix, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, 3, 1, 2, 3, 1, 3, 2, 1, 3, 2, and then 1, 4, 1."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And we can do that. We can represent this by an augmented matrix and then put that in reduced row echelon form. So let's do that. So I can represent this problem as the augmented matrix, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, 3, 1, 2, 3, 1, 3, 2, 1, 3, 2, and then 1, 4, 1. And then I augment that with the zero vector. And the immediate thing you should notice is, look, we took the pain of multiplying this times this to equal that, and we wrote this as a system of equations. But now that we want to solve the system of equations, we're kind of going back to the augmented matrix world."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So I can represent this problem as the augmented matrix, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, 3, 1, 2, 3, 1, 3, 2, 1, 3, 2, and then 1, 4, 1. And then I augment that with the zero vector. And the immediate thing you should notice is, look, we took the pain of multiplying this times this to equal that, and we wrote this as a system of equations. But now that we want to solve the system of equations, we're kind of going back to the augmented matrix world. And what does this augmented matrix look like? Well, this is just our matrix A right there. And that's just the zero vector right there."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "But now that we want to solve the system of equations, we're kind of going back to the augmented matrix world. And what does this augmented matrix look like? Well, this is just our matrix A right there. And that's just the zero vector right there. And to solve this, and we've done this before, we're just going to put this augmented matrix into row echelon form. And what you're going to find is when you put it in a row echelon form, this right side's not going to change at all, because no matter what you multiply or subtract by, you're just doing it all times 0. So you just keep ending up with 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And that's just the zero vector right there. And to solve this, and we've done this before, we're just going to put this augmented matrix into row echelon form. And what you're going to find is when you put it in a row echelon form, this right side's not going to change at all, because no matter what you multiply or subtract by, you're just doing it all times 0. So you just keep ending up with 0. So as we put this into reduced row echelon form, we're actually just putting matrix A into reduced echelon form. So let me do that instead of just talking about it. So let me start off by keeping row 1 the same."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So you just keep ending up with 0. So as we put this into reduced row echelon form, we're actually just putting matrix A into reduced echelon form. So let me do that instead of just talking about it. So let me start off by keeping row 1 the same. Row 1 is 1, 1, 1, 1, 0. And then I want to eliminate this one right here. So let me just subtract."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let me start off by keeping row 1 the same. Row 1 is 1, 1, 1, 1, 0. And then I want to eliminate this one right here. So let me just subtract. Let me replace row 2 with row 2 minus row 1. So 1 minus 1 is 0. 2 minus 1 is 1."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So let me just subtract. Let me replace row 2 with row 2 minus row 1. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. 4 minus 1 is 3. 0 minus 0 is 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 1 is 1. 3 minus 1 is 2. 4 minus 1 is 3. 0 minus 0 is 0. You can see the 0's aren't going to change. And then let me replace this guy with 4 times this guy minus this guy. So I want to get rid of this."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "0 minus 0 is 0. You can see the 0's aren't going to change. And then let me replace this guy with 4 times this guy minus this guy. So I want to get rid of this. So 4 times 1 minus 4 is 0. 4 times 1 minus 3 is 1. 4 times 1 minus 2 is 2."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So I want to get rid of this. So 4 times 1 minus 4 is 0. 4 times 1 minus 3 is 1. 4 times 1 minus 2 is 2. 4 times 1 minus 1 is 3. 4 times 0 minus 0 is 0. Now I want to get rid of, if I want to put this in reduced row echelon form, I want to get rid of that term and that term."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "4 times 1 minus 2 is 2. 4 times 1 minus 1 is 3. 4 times 0 minus 0 is 0. Now I want to get rid of, if I want to put this in reduced row echelon form, I want to get rid of that term and that term. So let me keep my middle row the same. My middle row is 0, 1, 2, 3. Middle row the same."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Now I want to get rid of, if I want to put this in reduced row echelon form, I want to get rid of that term and that term. So let me keep my middle row the same. My middle row is 0, 1, 2, 3. Middle row the same. So that's 0 on the augmented side of it, although these 0's are never going to change. So it's really just a little bit of an exercise just to keep writing them. And then my first row, let me replace it with the first row minus the second row so I can get rid of this 1."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Middle row the same. So that's 0 on the augmented side of it, although these 0's are never going to change. So it's really just a little bit of an exercise just to keep writing them. And then my first row, let me replace it with the first row minus the second row so I can get rid of this 1. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then my first row, let me replace it with the first row minus the second row so I can get rid of this 1. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1. 1 minus 3 is minus 2. And 0 minus 0 is 0. And let me replace this last row with the last row minus the middle row."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "1 minus 2 is minus 1. 1 minus 3 is minus 2. And 0 minus 0 is 0. And let me replace this last row with the last row minus the middle row. So 0 minus 0 is 0. 1 minus 1 is 0. 2 minus 2 is 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And let me replace this last row with the last row minus the middle row. So 0 minus 0 is 0. 1 minus 1 is 0. 2 minus 2 is 0. I think you see where this is going. 3 minus 3 is 0. And obviously 0 minus 0 is 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 2 is 0. I think you see where this is going. 3 minus 3 is 0. And obviously 0 minus 0 is 0. So this system of equations has been reduced to, just by doing reduced row echelon form, this problem. If I just rewrite this right here, this can be written as a system of equations of x1 minus x3 minus x4, the 0 x2's, is equal to 0. And then this second row right here, it could be written as there's no x1, so you just have an x2 plus 2x3 plus 3x2 is equal to 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And obviously 0 minus 0 is 0. So this system of equations has been reduced to, just by doing reduced row echelon form, this problem. If I just rewrite this right here, this can be written as a system of equations of x1 minus x3 minus x4, the 0 x2's, is equal to 0. And then this second row right here, it could be written as there's no x1, so you just have an x2 plus 2x3 plus 3x2 is equal to 0. And this obviously gives me no information whatsoever. And so I can solve this for x1 and x2. And what do I get?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And then this second row right here, it could be written as there's no x1, so you just have an x2 plus 2x3 plus 3x2 is equal to 0. And this obviously gives me no information whatsoever. And so I can solve this for x1 and x2. And what do I get? I get x1 is equal to x3 plus x4. Actually, I made a mistake here. This is x1 minus x3 minus 2 times x4 is equal to 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And what do I get? I get x1 is equal to x3 plus x4. Actually, I made a mistake here. This is x1 minus x3 minus 2 times x4 is equal to 0. So if I rewrite this, I get x1 is equal to x3 plus 2x4. And then I get x2 is equal to minus 2x3 minus 3x2. And so if I wanted to write the solution set to this equation, if I wanted to write it in terms of this, I could write x1, x2, x3, x4 is equal to, what's x1 equal to?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "This is x1 minus x3 minus 2 times x4 is equal to 0. So if I rewrite this, I get x1 is equal to x3 plus 2x4. And then I get x2 is equal to minus 2x3 minus 3x2. And so if I wanted to write the solution set to this equation, if I wanted to write it in terms of this, I could write x1, x2, x3, x4 is equal to, what's x1 equal to? It's equal to x3 times 1 plus x4 times 2. I just got this right here from this equation right here. x1 is equal to 1 times x3 plus 2 times x4."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And so if I wanted to write the solution set to this equation, if I wanted to write it in terms of this, I could write x1, x2, x3, x4 is equal to, what's x1 equal to? It's equal to x3 times 1 plus x4 times 2. I just got this right here from this equation right here. x1 is equal to 1 times x3 plus 2 times x4. That's just that right there. Now x2 is equal to x3 times minus 2 plus x4 times minus 3. What am I doing?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "x1 is equal to 1 times x3 plus 2 times x4. That's just that right there. Now x2 is equal to x3 times minus 2 plus x4 times minus 3. What am I doing? I'm losing track of things. This x2 right here is x2 plus 2x3 plus 3x4 is equal to 0. So x2 is equal to minus 2x3 minus 3x4."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "What am I doing? I'm losing track of things. This x2 right here is x2 plus 2x3 plus 3x4 is equal to 0. So x2 is equal to minus 2x3 minus 3x4. Write it like that. Sorry, my brain isn't completely in the problem. I'm making these silly mistakes."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So x2 is equal to minus 2x3 minus 3x4. Write it like that. Sorry, my brain isn't completely in the problem. I'm making these silly mistakes. But I think you understand this now. And then what is x3 equal to? Well, it's just equal to 1 times x3 plus 0 times x4."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "I'm making these silly mistakes. But I think you understand this now. And then what is x3 equal to? Well, it's just equal to 1 times x3 plus 0 times x4. Right? x3 is equal to x3. And what's x4 equal to?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's just equal to 1 times x3 plus 0 times x4. Right? x3 is equal to x3. And what's x4 equal to? It's equal to 0 times x3 plus 1 times x4. So all of the vectors in R4, these are a member of R4, which satisfy the equation, our original equation, Ax is equal to 0, can be represented as a linear combination of these two vectors, of those two vectors. Right?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And what's x4 equal to? It's equal to 0 times x3 plus 1 times x4. So all of the vectors in R4, these are a member of R4, which satisfy the equation, our original equation, Ax is equal to 0, can be represented as a linear combination of these two vectors, of those two vectors. Right? These are just random scalars that are a member of, we can pick any real number for x3, and we could pick any real number for x4. So our solution set is just a linear combination of those two vectors. Well, what's another way of saying a linear combination of two vectors?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Right? These are just random scalars that are a member of, we can pick any real number for x3, and we could pick any real number for x4. So our solution set is just a linear combination of those two vectors. Well, what's another way of saying a linear combination of two vectors? So let me write this. The null space of A, which is just a solution set of this equation, it's just all of the x's that satisfy this equation, it equals all of the linear combinations of this vector and that vector. Well, what do we call all the linear combinations of two vectors?"}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well, what's another way of saying a linear combination of two vectors? So let me write this. The null space of A, which is just a solution set of this equation, it's just all of the x's that satisfy this equation, it equals all of the linear combinations of this vector and that vector. Well, what do we call all the linear combinations of two vectors? It's the span of those two vectors. So it equals the span of that vector and that vector, of the vector 1, minus 2, 1, 0, and the vector 2, minus 3, 0, 1. And this is our null space."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "Well, what do we call all the linear combinations of two vectors? It's the span of those two vectors. So it equals the span of that vector and that vector, of the vector 1, minus 2, 1, 0, and the vector 2, minus 3, 0, 1. And this is our null space. And before letting it go, let me just point out one interesting thing right here. We represented our system of equation like this, and we put it into reduced row echelon form. So this is A and this is 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "And this is our null space. And before letting it go, let me just point out one interesting thing right here. We represented our system of equation like this, and we put it into reduced row echelon form. So this is A and this is 0. This right here is the reduced row echelon form of A. And so we're essentially, this equation, this is a linear equation that's trying to solve this problem. The reduced row echelon form of A times our vector x is equal to 0."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So this is A and this is 0. This right here is the reduced row echelon form of A. And so we're essentially, this equation, this is a linear equation that's trying to solve this problem. The reduced row echelon form of A times our vector x is equal to 0. So all the solutions to this are also the solutions to our original problem, to our original Ax is equal to 0. So what's the solution to this? All the x's that satisfy this, these are the null space of the reduced row echelon form of A."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "The reduced row echelon form of A times our vector x is equal to 0. So all the solutions to this are also the solutions to our original problem, to our original Ax is equal to 0. So what's the solution to this? All the x's that satisfy this, these are the null space of the reduced row echelon form of A. So here, all of the x's, this is the null space, this problem, if we find all the solutions that all of the x's here, this is the null space of the reduced row echelon form of our matrix A. But we're saying that this problem is the same problem as this one, right? So we can write that the null space of A is equal to the null space of the reduced row echelon form of A."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "All the x's that satisfy this, these are the null space of the reduced row echelon form of A. So here, all of the x's, this is the null space, this problem, if we find all the solutions that all of the x's here, this is the null space of the reduced row echelon form of our matrix A. But we're saying that this problem is the same problem as this one, right? So we can write that the null space of A is equal to the null space of the reduced row echelon form of A. And that might seem a little bit confusing. Hey, why are you even writing this out? But it's actually very useful when you're actually trying to calculate null spaces."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "So we can write that the null space of A is equal to the null space of the reduced row echelon form of A. And that might seem a little bit confusing. Hey, why are you even writing this out? But it's actually very useful when you're actually trying to calculate null spaces. Because we didn't even have to write a big augmented matrix here, we could just say, OK, take our matrix A, put it in reduced row echelon form, and then figure out its null space. So we would have gone straight to this point right here. This is the reduced row echelon form of A."}, {"video_title": "Null space 2 Calculating the null space of a matrix Linear Algebra Khan Academy.mp3", "Sentence": "But it's actually very useful when you're actually trying to calculate null spaces. Because we didn't even have to write a big augmented matrix here, we could just say, OK, take our matrix A, put it in reduced row echelon form, and then figure out its null space. So we would have gone straight to this point right here. This is the reduced row echelon form of A. And then I could have immediately solved these equations, right? I would have just taken the dot product of the reduced row echelon form, or not the dot product, the matrix vector product of the reduced row echelon form of A with this vector, and I would have gotten these equations. And then these equations would have immediately, I can just kind of rewrite them in this form, and I would have gotten our result."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's got n components in it. So v1, v2, all the way down to vn. I've touched on the idea before, but now that we've seen what a transpose is and we've taken transposes of matrices, there's no reason why we can't take the transpose of a vector, or a column vector in this case. So what would v transpose look like? Well, if you think of this as a n by 1 matrix, which it is, it has n rows and 1 column, then what are we going to get? We're going to have a 1 by n matrix when you take the transpose of it. And this 1 column is going to turn into the 1 row."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what would v transpose look like? Well, if you think of this as a n by 1 matrix, which it is, it has n rows and 1 column, then what are we going to get? We're going to have a 1 by n matrix when you take the transpose of it. And this 1 column is going to turn into the 1 row. So you're going to have it be equal to v1, v2, all the way to vn. And you might remember, we've already touched on this in a lot of matrices before. Let's say that's some matrix A."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this 1 column is going to turn into the 1 row. So you're going to have it be equal to v1, v2, all the way to vn. And you might remember, we've already touched on this in a lot of matrices before. Let's say that's some matrix A. We called the row vectors of those matrix, we call them the transpose of some column vectors. A transpose, A1 transpose, A2 transpose, all the way down to An transpose. In fact, not so many videos ago I had those row vectors, and I could have just called them the transpose of column vectors just like that."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that's some matrix A. We called the row vectors of those matrix, we call them the transpose of some column vectors. A transpose, A1 transpose, A2 transpose, all the way down to An transpose. In fact, not so many videos ago I had those row vectors, and I could have just called them the transpose of column vectors just like that. And that would have been, in some ways, a better way to do it, because we've defined all these operations around column vectors, so you could always refer to the transpose of the transpose and then do some operations on them. But anyway, I don't want to get too diverted, but let's think a little bit of what happens when you operate this vector, or you take some operation of this vector with some other vectors. So let's say I have another vector here that's w, and it's also a member of Rn."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "In fact, not so many videos ago I had those row vectors, and I could have just called them the transpose of column vectors just like that. And that would have been, in some ways, a better way to do it, because we've defined all these operations around column vectors, so you could always refer to the transpose of the transpose and then do some operations on them. But anyway, I don't want to get too diverted, but let's think a little bit of what happens when you operate this vector, or you take some operation of this vector with some other vectors. So let's say I have another vector here that's w, and it's also a member of Rn. So you have w1, w2, all the way down to wn. There's a couple of things that were already, I think, reasonably familiar with. You could take the dot product of v and w. v dot w is equal to what?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have another vector here that's w, and it's also a member of Rn. So you have w1, w2, all the way down to wn. There's a couple of things that were already, I think, reasonably familiar with. You could take the dot product of v and w. v dot w is equal to what? It is equal to v1 times w1, v1 w1, plus v2 w2, plus v2 w2, and you just keep going all the way to vn wn. This is the definition of the dot product of two column vectors. Now, how can we relate that to maybe the transpose of v?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could take the dot product of v and w. v dot w is equal to what? It is equal to v1 times w1, v1 w1, plus v2 w2, plus v2 w2, and you just keep going all the way to vn wn. This is the definition of the dot product of two column vectors. Now, how can we relate that to maybe the transpose of v? Well, we could take the transpose of v. Let me write it this way. What is, if I did a matrix multiplication, so I did v1, v2, all the way to vn, so this is v transpose. That's v transpose."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, how can we relate that to maybe the transpose of v? Well, we could take the transpose of v. Let me write it this way. What is, if I did a matrix multiplication, so I did v1, v2, all the way to vn, so this is v transpose. That's v transpose. And I take the product of that with w. I take the product of that with w. So I have w1, w2, all the way down to wn. Now, if I view these as just matrices, this is w right here. If I viewed these just as matrices, is this matrix product well-defined?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's v transpose. And I take the product of that with w. I take the product of that with w. So I have w1, w2, all the way down to wn. Now, if I view these as just matrices, this is w right here. If I viewed these just as matrices, is this matrix product well-defined? See, over here I have a n by 1 matrix. Here I have a 1, sorry, here the first one I have is a 1 by n matrix. I have one row and n columns."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I viewed these just as matrices, is this matrix product well-defined? See, over here I have a n by 1 matrix. Here I have a 1, sorry, here the first one I have is a 1 by n matrix. I have one row and n columns. And here I have an n by 1 matrix. I have n rows and only one column. So this is well-defined."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I have one row and n columns. And here I have an n by 1 matrix. I have n rows and only one column. So this is well-defined. I have the same number of columns here as I have rows here. And this is going to result in a 1 by 1 matrix. And what's it going to look like?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is well-defined. I have the same number of columns here as I have rows here. And this is going to result in a 1 by 1 matrix. And what's it going to look like? It's going to be equal to v1 times w1. Let me write it like this. v1 w1 plus v2 w2."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what's it going to look like? It's going to be equal to v1 times w1. Let me write it like this. v1 w1 plus v2 w2. It's only going to have one entry. We could write it as just a 1 by 1 matrix like that. Let me just do it."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "v1 w1 plus v2 w2. It's only going to have one entry. We could write it as just a 1 by 1 matrix like that. Let me just do it. 1 by 1 matrix like that. v1 w1 plus v2 w2. Let me just, I could write v2 there."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me just do it. 1 by 1 matrix like that. v1 w1 plus v2 w2. Let me just, I could write v2 there. Plus all the way to vn wn. That's what it'll be. It'll just be a 1 by 1 matrix that looks like that."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me just, I could write v2 there. Plus all the way to vn wn. That's what it'll be. It'll just be a 1 by 1 matrix that looks like that. But you might notice that these two things are equivalent. So we can make the statement that v dot w, which is the same thing as w dot d, these things are equivalent to v dot w is the equivalent of, let me just write it once over here. v dot w is the same thing as the transpose of v times w. As just a matrix-matrix product."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It'll just be a 1 by 1 matrix that looks like that. But you might notice that these two things are equivalent. So we can make the statement that v dot w, which is the same thing as w dot d, these things are equivalent to v dot w is the equivalent of, let me just write it once over here. v dot w is the same thing as the transpose of v times w. As just a matrix-matrix product. So if you view v as a matrix, take its transpose and then just take that matrix and take the product of that with w. It's the same thing as v dot w. So that's an interesting takeaway. I guess you could argue somewhat obvious. And we've already been referring this, when I define matrix-matrix products, I kind of said you're taking the dot product of each row with each column."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "v dot w is the same thing as the transpose of v times w. As just a matrix-matrix product. So if you view v as a matrix, take its transpose and then just take that matrix and take the product of that with w. It's the same thing as v dot w. So that's an interesting takeaway. I guess you could argue somewhat obvious. And we've already been referring this, when I define matrix-matrix products, I kind of said you're taking the dot product of each row with each column. And you can see that it really is. It's really the dot product of the transpose of that row with each column. But you got the general idea."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we've already been referring this, when I define matrix-matrix products, I kind of said you're taking the dot product of each row with each column. And you can see that it really is. It's really the dot product of the transpose of that row with each column. But you got the general idea. But let's see if we can build on this a little bit. Let's say I have some matrix A. Let me save our little outcome that I have there."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But you got the general idea. But let's see if we can build on this a little bit. Let's say I have some matrix A. Let me save our little outcome that I have there. Let's say I have some, let me get a good color here. Let's say I have some matrix A. And it's an m by n matrix."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me save our little outcome that I have there. Let's say I have some, let me get a good color here. Let's say I have some matrix A. And it's an m by n matrix. Now, if I were to multiply that times a vector x, so I'm going to multiply it by some vector x. And let's say that x is a member of Rn. So it has n elements."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And it's an m by n matrix. Now, if I were to multiply that times a vector x, so I'm going to multiply it by some vector x. And let's say that x is a member of Rn. So it has n elements. Or another way you could view it is it's an n by 1 matrix. Now, when I take the product of these, what am I going to get, or another way to say it is, what is the vector Ax? When I take this product, I'm just going to get another vector, and what's it going to be?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it has n elements. Or another way you could view it is it's an n by 1 matrix. Now, when I take the product of these, what am I going to get, or another way to say it is, what is the vector Ax? When I take this product, I'm just going to get another vector, and what's it going to be? It's going to be an m by 1 vector. This is going to be an m by 1 vector. So we could say that Ax is a member of Rm."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When I take this product, I'm just going to get another vector, and what's it going to be? It's going to be an m by 1 vector. This is going to be an m by 1 vector. So we could say that Ax is a member of Rm. It's going to have m elements. If this was equal to, if you said that Ax is equal to, I don't know, let's say it's equal to z, z would have m elements. You would have z1, z2, all the way down to zm."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that Ax is a member of Rm. It's going to have m elements. If this was equal to, if you said that Ax is equal to, I don't know, let's say it's equal to z, z would have m elements. You would have z1, z2, all the way down to zm. And I know that because you have m rows in A, and you have only one, well, you could say this is m by n. This is n by 1. The resulting product will be m by 1, or it'll be a vector that is a member of Rm. It'll have exactly m elements."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You would have z1, z2, all the way down to zm. And I know that because you have m rows in A, and you have only one, well, you could say this is m by n. This is n by 1. The resulting product will be m by 1, or it'll be a vector that is a member of Rm. It'll have exactly m elements. Now, if that's a vector of Rm, then the idea of dotting this with another member of Rm is well-defined. So let's say that I have another member of Rm. Let's say I have a vector y."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It'll have exactly m elements. Now, if that's a vector of Rm, then the idea of dotting this with another member of Rm is well-defined. So let's say that I have another member of Rm. Let's say I have a vector y. Let's say y is also a member of Rm. This has, the vector Ax, the vector that you get when you take this product, has m elements. This has m elements."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have a vector y. Let's say y is also a member of Rm. This has, the vector Ax, the vector that you get when you take this product, has m elements. This has m elements. So the idea of taking their dot product is well-defined. Let me write that. So we could take Ax, that's a vector, and now we are dotting it with this vector right here."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This has m elements. So the idea of taking their dot product is well-defined. Let me write that. So we could take Ax, that's a vector, and now we are dotting it with this vector right here. And we'll get a number. We just take each of their terms, multiply the corresponding terms, add them all up, and you get their dot product. But what is this equal to?"}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could take Ax, that's a vector, and now we are dotting it with this vector right here. And we'll get a number. We just take each of their terms, multiply the corresponding terms, add them all up, and you get their dot product. But what is this equal to? We can just use this little, I guess you could call it a result that we got earlier on in this video. Using this result, the dot product of two vectors is equal to the transpose of the first vector as kind of a matrix, so you can view this as Ax transpose. This is a m by 1."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But what is this equal to? We can just use this little, I guess you could call it a result that we got earlier on in this video. Using this result, the dot product of two vectors is equal to the transpose of the first vector as kind of a matrix, so you can view this as Ax transpose. This is a m by 1. This is m by 1. Now this is now a 1 by m matrix. And now we can multiply 1 by m matrix times y."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a m by 1. This is m by 1. Now this is now a 1 by m matrix. And now we can multiply 1 by m matrix times y. Just like that. Now, what is this thing equal to? We saw a while ago, I think it was two or three videos ago, we saw that if we take the product of two matrices and take its transpose, that's equal to the reverse product of the transposes."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And now we can multiply 1 by m matrix times y. Just like that. Now, what is this thing equal to? We saw a while ago, I think it was two or three videos ago, we saw that if we take the product of two matrices and take its transpose, that's equal to the reverse product of the transposes. You just switch the order and then take the transposes. So this is going to be equal to, this purple part is going to be equal to x transpose times A transpose times y. And this is just matrix products."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We saw a while ago, I think it was two or three videos ago, we saw that if we take the product of two matrices and take its transpose, that's equal to the reverse product of the transposes. You just switch the order and then take the transposes. So this is going to be equal to, this purple part is going to be equal to x transpose times A transpose times y. And this is just matrix products. These are matrix products. These aren't necessarily vector operations. We're treating all of these vectors as matrices."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is just matrix products. These are matrix products. These aren't necessarily vector operations. We're treating all of these vectors as matrices. And of course, we're treating the matrix as a matrix. So what is this equal to? Well, we know that matrix products are associative."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're treating all of these vectors as matrices. And of course, we're treating the matrix as a matrix. So what is this equal to? Well, we know that matrix products are associative. You could put a parentheses, right now we have a parentheses around there from there, but we could just take another association. We could say that that is equal to x transpose times these two matrices times each other. This is a vector, but you can represent it as an m by 1 matrix, so times A transpose y."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we know that matrix products are associative. You could put a parentheses, right now we have a parentheses around there from there, but we could just take another association. We could say that that is equal to x transpose times these two matrices times each other. This is a vector, but you can represent it as an m by 1 matrix, so times A transpose y. Just like that. Now let's think about what A transpose y is. Let's think about it."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is a vector, but you can represent it as an m by 1 matrix, so times A transpose y. Just like that. Now let's think about what A transpose y is. Let's think about it. A transpose, we have here, A is m by n. What is A transpose? A transpose is going to be n by m. It's going to be an n by m. So this is an n by m. And then what is this vector y going to be? This is an m by 1."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's think about it. A transpose, we have here, A is m by n. What is A transpose? A transpose is going to be n by m. It's going to be an n by m. So this is an n by m. And then what is this vector y going to be? This is an m by 1. So when you take this product, you're going to get an n by 1 matrix, or you could imagine this as a vector that is a member of Rn. So this is a member of Rn. The entire product is going to result with a vector that's a member of Rn."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is an m by 1. So when you take this product, you're going to get an n by 1 matrix, or you could imagine this as a vector that is a member of Rn. So this is a member of Rn. The entire product is going to result with a vector that's a member of Rn. And of course, it's well-defined, because this is a 1 by n vector right there. Now we can go back to our identity. We have the transpose of some vector times some other vector, they have the same, well I guess you could say this has as many horizontal entries as this guy has vertical entries, just like that."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The entire product is going to result with a vector that's a member of Rn. And of course, it's well-defined, because this is a 1 by n vector right there. Now we can go back to our identity. We have the transpose of some vector times some other vector, they have the same, well I guess you could say this has as many horizontal entries as this guy has vertical entries, just like that. So what is this equal to? We just use that identity. This is equal to, though just regular x in this case, instead of x transpose, we'll just have x."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have the transpose of some vector times some other vector, they have the same, well I guess you could say this has as many horizontal entries as this guy has vertical entries, just like that. So what is this equal to? We just use that identity. This is equal to, though just regular x in this case, instead of x transpose, we'll just have x. So this is equal to x dot, remember we just untranspose it, I guess you can view it that way, dot a transpose y. Which is a pretty neat outcome. We got this being equal to that."}, {"video_title": "Transpose of a vector Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to, though just regular x in this case, instead of x transpose, we'll just have x. So this is equal to x dot, remember we just untranspose it, I guess you can view it that way, dot a transpose y. Which is a pretty neat outcome. We got this being equal to that. We can kind of change the associativity, although we have to essentially change the order a bit and take the transpose of our matrix. So let me rewrite that, just so that you can remember the outcome. So the two big outcomes of this video are, I'll rewrite this one up here, v dot w is equal to the matrix product of v transpose times w. And if I have some matrix, assume all of these matrix vector products are well-defined and all the dot products are well-defined."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "I've talked a lot about the idea that eigenvectors could make for good bases, for good basis vectors. So let's explore that idea a little bit more. Let's say I have some transformation. Let's say it's a transformation from Rn to Rn. And it can be represented by the matrix A. So the transformation of x is equal to the n by n matrix A, it's n by n, times x. Now, let's say that we have n linearly independent eigenvectors of A."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's a transformation from Rn to Rn. And it can be represented by the matrix A. So the transformation of x is equal to the n by n matrix A, it's n by n, times x. Now, let's say that we have n linearly independent eigenvectors of A. So let's say, and this isn't always going to be the case, but it can often be the case. It's definitely possible. Let's say, let's assume that A has n linearly independent eigenvectors."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's say that we have n linearly independent eigenvectors of A. So let's say, and this isn't always going to be the case, but it can often be the case. It's definitely possible. Let's say, let's assume that A has n linearly independent eigenvectors. So I'm going to call them v1, v2, all the way through vn. Now, n linearly independent vectors in Rn can definitely be a basis for Rn. We've seen that multiple times."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Let's say, let's assume that A has n linearly independent eigenvectors. So I'm going to call them v1, v2, all the way through vn. Now, n linearly independent vectors in Rn can definitely be a basis for Rn. We've seen that multiple times. And what I want to show you in this video is that this makes a particularly good basis for this transformation. So let's explore that. So the transformation of each of these vectors, I'm going to write it over here, the transformation of vector 1 is equal to A times vector 1."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "We've seen that multiple times. And what I want to show you in this video is that this makes a particularly good basis for this transformation. So let's explore that. So the transformation of each of these vectors, I'm going to write it over here, the transformation of vector 1 is equal to A times vector 1. And since vector 1 is an eigenvector of A, that's going to be equal to some eigenvalue lambda 1 times vector 1. We could do that for all of them. The transformation of vector 2 is equal to A times v2, which is equal to some eigenvalue lambda 2 times v2."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So the transformation of each of these vectors, I'm going to write it over here, the transformation of vector 1 is equal to A times vector 1. And since vector 1 is an eigenvector of A, that's going to be equal to some eigenvalue lambda 1 times vector 1. We could do that for all of them. The transformation of vector 2 is equal to A times v2, which is equal to some eigenvalue lambda 2 times v2. And I'm just going to skip all of them and just go straight to the nth one. We have n of these eigenvectors. You might have a lot more."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "The transformation of vector 2 is equal to A times v2, which is equal to some eigenvalue lambda 2 times v2. And I'm just going to skip all of them and just go straight to the nth one. We have n of these eigenvectors. You might have a lot more. We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "You might have a lot more. We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see. So the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector vn. Now, what are these also equal to?"}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Let's see. So the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector vn. Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times v1 plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 plus 0 times v2 plus 0 times all of these basis vectors, these eigenvectors, but lambda n times vn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times v1 plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 plus 0 times v2 plus 0 times all of these basis vectors, these eigenvectors, but lambda n times vn. I mean, this is almost stunningly obvious. I just rewrote this as this plus a bunch of 0 vectors. But the reason why I wrote that is because in a second we're going to take this as a basis, and we're going to find coordinates with respect to that basis."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "And then this guy down here, this is going to be 0 times v1 plus 0 times v2 plus 0 times all of these basis vectors, these eigenvectors, but lambda n times vn. I mean, this is almost stunningly obvious. I just rewrote this as this plus a bunch of 0 vectors. But the reason why I wrote that is because in a second we're going to take this as a basis, and we're going to find coordinates with respect to that basis. And so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "But the reason why I wrote that is because in a second we're going to take this as a basis, and we're going to find coordinates with respect to that basis. And so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis. So B is equal to the set of, actually I don't even have to write it that way. Let's say I say that B, I have some basis B that's equal to that. What I want to show you is that when I do a change of basis, so we've seen this before."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that we define this as some basis. So B is equal to the set of, actually I don't even have to write it that way. Let's say I say that B, I have some basis B that's equal to that. What I want to show you is that when I do a change of basis, so we've seen this before. In my standard coordinates, or in coordinates with respect to the standard basis, you give me some vector in Rn. I'm going to multiply it times a, and you're going to have the transformation of it. It's also going to be in Rn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "What I want to show you is that when I do a change of basis, so we've seen this before. In my standard coordinates, or in coordinates with respect to the standard basis, you give me some vector in Rn. I'm going to multiply it times a, and you're going to have the transformation of it. It's also going to be in Rn. Now we know we could do a change of basis. And in a change of basis, if you want to go in that way, you multiply by C inverse, which is, and remember, the change of basis matrix, C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's also going to be in Rn. Now we know we could do a change of basis. And in a change of basis, if you want to go in that way, you multiply by C inverse, which is, and remember, the change of basis matrix, C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct. But if you change your basis from x to our new basis, you multiply by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as a transpose."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's very easy to construct. But if you change your basis from x to our new basis, you multiply by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as a transpose. We can't assume that, though. And so this is going to be x in our new basis. And if we want to find the transformation matrix for t with respect to our new basis, it's going to be some matrix d. And if you multiply d times x, you're going to get this guy, but you're going to get the b representation of that guy."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "If they're all orthonormal, then this is the same thing as a transpose. We can't assume that, though. And so this is going to be x in our new basis. And if we want to find the transformation matrix for t with respect to our new basis, it's going to be some matrix d. And if you multiply d times x, you're going to get this guy, but you're going to get the b representation of that guy. The transformation of the vector x is b representation. And if we want to go back and forth between that guy and that guy, if we want to go in this direction, you can multiply this times C, and you'll just get the transformation of x. And if you want to go in that direction, you can multiply by the inverse of your change of basis matrix."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "And if we want to find the transformation matrix for t with respect to our new basis, it's going to be some matrix d. And if you multiply d times x, you're going to get this guy, but you're going to get the b representation of that guy. The transformation of the vector x is b representation. And if we want to go back and forth between that guy and that guy, if we want to go in this direction, you can multiply this times C, and you'll just get the transformation of x. And if you want to go in that direction, you can multiply by the inverse of your change of basis matrix. We've seen this multiple times already. But what I've claimed, or I've kind of hinted at, is that if I have a basis that's defined by eigenvectors of A, that this will be a very nice matrix. So this might be the coordinate system that you want to operate in, especially if you're going to apply this matrix a lot."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "And if you want to go in that direction, you can multiply by the inverse of your change of basis matrix. We've seen this multiple times already. But what I've claimed, or I've kind of hinted at, is that if I have a basis that's defined by eigenvectors of A, that this will be a very nice matrix. So this might be the coordinate system that you want to operate in, especially if you're going to apply this matrix a lot. If you're going to do this transformation on a lot of different things, you're going to do it over and over and over again, maybe to the same set, then it maybe is worth the overhead to do the conversion and just use this as your coordinate system. So let's see that this will actually be a nice looking, easy to compute with, and actually diagonal matrix. So we know that the transformation of, let's write this in a bunch of different formats."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So this might be the coordinate system that you want to operate in, especially if you're going to apply this matrix a lot. If you're going to do this transformation on a lot of different things, you're going to do it over and over and over again, maybe to the same set, then it maybe is worth the overhead to do the conversion and just use this as your coordinate system. So let's see that this will actually be a nice looking, easy to compute with, and actually diagonal matrix. So we know that the transformation of, let's write this in a bunch of different formats. Let me scroll down a little bit. So if I wanted to write the transformation of v1 in b coordinates, what would it be? It's just going to be equal to, well, these are the basis vectors, right?"}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So we know that the transformation of, let's write this in a bunch of different formats. Let me scroll down a little bit. So if I wanted to write the transformation of v1 in b coordinates, what would it be? It's just going to be equal to, well, these are the basis vectors, right? So it's the coefficient on the basis vector. So it's going to be equal to lambda 1, and then there's a bunch of 0's. It's lambda 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's just going to be equal to, well, these are the basis vectors, right? So it's the coefficient on the basis vector. So it's going to be equal to lambda 1, and then there's a bunch of 0's. It's lambda 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn. That's what it's equal to. But it's also equal to d, and we can write d like this. d is also a transformation between rn and rn, just a different coordinate system."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's lambda 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn. That's what it's equal to. But it's also equal to d, and we can write d like this. d is also a transformation between rn and rn, just a different coordinate system. So d is going to just be a bunch of column vectors, d1, d2, all the way through dn, times, this is the same thing as d times our b representation of the vector v1. But what is our b representation of the vector v1? Well, the vector v1 is just 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "d is also a transformation between rn and rn, just a different coordinate system. So d is going to just be a bunch of column vectors, d1, d2, all the way through dn, times, this is the same thing as d times our b representation of the vector v1. But what is our b representation of the vector v1? Well, the vector v1 is just 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn. v1 is a basis vector, so it's just 1 times itself plus 0 times everything else. So this is what its representation is in the b coordinate system. Now, what is this going to be equal to?"}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Well, the vector v1 is just 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn. v1 is a basis vector, so it's just 1 times itself plus 0 times everything else. So this is what its representation is in the b coordinate system. Now, what is this going to be equal to? And we've seen this before. This is all a bit of review. I might even be boring you."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is this going to be equal to? And we've seen this before. This is all a bit of review. I might even be boring you. This is just equal to 1 times d1 plus 0 times d2 plus 0 times all the other columns. So this is just equal to d1. So just like that, we have our first column of our matrix d. We could just keep doing that."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "I might even be boring you. This is just equal to 1 times d1 plus 0 times d2 plus 0 times all the other columns. So this is just equal to d1. So just like that, we have our first column of our matrix d. We could just keep doing that. I'll do it multiple times. The transformation of v2 in our new coordinate system with respect to our new basis is going to be equal to, well, we know what the transformation of v2 is. It's 0 times v1 plus lambda2 times v2, and then plus 0 times everything else."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So just like that, we have our first column of our matrix d. We could just keep doing that. I'll do it multiple times. The transformation of v2 in our new coordinate system with respect to our new basis is going to be equal to, well, we know what the transformation of v2 is. It's 0 times v1 plus lambda2 times v2, and then plus 0 times everything else. And that's the same thing as d, d1, d2, all the way through dn, times our b representation of vector 2. Well, vector 2 is one of the basis vectors. It's just 0 times v1 plus 1 times v2 plus 0 times v3 all the way, the rest is 0."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's 0 times v1 plus lambda2 times v2, and then plus 0 times everything else. And that's the same thing as d, d1, d2, all the way through dn, times our b representation of vector 2. Well, vector 2 is one of the basis vectors. It's just 0 times v1 plus 1 times v2 plus 0 times v3 all the way, the rest is 0. So what's this going to be equal to? This is 0 times d1 plus 1 times d2 and 0 times d2. 1 times d2 and 0 times everything else."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's just 0 times v1 plus 1 times v2 plus 0 times v3 all the way, the rest is 0. So what's this going to be equal to? This is 0 times d1 plus 1 times d2 and 0 times d2. 1 times d2 and 0 times everything else. So it's equal to d2. I think you get the general idea. I'll do it one more time just to really hammer the point home, the transformation of the nth basis vector, which is also an eigenvector of our original matrix A, or of our transformation in standard coordinates, in b coordinates, is going to be equal to what?"}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "1 times d2 and 0 times everything else. So it's equal to d2. I think you get the general idea. I'll do it one more time just to really hammer the point home, the transformation of the nth basis vector, which is also an eigenvector of our original matrix A, or of our transformation in standard coordinates, in b coordinates, is going to be equal to what? Well, we wrote it right up here. It's going to be a bunch of 0's. It's 0 times all of these guys plus lambda n times vn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it one more time just to really hammer the point home, the transformation of the nth basis vector, which is also an eigenvector of our original matrix A, or of our transformation in standard coordinates, in b coordinates, is going to be equal to what? Well, we wrote it right up here. It's going to be a bunch of 0's. It's 0 times all of these guys plus lambda n times vn. And this is going to be this guy, d1, d2, all the way to dn, times the b representation of the nth basis vector, which is just 0 times v1, 0 times v2, and 0 times all of them except for 1 times vn. And so this is going to be equal to 0 times d1 plus 0 times d2 plus 0 times all of these guys all the way to 1 times dn. So that's going to be equal to dn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's 0 times all of these guys plus lambda n times vn. And this is going to be this guy, d1, d2, all the way to dn, times the b representation of the nth basis vector, which is just 0 times v1, 0 times v2, and 0 times all of them except for 1 times vn. And so this is going to be equal to 0 times d1 plus 0 times d2 plus 0 times all of these guys all the way to 1 times dn. So that's going to be equal to dn. And just like that, we know what our transformation matrix is going to look like with respect to this new basis, where this basis was defined, or it's made up of n linearly independent eigenvectors of our original matrix A. So what does D look like? Our matrix D is going to look like its first column is right there."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be equal to dn. And just like that, we know what our transformation matrix is going to look like with respect to this new basis, where this basis was defined, or it's made up of n linearly independent eigenvectors of our original matrix A. So what does D look like? Our matrix D is going to look like its first column is right there. We figured that one out. Lambda 1, and then we just have a bunch of 0's. Its second column is right here."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Our matrix D is going to look like its first column is right there. We figured that one out. Lambda 1, and then we just have a bunch of 0's. Its second column is right here. d2 is equal to this. It's 0, lambda 2, and then a bunch of 0's. And then this is in general the case."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "Its second column is right here. d2 is equal to this. It's 0, lambda 2, and then a bunch of 0's. And then this is in general the case. The nth column is going to have a 0 everywhere except along the diagonal. It's going to be lambda n. It's going to be the eigenvalue for the nth eigenvector. And so the diagonal is going to look like you're going to have lambda 3 all the way down to lambda n. And our nth column is lambda n with just a bunch of 0's everywhere."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "And then this is in general the case. The nth column is going to have a 0 everywhere except along the diagonal. It's going to be lambda n. It's going to be the eigenvalue for the nth eigenvector. And so the diagonal is going to look like you're going to have lambda 3 all the way down to lambda n. And our nth column is lambda n with just a bunch of 0's everywhere. So D, when we picked, this is a neat result. When you, if A has n linearly independent eigenvectors, and this isn't always the case, but you can figure out the eigenvectors and say, hey, I can take a collection of n of these that are linearly independent, then those will be a basis for Rn. n linearly independent vectors in Rn are a basis for Rn."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "And so the diagonal is going to look like you're going to have lambda 3 all the way down to lambda n. And our nth column is lambda n with just a bunch of 0's everywhere. So D, when we picked, this is a neat result. When you, if A has n linearly independent eigenvectors, and this isn't always the case, but you can figure out the eigenvectors and say, hey, I can take a collection of n of these that are linearly independent, then those will be a basis for Rn. n linearly independent vectors in Rn are a basis for Rn. But when you use that basis, when you use the linearly independent eigenvectors of A as a basis, we call this an eigenbasis, the transformation matrix with respect to that eigenbasis, it becomes a very, very nice matrix. This is super easy to multiply. It's super easy to invert."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "n linearly independent vectors in Rn are a basis for Rn. But when you use that basis, when you use the linearly independent eigenvectors of A as a basis, we call this an eigenbasis, the transformation matrix with respect to that eigenbasis, it becomes a very, very nice matrix. This is super easy to multiply. It's super easy to invert. It's super easy to take the determinant of. We've seen it multiple times. It just has a ton of neat properties."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It's super easy to invert. It's super easy to take the determinant of. We've seen it multiple times. It just has a ton of neat properties. It's just a good basis to be dealing with. So that's kind of the big takeaway. In all of linear algebra, we did all the stuff with spaces and vectors and all of that."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "It just has a ton of neat properties. It's just a good basis to be dealing with. So that's kind of the big takeaway. In all of linear algebra, we did all the stuff with spaces and vectors and all of that. But in general, vectors are abstract representations of real world things. You could represent a vector as the stock returns, or it could be a vector of weather in a certain part of the country, and you can create these spaces based on the number of dimensions and all of that. And then you're going to have transformations."}, {"video_title": "Showing that an eigenbasis makes for good coordinate systems Linear Algebra Khan Academy.mp3", "Sentence": "In all of linear algebra, we did all the stuff with spaces and vectors and all of that. But in general, vectors are abstract representations of real world things. You could represent a vector as the stock returns, or it could be a vector of weather in a certain part of the country, and you can create these spaces based on the number of dimensions and all of that. And then you're going to have transformations. Sometimes when you learn about Markov chains, your transformations are essentially what's the probability of after one time increment that something's state will change to something else, and then you'll want to apply that matrix many, many, many, many times to see what the stable state is for a lot of things. And I know I'm not explaining any of this to you well, but I wanted to tell you that all of linear algebra is really just a very general way to solve a whole universe of problems. And what's useful about this is you can have transformation matrices that define these functions, essentially on data sets."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I've got a handful of matrices here. I have the matrix A that's an m by n matrix. You can see it has n columns and m rows. And actually, let me throw in one entry there. It might be useful. This is the j-th column. So the m-th row is going to look like this."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And actually, let me throw in one entry there. It might be useful. This is the j-th column. So the m-th row is going to look like this. a, m, j. That's that entry right there. And then I have matrix B defined similarly."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the m-th row is going to look like this. a, m, j. That's that entry right there. And then I have matrix B defined similarly. But instead of being an m by n matrix, B is an n by m matrix. And so this entry right here, I realize this might be useful. This is going to be my n-th row."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then I have matrix B defined similarly. But instead of being an m by n matrix, B is an n by m matrix. And so this entry right here, I realize this might be useful. This is going to be my n-th row. And it's going to be my j-th column. And then I also wrote out their transposes. So if you look at the transpose of B, B was an n by m matrix."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be my n-th row. And it's going to be my j-th column. And then I also wrote out their transposes. So if you look at the transpose of B, B was an n by m matrix. Now the transpose is going to be an m by n matrix. And each of its rows become its columns. And the same thing I did for A."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you look at the transpose of B, B was an n by m matrix. Now the transpose is going to be an m by n matrix. And each of its rows become its columns. And the same thing I did for A. Its transpose is right there. A was m by n. The transpose is n by m. And each of these rows become each of these columns. Now fair enough."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the same thing I did for A. Its transpose is right there. A was m by n. The transpose is n by m. And each of these rows become each of these columns. Now fair enough. Let's define two new matrices right now. Let's define the matrix C. Let me do it over here. I think the real estate will be valuable in this video."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now fair enough. Let's define two new matrices right now. Let's define the matrix C. Let me do it over here. I think the real estate will be valuable in this video. Let's define my matrix C as being equal to the product of A and B. So what are going to be the dimensions of C? Well, m by n matrix times an n by m matrix."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think the real estate will be valuable in this video. Let's define my matrix C as being equal to the product of A and B. So what are going to be the dimensions of C? Well, m by n matrix times an n by m matrix. These two have to be equal even for the matrix product to be defined, and it's going to result with an m by m matrix. So it's going to be an m by m matrix. Now let's define another matrix."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, m by n matrix times an n by m matrix. These two have to be equal even for the matrix product to be defined, and it's going to result with an m by m matrix. So it's going to be an m by m matrix. Now let's define another matrix. Let's call it D. And it's equal to B transpose times A transpose. And the dimensions are going to be the same, because this is an m by n times an n by m. So these are the same, which is a requirement for this product to be defined. And so the dimensions of B are going to be m by m. So let's explore a little bit what the different entries of C are going to look like."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let's define another matrix. Let's call it D. And it's equal to B transpose times A transpose. And the dimensions are going to be the same, because this is an m by n times an n by m. So these are the same, which is a requirement for this product to be defined. And so the dimensions of B are going to be m by m. So let's explore a little bit what the different entries of C are going to look like. So let me write my matrix C right here. So it's just going to have a bunch of entries. C1 1, C2 2, C1 2, all the way to C1 m. You can imagine, because it's an m by m matrix, you're going to have Cmm over here."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so the dimensions of B are going to be m by m. So let's explore a little bit what the different entries of C are going to look like. So let me write my matrix C right here. So it's just going to have a bunch of entries. C1 1, C2 2, C1 2, all the way to C1 m. You can imagine, because it's an m by m matrix, you're going to have Cmm over here. You know how this drill goes. But what I'm curious about is how do we figure out what C, the general Cij is? How do we figure out what a particular entry is?"}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "C1 1, C2 2, C1 2, all the way to C1 m. You can imagine, because it's an m by m matrix, you're going to have Cmm over here. You know how this drill goes. But what I'm curious about is how do we figure out what C, the general Cij is? How do we figure out what a particular entry is? We know that C is the products of A and B. So to get to a particular entry in C, and we've seen this before, so it's a particular entry in C, so Cij, it's going to be, you can view it as the dot product of the i-th row in A with the j-th column in B, just like that. And what's that going to be equal to?"}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How do we figure out what a particular entry is? We know that C is the products of A and B. So to get to a particular entry in C, and we've seen this before, so it's a particular entry in C, so Cij, it's going to be, you can view it as the dot product of the i-th row in A with the j-th column in B, just like that. And what's that going to be equal to? It's going to be equal to Ai1 times B1j plus Ai2 times B2j. And you're just going to keep going until you get to the last term here, Ain times the last term here, Bnj. Fair enough."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what's that going to be equal to? It's going to be equal to Ai1 times B1j plus Ai2 times B2j. And you're just going to keep going until you get to the last term here, Ain times the last term here, Bnj. Fair enough. Now, what about our matrix D? What are its entries going to look like? So D, similarly, it's going to look like you're going to have D11, D12, all the way to D1m."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now, what about our matrix D? What are its entries going to look like? So D, similarly, it's going to look like you're going to have D11, D12, all the way to D1m. You're going to have Dmm. I could keep putting entries here, but I'm curious about just some general entry here. Let's say I want to find Dji."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So D, similarly, it's going to look like you're going to have D11, D12, all the way to D1m. You're going to have Dmm. I could keep putting entries here, but I'm curious about just some general entry here. Let's say I want to find Dji. Dji. That's what I want to find. So I want to find a general way for any particular entry of D, the j-th row and i-th column, which is a little bit different than the convention we normally use for these letters, but it's fine."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I want to find Dji. Dji. That's what I want to find. So I want to find a general way for any particular entry of D, the j-th row and i-th column, which is a little bit different than the convention we normally use for these letters, but it's fine. The first one is D's row. The second one is this entry's column. So how do we figure that out?"}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I want to find a general way for any particular entry of D, the j-th row and i-th column, which is a little bit different than the convention we normally use for these letters, but it's fine. The first one is D's row. The second one is this entry's column. So how do we figure that out? So D sub ji, it's going to be equal to D is the product of these two guys. So to get the j-th row and i-th column entry here, we essentially take the dot product of the j-th row here, which is that right there, with the i-th column of A, which is that right there. So I'm going to take the dot product of that."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So how do we figure that out? So D sub ji, it's going to be equal to D is the product of these two guys. So to get the j-th row and i-th column entry here, we essentially take the dot product of the j-th row here, which is that right there, with the i-th column of A, which is that right there. So I'm going to take the dot product of that. And you might already see something interesting here. This thing right here is equivalent to that thing right there. And this thing right here is equivalent to that thing right there, because we took the transposes."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to take the dot product of that. And you might already see something interesting here. This thing right here is equivalent to that thing right there. And this thing right here is equivalent to that thing right there, because we took the transposes. But let's actually just write it out. So what is this dot product going to be equal to? That's going to be bij."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this thing right here is equivalent to that thing right there, because we took the transposes. But let's actually just write it out. So what is this dot product going to be equal to? That's going to be bij. Let me write it this way. It's going to be bij times Ai1. Or we could write it as Ai1 times b1j."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be bij. Let me write it this way. It's going to be bij times Ai1. Or we could write it as Ai1 times b1j. And that's going to be plus b2j times Ai2, which is the same thing as Ai2 times b2j. And you're going to keep going until you get bnj times Ain. Or you could write that as Ain times bnj."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or we could write it as Ai1 times b1j. And that's going to be plus b2j times Ai2, which is the same thing as Ai2 times b2j. And you're going to keep going until you get bnj times Ain. Or you could write that as Ain times bnj. Now notice something. These two things are equivalent. They're completely equivalent statement."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or you could write that as Ain times bnj. Now notice something. These two things are equivalent. They're completely equivalent statement. The d sub ji is equivalent to c sub ij. Let me write that. So d, or I could write c sub ij, is equivalent to d sub ji."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "They're completely equivalent statement. The d sub ji is equivalent to c sub ij. Let me write that. So d, or I could write c sub ij, is equivalent to d sub ji. Or another way you could say it is, anything that's at row, all the entries that's at row i, column j in C, is now in row j, column i in D. And this is true for all the entries. I stayed as general as possible. So what does this mean?"}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So d, or I could write c sub ij, is equivalent to d sub ji. Or another way you could say it is, anything that's at row, all the entries that's at row i, column j in C, is now in row j, column i in D. And this is true for all the entries. I stayed as general as possible. So what does this mean? This is the definition of a transpose. So we now get that c transpose is equal to d. Or you could say that c is equal to d transpose. Now this is pretty interesting, because how did we define these two?"}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what does this mean? This is the definition of a transpose. So we now get that c transpose is equal to d. Or you could say that c is equal to d transpose. Now this is pretty interesting, because how did we define these two? We said that our matrix C is equal to the matrix product A and B. And we said that d is equal to our matrix product B transpose times A transpose. I did those definitions right there."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now this is pretty interesting, because how did we define these two? We said that our matrix C is equal to the matrix product A and B. And we said that d is equal to our matrix product B transpose times A transpose. I did those definitions right there. Here are the definitions. Now we just found out that d is equal to the transpose of C. So we could write that c transpose, which is the same thing as A times B transpose, is equal to d. So it is equal to d, which is just B transpose, A transpose. And this is a pretty neat takeaway."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I did those definitions right there. Here are the definitions. Now we just found out that d is equal to the transpose of C. So we could write that c transpose, which is the same thing as A times B transpose, is equal to d. So it is equal to d, which is just B transpose, A transpose. And this is a pretty neat takeaway. If I take the product of two matrices and then transpose it, it's equivalent to switching the order, or transposing them, and then taking the product of the reversed order, B transpose, A transpose. Which is a pretty, pretty neat takeaway. And you can actually extend this to an arbitrary number of matrices that you're taking the product of."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is a pretty neat takeaway. If I take the product of two matrices and then transpose it, it's equivalent to switching the order, or transposing them, and then taking the product of the reversed order, B transpose, A transpose. Which is a pretty, pretty neat takeaway. And you can actually extend this to an arbitrary number of matrices that you're taking the product of. I'm not proving it here, but it's actually a very simple extension from this right now. If you take the matrices, let's say, A, let me do a different letter, x, y, z. Take their product and then transpose it, it's equal to z transpose, y transpose, x transpose."}, {"video_title": "Transpose of a matrix product Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you can actually extend this to an arbitrary number of matrices that you're taking the product of. I'm not proving it here, but it's actually a very simple extension from this right now. If you take the matrices, let's say, A, let me do a different letter, x, y, z. Take their product and then transpose it, it's equal to z transpose, y transpose, x transpose. I haven't proven this general case, and you could keep doing it with 4 or 5 or n matrices multiplied by each other, but it generally works. And you could essentially prove it using what we proved in this video right here. That you take the product of two matrices, take their transpose, it's equal to the product of their transposes in reverse order."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so any entry in C I can denote with a lowercase cij. So if I want the ith row and jth column, it'd be cij. And so each of its entries are going to be the sum of the corresponding columns in our matrices A and B. So our ith ij entry in C is going to be equal to the ij entry in A plus the ij entry in B. That's our definition of matrix addition. You just get the corresponding entry in the same row and column. Add them up and you get your entry in the same row and column in your new matrix."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So our ith ij entry in C is going to be equal to the ij entry in A plus the ij entry in B. That's our definition of matrix addition. You just get the corresponding entry in the same row and column. Add them up and you get your entry in the same row and column in your new matrix. It's the sum of the other two. Now, let's think a little bit about the transposes of these guys right here. So if A looks like this, I won't draw all of the entries, it takes forever."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Add them up and you get your entry in the same row and column in your new matrix. It's the sum of the other two. Now, let's think a little bit about the transposes of these guys right here. So if A looks like this, I won't draw all of the entries, it takes forever. But each of its entries are ij, just like that. Let's say that A transpose looks like this. Each of its entries, we would call it, let's say if you've got that same entry, we're going to call it A prime ij."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if A looks like this, I won't draw all of the entries, it takes forever. But each of its entries are ij, just like that. Let's say that A transpose looks like this. Each of its entries, we would call it, let's say if you've got that same entry, we're going to call it A prime ij. And these things aren't probably going to be the same. There's some chance they are, but they're probably not going to be the same. But that's its ij-th entry in the i-th row, j-th column in A transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Each of its entries, we would call it, let's say if you've got that same entry, we're going to call it A prime ij. And these things aren't probably going to be the same. There's some chance they are, but they're probably not going to be the same. But that's its ij-th entry in the i-th row, j-th column in A transpose. Now, the fact that this is a transpose of that means that everything that's in some row and column here is going to be in that column and row over here, that the rows and columns get switched. So we know that we could write that A prime ij, we're going to have the same entry that was in A ji. Maybe A ji is over here."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But that's its ij-th entry in the i-th row, j-th column in A transpose. Now, the fact that this is a transpose of that means that everything that's in some row and column here is going to be in that column and row over here, that the rows and columns get switched. So we know that we could write that A prime ij, we're going to have the same entry that was in A ji. Maybe A ji is over here. So this thing over here, which is in the same position as this one, is going to be equal to this guy over here if you switched the rows and columns. I think you can accept that. And you can make the same argument for B."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe A ji is over here. So this thing over here, which is in the same position as this one, is going to be equal to this guy over here if you switched the rows and columns. I think you can accept that. And you can make the same argument for B. Let me actually draw it out. So if I make B transpose, the entry in the i-th row and j-th column, I'll call it B prime ij, just like that, just like I did for A. So we could say that B prime ij is equal to, you take the matrix B, what's going to be the entry that's in the j-th row and i-th column."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you can make the same argument for B. Let me actually draw it out. So if I make B transpose, the entry in the i-th row and j-th column, I'll call it B prime ij, just like that, just like I did for A. So we could say that B prime ij is equal to, you take the matrix B, what's going to be the entry that's in the j-th row and i-th column. These are, you could almost say, the definition of the transpose. If I'm in the third row and second column now, it's going to be what was in the second row and third column. Fair enough."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could say that B prime ij is equal to, you take the matrix B, what's going to be the entry that's in the j-th row and i-th column. These are, you could almost say, the definition of the transpose. If I'm in the third row and second column now, it's going to be what was in the second row and third column. Fair enough. So what is, so we already have what Cij is equal to. What's the transpose of Cij going to be equal to? Let me write that down."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. So what is, so we already have what Cij is equal to. What's the transpose of Cij going to be equal to? Let me write that down. So C transpose, let me write it over here. C transpose is equal to, I'll use the same notation. The prime means that we're taking entries in the transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. So C transpose, let me write it over here. C transpose is equal to, I'll use the same notation. The prime means that we're taking entries in the transpose. So C transpose is just going to be a bunch of entries, i, j. I'll put a little prime there showing that that's entries in the matrix of the transpose and not in C itself. And we know that C prime ij is equal to Cji. Nothing new at all."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The prime means that we're taking entries in the transpose. So C transpose is just going to be a bunch of entries, i, j. I'll put a little prime there showing that that's entries in the matrix of the transpose and not in C itself. And we know that C prime ij is equal to Cji. Nothing new at all. We've just expressed kind of the definition of the transpose for these three matrices. Now what is Cji equal to? So let's focus on this a little bit."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Nothing new at all. We've just expressed kind of the definition of the transpose for these three matrices. Now what is Cji equal to? So let's focus on this a little bit. What is Cji equal to? We know that Cij is equal to A sub ij plus B sub ij. So if you swap them around, this is going to be equal to, you just swap the j's and the i's."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's focus on this a little bit. What is Cji equal to? We know that Cij is equal to A sub ij plus B sub ij. So if you swap them around, this is going to be equal to, you just swap the j's and the i's. A sub ji plus B sub ji. I just used this information here. You could almost view it as this assumption or this definition to go from this to this."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if you swap them around, this is going to be equal to, you just swap the j's and the i's. A sub ji plus B sub ji. I just used this information here. You could almost view it as this assumption or this definition to go from this to this. If I had an x and a y here, I'd have an x and a y here and an x and a y here. I have a j and an i here, so I have a j and an i there. And a j and an i right there."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could almost view it as this assumption or this definition to go from this to this. If I had an x and a y here, I'd have an x and a y here and an x and a y here. I have a j and an i here, so I have a j and an i there. And a j and an i right there. Now what are these equal to? This is equal to, this guy right here is equal to, let me do it in the green, is equal to the same entry for the transpose of A at ij. And this is equal to the same entry for the transpose of B at ij."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And a j and an i right there. Now what are these equal to? This is equal to, this guy right here is equal to, let me do it in the green, is equal to the same entry for the transpose of A at ij. And this is equal to the same entry for the transpose of B at ij. Now what is this telling us? What is this telling us? It's telling us that the transpose of C, which is the same thing as A plus B, so it's saying that A plus B transpose is the same thing as C transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this is equal to the same entry for the transpose of B at ij. Now what is this telling us? What is this telling us? It's telling us that the transpose of C, which is the same thing as A plus B, so it's saying that A plus B transpose is the same thing as C transpose. Let me write that. C transpose is the same thing as A plus B transpose. So these are the entries in A plus B transpose right here."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's telling us that the transpose of C, which is the same thing as A plus B, so it's saying that A plus B transpose is the same thing as C transpose. Let me write that. C transpose is the same thing as A plus B transpose. So these are the entries in A plus B transpose right here. And what is this over here? What are these? These are the entries right there."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these are the entries in A plus B transpose right here. And what is this over here? What are these? These are the entries right there. We'll do the equal sign over here. What are these? These are the entries in A transpose plus B transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are the entries right there. We'll do the equal sign over here. What are these? These are the entries in A transpose plus B transpose. These are the entries in A transpose. These are the entries in B transpose. If you take the sum of the two, you're just adding up the corresponding entries."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "These are the entries in A transpose plus B transpose. These are the entries in A transpose. These are the entries in B transpose. If you take the sum of the two, you're just adding up the corresponding entries. So that's straightforward to show that if you take the sum of two matrices and then transpose it, it's equivalent to transposing them first and then taking their sum, which is a reasonably neat outcome. Let's do one more. And I think we'll finish up all of our major transpose properties."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you take the sum of the two, you're just adding up the corresponding entries. So that's straightforward to show that if you take the sum of two matrices and then transpose it, it's equivalent to transposing them first and then taking their sum, which is a reasonably neat outcome. Let's do one more. And I think we'll finish up all of our major transpose properties. Let's say that A inverse, this is going to be a slightly different take on things. We're still going to take the transpose. So if we know that A inverse is the inverse of A, that means that A times A inverse is equal to the identity matrix, assuming that these are n by n matrices."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I think we'll finish up all of our major transpose properties. Let's say that A inverse, this is going to be a slightly different take on things. We're still going to take the transpose. So if we know that A inverse is the inverse of A, that means that A times A inverse is equal to the identity matrix, assuming that these are n by n matrices. So it's the n-dimensional identity matrix. And that A inverse times A is also going to be equal to the identity matrix. Now, let's take the transpose of both sides of this equation."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we know that A inverse is the inverse of A, that means that A times A inverse is equal to the identity matrix, assuming that these are n by n matrices. So it's the n-dimensional identity matrix. And that A inverse times A is also going to be equal to the identity matrix. Now, let's take the transpose of both sides of this equation. Let me start with, I'll do them both simultaneously. So if you take the transpose of both sides of the equation, you get A times A inverse transpose is equal to the identity matrix transpose. Now what's the transpose of the identity matrix?"}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let's take the transpose of both sides of this equation. Let me start with, I'll do them both simultaneously. So if you take the transpose of both sides of the equation, you get A times A inverse transpose is equal to the identity matrix transpose. Now what's the transpose of the identity matrix? Let's draw it out. The identity matrix looks like this. You have just 1's all the way down the diagonal and everything else is 0."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what's the transpose of the identity matrix? Let's draw it out. The identity matrix looks like this. You have just 1's all the way down the diagonal and everything else is 0. And you could view this as I11, I22, all the way down to INN. Everything else is 0. So when you take the transpose, you're just swapping up the 0's."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have just 1's all the way down the diagonal and everything else is 0. And you could view this as I11, I22, all the way down to INN. Everything else is 0. So when you take the transpose, you're just swapping up the 0's. These guys don't change. The diagonal does not change when you take the transpose. So the transpose of the identity matrix is equal to the identity matrix."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So when you take the transpose, you're just swapping up the 0's. These guys don't change. The diagonal does not change when you take the transpose. So the transpose of the identity matrix is equal to the identity matrix. And so we can apply that same thing here. Let's take the transpose for this statement. So we know that A inverse times A transpose is equal to the identity matrix transpose, which is equal to the identity matrix."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the transpose of the identity matrix is equal to the identity matrix. And so we can apply that same thing here. Let's take the transpose for this statement. So we know that A inverse times A transpose is equal to the identity matrix transpose, which is equal to the identity matrix. And then we know what happens when you take the transpose of a product. It's equal to the product of the transposes in reverse order. So this thing right here, we can rewrite as A inverse transpose times A transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we know that A inverse times A transpose is equal to the identity matrix transpose, which is equal to the identity matrix. And then we know what happens when you take the transpose of a product. It's equal to the product of the transposes in reverse order. So this thing right here, we can rewrite as A inverse transpose times A transpose. Transpose times A transpose, which is going to be equal to the identity matrix. We could do the same thing over here. This thing is going to be equal to A transpose times A inverse transpose, which is also going to be equal to the identity matrix."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here, we can rewrite as A inverse transpose times A transpose. Transpose times A transpose, which is going to be equal to the identity matrix. We could do the same thing over here. This thing is going to be equal to A transpose times A inverse transpose, which is also going to be equal to the identity matrix. Now, this is an interesting statement. The fact that if I have this guy right here, times the transpose of A is equal to the identity matrix. And the transpose of A times that same guy is equal to identity matrix, this implies that A inverse transpose is the inverse of A transpose."}, {"video_title": "Transposes of sums and inverses Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This thing is going to be equal to A transpose times A inverse transpose, which is also going to be equal to the identity matrix. Now, this is an interesting statement. The fact that if I have this guy right here, times the transpose of A is equal to the identity matrix. And the transpose of A times that same guy is equal to identity matrix, this implies that A inverse transpose is the inverse of A transpose. Or another way of writing that is if I take A transpose and if I take its inverse, that is going to be equal to this guy. Going to be equal to A inverse transpose. So another neat outcome dealing with transposes."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Let's figure out all of the interesting things that we can figure out about this matrix and this vector. So the first thing of interest, and this is all going to essentially help us visualize what we learned in the last video, is the null space of A. So to figure out the null space of A, we know that the null space of A is equal to the null space of the reduced row echelon form of A. So let's just find out the reduced row echelon form of A. If we, let's say we leave the first row the same, we get 3 minus 2. And let's replace the second row with the second row minus 2 times the first. So 6 minus 2 times 3 is 0, and minus 4 minus 2 times minus 1 is 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So let's just find out the reduced row echelon form of A. If we, let's say we leave the first row the same, we get 3 minus 2. And let's replace the second row with the second row minus 2 times the first. So 6 minus 2 times 3 is 0, and minus 4 minus 2 times minus 1 is 0. And then let's replace the first row with the first row divided by 3. So then that becomes 1 minus 2 thirds, and the second row is still 0. So that's the reduced row echelon form of A."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So 6 minus 2 times 3 is 0, and minus 4 minus 2 times minus 1 is 0. And then let's replace the first row with the first row divided by 3. So then that becomes 1 minus 2 thirds, and the second row is still 0. So that's the reduced row echelon form of A. And we want to find its null space. So we want to find all of the vectors that we can multiply it by, so this is a vector x1, x2, that is equal to the 0 vector in R2. So the second row gives us no information."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So that's the reduced row echelon form of A. And we want to find its null space. So we want to find all of the vectors that we can multiply it by, so this is a vector x1, x2, that is equal to the 0 vector in R2. So the second row gives us no information. 0 times x1 plus 0 times x2 is equal to 0. No information there, so our only constraint is the first row, 1 times x1. Let me write it here."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So the second row gives us no information. 0 times x1 plus 0 times x2 is equal to 0. No information there, so our only constraint is the first row, 1 times x1. Let me write it here. So 1 times x1 plus, let me write it this way, minus 2 thirds times x2 is equal to that 0 right there. Or we could write that x1 is equal to 2 thirds x2. So if we wanted to write the null space of A, and let me just make a, actually before I write that, just to simplify things, and just to show you that x2 isn't some special number, let's just say that x2 is equal to t, where t is some real number."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it here. So 1 times x1 plus, let me write it this way, minus 2 thirds times x2 is equal to that 0 right there. Or we could write that x1 is equal to 2 thirds x2. So if we wanted to write the null space of A, and let me just make a, actually before I write that, just to simplify things, and just to show you that x2 isn't some special number, let's just say that x2 is equal to t, where t is some real number. And then we would have x1 is equal to 2 thirds times t. So the null space of our matrix, the null space of A, is going to be equal to the set of all x1, x2's that are equal to some real number t times the vector. Let's see, we have x2 is equal to t times 1, and x1 is equal to 2 thirds t. So 2 thirds 1, just like that. That is our null space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So if we wanted to write the null space of A, and let me just make a, actually before I write that, just to simplify things, and just to show you that x2 isn't some special number, let's just say that x2 is equal to t, where t is some real number. And then we would have x1 is equal to 2 thirds times t. So the null space of our matrix, the null space of A, is going to be equal to the set of all x1, x2's that are equal to some real number t times the vector. Let's see, we have x2 is equal to t times 1, and x1 is equal to 2 thirds t. So 2 thirds 1, just like that. That is our null space. All of the multiples of the vector 2 thirds 1. And actually, just to make this a little bit simpler, we can pick t to be equal to 3c. So then what do you have?"}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "That is our null space. All of the multiples of the vector 2 thirds 1. And actually, just to make this a little bit simpler, we can pick t to be equal to 3c. So then what do you have? If this is equal to 3c, if you multiply the 3 times these, we could rewrite this as being equal to the x1's, x2's that are equal to the scalar c, where that's some other real number, times 3 times 2 thirds is 2, 3 times 1 is 3. And the whole reason why I did that is just to write the null space with a slightly simpler basis vector, the one that didn't have any fractions in it. So we could also write the null space is equal to the span of the vector 2, 3."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So then what do you have? If this is equal to 3c, if you multiply the 3 times these, we could rewrite this as being equal to the x1's, x2's that are equal to the scalar c, where that's some other real number, times 3 times 2 thirds is 2, 3 times 1 is 3. And the whole reason why I did that is just to write the null space with a slightly simpler basis vector, the one that didn't have any fractions in it. So we could also write the null space is equal to the span of the vector 2, 3. All of these statements are equivalent. So we figured out the null space. Maybe another interesting thing, especially if we're going to make this relate to what we did in the last video, is find the solution set to the equation Ax is equal to b."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So we could also write the null space is equal to the span of the vector 2, 3. All of these statements are equivalent. So we figured out the null space. Maybe another interesting thing, especially if we're going to make this relate to what we did in the last video, is find the solution set to the equation Ax is equal to b. To do that, we just set up an augmented matrix. So we set up an augmented matrix, 3 minus 2, 6 minus 4, and we augment it with b, 9, 18. And then we put the left-hand side in reduced row echelon form, and we get, let's keep the first row the same, we get 3 minus 2, and then 9."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Maybe another interesting thing, especially if we're going to make this relate to what we did in the last video, is find the solution set to the equation Ax is equal to b. To do that, we just set up an augmented matrix. So we set up an augmented matrix, 3 minus 2, 6 minus 4, and we augment it with b, 9, 18. And then we put the left-hand side in reduced row echelon form, and we get, let's keep the first row the same, we get 3 minus 2, and then 9. Let's replace the second row with the second row minus 2 times the first. So we get 6 minus 2 times 3 is 0, minus 4 minus 2 times minus 2, that's 0 as well, minus 4 plus 4. 18 minus 2 times 9, that's 18 minus 18, that's also 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And then we put the left-hand side in reduced row echelon form, and we get, let's keep the first row the same, we get 3 minus 2, and then 9. Let's replace the second row with the second row minus 2 times the first. So we get 6 minus 2 times 3 is 0, minus 4 minus 2 times minus 2, that's 0 as well, minus 4 plus 4. 18 minus 2 times 9, that's 18 minus 18, that's also 0. Almost there. Now let's just replace the first row with the first row divided by 3. So the second row is going to stay the same."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "18 minus 2 times 9, that's 18 minus 18, that's also 0. Almost there. Now let's just replace the first row with the first row divided by 3. So the second row is going to stay the same. And the first row is going to be 1 minus 2 thirds, 1 minus 2 thirds, and then we get a 3 there. So we can write, so if we wanted to rewrite this as a kind of an equation, we would write that the matrix 1 minus 2 thirds, 0, 0, times the vector that's in our solution set, x1, x2, is equal to the vector 3, 0. Or another way to say it, the second row gives us no real constraints, so we don't have to pay attention to it."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So the second row is going to stay the same. And the first row is going to be 1 minus 2 thirds, 1 minus 2 thirds, and then we get a 3 there. So we can write, so if we wanted to rewrite this as a kind of an equation, we would write that the matrix 1 minus 2 thirds, 0, 0, times the vector that's in our solution set, x1, x2, is equal to the vector 3, 0. Or another way to say it, the second row gives us no real constraints, so we don't have to pay attention to it. The first row tells us that x1 minus 2 thirds times x2 is equal to 3. Or that x1 is equal to 3 plus 2 thirds x2. Let's do the same exercise here."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Or another way to say it, the second row gives us no real constraints, so we don't have to pay attention to it. The first row tells us that x1 minus 2 thirds times x2 is equal to 3. Or that x1 is equal to 3 plus 2 thirds x2. Let's do the same exercise here. We could say that if we say that x2 is equal to t, then x1 is equal to 3 plus 2 thirds t. Or we could say that the solution set to Ax is equal to b is equal to the set of all x1, x2's that are equal to, let's see, x1 is equal to 3 plus 2 thirds, so plus t times 2 thirds. Let's write it like that. It's a plus right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Let's do the same exercise here. We could say that if we say that x2 is equal to t, then x1 is equal to 3 plus 2 thirds t. Or we could say that the solution set to Ax is equal to b is equal to the set of all x1, x2's that are equal to, let's see, x1 is equal to 3 plus 2 thirds, so plus t times 2 thirds. Let's write it like that. It's a plus right there. And x2 is just equal to t. So it's just equal to 1 times t plus, well, let's just say 0. So that's our solution set right there. And you might immediately recognize that it's some particular solution plus some scaled up version of our null space right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "It's a plus right there. And x2 is just equal to t. So it's just equal to 1 times t plus, well, let's just say 0. So that's our solution set right there. And you might immediately recognize that it's some particular solution plus some scaled up version of our null space right there. That's our solution set. I could do that same substitution that we did over here. I could say that let's replace t with 3c."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And you might immediately recognize that it's some particular solution plus some scaled up version of our null space right there. That's our solution set. I could do that same substitution that we did over here. I could say that let's replace t with 3c. And so we could rewrite this as being equal to, let's rewrite it neatly, x1, x2 is equal to the vector. I want to run a space 3, 0. Plus c times, let me scroll over a little bit, plus c times the vector."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "I could say that let's replace t with 3c. And so we could rewrite this as being equal to, let's rewrite it neatly, x1, x2 is equal to the vector. I want to run a space 3, 0. Plus c times, let me scroll over a little bit, plus c times the vector. 3 times 2 thirds is 2. 3 times 1 is 3. So it's equal to the vector 3, 0 plus some member of our null space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Plus c times, let me scroll over a little bit, plus c times the vector. 3 times 2 thirds is 2. 3 times 1 is 3. So it's equal to the vector 3, 0 plus some member of our null space. The null space was a span of 2, 3, or all the scalar multiples of our null space. And this is where c is any real number. So that's our solution set."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to the vector 3, 0 plus some member of our null space. The null space was a span of 2, 3, or all the scalar multiples of our null space. And this is where c is any real number. So that's our solution set. That's our null space. Now one other thing that might be of interest, just because it's the orthogonal complement to the null space, and this relates to what we were doing in the last video, is what the row space of A is. So the row space of A is just the column space of A transpose."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So that's our solution set. That's our null space. Now one other thing that might be of interest, just because it's the orthogonal complement to the null space, and this relates to what we were doing in the last video, is what the row space of A is. So the row space of A is just the column space of A transpose. This is the row space. And what is this equal to? This is equal to the span of the row vectors of A."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So the row space of A is just the column space of A transpose. This is the row space. And what is this equal to? This is equal to the span of the row vectors of A. So we have 3 minus 2. And we have 6 minus 4. But this guy is just 2 times that vector right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to the span of the row vectors of A. So we have 3 minus 2. And we have 6 minus 4. But this guy is just 2 times that vector right there. So we could just ignore it. So it's just the span of the vector 3 minus 2. Now let's graph all of these."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "But this guy is just 2 times that vector right there. So we could just ignore it. So it's just the span of the vector 3 minus 2. Now let's graph all of these. So we have the row space, we have the solution set, and we have the null space. Let's see if we can graph it. We can make a reasonable graph of this thing right here."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Now let's graph all of these. So we have the row space, we have the solution set, and we have the null space. Let's see if we can graph it. We can make a reasonable graph of this thing right here. So I have my vertical axis right there. I have my horizontal axis right there. Now what does the null space look like?"}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "We can make a reasonable graph of this thing right here. So I have my vertical axis right there. I have my horizontal axis right there. Now what does the null space look like? It's all the multiples of 2, 3. So if I go out 1, 2, and then I go up 1, 2, 3, the vector 2, 3 looks like this in its standard position. And the null space is all of the multiples of this."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Now what does the null space look like? It's all the multiples of 2, 3. So if I go out 1, 2, and then I go up 1, 2, 3, the vector 2, 3 looks like this in its standard position. And the null space is all of the multiples of this. So all the multiples of that. So if you do all of the multiples of that, you get a line that looks something like this. If you do all of the multiples of that vector, you get a bunch of vectors that point to every point on this green line right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And the null space is all of the multiples of this. So all the multiples of that. So if you do all of the multiples of that, you get a line that looks something like this. If you do all of the multiples of that vector, you get a bunch of vectors that point to every point on this green line right there. So that right there is the null space of A. Now what is the solution set? The solution set is the vector 3, 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "If you do all of the multiples of that vector, you get a bunch of vectors that point to every point on this green line right there. So that right there is the null space of A. Now what is the solution set? The solution set is the vector 3, 0. So it is the vector 3, 0. So 1, 2, 3. So it is the vector 3, 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "The solution set is the vector 3, 0. So it is the vector 3, 0. So 1, 2, 3. So it is the vector 3, 0. Plus members of the null space. So if you take any of these, let's say if you add 2, 3, 2, 3 would look like this. But you're adding any multiple of 2, 3."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So it is the vector 3, 0. Plus members of the null space. So if you take any of these, let's say if you add 2, 3, 2, 3 would look like this. But you're adding any multiple of 2, 3. So when you add all the different multiples of 2, 3, you get this line right here. It's shifted essentially by that vector 3, 0. So this right here is the solution set."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "But you're adding any multiple of 2, 3. So when you add all the different multiples of 2, 3, you get this line right here. It's shifted essentially by that vector 3, 0. So this right here is the solution set. So solution 2, Ax is equal to b. That's the solution set right there. Now what's our row space?"}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is the solution set. So solution 2, Ax is equal to b. That's the solution set right there. Now what's our row space? The row space is all of the multiples of the vector 3 minus 2. So what does 3 minus 2 look like? The vector 3 minus 2, we go 1, 2, 3."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Now what's our row space? The row space is all of the multiples of the vector 3 minus 2. So what does 3 minus 2 look like? The vector 3 minus 2, we go 1, 2, 3. And then we go down 2. So you go 1, 2. And it's going to look like this."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "The vector 3 minus 2, we go 1, 2, 3. And then we go down 2. So you go 1, 2. And it's going to look like this. It's going to look something like this. And then if we were to graph it, and you could see, well, it's going to look something like, I want to do it relatively neatly. So it's going to look something like this."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And it's going to look like this. It's going to look something like this. And then if we were to graph it, and you could see, well, it's going to look something like, I want to do it relatively neatly. So it's going to look something like this. Let me do it in another color actually. It's going to look like that. I can't do it at all straight."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to look something like this. Let me do it in another color actually. It's going to look like that. I can't do it at all straight. Let me do it a little bit. Now this is horrible. Something about the bottom right-hand side of my, that's a pretty decent attempt."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "I can't do it at all straight. Let me do it a little bit. Now this is horrible. Something about the bottom right-hand side of my, that's a pretty decent attempt. So that right there is my row space. Because if you have 3 minus 2, you get roughly to that point. And then you want all of the multiples of that vector."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Something about the bottom right-hand side of my, that's a pretty decent attempt. So that right there is my row space. Because if you have 3 minus 2, you get roughly to that point. And then you want all of the multiples of that vector. 3 minus 2. If you multiplied it by minus 1, you would get the vector minus 3, 2. So it would look like this."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And then you want all of the multiples of that vector. 3 minus 2. If you multiplied it by minus 1, you would get the vector minus 3, 2. So it would look like this. It would be minus 3, 2. It would look like this. Minus 2, 3."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So it would look like this. It would be minus 3, 2. It would look like this. Minus 2, 3. And then up 2. It would be like that. So all of the scalar multiples of that as well."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2, 3. And then up 2. It would be like that. So all of the scalar multiples of that as well. That is our row space. And notice that our row space is orthogonal to our null space. Now this is a nice visual representation of everything we can do essentially with this matrix right here, which is our matrix A."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So all of the scalar multiples of that as well. That is our row space. And notice that our row space is orthogonal to our null space. Now this is a nice visual representation of everything we can do essentially with this matrix right here, which is our matrix A. But in the last video, we had a very interesting result. We said that if we have some B, let me do it in a new color, we found out that if we had some B that is a member of the column space of A, then the solution with the smallest length, or we could say the smallest or the shortest solution to Ax is equal to B is a unique member of the row space of A. Now let's see."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Now this is a nice visual representation of everything we can do essentially with this matrix right here, which is our matrix A. But in the last video, we had a very interesting result. We said that if we have some B, let me do it in a new color, we found out that if we had some B that is a member of the column space of A, then the solution with the smallest length, or we could say the smallest or the shortest solution to Ax is equal to B is a unique member of the row space of A. Now let's see. This is our big takeaway from the last video. And it might have seemed a little hard to visualize before, but now that we've graphed it all out, I think we can visualize it. This blue line right here is a solution set."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Now let's see. This is our big takeaway from the last video. And it might have seemed a little hard to visualize before, but now that we've graphed it all out, I think we can visualize it. This blue line right here is a solution set. The row space right here is this line that's perpendicular to the solution set. But notice, one of the vectors on it is both pointing to a position on the solution set, and it is in my row space. This vector right here that I'm doing in a bold green."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "This blue line right here is a solution set. The row space right here is this line that's perpendicular to the solution set. But notice, one of the vectors on it is both pointing to a position on the solution set, and it is in my row space. This vector right here that I'm doing in a bold green. That vector right there, we could call that vector r, because that vector is both a member of my row space. If you view it as a position vector, it points to a point on the line that represents our row space. All of the members of our row space point to points on this line."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "This vector right here that I'm doing in a bold green. That vector right there, we could call that vector r, because that vector is both a member of my row space. If you view it as a position vector, it points to a point on the line that represents our row space. All of the members of our row space point to points on this line. But at the same time, it also points to a point right there that is a member of our solution set. And notice, it is the only vector in our row space that points to a point that is a member of our solution set. And if you just look at it from a geometric point of view, all of the other solutions point to all of the other points on that line."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "All of the members of our row space point to points on this line. But at the same time, it also points to a point right there that is a member of our solution set. And notice, it is the only vector in our row space that points to a point that is a member of our solution set. And if you just look at it from a geometric point of view, all of the other solutions point to all of the other points on that line. So that's a solution right there. This is a solution right there. That is a solution right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And if you just look at it from a geometric point of view, all of the other solutions point to all of the other points on that line. So that's a solution right there. This is a solution right there. That is a solution right there. All of the vectors that point to things on this solution line right there, there's all solutions. But the very shortest one is this green one. This green one is essentially orthogonal to it, because it's a member of the row space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "That is a solution right there. All of the vectors that point to things on this solution line right there, there's all solutions. But the very shortest one is this green one. This green one is essentially orthogonal to it, because it's a member of the row space. And this essentially has the same slope as our null space. It's orthogonal, so it's kind of the shortest path distance to getting to our solution set. And for this exact example, if we wanted to, we can actually figure out what this shortest vector r is."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "This green one is essentially orthogonal to it, because it's a member of the row space. And this essentially has the same slope as our null space. It's orthogonal, so it's kind of the shortest path distance to getting to our solution set. And for this exact example, if we wanted to, we can actually figure out what this shortest vector r is. We've got the vector 3, 0 over here, right? The vector 3, 0 is this vector right here. So if you take 3, 0 minus your vector, so let me say r, let me write it this way."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And for this exact example, if we wanted to, we can actually figure out what this shortest vector r is. We've got the vector 3, 0 over here, right? The vector 3, 0 is this vector right here. So if you take 3, 0 minus your vector, so let me say r, let me write it this way. Our vector r, the special shortest solution, it's going to be a member of our column space. Our column space is a span of 3 minus 2. So it's going to be equal to some scalar multiple of 3 minus 2."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So if you take 3, 0 minus your vector, so let me say r, let me write it this way. Our vector r, the special shortest solution, it's going to be a member of our column space. Our column space is a span of 3 minus 2. So it's going to be equal to some scalar multiple of 3 minus 2. Now, we know that 3, 0 points to another solution on our solution set. But we know that if we take the difference of these two vectors, if I take 3, 0, if I take the vector 3, 0, and from that I subtract the vector r, what do I get? I get this vector right there."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be equal to some scalar multiple of 3 minus 2. Now, we know that 3, 0 points to another solution on our solution set. But we know that if we take the difference of these two vectors, if I take 3, 0, if I take the vector 3, 0, and from that I subtract the vector r, what do I get? I get this vector right there. Let me do that in another color. I get this pink vector right there. That pink vector, it's not in the standard position, but that pink vector is going to be a member of our null space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "I get this vector right there. Let me do that in another color. I get this pink vector right there. That pink vector, it's not in the standard position, but that pink vector is going to be a member of our null space. So if I take 3, 0, that minus my r, so minus c times 3 minus 2, that's going to be this pink vector right there. So if I were to take the dot product, so this vector right here, let me make this clear, that vector is this pink vector right there. That's in my null space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "That pink vector, it's not in the standard position, but that pink vector is going to be a member of our null space. So if I take 3, 0, that minus my r, so minus c times 3 minus 2, that's going to be this pink vector right there. So if I were to take the dot product, so this vector right here, let me make this clear, that vector is this pink vector right there. That's in my null space. I could shift it over. It's not in the standard position there, but if I do it in the standard position, it points to something on my null space line. So that's in my null space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "That's in my null space. I could shift it over. It's not in the standard position there, but if I do it in the standard position, it points to something on my null space line. So that's in my null space. So if I were to dot it with any member of my row space, it's going to be equal to 0. The row space is the orthogonal complement of the null space. So let me take the dot product of that with some member of my row space."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So that's in my null space. So if I were to dot it with any member of my row space, it's going to be equal to 0. The row space is the orthogonal complement of the null space. So let me take the dot product of that with some member of my row space. Might as well just take the basis vector for my row space. That's a member of my row space. So if I dot it with 3 minus 2, that will be equal to 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So let me take the dot product of that with some member of my row space. Might as well just take the basis vector for my row space. That's a member of my row space. So if I dot it with 3 minus 2, that will be equal to 0. So let's see if we can solve for c. So let's see, if we do this, we get this inside part, this pink vector right there is going to be 3 minus 3c. And then you're going to have 0 minus 2c. So that's just 2c."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So if I dot it with 3 minus 2, that will be equal to 0. So let's see if we can solve for c. So let's see, if we do this, we get this inside part, this pink vector right there is going to be 3 minus 3c. And then you're going to have 0 minus 2c. So that's just 2c. So this part right here just simplifies to that. I just performed the scalar multiplication and then the subtraction. And then if we dot that with my basis vector for my row space, 3 minus 2, this should be equal to 0."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So that's just 2c. So this part right here just simplifies to that. I just performed the scalar multiplication and then the subtraction. And then if we dot that with my basis vector for my row space, 3 minus 2, this should be equal to 0. So what do we get? We get 3 times 3, which is 9, minus 3c times minus 2. Let me write it this way."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And then if we dot that with my basis vector for my row space, 3 minus 2, this should be equal to 0. So what do we get? We get 3 times 3, which is 9, minus 3c times minus 2. Let me write it this way. This is probably the easiest way to do it. Well, sorry, I was actually doing it wrong. This is the first entry."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it this way. This is probably the easiest way to do it. Well, sorry, I was actually doing it wrong. This is the first entry. So you have 3 minus 3c times 3. 3 minus 3c, the first entry, times the first entry here, plus 2c times minus 2 is equal to 0. I was performing that dot product in a strange way."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "This is the first entry. So you have 3 minus 3c times 3. 3 minus 3c, the first entry, times the first entry here, plus 2c times minus 2 is equal to 0. I was performing that dot product in a strange way. I think I got a little caught off guard by these two terms. But anyway, the first term times the first term plus the second term times the second term. That's what our dot product is."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "I was performing that dot product in a strange way. I think I got a little caught off guard by these two terms. But anyway, the first term times the first term plus the second term times the second term. That's what our dot product is. This isn't a matrix. This is just the first and the second term right there. So then what do we have here?"}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "That's what our dot product is. This isn't a matrix. This is just the first and the second term right there. So then what do we have here? We have 3 times 3 is 9, minus 9c, minus 4c is equal to 0. If we add, well, this is 9 minus 13c is equal to 0. Or we get 13."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So then what do we have here? We have 3 times 3 is 9, minus 9c, minus 4c is equal to 0. If we add, well, this is 9 minus 13c is equal to 0. Or we get 13. Let's see, 9 is equal to 13c. Or c is equal to 9 over 13. So just like that, we've solved for our special vector r. We said, hey, look."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Or we get 13. Let's see, 9 is equal to 13c. Or c is equal to 9 over 13. So just like that, we've solved for our special vector r. We said, hey, look. If you take this 3, 0 vector and the difference between that and our vector r, you're going to get some vector right here that is in our null space. And if you take the dot product of that with some member in our row space, you're going to get 0. And the dot product is that entry times that entry plus that entry times that entry."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "So just like that, we've solved for our special vector r. We said, hey, look. If you take this 3, 0 vector and the difference between that and our vector r, you're going to get some vector right here that is in our null space. And if you take the dot product of that with some member in our row space, you're going to get 0. And the dot product is that entry times that entry plus that entry times that entry. And so you get c is equal to 9 over 13. So our special vector r, the unique shortest solution to the equation ax is equal to 0 is 9 over 13 times the vector, times our basis vector, 3 minus 2. Or we could write that as 27 over 13."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "And the dot product is that entry times that entry plus that entry times that entry. And so you get c is equal to 9 over 13. So our special vector r, the unique shortest solution to the equation ax is equal to 0 is 9 over 13 times the vector, times our basis vector, 3 minus 2. Or we could write that as 27 over 13. And then we have 3 times 9, and then 2 times 9 is minus 18 over 13. And that right there is our shortest basis vector. The shortest vector in our row space that satisfies this equation."}, {"video_title": "Rowspace solution to Ax = b example Linear Algebra Khan Academy.mp3", "Sentence": "Or we could write that as 27 over 13. And then we have 3 times 9, and then 2 times 9 is minus 18 over 13. And that right there is our shortest basis vector. The shortest vector in our row space that satisfies this equation. Let me write it better. It's the unique member of our row space that also happens to be the shortest solution to ax is equal to b. So this is an example of what we talked about in the last video, and hopefully you see visually why this is so."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this topic is going to be useful when we do some types of transformations, actually the projections that I'll do in the next video. And the notion that I forgot to do is the notion of a unit vector. And all this is is a vector that has a length of 1. So let me see, has length, and we've defined length, has a length of 1. So if something is a unit vector, let's say that u right here is a unit vector, and it's a member of Rn, then that means that if we have u, u looks like this, has n components, u2, all the way to un. We know what the length of this is, right? The length of this, the definition of the length, we know that the length of u, sometimes called the norm of u, it's just equal to the square root of the squared sums of all of its components."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me see, has length, and we've defined length, has a length of 1. So if something is a unit vector, let's say that u right here is a unit vector, and it's a member of Rn, then that means that if we have u, u looks like this, has n components, u2, all the way to un. We know what the length of this is, right? The length of this, the definition of the length, we know that the length of u, sometimes called the norm of u, it's just equal to the square root of the squared sums of all of its components. And if you think about it, this is just an extension of the Pythagorean theorem to some degree. But so it's u1 squared plus u2 squared, all the way to un squared, and it's the square root of that. If this is a unit vector, that implies that the length of u will be equal to 1."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The length of this, the definition of the length, we know that the length of u, sometimes called the norm of u, it's just equal to the square root of the squared sums of all of its components. And if you think about it, this is just an extension of the Pythagorean theorem to some degree. But so it's u1 squared plus u2 squared, all the way to un squared, and it's the square root of that. If this is a unit vector, that implies that the length of u will be equal to 1. And that doesn't matter in what dimension space we are. This could be R100, this could be R2. For it to have a unit vector in any of those spaces, their length is 1."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If this is a unit vector, that implies that the length of u will be equal to 1. And that doesn't matter in what dimension space we are. This could be R100, this could be R2. For it to have a unit vector in any of those spaces, their length is 1. So the next obvious question is, how do you construct a unit vector? So let's say that I have some vector v. And let's say it's not a unit vector. So it's v1, v2, all the way to vn."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "For it to have a unit vector in any of those spaces, their length is 1. So the next obvious question is, how do you construct a unit vector? So let's say that I have some vector v. And let's say it's not a unit vector. So it's v1, v2, all the way to vn. And I want to turn it into some vector u that is a unit vector that just goes in the same direction. So u will go in the same direction as v, but just has a length of u is going to be equal to 1. How do I construct this vector u here?"}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's v1, v2, all the way to vn. And I want to turn it into some vector u that is a unit vector that just goes in the same direction. So u will go in the same direction as v, but just has a length of u is going to be equal to 1. How do I construct this vector u here? Well, what I could do is I could take the length of v. I could find out what the length of v is, and we know how to do that. We just apply this definition of vector length. And what happens if I figure out the length of v, and then I multiply the vector v times that?"}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How do I construct this vector u here? Well, what I could do is I could take the length of v. I could find out what the length of v is, and we know how to do that. We just apply this definition of vector length. And what happens if I figure out the length of v, and then I multiply the vector v times that? So what if I say u is equal to 1 over the length of v times v itself? What happens here? If I take the length of this thing right here, what do I get?"}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And what happens if I figure out the length of v, and then I multiply the vector v times that? So what if I say u is equal to 1 over the length of v times v itself? What happens here? If I take the length of this thing right here, what do I get? So the length of u is equal to the length of this scalar. Remember, this is just some number. It's equal to this scalar, and I'm assuming v is a non-zero vector."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I take the length of this thing right here, what do I get? So the length of u is equal to the length of this scalar. Remember, this is just some number. It's equal to this scalar, and I'm assuming v is a non-zero vector. So the length of this scalar number is times v. And we know that we can take this scalar out of the formula. We could show that. I think I've showed it in a previous video, that the length of c times v is equal to c times the length of v. Let me write that down."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to this scalar, and I'm assuming v is a non-zero vector. So the length of this scalar number is times v. And we know that we can take this scalar out of the formula. We could show that. I think I've showed it in a previous video, that the length of c times v is equal to c times the length of v. Let me write that down. And that's essentially what I'm assuming right here, that if I take the length of c times some vector v, that is equal to c times the length of v. I think we showed this when we first were introduced to the idea of length. So we know that this is going to be equal to 1 over the length of vector v, that's my c, times this thing right here, times the length of vector v. Well, what's this going to be equal to? 1 over something times that something."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I think I've showed it in a previous video, that the length of c times v is equal to c times the length of v. Let me write that down. And that's essentially what I'm assuming right here, that if I take the length of c times some vector v, that is equal to c times the length of v. I think we showed this when we first were introduced to the idea of length. So we know that this is going to be equal to 1 over the length of vector v, that's my c, times this thing right here, times the length of vector v. Well, what's this going to be equal to? 1 over something times that something. Well, this is just going to be equal to 1. So that's all a unit vector is. If you want to find a unit vector, or sometimes it's called a normalized vector, that goes in the same direction as some vector v, you just figure out the length of v, just using the definition of vector length in Rn, and then multiply 1 over that length times the vector v, and this is just a scalar, and then you get your vector u."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1 over something times that something. Well, this is just going to be equal to 1. So that's all a unit vector is. If you want to find a unit vector, or sometimes it's called a normalized vector, that goes in the same direction as some vector v, you just figure out the length of v, just using the definition of vector length in Rn, and then multiply 1 over that length times the vector v, and this is just a scalar, and then you get your vector u. Let me do an example just to make sure you get the idea. So let's say I have some vector v, and it's in R3. Let's say it's 1, 2, minus 1."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If you want to find a unit vector, or sometimes it's called a normalized vector, that goes in the same direction as some vector v, you just figure out the length of v, just using the definition of vector length in Rn, and then multiply 1 over that length times the vector v, and this is just a scalar, and then you get your vector u. Let me do an example just to make sure you get the idea. So let's say I have some vector v, and it's in R3. Let's say it's 1, 2, minus 1. What is the length of v? The length of v is equal to the square root of 1 squared plus 2 squared plus minus 1 squared, and that is equal to the square root of 1 plus 1 plus 4, square root of 6. So that is the length of v. So if I want to construct a normalized vector u that goes in the same direction as v, I can just define my vector u as being equal to 1 over the length of v, 1 over the square root of 6, times v. So times 1, 2, minus 1, which is equal to 1 over the square root of 6, 2 over the square root of 6, and minus 1 over the square root of 6."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say it's 1, 2, minus 1. What is the length of v? The length of v is equal to the square root of 1 squared plus 2 squared plus minus 1 squared, and that is equal to the square root of 1 plus 1 plus 4, square root of 6. So that is the length of v. So if I want to construct a normalized vector u that goes in the same direction as v, I can just define my vector u as being equal to 1 over the length of v, 1 over the square root of 6, times v. So times 1, 2, minus 1, which is equal to 1 over the square root of 6, 2 over the square root of 6, and minus 1 over the square root of 6. And I'll leave it for you to verify that the length of u is going to be equal to 1. Now I'll just throw out one other idea here that you'll often see. When something is a unit vector, instead of using this little arrow on top of the vector, they'll often write a unit vector with a little hat on top of it like that."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that is the length of v. So if I want to construct a normalized vector u that goes in the same direction as v, I can just define my vector u as being equal to 1 over the length of v, 1 over the square root of 6, times v. So times 1, 2, minus 1, which is equal to 1 over the square root of 6, 2 over the square root of 6, and minus 1 over the square root of 6. And I'll leave it for you to verify that the length of u is going to be equal to 1. Now I'll just throw out one other idea here that you'll often see. When something is a unit vector, instead of using this little arrow on top of the vector, they'll often write a unit vector with a little hat on top of it like that. That signifies that we're dealing with a unit vector. And for those of you who've taken your vector calculus or have done a little bit of engineering, you're probably familiar with the vectors i, j, and k. And the reason why they have this little hat here is because these are all unit vectors in R3. They're members of R3, and they're all unit vectors."}, {"video_title": "Unit vectors Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When something is a unit vector, instead of using this little arrow on top of the vector, they'll often write a unit vector with a little hat on top of it like that. That signifies that we're dealing with a unit vector. And for those of you who've taken your vector calculus or have done a little bit of engineering, you're probably familiar with the vectors i, j, and k. And the reason why they have this little hat here is because these are all unit vectors in R3. They're members of R3, and they're all unit vectors. These are actually the basis vectors in R3. And for those of you all who've been watching my transformation videos, these are equivalent to the vectors e1, which I could write with a hat on it, really, e2, and e3, which are the standard basis vectors in R3. Anyway, now that you've been exposed to it, now I can start to use the idea of a unit vector in future videos."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that the first vector is 1, 0, 0, 1. And the second vector is 0, 1, 0, 1. That is my subspace V. And you can see that these are going to be a basis, that these are linearly independent. Two vectors that are linear, or a set of vectors that are linearly independent and that span a subspace, are a basis for that subspace. You can see they're linearly independent. This guy has a 1 here. There's no way that you can take some combination of this guy to somehow get a 1 there."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Two vectors that are linear, or a set of vectors that are linearly independent and that span a subspace, are a basis for that subspace. You can see they're linearly independent. This guy has a 1 here. There's no way that you can take some combination of this guy to somehow get a 1 there. And then this guy has a 1 here. There's no way you can get some linear combination of these 0's here to get a 1 there. So they're linearly independent."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "There's no way that you can take some combination of this guy to somehow get a 1 there. And then this guy has a 1 here. There's no way you can get some linear combination of these 0's here to get a 1 there. So they're linearly independent. So you could also call this a basis for V. Now, given that, let's see if we can find out the transformation matrix for the projection of any arbitrary vector onto the subspace. So let's say that x, we're dealing in R4 here, right? So let's say that x is a member of R4."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So they're linearly independent. So you could also call this a basis for V. Now, given that, let's see if we can find out the transformation matrix for the projection of any arbitrary vector onto the subspace. So let's say that x, we're dealing in R4 here, right? So let's say that x is a member of R4. And I want to figure out what the projection, I want to figure out a transformation matrix for the projection onto V of x. Now, in the last video, we came up with a general way to figure this out. We said if A is a transformation matrix, sorry, if A is a matrix whose columns are the basis for the subspace, so let's say A is equal to 1, 0, 0, 1, 0, 1, 0, 1, so A is a matrix whose columns are the basis for a subspace."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that x is a member of R4. And I want to figure out what the projection, I want to figure out a transformation matrix for the projection onto V of x. Now, in the last video, we came up with a general way to figure this out. We said if A is a transformation matrix, sorry, if A is a matrix whose columns are the basis for the subspace, so let's say A is equal to 1, 0, 0, 1, 0, 1, 0, 1, so A is a matrix whose columns are the basis for a subspace. Then the projection of x onto V would be equal to, and this is kind of hard, the first time you look at it, it gives you a headache, but there's a certain pattern or symmetry or a way of, you could say it's A times, you're going to have something in the middle, and then you have A transpose times your vector x. And the way I remember it is, in the middle you have these two guys switched around, so then you have A transpose A, and you take the inverse of it. So you probably won't be using this in your everyday life five or 10 years from now, so it's OK if you don't memorize it."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "We said if A is a transformation matrix, sorry, if A is a matrix whose columns are the basis for the subspace, so let's say A is equal to 1, 0, 0, 1, 0, 1, 0, 1, so A is a matrix whose columns are the basis for a subspace. Then the projection of x onto V would be equal to, and this is kind of hard, the first time you look at it, it gives you a headache, but there's a certain pattern or symmetry or a way of, you could say it's A times, you're going to have something in the middle, and then you have A transpose times your vector x. And the way I remember it is, in the middle you have these two guys switched around, so then you have A transpose A, and you take the inverse of it. So you probably won't be using this in your everyday life five or 10 years from now, so it's OK if you don't memorize it. But temporarily put this in your medium term memory, because it's a good thing to know for doing these projection problems. So if we want to find the general matrix for this transformation, we just have to determine what this matrix is equal to. And that's just a bunch of matrix operations."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So you probably won't be using this in your everyday life five or 10 years from now, so it's OK if you don't memorize it. But temporarily put this in your medium term memory, because it's a good thing to know for doing these projection problems. So if we want to find the general matrix for this transformation, we just have to determine what this matrix is equal to. And that's just a bunch of matrix operations. So that's A. What is A transpose? A transpose is going to be equal to just all the rows turned to columns."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And that's just a bunch of matrix operations. So that's A. What is A transpose? A transpose is going to be equal to just all the rows turned to columns. So the first column becomes the first row, so it becomes 1, 0, 0, 1. Second column becomes the second row, 0, 1, 0, 1. That's what A transpose is."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "A transpose is going to be equal to just all the rows turned to columns. So the first column becomes the first row, so it becomes 1, 0, 0, 1. Second column becomes the second row, 0, 1, 0, 1. That's what A transpose is. Now what's A transpose A? To figure out that, I want to figure out what A transpose times A is. So let me multiply A transpose times A."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That's what A transpose is. Now what's A transpose A? To figure out that, I want to figure out what A transpose times A is. So let me multiply A transpose times A. So I'll rewrite A right here. 1, 0, 0, 1. 0, 1, 0, 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let me multiply A transpose times A. So I'll rewrite A right here. 1, 0, 0, 1. 0, 1, 0, 1. This will give us some good practice on matrix-matrix products. This is going to be equal to what? Well, first of all, this is a 2 by 4 matrix, and I'm multiplying it by a 4 by 2 matrix."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "0, 1, 0, 1. This will give us some good practice on matrix-matrix products. This is going to be equal to what? Well, first of all, this is a 2 by 4 matrix, and I'm multiplying it by a 4 by 2 matrix. So it's going to be a 2 by 2 matrix. So the first entry is essentially the dot product of that row with that column. So it's 1 times 1 plus 0 times 0 plus 0 times 0 plus 1 times 1, so it's just going to be 2 for that first entry right there."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Well, first of all, this is a 2 by 4 matrix, and I'm multiplying it by a 4 by 2 matrix. So it's going to be a 2 by 2 matrix. So the first entry is essentially the dot product of that row with that column. So it's 1 times 1 plus 0 times 0 plus 0 times 0 plus 1 times 1, so it's just going to be 2 for that first entry right there. And then you take the dot product of this guy with this guy right here. So it's 1 times 0, which is 0, plus 0 times 1, which is 0, plus 0 times 0, which is 0, plus 1 times 1, which is 1. Now we do this guy."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So it's 1 times 1 plus 0 times 0 plus 0 times 0 plus 1 times 1, so it's just going to be 2 for that first entry right there. And then you take the dot product of this guy with this guy right here. So it's 1 times 0, which is 0, plus 0 times 1, which is 0, plus 0 times 0, which is 0, plus 1 times 1, which is 1. Now we do this guy. This guy dotted with this column right there. 0 times 1 is 0, plus 1 times 0 is 0, plus 0 times 0 is 0, plus 1 times 1 is 1. And then finally, this row dotted with this second column, second row with second column."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Now we do this guy. This guy dotted with this column right there. 0 times 1 is 0, plus 1 times 0 is 0, plus 0 times 0 is 0, plus 1 times 1 is 1. And then finally, this row dotted with this second column, second row with second column. 0 times 0 is 0, 1 times 1 is 1, 0 times 0 is 0, 1 times 1 is 1, so we have 1 times 1 plus 1 times 1, so it's going to be 2, it's going to be equal to 2. So this right here, that right there is A transpose A. But that's not good enough."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, this row dotted with this second column, second row with second column. 0 times 0 is 0, 1 times 1 is 1, 0 times 0 is 0, 1 times 1 is 1, so we have 1 times 1 plus 1 times 1, so it's going to be 2, it's going to be equal to 2. So this right here, that right there is A transpose A. But that's not good enough. We need to figure out what the inverse of A transpose A is. This is A transpose A, but we need to figure out A transpose A inverse. So what's the inverse of this?"}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "But that's not good enough. We need to figure out what the inverse of A transpose A is. This is A transpose A, but we need to figure out A transpose A inverse. So what's the inverse of this? So let me write it here. The inverse, A transpose A inverse, is going to be equal to what? It's 1 over the determinant of this guy."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So what's the inverse of this? So let me write it here. The inverse, A transpose A inverse, is going to be equal to what? It's 1 over the determinant of this guy. And what's the determinant here? It's going to be 1 over the determinant of this. The determinant is 2 times 2, which is 4, minus 1 times 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "It's 1 over the determinant of this guy. And what's the determinant here? It's going to be 1 over the determinant of this. The determinant is 2 times 2, which is 4, minus 1 times 1. So it's 4 minus 1, which is 3. So 1 over the determinant times this guy, where if I swap these two, so I swap the 1's, sorry, I swap the 2's, so this 2 goes here, and then this orange 2 goes over here. And then I make these 1's negative."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "The determinant is 2 times 2, which is 4, minus 1 times 1. So it's 4 minus 1, which is 3. So 1 over the determinant times this guy, where if I swap these two, so I swap the 1's, sorry, I swap the 2's, so this 2 goes here, and then this orange 2 goes over here. And then I make these 1's negative. So this becomes a minus 1, and this becomes a minus 1. We learned that this is a general solution for the inverse of a 2 by 2 matrix. I think it was 10 or 11 videos ago, and you probably learned this in your Algebra 2 class, frankly."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then I make these 1's negative. So this becomes a minus 1, and this becomes a minus 1. We learned that this is a general solution for the inverse of a 2 by 2 matrix. I think it was 10 or 11 videos ago, and you probably learned this in your Algebra 2 class, frankly. But there you go. We have A transpose A inverse. So we have this guy."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I think it was 10 or 11 videos ago, and you probably learned this in your Algebra 2 class, frankly. But there you go. We have A transpose A inverse. So we have this guy. We have this whole guy here. It's just this matrix. I could multiply the 1 third into it, but I don't have to do that just yet."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So we have this guy. We have this whole guy here. It's just this matrix. I could multiply the 1 third into it, but I don't have to do that just yet. But let's figure out the whole matrix now. The whole A times this guy, A transpose A inverse times A transpose. So let me write it this way."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I could multiply the 1 third into it, but I don't have to do that just yet. But let's figure out the whole matrix now. The whole A times this guy, A transpose A inverse times A transpose. So let me write it this way. So the projection onto the subspace V of x is going to be equal to A 1, 0, 0, 1, 0. Let me do it a little bit bigger like this. So 1, 0, 0, 1, 0, 1, 0, 1 times A transpose A inverse."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let me write it this way. So the projection onto the subspace V of x is going to be equal to A 1, 0, 0, 1, 0. Let me do it a little bit bigger like this. So 1, 0, 0, 1, 0, 1, 0, 1 times A transpose A inverse. A times A transpose A inverse, which is this guy right here. And let's just put the 1 third out front, just because that's just a scalar. So I'll put the 1 third out front times this guy."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So 1, 0, 0, 1, 0, 1, 0, 1 times A transpose A inverse. A times A transpose A inverse, which is this guy right here. And let's just put the 1 third out front, just because that's just a scalar. So I'll put the 1 third out front times this guy. This A transpose A inverse is 1 third times 2 minus 1 minus 1, 2, and then I'm going to multiply it times A transpose. And all of that times our vector x. So A transpose is right there."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So I'll put the 1 third out front times this guy. This A transpose A inverse is 1 third times 2 minus 1 minus 1, 2, and then I'm going to multiply it times A transpose. And all of that times our vector x. So A transpose is right there. It is 1, 0, 0, 1, 0, 1, 0, 1, and then all of that's going to be times your vector x. So we still have some nice matrix-matrix products ahead of us. Let's see if we can do these."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So A transpose is right there. It is 1, 0, 0, 1, 0, 1, 0, 1, and then all of that's going to be times your vector x. So we still have some nice matrix-matrix products ahead of us. Let's see if we can do these. So the first one, let's just multiply these two guys. Let's multiply those two guys. I don't think there's any simple way to do it."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if we can do these. So the first one, let's just multiply these two guys. Let's multiply those two guys. I don't think there's any simple way to do it. So this is a 2 by 2 matrix, and this is a 2 by 4 matrix. So when I multiply them, I'm going to end up with a 2 by 4 matrix. So let me write that 2 by 4 matrix right here."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I don't think there's any simple way to do it. So this is a 2 by 2 matrix, and this is a 2 by 4 matrix. So when I multiply them, I'm going to end up with a 2 by 4 matrix. So let me write that 2 by 4 matrix right here. And then I can write this guy right here. 1, 0, 0, 1, 0, 1, 0, 1, and then I have the 1 third that was from A transpose A inverse, but I put the scaling factor out there. And all of this is equal to the projection of x onto v. So let's do this product."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So let me write that 2 by 4 matrix right here. And then I can write this guy right here. 1, 0, 0, 1, 0, 1, 0, 1, and then I have the 1 third that was from A transpose A inverse, but I put the scaling factor out there. And all of this is equal to the projection of x onto v. So let's do this product. So this first entry is going to be 2 times 1 plus minus 1 times 0. So that is just 2. And you're going to have 2 times 0 plus minus 1 times 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And all of this is equal to the projection of x onto v. So let's do this product. So this first entry is going to be 2 times 1 plus minus 1 times 0. So that is just 2. And you're going to have 2 times 0 plus minus 1 times 1. Well, that's minus 1. Then you're going to have 2 times 0 plus minus 1 times 0. Well, that's just 0."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And you're going to have 2 times 0 plus minus 1 times 1. Well, that's minus 1. Then you're going to have 2 times 0 plus minus 1 times 0. Well, that's just 0. And then you're going to have 2 times 1 plus minus 1 times 1. That's 2 minus 1. That's just 1, right?"}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Well, that's just 0. And then you're going to have 2 times 1 plus minus 1 times 1. That's 2 minus 1. That's just 1, right? 2 times 1 plus minus 1 times 1. Fair enough. Now let's do the second row."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That's just 1, right? 2 times 1 plus minus 1 times 1. Fair enough. Now let's do the second row. Minus 1 times 1 plus 2 times 0. So that's just minus 1. Minus 1 times 0 plus 2 times 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Now let's do the second row. Minus 1 times 1 plus 2 times 0. So that's just minus 1. Minus 1 times 0 plus 2 times 1. Well, that's just 2. Minus 1 times 0 plus 2 times 0. That's just 0."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 times 0 plus 2 times 1. Well, that's just 2. Minus 1 times 0 plus 2 times 0. That's just 0. Minus 1 times 1 plus 2 times 1. Well, that's minus 1 plus 2. So that is 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That's just 0. Minus 1 times 1 plus 2 times 1. Well, that's minus 1 plus 2. So that is 1. Almost there. And of course, we have to multiply it times x at the end. That's what the transformation is."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So that is 1. Almost there. And of course, we have to multiply it times x at the end. That's what the transformation is. But this right here is our transformation matrix. One more left to do. Let's hope I haven't made any careless mistakes and that I won't make any when doing this product."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That's what the transformation is. But this right here is our transformation matrix. One more left to do. Let's hope I haven't made any careless mistakes and that I won't make any when doing this product. Because this is going to be a little more complicated. Because this is a 4 by 2 times a 2 by 4. I'm going to end up with a 4 by 4 matrix."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "Let's hope I haven't made any careless mistakes and that I won't make any when doing this product. Because this is going to be a little more complicated. Because this is a 4 by 2 times a 2 by 4. I'm going to end up with a 4 by 4 matrix. So let me give myself some breathing room here. Because I'm going to generate a 4 by 4 matrix right there. And so what am I going to get?"}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to end up with a 4 by 4 matrix. So let me give myself some breathing room here. Because I'm going to generate a 4 by 4 matrix right there. And so what am I going to get? So this first entry is going to be 1 times 2 plus 0 times minus 1. So it's just going to be equal to 2. The next entry, this row times any column here is just going to be the first entry in the column."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And so what am I going to get? So this first entry is going to be 1 times 2 plus 0 times minus 1. So it's just going to be equal to 2. The next entry, this row times any column here is just going to be the first entry in the column. Because it's zeroed out. So 1 times 2 plus 0 times minus 1 is just 2. 1 times minus 1 plus 0 times 2 is just minus 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "The next entry, this row times any column here is just going to be the first entry in the column. Because it's zeroed out. So 1 times 2 plus 0 times minus 1 is just 2. 1 times minus 1 plus 0 times 2 is just minus 1. 1 times 0 plus 0 times 0 is 0. 1 times 1 plus 0 times 1 is just 1. When you take this row and you multiply it times these columns, you literally just got your first row there."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "1 times minus 1 plus 0 times 2 is just minus 1. 1 times 0 plus 0 times 0 is 0. 1 times 1 plus 0 times 1 is just 1. When you take this row and you multiply it times these columns, you literally just got your first row there. Now let's do this row times these columns. Now you've got a 0 here, so you're going to have a 0 times the first entry of all of these and a 1 times the second one. So 0 times 2 plus 1 times minus 1 is minus 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "When you take this row and you multiply it times these columns, you literally just got your first row there. Now let's do this row times these columns. Now you've got a 0 here, so you're going to have a 0 times the first entry of all of these and a 1 times the second one. So 0 times 2 plus 1 times minus 1 is minus 1. 0 times minus 1 plus 1 times 2 is 2. You're just going to get the second row here. 2, 0, 1."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So 0 times 2 plus 1 times minus 1 is minus 1. 0 times minus 1 plus 1 times 2 is 2. You're just going to get the second row here. 2, 0, 1. That actually makes sense. Because if you just look at this part of the matrix, it's the 2 by 2 identity matrix. So anyway, that's a little hint why this looks very much like that, but we're just going to go through this matrix product."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "2, 0, 1. That actually makes sense. Because if you just look at this part of the matrix, it's the 2 by 2 identity matrix. So anyway, that's a little hint why this looks very much like that, but we're just going to go through this matrix product. Now you multiply this. Let me do it in a different color. You multiply this guy times each of these columns."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So anyway, that's a little hint why this looks very much like that, but we're just going to go through this matrix product. Now you multiply this. Let me do it in a different color. You multiply this guy times each of these columns. That guy dotted with that is just going to be 0, because this guy is essentially the 0 row vector. So you're just going to get a bunch of 0's. And then finally, this last row, it's 1 times the first entry plus 1 times the second entry."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "You multiply this guy times each of these columns. That guy dotted with that is just going to be 0, because this guy is essentially the 0 row vector. So you're just going to get a bunch of 0's. And then finally, this last row, it's 1 times the first entry plus 1 times the second entry. So this guy's going to be 2 plus minus 1, which is 1. Minus 1 plus 2, which is 1. 0 plus 0, which is 0."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, this last row, it's 1 times the first entry plus 1 times the second entry. So this guy's going to be 2 plus minus 1, which is 1. Minus 1 plus 2, which is 1. 0 plus 0, which is 0. And then 1 plus 1, which is 2. And all of that times x. And there you have it."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "0 plus 0, which is 0. And then 1 plus 1, which is 2. And all of that times x. And there you have it. This is exciting. The projection onto v of x is equal to this whole matrix times x. So this thing right here, I could multiply the 1 third into it."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "And there you have it. This is exciting. The projection onto v of x is equal to this whole matrix times x. So this thing right here, I could multiply the 1 third into it. But we don't have to do that. That'll just make it a little bit more messy. This thing right here is the transformation matrix."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "So this thing right here, I could multiply the 1 third into it. But we don't have to do that. That'll just make it a little bit more messy. This thing right here is the transformation matrix. And as you can see, since we're transforming, remember this projection onto v, this is a linear transformation from R4 to R4. You give me some member of R4, and I'll give you another member of R4 that's in my subspace. That is the projection."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "This thing right here is the transformation matrix. And as you can see, since we're transforming, remember this projection onto v, this is a linear transformation from R4 to R4. You give me some member of R4, and I'll give you another member of R4 that's in my subspace. That is the projection. So this is going to be a 4 by 4 matrix. You can see it right there. Anyway, hopefully you found that useful to actually see a tangible result."}, {"video_title": "Subspace projection matrix example Linear Algebra Khan Academy.mp3", "Sentence": "That is the projection. So this is going to be a 4 by 4 matrix. You can see it right there. Anyway, hopefully you found that useful to actually see a tangible result. R4 is very abstract. So this would even be beyond our three-dimensional programming example. This would be, we're dealing with a more abstract data set where we're interested in finding a projection."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the first question I'm going to ask about the set of vectors s, and they're all three-dimensional vectors. They have three components, is the span of s equal to r3? And it seems like it might be. If each of these add new information, it seems like maybe I could describe any vector in r3 by these three vectors, by some combination of these three vectors. And the second question I'm going to ask is are they linearly independent? And maybe I'll be able to answer them at the same time. So let's answer the first one."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If each of these add new information, it seems like maybe I could describe any vector in r3 by these three vectors, by some combination of these three vectors. And the second question I'm going to ask is are they linearly independent? And maybe I'll be able to answer them at the same time. So let's answer the first one. Do they span r3? And to span r3, that means some linear combination of these three vectors should be able to construct any vector in r3. So let me give you a linear combination of these vectors."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's answer the first one. Do they span r3? And to span r3, that means some linear combination of these three vectors should be able to construct any vector in r3. So let me give you a linear combination of these vectors. So I could have c1 times the first vector, 1 minus 1, 2, plus some other arbitrary constant c2, some scalar, times the second vector, 2, 1, 3, plus some third scaling factor, times the third vector, minus 1, 0, 2. I should be able to, using some arbitrary constants, take a combination of these vectors that sum up to any vector in r3, and I'm going to represent any vector in r3 by the vector a, b, and c, where a, b, and c are any real numbers. So if you give me any a, b, and c, and I can give you a formula for telling you what your c3's, your c2's, and your c1's are, then that essentially means that it spans r3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me give you a linear combination of these vectors. So I could have c1 times the first vector, 1 minus 1, 2, plus some other arbitrary constant c2, some scalar, times the second vector, 2, 1, 3, plus some third scaling factor, times the third vector, minus 1, 0, 2. I should be able to, using some arbitrary constants, take a combination of these vectors that sum up to any vector in r3, and I'm going to represent any vector in r3 by the vector a, b, and c, where a, b, and c are any real numbers. So if you give me any a, b, and c, and I can give you a formula for telling you what your c3's, your c2's, and your c1's are, then that essentially means that it spans r3. Because if you give me a vector, I can always tell you how to construct that vector with these three. So let's see if I can do that. So just from our definition of scalar multiplication of a vector, we know that c1 times this vector, I could rewrite it if I want."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you give me any a, b, and c, and I can give you a formula for telling you what your c3's, your c2's, and your c1's are, then that essentially means that it spans r3. Because if you give me a vector, I can always tell you how to construct that vector with these three. So let's see if I can do that. So just from our definition of scalar multiplication of a vector, we know that c1 times this vector, I could rewrite it if I want. I normally skip this step, but I really want to make it clear. So c1 times this, I could just rewrite it as each of the terms times c1. Similarly, c2 times this is the same thing as each of the terms times c2."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So just from our definition of scalar multiplication of a vector, we know that c1 times this vector, I could rewrite it if I want. I normally skip this step, but I really want to make it clear. So c1 times this, I could just rewrite it as each of the terms times c1. Similarly, c2 times this is the same thing as each of the terms times c2. And c3 times this is the same thing as each of the terms times c3. I want to show you that everything we do, it just formally comes from our definition of multiplication of a vector times a scalar, which is what we just did. Or vector addition, which is what we're about to do."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Similarly, c2 times this is the same thing as each of the terms times c2. And c3 times this is the same thing as each of the terms times c3. I want to show you that everything we do, it just formally comes from our definition of multiplication of a vector times a scalar, which is what we just did. Or vector addition, which is what we're about to do. So vector addition tells us that this term plus this term plus this term needs to equal that term. So let me write that down. We get c1 plus 2c2 minus c3 will be equal to a."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or vector addition, which is what we're about to do. So vector addition tells us that this term plus this term plus this term needs to equal that term. So let me write that down. We get c1 plus 2c2 minus c3 will be equal to a. Likewise, we can do the same thing with the next row. Minus c1 plus c2 plus 0c3 must be equal to b. So we get minus c1 plus c2 plus 0c3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We get c1 plus 2c2 minus c3 will be equal to a. Likewise, we can do the same thing with the next row. Minus c1 plus c2 plus 0c3 must be equal to b. So we get minus c1 plus c2 plus 0c3. So we only have to write that. It's going to be equal to b. And then finally, let's just do that last row."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we get minus c1 plus c2 plus 0c3. So we only have to write that. It's going to be equal to b. And then finally, let's just do that last row. 2c1 plus 3c2 plus 2c3 is going to be equal to c. Right there. Now let's see if we can solve for our different constants. So I'm going to do it by elimination."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then finally, let's just do that last row. 2c1 plus 3c2 plus 2c3 is going to be equal to c. Right there. Now let's see if we can solve for our different constants. So I'm going to do it by elimination. I think you might be familiar with this process. I think I've done it in some of the earlier linear algebra videos before I started doing a formal presentation of it. And I'm going to review it again in a few videos from now."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to do it by elimination. I think you might be familiar with this process. I think I've done it in some of the earlier linear algebra videos before I started doing a formal presentation of it. And I'm going to review it again in a few videos from now. But I think you understand how to solve it this way. What I'm going to do is I'm going to first eliminate these two terms, and then I'm going to eliminate this term. And then I can solve for my various constants."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I'm going to review it again in a few videos from now. But I think you understand how to solve it this way. What I'm going to do is I'm going to first eliminate these two terms, and then I'm going to eliminate this term. And then I can solve for my various constants. So if I want to eliminate this term right here, what I could do is I could add this equation to that equation. Or even better, I can replace this equation with the sum of these two equations. So let me do that."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I can solve for my various constants. So if I want to eliminate this term right here, what I could do is I could add this equation to that equation. Or even better, I can replace this equation with the sum of these two equations. So let me do that. So I'm just going to add these two equations to each other and replace this one with that sum. So minus c1 plus c1, that just gives you 0. I can ignore it."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me do that. So I'm just going to add these two equations to each other and replace this one with that sum. So minus c1 plus c1, that just gives you 0. I can ignore it. Then c2 plus 2c2, that's 3c2. And then 0 plus minus c3 is equal to minus c3. And I'm replacing this with the sum of these two."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I can ignore it. Then c2 plus 2c2, that's 3c2. And then 0 plus minus c3 is equal to minus c3. And I'm replacing this with the sum of these two. So b plus a. It equals b plus a. Let me write down that first equation on the top."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I'm replacing this with the sum of these two. So b plus a. It equals b plus a. Let me write down that first equation on the top. So the first equation, I'm not doing anything to it. So I get c1 plus 2c2 minus c3 is equal to a. Now in this last equation, I want to eliminate this term."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write down that first equation on the top. So the first equation, I'm not doing anything to it. So I get c1 plus 2c2 minus c3 is equal to a. Now in this last equation, I want to eliminate this term. So let's take this equation and subtract from it 2 times this top equation. So you could also view it as, let's add this to minus 2 times this top equation. And since we're almost done using this, we actually even wrote it, let's just multiply this times minus 2."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now in this last equation, I want to eliminate this term. So let's take this equation and subtract from it 2 times this top equation. So you could also view it as, let's add this to minus 2 times this top equation. And since we're almost done using this, we actually even wrote it, let's just multiply this times minus 2. So this becomes a minus 2c1 minus 4c2 plus 2c3. So plus 2c3 is equal to minus 2a. If you just multiply each of these terms, I want to be very careful, I don't want to make a careless mistake."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And since we're almost done using this, we actually even wrote it, let's just multiply this times minus 2. So this becomes a minus 2c1 minus 4c2 plus 2c3. So plus 2c3 is equal to minus 2a. If you just multiply each of these terms, I want to be very careful, I don't want to make a careless mistake. Minus 2 times c1 minus 4 plus 2 and then minus 2. And now we can add these two together. And what do we get?"}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you just multiply each of these terms, I want to be very careful, I don't want to make a careless mistake. Minus 2 times c1 minus 4 plus 2 and then minus 2. And now we can add these two together. And what do we get? The 2c1 minus 2c1, that's a 0. Don't have to write it. 3c2 minus 4c2, that's a minus c2."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what do we get? The 2c1 minus 2c1, that's a 0. Don't have to write it. 3c2 minus 4c2, that's a minus c2. And then you have your 2c3 plus another 2c3. So that is equal to plus 4c3 is equal to c minus 2a. All I did is I replaced this with this minus 2 times that."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "3c2 minus 4c2, that's a minus c2. And then you have your 2c3 plus another 2c3. So that is equal to plus 4c3 is equal to c minus 2a. All I did is I replaced this with this minus 2 times that. And I got this. Now I want to eliminate, now I'm going to keep my top equation constant again. I'm not going to do anything to it, so I'm just going to move it to the right."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I replaced this with this minus 2 times that. And I got this. Now I want to eliminate, now I'm going to keep my top equation constant again. I'm not going to do anything to it, so I'm just going to move it to the right. So I get c1 plus 2c2 minus c3 is equal to a. I'm also going to keep my second equation the same. So I get 3c2 minus c3 is equal to b plus a. Let me scroll over a good bit."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm not going to do anything to it, so I'm just going to move it to the right. So I get c1 plus 2c2 minus c3 is equal to a. I'm also going to keep my second equation the same. So I get 3c2 minus c3 is equal to b plus a. Let me scroll over a good bit. And then this last equation I want to eliminate. My goal is to eliminate this term right here. So what I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll over a good bit. And then this last equation I want to eliminate. My goal is to eliminate this term right here. So what I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here. So if I multiply this bottom equation times 3, let me just do, well, actually, I don't want to make things messier. So this becomes a minus 3 plus a 3, so those cancel out. This becomes a 12 minus a 1, right?"}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here. So if I multiply this bottom equation times 3, let me just do, well, actually, I don't want to make things messier. So this becomes a minus 3 plus a 3, so those cancel out. This becomes a 12 minus a 1, right? So this becomes 12c3 minus c3, which is 11c3. And then this becomes a, oh, sorry. I was already done."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This becomes a 12 minus a 1, right? So this becomes 12c3 minus c3, which is 11c3. And then this becomes a, oh, sorry. I was already done. When I do 3 times this plus that, those canceled out. And when I multiply 3 times this, I get 12c3 minus a c3, so that's 11c3. And I multiply this times 3 plus this."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I was already done. When I do 3 times this plus that, those canceled out. And when I multiply 3 times this, I get 12c3 minus a c3, so that's 11c3. And I multiply this times 3 plus this. So I get 3c minus 6a, right? I'm just multiplying this times 3, plus this. Plus b plus a."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I multiply this times 3 plus this. So I get 3c minus 6a, right? I'm just multiplying this times 3, plus this. Plus b plus a. And so what can I rewrite this by? Actually, I want to make something very clear. This c is different than these c1's, c2's, and c3's that I had up here."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Plus b plus a. And so what can I rewrite this by? Actually, I want to make something very clear. This c is different than these c1's, c2's, and c3's that I had up here. I think you realize that. But I just realized that I used the letter c twice, and I just didn't want any confusion here. So this c that doesn't have any subscript is a different constant than all of these things over here."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This c is different than these c1's, c2's, and c3's that I had up here. I think you realize that. But I just realized that I used the letter c twice, and I just didn't want any confusion here. So this c that doesn't have any subscript is a different constant than all of these things over here. So let's see if we can simplify this. So we have an a and a minus 6a, so let's just add them. So let's go to that a, and then this becomes minus 5a."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this c that doesn't have any subscript is a different constant than all of these things over here. So let's see if we can simplify this. So we have an a and a minus 6a, so let's just add them. So let's go to that a, and then this becomes minus 5a. So we get, and then if we divide both sides of this equation by 11, what do we get? We get c3 is equal to 1 over 11 times 3c minus 5a. So you give me any a or c, and I'll already tell you what c3 is."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's go to that a, and then this becomes minus 5a. So we get, and then if we divide both sides of this equation by 11, what do we get? We get c3 is equal to 1 over 11 times 3c minus 5a. So you give me any a or c, and I'll already tell you what c3 is. Now what is c2? c2 is equal to, let me simplify this equation right here, let me do it right there. So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you give me any a or c, and I'll already tell you what c3 is. Now what is c2? c2 is equal to, let me simplify this equation right here, let me do it right there. So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3. Well, I know, and if I divide both sides of this by 3, I get c2 is equal to 1 over 3 times b plus a plus c3. I'll just leave it like that for now. And then what is c1 is equal to?"}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3. Well, I know, and if I divide both sides of this by 3, I get c2 is equal to 1 over 3 times b plus a plus c3. I'll just leave it like that for now. And then what is c1 is equal to? Well, I could just rewrite this top equation as, if I subtract 2c2 and add c3 to both sides, I get c1 is equal to a minus 2c2 plus c3. So what have I just shown you? You can give me any vector in R3 that you want to find."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then what is c1 is equal to? Well, I could just rewrite this top equation as, if I subtract 2c2 and add c3 to both sides, I get c1 is equal to a minus 2c2 plus c3. So what have I just shown you? You can give me any vector in R3 that you want to find. So you can give me any real number for a, any real number for b, any real number for c. And if you give me those numbers, I'm claiming now that I can always tell you some combination of these three vectors that will add up to those. And I've actually already solved for what I have to multiply each of those vectors by to add up to this third vector. So you give me your a's, b's, and c's."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can give me any vector in R3 that you want to find. So you can give me any real number for a, any real number for b, any real number for c. And if you give me those numbers, I'm claiming now that I can always tell you some combination of these three vectors that will add up to those. And I've actually already solved for what I have to multiply each of those vectors by to add up to this third vector. So you give me your a's, b's, and c's. I just have to substitute into the a's and the c's right here. And oh, sorry, I forgot this b over here. There's also a b."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you give me your a's, b's, and c's. I just have to substitute into the a's and the c's right here. And oh, sorry, I forgot this b over here. There's also a b. It was suspicious that I didn't have to deal with a b. So this is a, there was a b right there. So this is 3c minus 5a plus b."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's also a b. It was suspicious that I didn't have to deal with a b. So this is a, there was a b right there. So this is 3c minus 5a plus b. Let me write that. Plus a b. There's a b right there in the parentheses."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is 3c minus 5a plus b. Let me write that. Plus a b. There's a b right there in the parentheses. But I think you get the general idea. You give me your a, b's, and c's. Any real numbers can apply."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There's a b right there in the parentheses. But I think you get the general idea. You give me your a, b's, and c's. Any real numbers can apply. I'm not dividing by, there's no division over here, so I don't have to worry about dividing by 0. So this is just a linear combination of any real numbers, so I can clearly get another real number. So you give me your a, b's, and c's."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Any real numbers can apply. I'm not dividing by, there's no division over here, so I don't have to worry about dividing by 0. So this is just a linear combination of any real numbers, so I can clearly get another real number. So you give me your a, b's, and c's. I'm going to give you a c3. Now, you give me a, b's, and c's. I got a c3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you give me your a, b's, and c's. I'm going to give you a c3. Now, you give me a, b's, and c's. I got a c3. This is just going to be another real number. And I'm just going to take that with your former a's and b's, and I'm going to be able to give you c2. And then I already give you, we already are able to solve for a c2 and a c3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I got a c3. This is just going to be another real number. And I'm just going to take that with your former a's and b's, and I'm going to be able to give you c2. And then I already give you, we already are able to solve for a c2 and a c3. And then I just use your a as well. And then I'm going to give you a c1. So hopefully you're seeing that no matter what a, b, and c you give me, I can give you a c1, c2, or c3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I already give you, we already are able to solve for a c2 and a c3. And then I just use your a as well. And then I'm going to give you a c1. So hopefully you're seeing that no matter what a, b, and c you give me, I can give you a c1, c2, or c3. There's no reason that any a, b's, or c's should break down these formulas. We're not doing any division, so it's not like a 0 would break it down. So I can say definitively that the set of vectors, of these three vectors, does indeed span R3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So hopefully you're seeing that no matter what a, b, and c you give me, I can give you a c1, c2, or c3. There's no reason that any a, b's, or c's should break down these formulas. We're not doing any division, so it's not like a 0 would break it down. So I can say definitively that the set of vectors, of these three vectors, does indeed span R3. Now let me ask you another question, where I already asked it. Are these vectors linearly independent? Are they linearly independent?"}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I can say definitively that the set of vectors, of these three vectors, does indeed span R3. Now let me ask you another question, where I already asked it. Are these vectors linearly independent? Are they linearly independent? Now we said that in order for them to be linearly independent, the only solution to c1 times my first vector, 1 minus 1, 2, plus c2 times my second vector, 2, 1, 3, plus c3 times my third vector, minus 1, 0, 2. So if something is linearly independent, that means that the only solution to this equation. So if I want to find some set of combinations of these vectors that add up to the 0 vector, and I did that in the previous video, if they are linearly dependent, there must be some non-zero solution."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Are they linearly independent? Now we said that in order for them to be linearly independent, the only solution to c1 times my first vector, 1 minus 1, 2, plus c2 times my second vector, 2, 1, 3, plus c3 times my third vector, minus 1, 0, 2. So if something is linearly independent, that means that the only solution to this equation. So if I want to find some set of combinations of these vectors that add up to the 0 vector, and I did that in the previous video, if they are linearly dependent, there must be some non-zero solution. So at least one of these constants would be non-zero for this solution. I mean, you can always make them 0 no matter what, but if they're linearly dependent, then one of these could be non-zero. If they're linearly independent, then all of these have to be, the only solution to this equation would be c1, c2, c3, all have to be equal to 0."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I want to find some set of combinations of these vectors that add up to the 0 vector, and I did that in the previous video, if they are linearly dependent, there must be some non-zero solution. So at least one of these constants would be non-zero for this solution. I mean, you can always make them 0 no matter what, but if they're linearly dependent, then one of these could be non-zero. If they're linearly independent, then all of these have to be, the only solution to this equation would be c1, c2, c3, all have to be equal to 0. Linear independence implies this, this implies linear independence. Now, this is the exact same thing we did here, but in this case, I'm just picking my a, b's, and c's to be 0. This is a, this is b, and this is c. I can pick any vector in R3 for my a's, b's, and c's."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If they're linearly independent, then all of these have to be, the only solution to this equation would be c1, c2, c3, all have to be equal to 0. Linear independence implies this, this implies linear independence. Now, this is the exact same thing we did here, but in this case, I'm just picking my a, b's, and c's to be 0. This is a, this is b, and this is c. I can pick any vector in R3 for my a's, b's, and c's. I'm now picking the 0 vector. So let's see what our c1's, c2's, and c3's are. So my a equals b is equal to c is equal to 0."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a, this is b, and this is c. I can pick any vector in R3 for my a's, b's, and c's. I'm now picking the 0 vector. So let's see what our c1's, c2's, and c3's are. So my a equals b is equal to c is equal to 0. I'm setting it equal to the 0 vector. What linear combination of these three vectors equal the 0 vector? Well, if a, b, and c are all equal to 0, that term is 0, that is 0, that is 0."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So my a equals b is equal to c is equal to 0. I'm setting it equal to the 0 vector. What linear combination of these three vectors equal the 0 vector? Well, if a, b, and c are all equal to 0, that term is 0, that is 0, that is 0. You have 1 11th times 0 minus 0 plus 0, that's just 0. So c3 is equal to 0. Now, c3 is equal to 0."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, if a, b, and c are all equal to 0, that term is 0, that is 0, that is 0. You have 1 11th times 0 minus 0 plus 0, that's just 0. So c3 is equal to 0. Now, c3 is equal to 0. We already know that a is equal to 0 and b is equal to 0. So c2 is 1 3rd times 0, so it equals 0. Now, what's c1?"}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, c3 is equal to 0. We already know that a is equal to 0 and b is equal to 0. So c2 is 1 3rd times 0, so it equals 0. Now, what's c1? Well, it's c3, which is 0. c2 is 0, so 2 times 0 is 0. So c1 is just going to be equal to a. What I just said, a is equal to 0."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, what's c1? Well, it's c3, which is 0. c2 is 0, so 2 times 0 is 0. So c1 is just going to be equal to a. What I just said, a is equal to 0. So the only solution to this equation right here, the only combination of, the only linear combination of these three vectors that result in the 0 vector are when you weight all of them by 0. So I just showed you that c1, c2, and c3 all have to be 0. And because they're all 0, we know that this is a linearly independent set of vectors."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What I just said, a is equal to 0. So the only solution to this equation right here, the only combination of, the only linear combination of these three vectors that result in the 0 vector are when you weight all of them by 0. So I just showed you that c1, c2, and c3 all have to be 0. And because they're all 0, we know that this is a linearly independent set of vectors. Or that none of these vectors can be represented as a combination of the other two. None of these vectors can be represented as a combination of another two. So this is interesting."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And because they're all 0, we know that this is a linearly independent set of vectors. Or that none of these vectors can be represented as a combination of the other two. None of these vectors can be represented as a combination of another two. So this is interesting. I have exactly three vectors that span R3, and they're linearly independent. And linearly independent, in my brain, that means that, look, I don't have any redundant vectors, anything that could have just been built with the other vectors. And I have exactly three vectors, and it's spanning R3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is interesting. I have exactly three vectors that span R3, and they're linearly independent. And linearly independent, in my brain, that means that, look, I don't have any redundant vectors, anything that could have just been built with the other vectors. And I have exactly three vectors, and it's spanning R3. So in general, and I haven't proven this to you, but I could, is that if you have exactly three vectors and they do span R3, they have to be linearly independent. If they weren't linearly independent, then one of these would be redundant. Then let's say that that guy was a redundant one."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I have exactly three vectors, and it's spanning R3. So in general, and I haven't proven this to you, but I could, is that if you have exactly three vectors and they do span R3, they have to be linearly independent. If they weren't linearly independent, then one of these would be redundant. Then let's say that that guy was a redundant one. I always pick the third one. But let's say this guy would be redundant, which means that the span of this would be equal to the span of these two, because if this guy is redundant, he could be part of the span of these two guys. And the span of two vectors could never span R3."}, {"video_title": "Span and linear independence example Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then let's say that that guy was a redundant one. I always pick the third one. But let's say this guy would be redundant, which means that the span of this would be equal to the span of these two, because if this guy is redundant, he could be part of the span of these two guys. And the span of two vectors could never span R3. Or in the other way, you could go, if you have three linear independent, three tuples, and they're all independent, then you could also say that that spans R3. But I haven't proven that to you, but hopefully you get the sense that each of these is contributing new directionality. One is going like that."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have a matrix where everything below the main diagonal is a 0. And I'll start, just for the sake of argument, let's start with a 2 by 2 matrix. So let's start with a 2 by 2 matrix. So I have the values a, b, 0, and d. So instead of a c, I have a 0 there. So everything below the main diagonal is a 0. So what is the determinant of this going to be? Let's call that matrix A."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I have the values a, b, 0, and d. So instead of a c, I have a 0 there. So everything below the main diagonal is a 0. So what is the determinant of this going to be? Let's call that matrix A. So the determinant of A is going to be equal to ad minus b times 0. So that's just 0. So we don't have to write it."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that matrix A. So the determinant of A is going to be equal to ad minus b times 0. So that's just 0. So we don't have to write it. So it's equal to a times d. Now let's say I have another matrix. Let's call it B. And let's say it's a 3 by 3 matrix."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we don't have to write it. So it's equal to a times d. Now let's say I have another matrix. Let's call it B. And let's say it's a 3 by 3 matrix. And let's say its entries are a, b, c. You've got a 0 here. Then let's say you have a d here, e. Then you have another 0 here, another 0 here, and you have an f. So once again, all of the entries below the main diagonal are 0. What's this guy's determinant?"}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say it's a 3 by 3 matrix. And let's say its entries are a, b, c. You've got a 0 here. Then let's say you have a d here, e. Then you have another 0 here, another 0 here, and you have an f. So once again, all of the entries below the main diagonal are 0. What's this guy's determinant? Well, we learned several videos ago that you can always pick the row and the column that has the most 0's on it. That simplifies your situation. So let's find the determinant along this column right here."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What's this guy's determinant? Well, we learned several videos ago that you can always pick the row and the column that has the most 0's on it. That simplifies your situation. So let's find the determinant along this column right here. So the determinant of B is going to be equal to A times the submatrix if you were to ignore A's row and column. So A times the determinant of d, e, 0, f. And then minus 0 times its submatrix. You cancel out."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's find the determinant along this column right here. So the determinant of B is going to be equal to A times the submatrix if you were to ignore A's row and column. So A times the determinant of d, e, 0, f. And then minus 0 times its submatrix. You cancel out. Or times the determinant of its submatrix. That row and that column. You get B, c, 0, f. And then you have plus 0 times, you get rid of that row, that column, you get B, c, d, e. Now obviously these two guys are going to be 0."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You cancel out. Or times the determinant of its submatrix. That row and that column. You get B, c, 0, f. And then you have plus 0 times, you get rid of that row, that column, you get B, c, d, e. Now obviously these two guys are going to be 0. I don't care what these 2 by 2 matrices, what their determinants end up evaluating to. So these are both going to be equal to 0 because we're multiplying by 0. We're left with A times the determinant of this."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You get B, c, 0, f. And then you have plus 0 times, you get rid of that row, that column, you get B, c, d, e. Now obviously these two guys are going to be 0. I don't care what these 2 by 2 matrices, what their determinants end up evaluating to. So these are both going to be equal to 0 because we're multiplying by 0. We're left with A times the determinant of this. And the determinant of this is pretty straightforward. We're going to have, it's going to be equal to A times, the determinant of this is df minus 0 times e. So it's just going to be df. So the determinant of B is A, d, f. Notice the determinant of A was just A and d. Now you might see a pattern."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're left with A times the determinant of this. And the determinant of this is pretty straightforward. We're going to have, it's going to be equal to A times, the determinant of this is df minus 0 times e. So it's just going to be df. So the determinant of B is A, d, f. Notice the determinant of A was just A and d. Now you might see a pattern. In both cases we had 0's below the main diagonal. This was the main diagonal right here. And when we took the determinants of the matrix, the determinant just ended up being the product of the entries along the main diagonal."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of B is A, d, f. Notice the determinant of A was just A and d. Now you might see a pattern. In both cases we had 0's below the main diagonal. This was the main diagonal right here. And when we took the determinants of the matrix, the determinant just ended up being the product of the entries along the main diagonal. And if you think that that's a general trend that always applies, you are correct. We can do it in the general case. Let's do the general case."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And when we took the determinants of the matrix, the determinant just ended up being the product of the entries along the main diagonal. And if you think that that's a general trend that always applies, you are correct. We can do it in the general case. Let's do the general case. So let's say we have some matrix A, and it is equal to A11, then you have A22, you're going to have a 0 right there. And then you just keep going all the way down to A and n. In this row, everything is going to be a 0, except for that last column. This is all a 0 right here."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's do the general case. So let's say we have some matrix A, and it is equal to A11, then you have A22, you're going to have a 0 right there. And then you just keep going all the way down to A and n. In this row, everything is going to be a 0, except for that last column. This is all a 0 right here. So everything below the main diagonal is a 0. Just like this one, but we're doing it in the general n by n case. And everything up here is, well, it doesn't have to be 0."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is all a 0 right here. So everything below the main diagonal is a 0. Just like this one, but we're doing it in the general n by n case. And everything up here is, well, it doesn't have to be 0. This is A12 all the way to A1n. This is A2n, keep going down. So everything at the main diagonal or above isn't necessarily equal to 0."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And everything up here is, well, it doesn't have to be 0. This is A12 all the way to A1n. This is A2n, keep going down. So everything at the main diagonal or above isn't necessarily equal to 0. So if you wanted to find the determinant of A, we could do the same thing we did here. We could go down that first row right there. So the determinant of our matrix A is equal to this guy, A11, times the determinant of its submatrix."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So everything at the main diagonal or above isn't necessarily equal to 0. So if you wanted to find the determinant of A, we could do the same thing we did here. We could go down that first row right there. So the determinant of our matrix A is equal to this guy, A11, times the determinant of its submatrix. So that's going to be A22 goes all the way to A2n, and then A33 all the way to Ann, and then everything down here, these are all 0's. So once again, we have another situation where all of the entries below the main diagonal are 0. So what's the determinant of this guy right here?"}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of our matrix A is equal to this guy, A11, times the determinant of its submatrix. So that's going to be A22 goes all the way to A2n, and then A33 all the way to Ann, and then everything down here, these are all 0's. So once again, we have another situation where all of the entries below the main diagonal are 0. So what's the determinant of this guy right here? And you might say, hey, what about the rest of that row? Well, the rest of the row is just a bunch of 0's, just like we had here. It's 0 times the determinant of its submatrix, and then that'd be a minus and a plus, 0 times the determinant of its submatrix, so on and so forth."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what's the determinant of this guy right here? And you might say, hey, what about the rest of that row? Well, the rest of the row is just a bunch of 0's, just like we had here. It's 0 times the determinant of its submatrix, and then that'd be a minus and a plus, 0 times the determinant of its submatrix, so on and so forth. So we just have to pay attention to this term right there. Now, the same argument we can do here. To find this determinant, we can just go down that row."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's 0 times the determinant of its submatrix, and then that'd be a minus and a plus, 0 times the determinant of its submatrix, so on and so forth. So we just have to pay attention to this term right there. Now, the same argument we can do here. To find this determinant, we can just go down that row. So the determinant of this is just going to be equal to, let's write out, let's not forget our A11 out there. A11, and the determinant of this is going to be A22 times the determinant of its submatrix. Get rid of its row and its column, and you're just left with A33 all the way down to Ann."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "To find this determinant, we can just go down that row. So the determinant of this is just going to be equal to, let's write out, let's not forget our A11 out there. A11, and the determinant of this is going to be A22 times the determinant of its submatrix. Get rid of its row and its column, and you're just left with A33 all the way down to Ann. Everything up here is non-zero. This is A3n. And then everything below the diagonal, once again, is just a bunch of 0's."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Get rid of its row and its column, and you're just left with A33 all the way down to Ann. Everything up here is non-zero. This is A3n. And then everything below the diagonal, once again, is just a bunch of 0's. Everything down here is a bunch of 0's. Another what we call upper triangular matrix. Let me write that down."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then everything below the diagonal, once again, is just a bunch of 0's. Everything down here is a bunch of 0's. Another what we call upper triangular matrix. Let me write that down. This whole class where you have 0's below the main diagonal, these are called upper triangular matrices. Just like that. Now, we keep doing the process over and over again."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. This whole class where you have 0's below the main diagonal, these are called upper triangular matrices. Just like that. Now, we keep doing the process over and over again. If you just keep following this pattern over again, now you're going to have the determinant of this is A33 times its submatrix. And every time the submatrix is getting smaller and smaller, you will eventually get to A11 times A22 times all the way to An minus 2 times a 2 by 2 matrix over here. There's just going to be An minus 1 An sub n. This is going to be A sub n minus 1 n. And then you're going to have a 0 right here."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we keep doing the process over and over again. If you just keep following this pattern over again, now you're going to have the determinant of this is A33 times its submatrix. And every time the submatrix is getting smaller and smaller, you will eventually get to A11 times A22 times all the way to An minus 2 times a 2 by 2 matrix over here. There's just going to be An minus 1 An sub n. This is going to be A sub n minus 1 n. And then you're going to have a 0 right here. So it's just the bottom right-hand corner of our original matrix is what you're going to be left with. And what is the determinant of this? Well, it's just the product of these two things."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "There's just going to be An minus 1 An sub n. This is going to be A sub n minus 1 n. And then you're going to have a 0 right here. So it's just the bottom right-hand corner of our original matrix is what you're going to be left with. And what is the determinant of this? Well, it's just the product of these two things. It's just this guy times this guy minus this guy times that guy, but that's just 0. So the determinant of A ends up becoming A11 times A22 all the way to Ann, or the product of all of the entries of the domain diagonal, which is a super important takeaway because it really simplifies finding the determinants of what would otherwise be really hard matrices to find the determinants of. You could imagine if this was a 100 by 100 matrix."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's just the product of these two things. It's just this guy times this guy minus this guy times that guy, but that's just 0. So the determinant of A ends up becoming A11 times A22 all the way to Ann, or the product of all of the entries of the domain diagonal, which is a super important takeaway because it really simplifies finding the determinants of what would otherwise be really hard matrices to find the determinants of. You could imagine if this was a 100 by 100 matrix. Now we can just multiply the diagonal. So just to make sure that things are clear, let me do an example. Let's say we find the determinant of 7, 3, 4, 2."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could imagine if this was a 100 by 100 matrix. Now we can just multiply the diagonal. So just to make sure that things are clear, let me do an example. Let's say we find the determinant of 7, 3, 4, 2. Let's say we have 0's here. Let's say it's a minus 2, 1, and a 3. We don't want 0's there."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we find the determinant of 7, 3, 4, 2. Let's say we have 0's here. Let's say it's a minus 2, 1, and a 3. We don't want 0's there. We don't need to have 0's there. 6, 7. We actually could have 0's there, but we don't need to have 0's there."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We don't want 0's there. We don't need to have 0's there. 6, 7. We actually could have 0's there, but we don't need to have 0's there. You have a 0 there, and you have 0's there. Just like that. So it's upper triangular matrix."}, {"video_title": "Upper triangular determinant Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We actually could have 0's there, but we don't need to have 0's there. You have a 0 there, and you have 0's there. Just like that. So it's upper triangular matrix. If you want to evaluate this determinant, you just multiply these entries right here. So the determinant is equal to 7 times minus 2 times 1 times 3, so it's 7 times minus 6, which is equal to minus 42. And it's that easy."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So it'll map it over here. And we could call that the mapping of T or the mapping of x or T of x. And since T is a linear transformation, we know that the mapping of x to its codomain is equivalent to x being multiplied by some matrix A. So we know that this thing right here is equal to some matrix A times x. You've seen all of this multiple, multiple times. And just to make sure we understand the wording properly, we've used the word that A is the matrix for T, or let's say it's the transformation matrix for T. Now, in the last couple of videos, we've learned that the same vector can be represented in different ways. It can be represented in different coordinate systems."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So we know that this thing right here is equal to some matrix A times x. You've seen all of this multiple, multiple times. And just to make sure we understand the wording properly, we've used the word that A is the matrix for T, or let's say it's the transformation matrix for T. Now, in the last couple of videos, we've learned that the same vector can be represented in different ways. It can be represented in different coordinate systems. And when I just write the vector x like that, we just assume that it's being represented in standard coordinates, or it's being represented with respect to the standard basis. So let's be a little bit more particular. A is the transformation for T only when x is represented in standard coordinates, or only when x is written in coordinates with respect to the standard basis."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "It can be represented in different coordinate systems. And when I just write the vector x like that, we just assume that it's being represented in standard coordinates, or it's being represented with respect to the standard basis. So let's be a little bit more particular. A is the transformation for T only when x is represented in standard coordinates, or only when x is written in coordinates with respect to the standard basis. So let me write a little qualifier here. A is the transformation matrix for T with respect to the standard basis. I never wrote this blue part before."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "A is the transformation for T only when x is represented in standard coordinates, or only when x is written in coordinates with respect to the standard basis. So let me write a little qualifier here. A is the transformation matrix for T with respect to the standard basis. I never wrote this blue part before. I never even said this blue part before, because the only coordinate system we were dealing with was the standard coordinate system, or the coordinates with respect to the standard basis. But now we know that there are multiple coordinate systems. There are multiple ways to represent this vector."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "I never wrote this blue part before. I never even said this blue part before, because the only coordinate system we were dealing with was the standard coordinate system, or the coordinates with respect to the standard basis. But now we know that there are multiple coordinate systems. There are multiple ways to represent this vector. There are multiple ways to represent that vector, because Rn has multiple spanning bases, or there's multiple bases that can represent Rn. And each of those bases essentially can generate a coordinate system where you can represent any vector in Rn by some, with coordinates with respect to any of those bases. So that last part I said was a bit of a mouthful, so let me make it a little bit more concrete."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "There are multiple ways to represent this vector. There are multiple ways to represent that vector, because Rn has multiple spanning bases, or there's multiple bases that can represent Rn. And each of those bases essentially can generate a coordinate system where you can represent any vector in Rn by some, with coordinates with respect to any of those bases. So that last part I said was a bit of a mouthful, so let me make it a little bit more concrete. Let's say that I have some basis B that's made up of n, it has to be linearly independent, that's the definition of a basis, of n vectors, v1, v2, all the way to vn. Now, these are n linearly independent vectors. Each of these are members of Rn, so B is a basis for Rn, which is just another way of saying that all of these vectors are linearly independent, and any vector in Rn can be represented as a linear combination of these guys."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So that last part I said was a bit of a mouthful, so let me make it a little bit more concrete. Let's say that I have some basis B that's made up of n, it has to be linearly independent, that's the definition of a basis, of n vectors, v1, v2, all the way to vn. Now, these are n linearly independent vectors. Each of these are members of Rn, so B is a basis for Rn, which is just another way of saying that all of these vectors are linearly independent, and any vector in Rn can be represented as a linear combination of these guys. Which is another way of saying that any vector in Rn can be represented with coordinates with respect to this basis right there. So the same vector x, the same dot here, when we represent it in standard coordinates, we write, it's just going to be that right there, that vector x. But what if we want to represent it in coordinates with respect to this new basis?"}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Each of these are members of Rn, so B is a basis for Rn, which is just another way of saying that all of these vectors are linearly independent, and any vector in Rn can be represented as a linear combination of these guys. Which is another way of saying that any vector in Rn can be represented with coordinates with respect to this basis right there. So the same vector x, the same dot here, when we represent it in standard coordinates, we write, it's just going to be that right there, that vector x. But what if we want to represent it in coordinates with respect to this new basis? Well, then we could call that same vector x will look like this. Or we would denote it by this. The same vector can be represented with respect to this basis."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "But what if we want to represent it in coordinates with respect to this new basis? Well, then we could call that same vector x will look like this. Or we would denote it by this. The same vector can be represented with respect to this basis. So this could be some set of coordinates, this would be some other set of coordinates, but it's still representing the same basis. Likewise, this vector right here, that vector right there, is also in Rn, so it can be represented by some linear combination of these guys. Or you can represent it with coordinates with respect to this basis."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "The same vector can be represented with respect to this basis. So this could be some set of coordinates, this would be some other set of coordinates, but it's still representing the same basis. Likewise, this vector right here, that vector right there, is also in Rn, so it can be represented by some linear combination of these guys. Or you can represent it with coordinates with respect to this basis. So that same point right there, I could represent it, so that point is this, but I could represent it with coordinates with respect to my basis, just like that. So this is an interesting question, or this should maybe bring an interesting question into your brain. If I start off with something that's in standard coordinates, and I apply the transformation T, that's like applying this matrix A to it, or multiplying that thing in standard coordinates times the matrix A, I then get the mapping of T in standard coordinates."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Or you can represent it with coordinates with respect to this basis. So that same point right there, I could represent it, so that point is this, but I could represent it with coordinates with respect to my basis, just like that. So this is an interesting question, or this should maybe bring an interesting question into your brain. If I start off with something that's in standard coordinates, and I apply the transformation T, that's like applying this matrix A to it, or multiplying that thing in standard coordinates times the matrix A, I then get the mapping of T in standard coordinates. Now, what if I start off with that thing in non-standard coordinates? If I have coordinates with respect to this other basis here? Well, T should still map it to this guy, right?"}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "If I start off with something that's in standard coordinates, and I apply the transformation T, that's like applying this matrix A to it, or multiplying that thing in standard coordinates times the matrix A, I then get the mapping of T in standard coordinates. Now, what if I start off with that thing in non-standard coordinates? If I have coordinates with respect to this other basis here? Well, T should still map it to this guy, right? The transformation, no matter what, should always map from that dot to that dot. It shouldn't care what your coordinates are. So T should still map to that same exact point."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Well, T should still map it to this guy, right? The transformation, no matter what, should always map from that dot to that dot. It shouldn't care what your coordinates are. So T should still map to that same exact point. T should still be a linear transformation, and it could map from x to T of x, but that's the same thing as mapping from this kind of way of labeling x to this way of labeling x. So we could say maybe T could be this guy right here could be some other matrix times this guy over here. So let me write this over here."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So T should still map to that same exact point. T should still be a linear transformation, and it could map from x to T of x, but that's the same thing as mapping from this kind of way of labeling x to this way of labeling x. So we could say maybe T could be this guy right here could be some other matrix times this guy over here. So let me write this over here. So maybe T, I mean, these are just different coordinate systems, so maybe I shouldn't just even say maybe. This guy should be able to be represented. So if I represent the mapping of x in our codomain in coordinates with respect to b, so that's what that guy is right there, so if I want to represent that dot with this other coordinate system, with coordinates with respect to this basis, I want to represent it, it should be equal to the product of some other matrix."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So let me write this over here. So maybe T, I mean, these are just different coordinate systems, so maybe I shouldn't just even say maybe. This guy should be able to be represented. So if I represent the mapping of x in our codomain in coordinates with respect to b, so that's what that guy is right there, so if I want to represent that dot with this other coordinate system, with coordinates with respect to this basis, I want to represent it, it should be equal to the product of some other matrix. Let me call that other matrix D. Some other matrix D times this representation of x times the coordinates of x with respect to my alternate non-standard coordinate system. I should be able to find some matrix D that does this. And then we would call D, we would say that D is the transformation matrix for T. And A was with respect, A assumes that you have x in terms in standard coordinates."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So if I represent the mapping of x in our codomain in coordinates with respect to b, so that's what that guy is right there, so if I want to represent that dot with this other coordinate system, with coordinates with respect to this basis, I want to represent it, it should be equal to the product of some other matrix. Let me call that other matrix D. Some other matrix D times this representation of x times the coordinates of x with respect to my alternate non-standard coordinate system. I should be able to find some matrix D that does this. And then we would call D, we would say that D is the transformation matrix for T. And A was with respect, A assumes that you have x in terms in standard coordinates. But now D assumes that you have x in this other, in coordinates with respect to this basis. So with respect to the basis b. No reason why we shouldn't be able to do this."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And then we would call D, we would say that D is the transformation matrix for T. And A was with respect, A assumes that you have x in terms in standard coordinates. But now D assumes that you have x in this other, in coordinates with respect to this basis. So with respect to the basis b. No reason why we shouldn't be able to do this. These things are just different ways of representing the exact same vector, the exact same dot in our sets here. So if I represent it one way, the standard way, I multiply it by a and I get ax. If I represent it in non-standard coordinates, I should be able to multiply it by some other matrix and get another non-standard coordinate representation of what it gets mapped to."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "No reason why we shouldn't be able to do this. These things are just different ways of representing the exact same vector, the exact same dot in our sets here. So if I represent it one way, the standard way, I multiply it by a and I get ax. If I represent it in non-standard coordinates, I should be able to multiply it by some other matrix and get another non-standard coordinate representation of what it gets mapped to. So let's see if we can find some relation between D and between A. So we learned a couple of videos ago that there's a change of basis matrix that we can generate from this basis. And it's pretty easy to generate."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "If I represent it in non-standard coordinates, I should be able to multiply it by some other matrix and get another non-standard coordinate representation of what it gets mapped to. So let's see if we can find some relation between D and between A. So we learned a couple of videos ago that there's a change of basis matrix that we can generate from this basis. And it's pretty easy to generate. The change of basis matrix is just a matrix whose columns are these basis vectors. So v1, v2, no I shouldn't put a comma there, these are just the columns, v2, all the way to vn. This is an n by n matrix."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And it's pretty easy to generate. The change of basis matrix is just a matrix whose columns are these basis vectors. So v1, v2, no I shouldn't put a comma there, these are just the columns, v2, all the way to vn. This is an n by n matrix. Each of these guys are members of Rn and we have n of them. This is an n by n matrix where all of the columns are linearly independent. So we know that c is invertible."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "This is an n by n matrix. Each of these guys are members of Rn and we have n of them. This is an n by n matrix where all of the columns are linearly independent. So we know that c is invertible. These are all column vectors right here. So we know that c is invertible. And we learned in the last two or three videos that if we have some vector x, if we have our vector x and it's being represented by coordinates with respect to our basis b, we can just multiply that by c and we'll get our vector x."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So we know that c is invertible. These are all column vectors right here. So we know that c is invertible. And we learned in the last two or three videos that if we have some vector x, if we have our vector x and it's being represented by coordinates with respect to our basis b, we can just multiply that by c and we'll get our vector x. This is essentially, it will tell us the linear combination of these guys that will give us x. And since c is invertible, we also saw that if we have just the standard format for x or the standard coordinates for x, we can multiply that by c inverse and then that will get us the coordinates for x with respect to b, with respect to the basis b. And these two things, if you just multiply both sides of this equation by c inverse on the left-hand side, you're going to get this equation right there."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And we learned in the last two or three videos that if we have some vector x, if we have our vector x and it's being represented by coordinates with respect to our basis b, we can just multiply that by c and we'll get our vector x. This is essentially, it will tell us the linear combination of these guys that will give us x. And since c is invertible, we also saw that if we have just the standard format for x or the standard coordinates for x, we can multiply that by c inverse and then that will get us the coordinates for x with respect to b, with respect to the basis b. And these two things, if you just multiply both sides of this equation by c inverse on the left-hand side, you're going to get this equation right there. Now, given that, let's see if we can find some type of relation. Now, we know, or we're saying, let's see what d times xb is equal to. So let's say we take d times xb."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And these two things, if you just multiply both sides of this equation by c inverse on the left-hand side, you're going to get this equation right there. Now, given that, let's see if we can find some type of relation. Now, we know, or we're saying, let's see what d times xb is equal to. So let's say we take d times xb. So this thing right here should be equal to d times the representation or the coordinates of x with respect to the basis b. That's what we're claiming. We're saying that this guy is equal to d times the representation of x with respect to the coordinates, with respect to the basis b."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So let's say we take d times xb. So this thing right here should be equal to d times the representation or the coordinates of x with respect to the basis b. That's what we're claiming. We're saying that this guy is equal to d times the representation of x with respect to the coordinates, with respect to the basis b. So let me write all of this down. Scroll down. I'll do it right here because I think it's nice to have this graphic up here."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "We're saying that this guy is equal to d times the representation of x with respect to the coordinates, with respect to the basis b. So let me write all of this down. Scroll down. I'll do it right here because I think it's nice to have this graphic up here. So we can say that d times xb is equal to this thing right here. It's the same thing as the transformation of x represented in coordinates with respect to b, or in these non-standard coordinates. So it's equal to the transformation of x represented in this coordinate system, represented in coordinates with respect to b."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it right here because I think it's nice to have this graphic up here. So we can say that d times xb is equal to this thing right here. It's the same thing as the transformation of x represented in coordinates with respect to b, or in these non-standard coordinates. So it's equal to the transformation of x represented in this coordinate system, represented in coordinates with respect to b. We see that right there. But what is the transformation of x? What is the transformation of x?"}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to the transformation of x represented in this coordinate system, represented in coordinates with respect to b. We see that right there. But what is the transformation of x? What is the transformation of x? Well, that's the same thing as a times x, right? That's kind of the standard transformation. If x was represented in standard coordinates."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "What is the transformation of x? Well, that's the same thing as a times x, right? That's kind of the standard transformation. If x was represented in standard coordinates. So this is equal to x in standard coordinates times the matrix A, and then that will get us to this dot in standard coordinates. But then we want to convert it to this non-standard coordinates. Just like that."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "If x was represented in standard coordinates. So this is equal to x in standard coordinates times the matrix A, and then that will get us to this dot in standard coordinates. But then we want to convert it to this non-standard coordinates. Just like that. Now, if we have this, how can we just figure out what the vector Ax should look like? What this vector should look like? Well, we can look at this equation right here."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. Now, if we have this, how can we just figure out what the vector Ax should look like? What this vector should look like? Well, we can look at this equation right here. We have this. This is the same thing as this. And so if we apply, or actually no, we're going to go the other way."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Well, we can look at this equation right here. We have this. This is the same thing as this. And so if we apply, or actually no, we're going to go the other way. We have this. We have that right there. That's this right there."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And so if we apply, or actually no, we're going to go the other way. We have this. We have that right there. That's this right there. And we want to get just this dot represented in regular standard coordinates. So what do we do? We multiply it by c. So let me write it this way."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "That's this right there. And we want to get just this dot represented in regular standard coordinates. So what do we do? We multiply it by c. So let me write it this way. If we multiply both sides of this equation times c, what do we get? We get this right here. Actually, no."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "We multiply it by c. So let me write it this way. If we multiply both sides of this equation times c, what do we get? We get this right here. Actually, no. I was looking at the right equation the first time. We have this right here, which is the same. Let me, first intuition is always right."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Actually, no. I was looking at the right equation the first time. We have this right here, which is the same. Let me, first intuition is always right. We have this, which is the same thing as this right here. So this can be rewritten as c inverse. We don't have an x here."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Let me, first intuition is always right. We have this, which is the same thing as this right here. So this can be rewritten as c inverse. We don't have an x here. We have an Ax here. So c inverse times Ax. The vector Ax represented in this non-standard coordinates is the same thing as multiplying the inverse of our change of basis matrix times the vector Ax."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "We don't have an x here. We have an Ax here. So c inverse times Ax. The vector Ax represented in this non-standard coordinates is the same thing as multiplying the inverse of our change of basis matrix times the vector Ax. This will essentially, if I have my vector Ax and I multiply it times the inverse of the change of basis matrix, I will then have a representation of the vector Ax in my non-standard basis. Now, what is the vector x equal to? The vector x is equal to our change of basis matrix times x represented in these non-standard coordinates."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "The vector Ax represented in this non-standard coordinates is the same thing as multiplying the inverse of our change of basis matrix times the vector Ax. This will essentially, if I have my vector Ax and I multiply it times the inverse of the change of basis matrix, I will then have a representation of the vector Ax in my non-standard basis. Now, what is the vector x equal to? The vector x is equal to our change of basis matrix times x represented in these non-standard coordinates. So this is going to be equal to c inverse a times x. x is just the same thing as c. x is just c times our non-standard coordinates for x. Just like that. Let me summarize it, just because I waffled a little bit on this point right there, just because I got a little bit confused."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "The vector x is equal to our change of basis matrix times x represented in these non-standard coordinates. So this is going to be equal to c inverse a times x. x is just the same thing as c. x is just c times our non-standard coordinates for x. Just like that. Let me summarize it, just because I waffled a little bit on this point right there, just because I got a little bit confused. If I start off with the non-standard representation of x, or x in coordinates with respect to b, I multiply them times d. So if I start with this, I multiply them times d, I get to that point right there. So this right there is the same thing as this point right there. That point right there should be the non-standard representation of the transformation of x."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "Let me summarize it, just because I waffled a little bit on this point right there, just because I got a little bit confused. If I start off with the non-standard representation of x, or x in coordinates with respect to b, I multiply them times d. So if I start with this, I multiply them times d, I get to that point right there. So this right there is the same thing as this point right there. That point right there should be the non-standard representation of the transformation of x. The non-standard representation, or the coordinates of the transformation of x with respect to b. Now the transformation of x, if x is in standard coordinates, is just a times x. So this is just a times x, but I want to represent it in these non-standard coordinates."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "That point right there should be the non-standard representation of the transformation of x. The non-standard representation, or the coordinates of the transformation of x with respect to b. Now the transformation of x, if x is in standard coordinates, is just a times x. So this is just a times x, but I want to represent it in these non-standard coordinates. Now a times x in non-standard coordinates is the same thing as c inverse times a times x. If you think this is the same thing as this. And so if you have this and you want to represent it in non-standard coordinates, you multiply it by c inverse."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So this is just a times x, but I want to represent it in these non-standard coordinates. Now a times x in non-standard coordinates is the same thing as c inverse times a times x. If you think this is the same thing as this. And so if you have this and you want to represent it in non-standard coordinates, you multiply it by c inverse. So then you'll get that representation in non-standard coordinates. And then finally we say, look, x is the same thing as c times the non-standard coordinate representation of x. So x we can replace with that right there."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "And so if you have this and you want to represent it in non-standard coordinates, you multiply it by c inverse. So then you'll get that representation in non-standard coordinates. And then finally we say, look, x is the same thing as c times the non-standard coordinate representation of x. So x we can replace with that right there. And so the big takeaway here is that d times the coordinates of x with respect to the basis b is equal to c inverse a times c times the coordinates of x with respect to the basis b. And just like that we have a version. So d must be equal to c inverse ac."}, {"video_title": "Transformation matrix with respect to a basis Linear Algebra Khan Academy.mp3", "Sentence": "So x we can replace with that right there. And so the big takeaway here is that d times the coordinates of x with respect to the basis b is equal to c inverse a times c times the coordinates of x with respect to the basis b. And just like that we have a version. So d must be equal to c inverse ac. So if d is the transformation matrix for t with respect to the basis b and c is the change of basis matrix for b, then, and we know a is the, let me write that down, and, might as well, because this is our big takeaway, and a is the transformation, I'll write it in shorthand, matrix for t with respect to the standard basis, then we can say, and this is the big takeaway, that d, our matrix d, is equal to c inverse times a times c. That's our big takeaway from this video. Which is really interesting. I don't want you to lose this point."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "We came up with a process for generating an orthonormal basis in the last video. And it wasn't a new discovery. It's called the Gram-Schmidt process. But let's apply that now to some real examples. And hopefully we'll see that it's a lot more concrete than it might have looked in the last video. Let's say I have the plane x1 plus x2 plus x3 is equal to 0. This is a plane in R3."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But let's apply that now to some real examples. And hopefully we'll see that it's a lot more concrete than it might have looked in the last video. Let's say I have the plane x1 plus x2 plus x3 is equal to 0. This is a plane in R3. So let me just say that subspace V is equal to the plane defined by this guy right here. x1 plus x2 plus x3 is equal to 0. All of the vectors in the subspace, if you take their entries and you add them up, you're going to get 0."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "This is a plane in R3. So let me just say that subspace V is equal to the plane defined by this guy right here. x1 plus x2 plus x3 is equal to 0. All of the vectors in the subspace, if you take their entries and you add them up, you're going to get 0. So first we need just any basis for V. So let's see if we could come up with that. So if we subtract x2 and x3 from both sides of this equation, we know that x1 is going to be equal to minus x2 minus x3. Or we could say that our subspace V is equal to the set of all of the vectors in R3, x1, x2, and x3, that satisfy the equation, let's say, minus, let me write it this way, let's say that x2 is equal to c1 and x3 is equal to c2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "All of the vectors in the subspace, if you take their entries and you add them up, you're going to get 0. So first we need just any basis for V. So let's see if we could come up with that. So if we subtract x2 and x3 from both sides of this equation, we know that x1 is going to be equal to minus x2 minus x3. Or we could say that our subspace V is equal to the set of all of the vectors in R3, x1, x2, and x3, that satisfy the equation, let's say, minus, let me write it this way, let's say that x2 is equal to c1 and x3 is equal to c2. Then this equation would be x1 is equal to minus c1 minus c2. So if we write it that way, then the subspace V is the set of all of the vectors in R3 such that c1 times some vector, let me write it this way, c1 times, let's see, x1 is, let me write it this way, plus c2 times some other vector, where c1 and c2 are any real numbers. So c1 and c2 are a member of the reals."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or we could say that our subspace V is equal to the set of all of the vectors in R3, x1, x2, and x3, that satisfy the equation, let's say, minus, let me write it this way, let's say that x2 is equal to c1 and x3 is equal to c2. Then this equation would be x1 is equal to minus c1 minus c2. So if we write it that way, then the subspace V is the set of all of the vectors in R3 such that c1 times some vector, let me write it this way, c1 times, let's see, x1 is, let me write it this way, plus c2 times some other vector, where c1 and c2 are any real numbers. So c1 and c2 are a member of the reals. And so what is x1? x1 is equal to minus c1 minus c2. So x1 is equal to minus 1 c1 minus 1 c2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So c1 and c2 are a member of the reals. And so what is x1? x1 is equal to minus c1 minus c2. So x1 is equal to minus 1 c1 minus 1 c2. x2 is just equal to c1. So x2 is equal to 1 times c1 plus 0 times c2. And then x3 is equal to c2, or 0 times c1 plus 1 times c2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So x1 is equal to minus 1 c1 minus 1 c2. x2 is just equal to c1. So x2 is equal to 1 times c1 plus 0 times c2. And then x3 is equal to c2, or 0 times c1 plus 1 times c2. So V is essentially the span of these two vectors, all of the linear combinations of these two vectors. That would represent that plane. So let me write it like this."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then x3 is equal to c2, or 0 times c1 plus 1 times c2. So V is essentially the span of these two vectors, all of the linear combinations of these two vectors. That would represent that plane. So let me write it like this. V is equal to the span of the vectors minus 1, 1, 0, and the vector minus 1, 0, 1. And you can see that these are linearly independent. Obviously, there's no linear combination of this guy that can give you a 1 over here."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let me write it like this. V is equal to the span of the vectors minus 1, 1, 0, and the vector minus 1, 0, 1. And you can see that these are linearly independent. Obviously, there's no linear combination of this guy that can give you a 1 over here. There's no linear combination of this guy that will give you a 1 right there. So this is what V is. But what we want, the whole reason why I'm making this video, is to find an orthonormal basis for V. This is just a basis."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Obviously, there's no linear combination of this guy that can give you a 1 over here. There's no linear combination of this guy that will give you a 1 right there. So this is what V is. But what we want, the whole reason why I'm making this video, is to find an orthonormal basis for V. This is just a basis. These guys right here are just a basis for V. Let's find an orthonormal basis. Let's call this vector up here. Let's call that V1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But what we want, the whole reason why I'm making this video, is to find an orthonormal basis for V. This is just a basis. These guys right here are just a basis for V. Let's find an orthonormal basis. Let's call this vector up here. Let's call that V1. And let's call this vector right here V2. So if we wanted to find an orthonormal basis for the span of V1, let me write this down. Let me define subspace V1 is equal to the span of just my vector V1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let's call that V1. And let's call this vector right here V2. So if we wanted to find an orthonormal basis for the span of V1, let me write this down. Let me define subspace V1 is equal to the span of just my vector V1. Well, we saw in the last video, if we just divide V1 by its length, then we'll have the span of that vector is going to be a unit vector. And it's going to be the same thing as this subspace V1, as this line in R3. So let's do that."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me define subspace V1 is equal to the span of just my vector V1. Well, we saw in the last video, if we just divide V1 by its length, then we'll have the span of that vector is going to be a unit vector. And it's going to be the same thing as this subspace V1, as this line in R3. So let's do that. What is the length of V1? The length of V1 is equal to the square root of minus 1 squared, which is 1, plus 1 squared, which is 1, plus 0 squared, which is 0. So it's equal to the square root of 2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So let's do that. What is the length of V1? The length of V1 is equal to the square root of minus 1 squared, which is 1, plus 1 squared, which is 1, plus 0 squared, which is 0. So it's equal to the square root of 2. So let's define some vector U1 is equal to 1 divided by the length of V1. So 1 over the square root of 2 times V1, times minus 1, 1, 0. Then the span of V1 is just the same thing as the span of U1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to the square root of 2. So let's define some vector U1 is equal to 1 divided by the length of V1. So 1 over the square root of 2 times V1, times minus 1, 1, 0. Then the span of V1 is just the same thing as the span of U1. And so this would be an orthonormal basis. Just this vector right here would be an orthonormal basis for just the span of V1. But we don't want just the span of V1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Then the span of V1 is just the same thing as the span of U1. And so this would be an orthonormal basis. Just this vector right here would be an orthonormal basis for just the span of V1. But we don't want just the span of V1. We want the span of V1 and V2. So let me draw it. So right now we have a basis, if I just do U1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But we don't want just the span of V1. We want the span of V1 and V2. So let me draw it. So right now we have a basis, if I just do U1. I'm not going to actually draw what this looks like. It looks something like this. And its span is this entire line in R3."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So right now we have a basis, if I just do U1. I'm not going to actually draw what this looks like. It looks something like this. And its span is this entire line in R3. So its span is this entire line. The span of just one vector in Rn is just going to be all the scalar multiples of it, or a line in Rn. So this right here is a subspace V1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And its span is this entire line in R3. So its span is this entire line. The span of just one vector in Rn is just going to be all the scalar multiples of it, or a line in Rn. So this right here is a subspace V1. Now we have V2 here, which is linearly independent from this guy, which means it's linearly independent from this guy, because he's just a scaled version of this guy. So V2 is going to maybe look like that. That is V2 right there."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is a subspace V1. Now we have V2 here, which is linearly independent from this guy, which means it's linearly independent from this guy, because he's just a scaled version of this guy. So V2 is going to maybe look like that. That is V2 right there. This, of course, was our U1. What we want to do is we want to define a subspace V. I'll call it V2 for now. V2 is equal to the span of V1 and V2, which is the same thing as the span."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That is V2 right there. This, of course, was our U1. What we want to do is we want to define a subspace V. I'll call it V2 for now. V2 is equal to the span of V1 and V2, which is the same thing as the span. Anything that's spanned by V1 is also spanned by U1, as the span of U1 and V2. So we want to see everything that can be generated by linear combinations of U1 and V2. And obviously this thing right here is our plane that we're talking about."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "V2 is equal to the span of V1 and V2, which is the same thing as the span. Anything that's spanned by V1 is also spanned by U1, as the span of U1 and V2. So we want to see everything that can be generated by linear combinations of U1 and V2. And obviously this thing right here is our plane that we're talking about. The span of these two guys, that is the whole subspace that we're talking about in this problem. So that is equal to V. So once we find this, if we find an orthonormal version for this span, we're done. So how can we do that?"}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And obviously this thing right here is our plane that we're talking about. The span of these two guys, that is the whole subspace that we're talking about in this problem. So that is equal to V. So once we find this, if we find an orthonormal version for this span, we're done. So how can we do that? Well, if I can find a vector that's orthogonal to all of the linear combinations of this, that if I add up some linear combination of this to that vector, I can get V2. I can replace V2 with that vector. So we could call that vector right there, Y2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So how can we do that? Well, if I can find a vector that's orthogonal to all of the linear combinations of this, that if I add up some linear combination of this to that vector, I can get V2. I can replace V2 with that vector. So we could call that vector right there, Y2. If I can determine a Y2, this Y2 is clearly orthogonal to everything over here. And I can take some vector in V1 in this line and add it to Y2, and I can get to V2. So the combinations of these guys are just as good as V2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So we could call that vector right there, Y2. If I can determine a Y2, this Y2 is clearly orthogonal to everything over here. And I can take some vector in V1 in this line and add it to Y2, and I can get to V2. So the combinations of these guys are just as good as V2. So this is going to be equal to the span of U1 and Y2. Now what is Y2 equal to? Well, we saw in the last video, this is just the projection of V2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So the combinations of these guys are just as good as V2. So this is going to be equal to the span of U1 and Y2. Now what is Y2 equal to? Well, we saw in the last video, this is just the projection of V2. This vector right here is the projection of V2 onto the subspace V1. And how do we figure out what that? And then what would Y2 be?"}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Well, we saw in the last video, this is just the projection of V2. This vector right here is the projection of V2 onto the subspace V1. And how do we figure out what that? And then what would Y2 be? Y2 would be V2 minus that. So Y2 is equal to V2 minus the projection of V2 onto V1. Or if we were to actually write it out, what is that going to be equal to?"}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And then what would Y2 be? Y2 would be V2 minus that. So Y2 is equal to V2 minus the projection of V2 onto V1. Or if we were to actually write it out, what is that going to be equal to? So it's going to be equal to V2 is this vector right here. So it's minus 1, 0, 1. That is V2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Or if we were to actually write it out, what is that going to be equal to? So it's going to be equal to V2 is this vector right here. So it's minus 1, 0, 1. That is V2. V2 minus the projection of V2 onto V1. Well, the projection of the vector V2 onto the subspace V1 is just V2 minus 1, 0, 1 dotted with the orthonormal basis for V1. The orthonormal basis for V1 is just U1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "That is V2. V2 minus the projection of V2 onto V1. Well, the projection of the vector V2 onto the subspace V1 is just V2 minus 1, 0, 1 dotted with the orthonormal basis for V1. The orthonormal basis for V1 is just U1. And we solved for U1 up here. So that's going to be that dotted with 1 over the square root of 2 times, let me do that in yellow actually, just so you can see that this is U1. So dotting it with U1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The orthonormal basis for V1 is just U1. And we solved for U1 up here. So that's going to be that dotted with 1 over the square root of 2 times, let me do that in yellow actually, just so you can see that this is U1. So dotting it with U1. So dotting it with 1 over the square root of 2 times minus 1, 1, 0. I like leaving the 1 over the square root of 2 out of there just to keep things simple. All of that divided by, actually not divided by anything, because if we were doing a projection of a line, it would be divided by the dot product of the orthonormal basis with itself, but its length is 1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So dotting it with U1. So dotting it with 1 over the square root of 2 times minus 1, 1, 0. I like leaving the 1 over the square root of 2 out of there just to keep things simple. All of that divided by, actually not divided by anything, because if we were doing a projection of a line, it would be divided by the dot product of the orthonormal basis with itself, but its length is 1. So we don't have that, and we saw that before. So actually, let me write this a little bit. I'll just move it down."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "All of that divided by, actually not divided by anything, because if we were doing a projection of a line, it would be divided by the dot product of the orthonormal basis with itself, but its length is 1. So we don't have that, and we saw that before. So actually, let me write this a little bit. I'll just move it down. Let's see if I can move it. So it's just equal to that guy. Let me make the numbers clear."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I'll just move it down. Let's see if I can move it. So it's just equal to that guy. Let me make the numbers clear. This is V2 minus the projection of V2 onto the subspace 1. So that's just V2 dotted with my orthonormal basis for V1. My first vector in my orthonormal basis."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Let me make the numbers clear. This is V2 minus the projection of V2 onto the subspace 1. So that's just V2 dotted with my orthonormal basis for V1. My first vector in my orthonormal basis. There's only 1, so I'm only going to have 1 term here. And then all of that times my orthonormal basis vector for V1. So 1 over the square root of 2 times the vector minus 1, 1, 0."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "My first vector in my orthonormal basis. There's only 1, so I'm only going to have 1 term here. And then all of that times my orthonormal basis vector for V1. So 1 over the square root of 2 times the vector minus 1, 1, 0. Now this looks really fancy. This right here is our orthonormal basis for the subspace V1. But what does this simplify to?"}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So 1 over the square root of 2 times the vector minus 1, 1, 0. Now this looks really fancy. This right here is our orthonormal basis for the subspace V1. But what does this simplify to? So this is going to be equal to, remember, this right here, this piece right there, that's the projection onto V1 of V2. That's what that was right there. So this is going to be equal to the vector minus 1, 0, 1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "But what does this simplify to? So this is going to be equal to, remember, this right here, this piece right there, that's the projection onto V1 of V2. That's what that was right there. So this is going to be equal to the vector minus 1, 0, 1. Minus, now if I could take the 1 over the square root of 2 on the outside, actually I could take both of these onto the outside. So 1 over the square root of 2 times 1 over the square root of 2 is just going to be 1 over 2. So this is going to be minus 1 half times these guys dotted with each other."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be equal to the vector minus 1, 0, 1. Minus, now if I could take the 1 over the square root of 2 on the outside, actually I could take both of these onto the outside. So 1 over the square root of 2 times 1 over the square root of 2 is just going to be 1 over 2. So this is going to be minus 1 half times these guys dotted with each other. So what is, let me just write it this way, so what is these guys dotted with each other? It's going to be, it's just going to be a number. Minus 1 times minus 1 is 1, plus 0 times 1, so plus 0, plus 1 times 0, so plus 0."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is going to be minus 1 half times these guys dotted with each other. So what is, let me just write it this way, so what is these guys dotted with each other? It's going to be, it's just going to be a number. Minus 1 times minus 1 is 1, plus 0 times 1, so plus 0, plus 1 times 0, so plus 0. All of that times, well we already used this part of it, so we just have this part left over, times minus 1, 1, and 0. This was the dot product, and well we took the two scaling factors out, when you multiply them you got a 1 half. So this is just going to be a 1, which simplifies things."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 times minus 1 is 1, plus 0 times 1, so plus 0, plus 1 times 0, so plus 0. All of that times, well we already used this part of it, so we just have this part left over, times minus 1, 1, and 0. This was the dot product, and well we took the two scaling factors out, when you multiply them you got a 1 half. So this is just going to be a 1, which simplifies things. So this is going to be equal to the vector minus 1, 0, 1, minus 1 half times this. Or we could just write, so 1 half times minus 1 is minus 1 half, we have 1 half, and then we have 0. And so this is going to be equal to minus 1 minus minus 1 half, that's plus 1 half, so that's going to be just minus 1 half."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this is just going to be a 1, which simplifies things. So this is going to be equal to the vector minus 1, 0, 1, minus 1 half times this. Or we could just write, so 1 half times minus 1 is minus 1 half, we have 1 half, and then we have 0. And so this is going to be equal to minus 1 minus minus 1 half, that's plus 1 half, so that's going to be just minus 1 half. 0 minus 1 half is minus 1 half. And then 1 minus 0 is just 1. So this right here is our vector y2."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "And so this is going to be equal to minus 1 minus minus 1 half, that's plus 1 half, so that's going to be just minus 1 half. 0 minus 1 half is minus 1 half. And then 1 minus 0 is just 1. So this right here is our vector y2. And if you combine u1 right here and y2, we are spanning our subspace V. But we don't have an orthonormal basis yet. These guys are orthogonal with respect to each other, but this guy does not have length 1 yet. So to make it equal to length 1, let's replace him, let's define another vector."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is our vector y2. And if you combine u1 right here and y2, we are spanning our subspace V. But we don't have an orthonormal basis yet. These guys are orthogonal with respect to each other, but this guy does not have length 1 yet. So to make it equal to length 1, let's replace him, let's define another vector. Let's define another vector u2 that's equal to 1 over the length of y2 times y2. So what is the length of y2? The length of y2 is equal to the square root of minus 1 half squared is 1 fourth plus 1 squared."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So to make it equal to length 1, let's replace him, let's define another vector. Let's define another vector u2 that's equal to 1 over the length of y2 times y2. So what is the length of y2? The length of y2 is equal to the square root of minus 1 half squared is 1 fourth plus 1 squared. So it's the square root of what is this? 1 and 1 half, or 3 halves. So it's equal to the square root of 3 halves."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "The length of y2 is equal to the square root of minus 1 half squared is 1 fourth plus 1 squared. So it's the square root of what is this? 1 and 1 half, or 3 halves. So it's equal to the square root of 3 halves. This is 1 half plus 1 is 1 and 1 half, which is 3 halves. So it's equal to the square root of 3 halves. So if I define u2, u2 is equal to 1 over the square root of 3 halves, or that's the same thing as the square root of 2 thirds times y2, which is this guy right here, minus 1 half and 1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to the square root of 3 halves. This is 1 half plus 1 is 1 and 1 half, which is 3 halves. So it's equal to the square root of 3 halves. So if I define u2, u2 is equal to 1 over the square root of 3 halves, or that's the same thing as the square root of 2 thirds times y2, which is this guy right here, minus 1 half and 1. And I already had defined u1 up here. u1 was right up here. I think I can just move it down."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "So if I define u2, u2 is equal to 1 over the square root of 3 halves, or that's the same thing as the square root of 2 thirds times y2, which is this guy right here, minus 1 half and 1. And I already had defined u1 up here. u1 was right up here. I think I can just move it down. So I found u1 right here. We now have 2 vectors that are orthogonal with respect to each other. So if I have the set of u1 and u2, these guys both have length 1."}, {"video_title": "Gram-Schmidt process example Alternate coordinate systems (bases) Linear Algebra Khan Academy.mp3", "Sentence": "I think I can just move it down. So I found u1 right here. We now have 2 vectors that are orthogonal with respect to each other. So if I have the set of u1 and u2, these guys both have length 1. They're orthogonal with respect to each other, and they span v. So this is an orthonormal basis for the plane that we started this video out with, for v. And we're done. We have done the Gram-Schmidt process. These are our new orthonormal basis vectors."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it fairly neatly. I want to do it neater than that. So that is my vertical coordinate. This is my horizontal coordinate. That's good enough. And let's say I have four points in R2 specified by four position vectors. So this is R2 right here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is my horizontal coordinate. That's good enough. And let's say I have four points in R2 specified by four position vectors. So this is R2 right here. So my first position vector is the 0 vector. So it just specifies the point right there. So it's 0, 0."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is R2 right here. So my first position vector is the 0 vector. So it just specifies the point right there. So it's 0, 0. My second point is specified by the vector, say, k1, 0. So some constant is times 0. Let me write it a little bit neater than that."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's 0, 0. My second point is specified by the vector, say, k1, 0. So some constant is times 0. Let me write it a little bit neater than that. A little bit bigger. So it's k1, 0. So that's my second position vector, which would specify that point right there."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write it a little bit neater than that. A little bit bigger. So it's k1, 0. So that's my second position vector, which would specify that point right there. So let me draw my vector right there. Let's say my third point is going to be up here. As you can see, I'm creating a rectangle."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's my second position vector, which would specify that point right there. So let me draw my vector right there. Let's say my third point is going to be up here. As you can see, I'm creating a rectangle. And let's say it's specified by the position vector. Draw the position vector like that. And that position vector is going to be the vector k1."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "As you can see, I'm creating a rectangle. And let's say it's specified by the position vector. Draw the position vector like that. And that position vector is going to be the vector k1. That's going to be its horizontal component. And then k2 is its vertical component. This is, if we wanted to graph it on the coordinates, this would be the k2 coordinate right there."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And that position vector is going to be the vector k1. That's going to be its horizontal component. And then k2 is its vertical component. This is, if we wanted to graph it on the coordinates, this would be the k2 coordinate right there. And let's say that the last point is this point right here, and it is specified by the position vector 0, k2. And let me define a set. Let me call that set rectangle."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is, if we wanted to graph it on the coordinates, this would be the k2 coordinate right there. And let's say that the last point is this point right here, and it is specified by the position vector 0, k2. And let me define a set. Let me call that set rectangle. Let's say R for rectangle. Let me just call it REC. Let's say that rectangle is equal to the set of all of the points when you connect the points specified by those position vectors."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me call that set rectangle. Let's say R for rectangle. Let me just call it REC. Let's say that rectangle is equal to the set of all of the points when you connect the points specified by those position vectors. So the rectangle formed by connecting points specified by these guys here. So if we call this vector a, this is vector b, this is vector c, and then this is vector d right here. So rectangle specified by connecting the points a, b, c, and d. Now, let's say, so let me draw what our set rectangle is going to be."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that rectangle is equal to the set of all of the points when you connect the points specified by those position vectors. So the rectangle formed by connecting points specified by these guys here. So if we call this vector a, this is vector b, this is vector c, and then this is vector d right here. So rectangle specified by connecting the points a, b, c, and d. Now, let's say, so let me draw what our set rectangle is going to be. It's going to be all of these points here, all of these points there, all of those points there, and then all of those points there, including our, I guess you could call them, our vertices. So that is my rectangle. Now, what is the area within this rectangle?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So rectangle specified by connecting the points a, b, c, and d. Now, let's say, so let me draw what our set rectangle is going to be. It's going to be all of these points here, all of these points there, all of those points there, and then all of those points there, including our, I guess you could call them, our vertices. So that is my rectangle. Now, what is the area within this rectangle? So this is the area of rectangle. What's the area within this rectangle? Well, it's just base times height."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, what is the area within this rectangle? So this is the area of rectangle. What's the area within this rectangle? Well, it's just base times height. The base has length k1, right? The distance from here to here is k1, so it's just k1. And then what's the height?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's just base times height. The base has length k1, right? The distance from here to here is k1, so it's just k1. And then what's the height? It's the distance from here to here, which is just k2. So it's just k1 times k2. That's fairly straightforward."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then what's the height? It's the distance from here to here, which is just k2. So it's just k1 times k2. That's fairly straightforward. I haven't shown you anything that fascinating just yet. But let's say we're going to transform this set. We're going to transform this rectangle, and we're going to transform it with the transformation t. Let me pick a nice color for t. Let's say we have the transformation t. It is a mapping from R2 to R2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's fairly straightforward. I haven't shown you anything that fascinating just yet. But let's say we're going to transform this set. We're going to transform this rectangle, and we're going to transform it with the transformation t. Let me pick a nice color for t. Let's say we have the transformation t. It is a mapping from R2 to R2. And it can be represented as, so when you apply the transformation t to some vector x, it is equal to the matrix ABCD times your vector x, just like that. Now, let's see what happens when we apply the transformation to each of these points. We saw many videos ago that if you want essentially the image of our rectangle under this transformation, you can essentially just get the transformation of each of the endpoints, and then connect the dots in your codomain, which is also going to be R2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're going to transform this rectangle, and we're going to transform it with the transformation t. Let me pick a nice color for t. Let's say we have the transformation t. It is a mapping from R2 to R2. And it can be represented as, so when you apply the transformation t to some vector x, it is equal to the matrix ABCD times your vector x, just like that. Now, let's see what happens when we apply the transformation to each of these points. We saw many videos ago that if you want essentially the image of our rectangle under this transformation, you can essentially just get the transformation of each of the endpoints, and then connect the dots in your codomain, which is also going to be R2. So we just have to figure out the transformation of each of these guys. So what is my transformation? I'll do it over here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We saw many videos ago that if you want essentially the image of our rectangle under this transformation, you can essentially just get the transformation of each of the endpoints, and then connect the dots in your codomain, which is also going to be R2. So we just have to figure out the transformation of each of these guys. So what is my transformation? I'll do it over here. I'll do it in the colors. What is the transformation applied to the 0 vector, applied to point A right there? Well, it's just going to be A times 0 plus B times 0, which is just 0."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it over here. I'll do it in the colors. What is the transformation applied to the 0 vector, applied to point A right there? Well, it's just going to be A times 0 plus B times 0, which is just 0. And then the second term is going to be C times 0 plus D times 0, which is just 0. That's vector A. Vector B. If I apply the transformation to vector B, vector B is K1 0, just like that."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, it's just going to be A times 0 plus B times 0, which is just 0. And then the second term is going to be C times 0 plus D times 0, which is just 0. That's vector A. Vector B. If I apply the transformation to vector B, vector B is K1 0, just like that. If I apply the transformation, it's going to be A times K. So the first entry in the transformed vector, or what we're mapping to in our codomain, is going to be A times K1 plus B times 0. So it's just A K1. And then the second entry is going to be C times K1 plus D times 0."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I apply the transformation to vector B, vector B is K1 0, just like that. If I apply the transformation, it's going to be A times K. So the first entry in the transformed vector, or what we're mapping to in our codomain, is going to be A times K1 plus B times 0. So it's just A K1. And then the second entry is going to be C times K1 plus D times 0. So it's going to be C times K1. All I'm doing is taking the matrix vector product of this guy and that guy. And then let's move on to point C. So the transformation of point C of K1 K2 is equal to, or the transformation of vector C, we could say, it's equal to what?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the second entry is going to be C times K1 plus D times 0. So it's going to be C times K1. All I'm doing is taking the matrix vector product of this guy and that guy. And then let's move on to point C. So the transformation of point C of K1 K2 is equal to, or the transformation of vector C, we could say, it's equal to what? The first entry is going to be A times K1 plus B times K2. A times K1 plus B times K2. And the second entry is going to be C times K1 plus D times K2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then let's move on to point C. So the transformation of point C of K1 K2 is equal to, or the transformation of vector C, we could say, it's equal to what? The first entry is going to be A times K1 plus B times K2. A times K1 plus B times K2. And the second entry is going to be C times K1 plus D times K2. C times K1 plus D times K2. We have one point left. The transformation, if we take the transformation of our vector 0 K2, what do we get?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the second entry is going to be C times K1 plus D times K2. C times K1 plus D times K2. We have one point left. The transformation, if we take the transformation of our vector 0 K2, what do we get? We get A times 0 plus B times K2, which is just B times K2. And then C times 0 plus D times K2, which is just D times K2. Just like that."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The transformation, if we take the transformation of our vector 0 K2, what do we get? We get A times 0 plus B times K2, which is just B times K2. And then C times 0 plus D times K2, which is just D times K2. Just like that. So let's draw the image of our rectangle under our transformations. Let me redraw my axes. So this is my vertical axis right there."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Just like that. So let's draw the image of our rectangle under our transformations. Let me redraw my axes. So this is my vertical axis right there. This is my horizontal axis right there. And so A gets mapped to the 0 vector right there. So this is the transformation of 0, 0."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is my vertical axis right there. This is my horizontal axis right there. And so A gets mapped to the 0 vector right there. So this is the transformation of 0, 0. Right like that. Now what does the vector B get transformed? The vector B gets transformed to A K1, C K1."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the transformation of 0, 0. Right like that. Now what does the vector B get transformed? The vector B gets transformed to A K1, C K1. So let me draw it like this. The vector B gets transformed to something that looks like this, where this is the vector. So this is the transformation of K1 and 0, which is equal to A K1, C K1."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The vector B gets transformed to A K1, C K1. So let me draw it like this. The vector B gets transformed to something that looks like this, where this is the vector. So this is the transformation of K1 and 0, which is equal to A K1, C K1. Saw that already. Now what does this yellow guy get transformed to? Let me do the blue one first."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is the transformation of K1 and 0, which is equal to A K1, C K1. Saw that already. Now what does this yellow guy get transformed to? Let me do the blue one first. So the blue one gets transformed here. Let me finish here. So if we come down here, this is going to be A K1 right there, and if we go to the left like that, that's going to be C K1."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do the blue one first. So the blue one gets transformed here. Let me finish here. So if we come down here, this is going to be A K1 right there, and if we go to the left like that, that's going to be C K1. That's how we graph it. Now this last guy, this blue guy, maybe he looks something like this. This is the transformation of that vector right there by this linear transformation, 0 K2, which is equal to the vector B K2 and D K2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we come down here, this is going to be A K1 right there, and if we go to the left like that, that's going to be C K1. That's how we graph it. Now this last guy, this blue guy, maybe he looks something like this. This is the transformation of that vector right there by this linear transformation, 0 K2, which is equal to the vector B K2 and D K2. So this point right here is B K2, and then this coordinate right there is D K2. Now this last guy, this yellow point, when I transformed it, what do I get? I got A K1 plus B K2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the transformation of that vector right there by this linear transformation, 0 K2, which is equal to the vector B K2 and D K2. So this point right here is B K2, and then this coordinate right there is D K2. Now this last guy, this yellow point, when I transformed it, what do I get? I got A K1 plus B K2. So A K1 plus B K2 will get you right about there. So that's going to be its x-coordinate. And its y-coordinate is C K1 plus D K2."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I got A K1 plus B K2. So A K1 plus B K2 will get you right about there. So that's going to be its x-coordinate. And its y-coordinate is C K1 plus D K2. C K1 plus D K2. So C K1, and then I'm going to add that much again to it. So it's going to get me someplace over here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And its y-coordinate is C K1 plus D K2. C K1 plus D K2. So C K1, and then I'm going to add that much again to it. So it's going to get me someplace over here. So I'm going to add this distance to this whole distance over here. So I'm going to get up here some point. So I'm going to get right over here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to get me someplace over here. So I'm going to add this distance to this whole distance over here. So I'm going to get up here some point. So I'm going to get right over here. So that vector is going to look something like that. That is the transformation of K1 K2. That's the transformation of that guy."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to get right over here. So that vector is going to look something like that. That is the transformation of K1 K2. That's the transformation of that guy. And then if we connect the dots, remember the transformation of this set, of this rectangle, or we could say the image of this rectangle under the transformation, we just take each of the points that define that rectangle, and then we connect the dots. We saw that a while ago. So let me draw it this way."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the transformation of that guy. And then if we connect the dots, remember the transformation of this set, of this rectangle, or we could say the image of this rectangle under the transformation, we just take each of the points that define that rectangle, and then we connect the dots. We saw that a while ago. So let me draw it this way. This line right here, that connects those two dots, will be transformed to this line that connects these two dots right there. This line, let me see, I was going to do that in a different color. This line that connects those two dots is going to be transformed to this line that connects those two dots."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me draw it this way. This line right here, that connects those two dots, will be transformed to this line that connects these two dots right there. This line, let me see, I was going to do that in a different color. This line that connects those two dots is going to be transformed to this line that connects those two dots. And then we have this line that connects those two dots is going to be transformed to this line that connects these two dots right there. And then finally, this line that connects these two dots is going to be transformed to this line that connects these two dots right there. Now, just out of curiosity, what is, so let me just write this right here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This line that connects those two dots is going to be transformed to this line that connects those two dots. And then we have this line that connects those two dots is going to be transformed to this line that connects these two dots right there. And then finally, this line that connects these two dots is going to be transformed to this line that connects these two dots right there. Now, just out of curiosity, what is, so let me just write this right here. So this rectangle right here is the image. We could write this as T of our rectangle, or if we wanted to write it in words, this is equal to the image of the rectangle under our transformation T. Now, what is the area of the image of our rectangle under our transformation? What is the area of this thing?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, just out of curiosity, what is, so let me just write this right here. So this rectangle right here is the image. We could write this as T of our rectangle, or if we wanted to write it in words, this is equal to the image of the rectangle under our transformation T. Now, what is the area of the image of our rectangle under our transformation? What is the area of this thing? What is this area right there? Well, we could view this parallelogram, essentially, as a parallelogram generated by this vector and that vector. Or if we want to write it another way, if we had a matrix, if we had some matrix whose column vectors are this guy and that guy, so let's say its first column vector is this guy."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What is the area of this thing? What is this area right there? Well, we could view this parallelogram, essentially, as a parallelogram generated by this vector and that vector. Or if we want to write it another way, if we had a matrix, if we had some matrix whose column vectors are this guy and that guy, so let's say its first column vector is this guy. So a k1, c k1, and this one right here, this was the transformation of k1, 0. And then its second column vector is this guy right here, which was b k2 and d k2. This parallelogram is a parallelogram generated by these two column vectors."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Or if we want to write it another way, if we had a matrix, if we had some matrix whose column vectors are this guy and that guy, so let's say its first column vector is this guy. So a k1, c k1, and this one right here, this was the transformation of k1, 0. And then its second column vector is this guy right here, which was b k2 and d k2. This parallelogram is a parallelogram generated by these two column vectors. This was the transformation right here. This was the transformation of 0, k2. And in the last video, we said that the area of this parallelogram is equivalent to the area of the parallelogram, which is the same thing as the area of the image of our original rectangle under our transformation."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This parallelogram is a parallelogram generated by these two column vectors. This was the transformation right here. This was the transformation of 0, k2. And in the last video, we said that the area of this parallelogram is equivalent to the area of the parallelogram, which is the same thing as the area of the image of our original rectangle under our transformation. The area of this parallelogram is equivalent to the absolute value, let me call this determinant right here, let me call this i for image. So it's equal to the absolute value of the determinant of the matrix whose column vectors generate that parallelogram. So it should be equal to the absolute value of the determinant of i."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And in the last video, we said that the area of this parallelogram is equivalent to the area of the parallelogram, which is the same thing as the area of the image of our original rectangle under our transformation. The area of this parallelogram is equivalent to the absolute value, let me call this determinant right here, let me call this i for image. So it's equal to the absolute value of the determinant of the matrix whose column vectors generate that parallelogram. So it should be equal to the absolute value of the determinant of i. We saw that in the last video. That's what the area of this thing is going to be. Well, what is the determinant of this?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it should be equal to the absolute value of the determinant of i. We saw that in the last video. That's what the area of this thing is going to be. Well, what is the determinant of this? Well, this is going to be equal to, we've got to keep our absolute value signs. So the determinant of this is a k1 times d k2. So we could write this as, let me switch colors, we could write it as k1, k2, a d. I just multiplied that term times that term minus this term times that term."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, what is the determinant of this? Well, this is going to be equal to, we've got to keep our absolute value signs. So the determinant of this is a k1 times d k2. So we could write this as, let me switch colors, we could write it as k1, k2, a d. I just multiplied that term times that term minus this term times that term. So minus, say, k1, k2, bc. That's just the definition of the determinant of a 2 by 2 matrix. We saw this in the last video."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we could write this as, let me switch colors, we could write it as k1, k2, a d. I just multiplied that term times that term minus this term times that term. So minus, say, k1, k2, bc. That's just the definition of the determinant of a 2 by 2 matrix. We saw this in the last video. This is nothing new here, or maybe it might be relatively new if you just watched the last video. But all I'm saying is, look, if we want to figure out this parallelogram, this is the parallelogram generated by that vector and that vector. And if we say that those two guys are the column vectors of some matrix right there, we saw in the last video that the area of this parallelogram is equal to the absolute value of the determinant of this matrix."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We saw this in the last video. This is nothing new here, or maybe it might be relatively new if you just watched the last video. But all I'm saying is, look, if we want to figure out this parallelogram, this is the parallelogram generated by that vector and that vector. And if we say that those two guys are the column vectors of some matrix right there, we saw in the last video that the area of this parallelogram is equal to the absolute value of the determinant of this matrix. Now what is this equal to? This is equal to, we can factor out a k1, k2, so it's the absolute value of k1, k2 times a d minus bc. Now what is this equal to?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if we say that those two guys are the column vectors of some matrix right there, we saw in the last video that the area of this parallelogram is equal to the absolute value of the determinant of this matrix. Now what is this equal to? This is equal to, we can factor out a k1, k2, so it's the absolute value of k1, k2 times a d minus bc. Now what is this equal to? Well, a d minus bc, that is the determinant of our transformation vector. So if we say that T of x, if we call that vector right, or our transformation matrix, if we call this matrix A, this is equal to k1, k2 times the determinant of A. Which is a pretty interesting takeaway."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what is this equal to? Well, a d minus bc, that is the determinant of our transformation vector. So if we say that T of x, if we call that vector right, or our transformation matrix, if we call this matrix A, this is equal to k1, k2 times the determinant of A. Which is a pretty interesting takeaway. What this is telling us is, look, if I have some region, or in this case I just have the rectangle, but if I have some region in my domain and I apply a transformation, let's say my region has some area. Let's say in this case it has some area A. Let me write this as area."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which is a pretty interesting takeaway. What this is telling us is, look, if I have some region, or in this case I just have the rectangle, but if I have some region in my domain and I apply a transformation, let's say my region has some area. Let's say in this case it has some area A. Let me write this as area. This was my original area in my original region, right? And then I apply some transformation to it, where the transformation is equal to some matrix A times any member of your domain. I apply some transformation to it, I get some new region."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this as area. This was my original area in my original region, right? And then I apply some transformation to it, where the transformation is equal to some matrix A times any member of your domain. I apply some transformation to it, I get some new region. I get the image of this set under my transformation. Your new area is going to be equal to the absolute value of the determinant of your transformation matrix. That's the determinant of the transformation matrix."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I apply some transformation to it, I get some new region. I get the image of this set under my transformation. Your new area is going to be equal to the absolute value of the determinant of your transformation matrix. That's the determinant of the transformation matrix. That's the transformation matrix. So it's equal to the determinant of the transformation matrix times the area of your original rectangle in this case. Times the area of your original rectangle in this case, right?"}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the determinant of the transformation matrix. That's the transformation matrix. So it's equal to the determinant of the transformation matrix times the area of your original rectangle in this case. Times the area of your original rectangle in this case, right? This was your original rectangle. You take the absolute value, just because sometimes if you flip-swap these vectors you might get a negative sign. So you take the absolute value."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times the area of your original rectangle in this case, right? This was your original rectangle. You take the absolute value, just because sometimes if you flip-swap these vectors you might get a negative sign. So you take the absolute value. But that's a really neat idea. The determinant of the transformation matrix is essentially a scaling factor on the area of a certain region. I'm not going to prove it to you here, but you can kind of imagine it."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you take the absolute value. But that's a really neat idea. The determinant of the transformation matrix is essentially a scaling factor on the area of a certain region. I'm not going to prove it to you here, but you can kind of imagine it. If I have, let's say I have some, let me go abstract now. Let's say I have some region in R2 that, well let me do a better, let me do a shape that you might recognize. Let's say I have something like that."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm not going to prove it to you here, but you can kind of imagine it. If I have, let's say I have some, let me go abstract now. Let's say I have some region in R2 that, well let me do a better, let me do a shape that you might recognize. Let's say I have something like that. Let's say it's a bit of a, let's say it's an ellipse. And this is our domain. This is my domain."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have something like that. Let's say it's a bit of a, let's say it's an ellipse. And this is our domain. This is my domain. It's R2. And let's say that the area here, the area of this region right here, is equal to a. So area is equal to a."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is my domain. It's R2. And let's say that the area here, the area of this region right here, is equal to a. So area is equal to a. And let's say I have some transformation, t, that is a mapping from R2 to R2. And t is defined, so t applied to some vector x in my domain is equal to, well I already used a, so let's say it's equal to some matrix b times any vector in my domain. If I take the image of this area under t, so this is my domain right here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So area is equal to a. And let's say I have some transformation, t, that is a mapping from R2 to R2. And t is defined, so t applied to some vector x in my domain is equal to, well I already used a, so let's say it's equal to some matrix b times any vector in my domain. If I take the image of this area under t, so this is my domain right here. Let me do it in the codomain. Let's say the image of this under b, let's say it looks something like, or my image under t, let's say it looks something like this. I don't know what it's going to look like, but let's say it looks something like this."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If I take the image of this area under t, so this is my domain right here. Let me do it in the codomain. Let's say the image of this under b, let's say it looks something like, or my image under t, let's say it looks something like this. I don't know what it's going to look like, but let's say it looks something like this. Let's say it looks something like that right there. So this right here. So if we call this set, let me call it e for ellipse."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I don't know what it's going to look like, but let's say it looks something like this. Let's say it looks something like that right there. So this right here. So if we call this set, let me call it e for ellipse. That's this whole thing right here. This is the ellipse right there. This is the image of my ellipse under the transformation t. If I take every point of this ellipse, I'll construct this right here."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we call this set, let me call it e for ellipse. That's this whole thing right here. This is the ellipse right there. This is the image of my ellipse under the transformation t. If I take every point of this ellipse, I'll construct this right here. Or I guess maybe just to keep the analogy straight for this example, let's say that the ellipse is the set of just the boundary, but it also applies actually if you were to fill up the whole region. So this boundary gets mapped to this boundary right here. But it would have worked just as well if we filled up the region, since I'm not proving either to you very rigorously."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is the image of my ellipse under the transformation t. If I take every point of this ellipse, I'll construct this right here. Or I guess maybe just to keep the analogy straight for this example, let's say that the ellipse is the set of just the boundary, but it also applies actually if you were to fill up the whole region. So this boundary gets mapped to this boundary right here. But it would have worked just as well if we filled up the region, since I'm not proving either to you very rigorously. You would have to accept it as a leaf of faith. So this is the image of this boundary under our transformation. And if the area surrounded by this boundary, or if you could say the area of this region is a, then the area here, the area of the image of our ellipse under our transformation, is going to be equal to the absolute value of our area, our original area, times the determinant of our transformation matrix."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But it would have worked just as well if we filled up the region, since I'm not proving either to you very rigorously. You would have to accept it as a leaf of faith. So this is the image of this boundary under our transformation. And if the area surrounded by this boundary, or if you could say the area of this region is a, then the area here, the area of the image of our ellipse under our transformation, is going to be equal to the absolute value of our area, our original area, times the determinant of our transformation matrix. And if you find it a little bit of a big leap to go from just a arbitrary rectangle transformation to maybe these more generalized shapes, you can imagine. So t is going from there to there. It's always nice to draw that arrow."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if the area surrounded by this boundary, or if you could say the area of this region is a, then the area here, the area of the image of our ellipse under our transformation, is going to be equal to the absolute value of our area, our original area, times the determinant of our transformation matrix. And if you find it a little bit of a big leap to go from just a arbitrary rectangle transformation to maybe these more generalized shapes, you can imagine. So t is going from there to there. It's always nice to draw that arrow. But if you view it as a bit of a leap to make that jump from rectangles to curves, you can imagine this shape to be able to be constructed by a bunch of arbitrary rectangles. So you could essentially fill this space with a bunch of arbitrary rectangles like this. And you could keep making them infinitely small so that you completely fill the space."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's always nice to draw that arrow. But if you view it as a bit of a leap to make that jump from rectangles to curves, you can imagine this shape to be able to be constructed by a bunch of arbitrary rectangles. So you could essentially fill this space with a bunch of arbitrary rectangles like this. And you could keep making them infinitely small so that you completely fill the space. You completely fill that space with rectangles. And if you were to map each of those rectangles using t, each of those rectangles might look something like this. Maybe they look something like this."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you could keep making them infinitely small so that you completely fill the space. You completely fill that space with rectangles. And if you were to map each of those rectangles using t, each of those rectangles might look something like this. Maybe they look something like this. I don't know what they'll look like. They'll look like a bunch of parallelograms. Maybe they'll look something like that."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe they look something like this. I don't know what they'll look like. They'll look like a bunch of parallelograms. Maybe they'll look something like that. I'm not drawing it that neatly. They'll look like a bunch of parallelograms. And they will essentially fill this space right there."}, {"video_title": "Determinant as scaling factor Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Maybe they'll look something like that. I'm not drawing it that neatly. They'll look like a bunch of parallelograms. And they will essentially fill this space right there. So that's a way you can get your head around the idea of kind of jumping from just the arbitrary rectangles to these arbitrary curves or shapes or regions. But this is a pretty neat outcome. And it's a very interesting way to view a determinant."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's see if taking the transpose of a matrix does anything to its determinant. So a good place to start is just with the 2 by 2 scenario. So let's do the 2 by 2 scenario. So if I start with some matrix here, let me just take its determinant. So A, B, C, D. And then let me take its determinant. So this is going to be equal to A, D, minus B, C. Now let me take the transpose of this and then take its determinant. So that would be the determinant of A, C, the columns turn into the rows, and then B, D, and the rows turn into the columns."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if I start with some matrix here, let me just take its determinant. So A, B, C, D. And then let me take its determinant. So this is going to be equal to A, D, minus B, C. Now let me take the transpose of this and then take its determinant. So that would be the determinant of A, C, the columns turn into the rows, and then B, D, and the rows turn into the columns. What is this going to be equal to? This is going to be equal to A, D, minus B, C again. The only thing that happened is these two guys got swapped and they multiply times each other anyway."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that would be the determinant of A, C, the columns turn into the rows, and then B, D, and the rows turn into the columns. What is this going to be equal to? This is going to be equal to A, D, minus B, C again. The only thing that happened is these two guys got swapped and they multiply times each other anyway. So these two things are equivalent. So at least for the 2 by 2 case, the determinant of some matrix is equal to the determinant of the transpose of that matrix. Now that's just the 2 by 2 case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The only thing that happened is these two guys got swapped and they multiply times each other anyway. So these two things are equivalent. So at least for the 2 by 2 case, the determinant of some matrix is equal to the determinant of the transpose of that matrix. Now that's just the 2 by 2 case. Now I'm going to make an inductive argument, or I guess I should say an argument by induction, to show that this works for all n by n, for all cases. And the way you construct an argument by induction is you assume that it's true for the n by n case. So let's assume that for n by n, let's say I have some matrix, let me call it matrix B."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now that's just the 2 by 2 case. Now I'm going to make an inductive argument, or I guess I should say an argument by induction, to show that this works for all n by n, for all cases. And the way you construct an argument by induction is you assume that it's true for the n by n case. So let's assume that for n by n, let's say I have some matrix, let me call it matrix B. Let's say it's an n by n matrix, we assume that the determinant of any matrix B, that's n by n, is equal to the determinant of B's transpose. That's where we start off with our inductive argument. And then we see if given that, is it true of n plus 1 by n plus 1 matrix?"}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's assume that for n by n, let's say I have some matrix, let me call it matrix B. Let's say it's an n by n matrix, we assume that the determinant of any matrix B, that's n by n, is equal to the determinant of B's transpose. That's where we start off with our inductive argument. And then we see if given that, is it true of n plus 1 by n plus 1 matrix? Because if we can, if we can say look, given that it's true for the n by n case, it's going to be true for the n plus 1 by n plus 1 case, then we're done. Because we know it's true for the base case, the 2 by 2 case, which you could say, well that's your first n by n. So if it's true for the 2 by 2 case, then it'll be true for the 3 by 3 case, because that's just one increment. But then if it's true for the 3 by 3 case, it'll be true for the 4 by 4 case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we see if given that, is it true of n plus 1 by n plus 1 matrix? Because if we can, if we can say look, given that it's true for the n by n case, it's going to be true for the n plus 1 by n plus 1 case, then we're done. Because we know it's true for the base case, the 2 by 2 case, which you could say, well that's your first n by n. So if it's true for the 2 by 2 case, then it'll be true for the 3 by 3 case, because that's just one increment. But then if it's true for the 3 by 3 case, it'll be true for the 4 by 4 case. But if it's true for the 4 by 4 case, it'll be true for the 5 by 5 case. And you just keep going up like that. So when you do a proof by induction, you prove a base case, and then you prove that if it's true for n, or in this case an n by n determinant, if you can prove that given it's true for an n by n determinant, it's going to be true for an n plus 1 by n plus 1 determinant, or an n plus 1 by n plus 1 matrix, then you have completed your proof."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But then if it's true for the 3 by 3 case, it'll be true for the 4 by 4 case. But if it's true for the 4 by 4 case, it'll be true for the 5 by 5 case. And you just keep going up like that. So when you do a proof by induction, you prove a base case, and then you prove that if it's true for n, or in this case an n by n determinant, if you can prove that given it's true for an n by n determinant, it's going to be true for an n plus 1 by n plus 1 determinant, or an n plus 1 by n plus 1 matrix, then you have completed your proof. So let's see if this is the case. So let me construct an n plus 1 by n plus 1 matrix. So let's say I have my matrix A, my favorite letter to use for matrices."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So when you do a proof by induction, you prove a base case, and then you prove that if it's true for n, or in this case an n by n determinant, if you can prove that given it's true for an n by n determinant, it's going to be true for an n plus 1 by n plus 1 determinant, or an n plus 1 by n plus 1 matrix, then you have completed your proof. So let's see if this is the case. So let me construct an n plus 1 by n plus 1 matrix. So let's say I have my matrix A, my favorite letter to use for matrices. I think the entire linear algebra is a favorite letter to use for matrices. And let's say it's an n plus 1 by n plus 1 matrix. And just to simplify my notation, let's just say that m is equal to n plus 1."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have my matrix A, my favorite letter to use for matrices. I think the entire linear algebra is a favorite letter to use for matrices. And let's say it's an n plus 1 by n plus 1 matrix. And just to simplify my notation, let's just say that m is equal to n plus 1. So we could call it an m by m matrix. And what is it going to look like? Let's draw its entries right here."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And just to simplify my notation, let's just say that m is equal to n plus 1. So we could call it an m by m matrix. And what is it going to look like? Let's draw its entries right here. So it's going to have, I'm going to draw more than the normal amount of entries. A11, this is our first row, A12, all the way to A1m. We have m columns, which is the same thing as n plus 1."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's draw its entries right here. So it's going to have, I'm going to draw more than the normal amount of entries. A11, this is our first row, A12, all the way to A1m. We have m columns, which is the same thing as n plus 1. That's not n times 1 columns. This is n plus 1 as well. And then we do our second row right here."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We have m columns, which is the same thing as n plus 1. That's not n times 1 columns. This is n plus 1 as well. And then we do our second row right here. A21, A22, A23, all the way to A2m. Then you have your third row right here. A31, A32, A33, all the way to A3m."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we do our second row right here. A21, A22, A23, all the way to A2m. Then you have your third row right here. A31, A32, A33, all the way to A3m. And then you go all the way down here at the end. You have your mth row, which you could also say is your n plus 1th row. Which is your mth row, first column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A31, A32, A33, all the way to A3m. And then you go all the way down here at the end. You have your mth row, which you could also say is your n plus 1th row. Which is your mth row, first column. And then you have A2m, and then A3m, all the way to A sub mm. Fair enough. Now let me draw the transpose of A."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Which is your mth row, first column. And then you have A2m, and then A3m, all the way to A sub mm. Fair enough. Now let me draw the transpose of A. So A transpose is also going to be an n plus 1 by n plus 1 matrix, which you could also write as an m by m matrix. I'm just going to have to take the transpose of this. So the transpose of that."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now let me draw the transpose of A. So A transpose is also going to be an n plus 1 by n plus 1 matrix, which you could also write as an m by m matrix. I'm just going to have to take the transpose of this. So the transpose of that. This row becomes a column, so it becomes A11. Then this entry right here is A12. It's this entry right there."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the transpose of that. This row becomes a column, so it becomes A11. Then this entry right here is A12. It's this entry right there. Then you can go all the way down to A1m. Then this pink row becomes a pink column here. A21, you have A, I wanted to do it in pink."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's this entry right there. Then you can go all the way down to A1m. Then this pink row becomes a pink column here. A21, you have A, I wanted to do it in pink. You have A21, A22, you have an A23. It goes all the way down to A2m. You have your green row right here."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "A21, you have A, I wanted to do it in pink. You have A21, A22, you have an A23. It goes all the way down to A2m. You have your green row right here. It's your third one. So it's A31, A32, A33, all the way down to A3m. And then we can just skip a bunch of rows in this case, but it's columns in this case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You have your green row right here. It's your third one. So it's A31, A32, A33, all the way down to A3m. And then we can just skip a bunch of rows in this case, but it's columns in this case. So you just draw some dots, and you have Am1, Am2. I'm just going down this guy, but this row is now going to become, which was the last row, is now going to become the last column. Am3, all the way down to Amm."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we can just skip a bunch of rows in this case, but it's columns in this case. So you just draw some dots, and you have Am1, Am2. I'm just going down this guy, but this row is now going to become, which was the last row, is now going to become the last column. Am3, all the way down to Amm. And I have my transpose. Now let's take the determinant of A. Let me do it in purple."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Am3, all the way down to Amm. And I have my transpose. Now let's take the determinant of A. Let me do it in purple. So the determinant of my matrix A, we could just go down this first row up here. It's going to be equal to A11 times the determinant of its submatrix. So it's the determinant of this submatrix right here."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me do it in purple. So the determinant of my matrix A, we could just go down this first row up here. It's going to be equal to A11 times the determinant of its submatrix. So it's the determinant of this submatrix right here. The determinant of that submatrix right there. We could call that the submatrix A sub 11. We've seen this notation before."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's the determinant of this submatrix right here. The determinant of that submatrix right there. We could call that the submatrix A sub 11. We've seen this notation before. So it's the determinant of A sub 11. And then it is minus A12 times the determinant of its submatrix. So you cross out that row and that column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've seen this notation before. So it's the determinant of A sub 11. And then it is minus A12 times the determinant of its submatrix. So you cross out that row and that column. And so that is going to be A sub 12. And you're going to go all the way to, and I don't know what the sign on that is, so we could call it negative 1 to the 1 plus m. That'll give us the right sign for the checkerboard pattern. Times the determinant of the submatrix for this guy."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you cross out that row and that column. And so that is going to be A sub 12. And you're going to go all the way to, and I don't know what the sign on that is, so we could call it negative 1 to the 1 plus m. That'll give us the right sign for the checkerboard pattern. Times the determinant of the submatrix for this guy. So we call it A1m. Where you cross out that guy's row, that guy's column, and you're just left with all of this stuff over here. Fair enough."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Times the determinant of the submatrix for this guy. So we call it A1m. Where you cross out that guy's row, that guy's column, and you're just left with all of this stuff over here. Fair enough. Now let's look at the determinant of A transpose. We learned earlier, you don't have to go down the first row or you don't have to even go down a row. You could go down a column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Fair enough. Now let's look at the determinant of A transpose. We learned earlier, you don't have to go down the first row or you don't have to even go down a row. You could go down a column. Let me be clear. So for our determinant of A, we went down this row. And our submatrix, this was my first submatrix, my second submatrix."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You could go down a column. Let me be clear. So for our determinant of A, we went down this row. And our submatrix, this was my first submatrix, my second submatrix. You know what it looks like. You would cross out the second column and that row and whatever's left over would be the second submatrix and so on and so forth. But for our determinant of A transpose, let's just go down this first column and get the submatrixes like that."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And our submatrix, this was my first submatrix, my second submatrix. You know what it looks like. You would cross out the second column and that row and whatever's left over would be the second submatrix and so on and so forth. But for our determinant of A transpose, let's just go down this first column and get the submatrixes like that. So this is going to be equal to, let's get our first term right here, A sub 11 times the determinant of its submatrix. So what's the determinant of its submatrix? It's going to be its submatrix."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But for our determinant of A transpose, let's just go down this first column and get the submatrixes like that. So this is going to be equal to, let's get our first term right here, A sub 11 times the determinant of its submatrix. So what's the determinant of its submatrix? It's going to be its submatrix. You cross out its row and its column and you're going to be left with this thing right here. Now, an interesting question is how does this thing that I've just squared off, this guy's submatrix, relate to this guy's submatrix? Well, if you look at it carefully, this row from A22 to A2m has now become a column from A22 to A2m."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be its submatrix. You cross out its row and its column and you're going to be left with this thing right here. Now, an interesting question is how does this thing that I've just squared off, this guy's submatrix, relate to this guy's submatrix? Well, if you look at it carefully, this row from A22 to A2m has now become a column from A22 to A2m. This row, which is the next one, from A32 to A3m, has now become a column from A32 to A3m. If you keep going down, this last row has now become this column. So this guy's submatrix, or the thing we're going to have to take the determinant of right here, is equal to the transpose of this guy."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, if you look at it carefully, this row from A22 to A2m has now become a column from A22 to A2m. This row, which is the next one, from A32 to A3m, has now become a column from A32 to A3m. If you keep going down, this last row has now become this column. So this guy's submatrix, or the thing we're going to have to take the determinant of right here, is equal to the transpose of this guy. So this is equal to A sub 11 transpose. And if you go through it, so then we go minus this guy, minus A12 times the determinant of his submatrix. And if we cross out this guy's row and that guy's column, what is that going to look like?"}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this guy's submatrix, or the thing we're going to have to take the determinant of right here, is equal to the transpose of this guy. So this is equal to A sub 11 transpose. And if you go through it, so then we go minus this guy, minus A12 times the determinant of his submatrix. And if we cross out this guy's row and that guy's column, what is that going to look like? His submatrix is going to look like this. It's going to have that there and it's going to have that right there. How does that compare to A12?"}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And if we cross out this guy's row and that guy's column, what is that going to look like? His submatrix is going to look like this. It's going to have that there and it's going to have that right there. How does that compare to A12? So A12 is if you crossed out this and this, and then you crossed out this and this, and you're left with all of this right here. All of this right here. So once again, you see that this row is the same thing as this column, that this row is the same thing as this column, that that row is the same thing as that column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "How does that compare to A12? So A12 is if you crossed out this and this, and then you crossed out this and this, and you're left with all of this right here. All of this right here. So once again, you see that this row is the same thing as this column, that this row is the same thing as this column, that that row is the same thing as that column. So once again, the submatrix we have to take the determinant of is equal to the transpose of this thing over here. So it's equal to A12 transpose. This thing is equal to the transpose of this thing."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So once again, you see that this row is the same thing as this column, that this row is the same thing as this column, that that row is the same thing as that column. So once again, the submatrix we have to take the determinant of is equal to the transpose of this thing over here. So it's equal to A12 transpose. This thing is equal to the transpose of this thing. So in general, each of these submatrices, when we go down this row, is equal to the transpose of each of these. So you keep going. So then you're going to go all the way to plus minus 1."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This thing is equal to the transpose of this thing. So in general, each of these submatrices, when we go down this row, is equal to the transpose of each of these. So you keep going. So then you're going to go all the way to plus minus 1. We're going to go all the way down to the minus 1 plus m times the determinant of this guy. It's going to be this guy transpose. You can even do it."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So then you're going to go all the way to plus minus 1. We're going to go all the way down to the minus 1 plus m times the determinant of this guy. It's going to be this guy transpose. You can even do it. If you go all the way to, if you cross that guy out and that guy out, you're left with everything else on this matrix. And that's equal to the transpose of if you cross this guy out and that guy out. This row turns into that column, that row turns into that column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "You can even do it. If you go all the way to, if you cross that guy out and that guy out, you're left with everything else on this matrix. And that's equal to the transpose of if you cross this guy out and that guy out. This row turns into that column, that row turns into that column. I think you see the point. I don't want to beat a dead horse. So that's going to be equal to A sub 1m transpose."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This row turns into that column, that row turns into that column. I think you see the point. I don't want to beat a dead horse. So that's going to be equal to A sub 1m transpose. Now, remember, going into this inductive proof, or proof by induction, I assumed that for, remember this is an n plus 1 by n plus 1 matrix. But going into it, I assumed that for an n by n matrix, the determinant of B is equal to the determinant of B transpose. Well, these guys right here, these are n by n matrices."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be equal to A sub 1m transpose. Now, remember, going into this inductive proof, or proof by induction, I assumed that for, remember this is an n plus 1 by n plus 1 matrix. But going into it, I assumed that for an n by n matrix, the determinant of B is equal to the determinant of B transpose. Well, these guys right here, these are n by n matrices. This guy right here is an n plus 1 by n plus 1. Same thing for this guy right here. But these guys right here are n by n. So if we assume for the n by n case that the determinant of a matrix is equal to the determinant of a transpose, this is the determinant of a matrix, this is the determinant of its transpose, these two things have to be equal."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, these guys right here, these are n by n matrices. This guy right here is an n plus 1 by n plus 1. Same thing for this guy right here. But these guys right here are n by n. So if we assume for the n by n case that the determinant of a matrix is equal to the determinant of a transpose, this is the determinant of a matrix, this is the determinant of its transpose, these two things have to be equal. So we can then say that the determinant of A transpose is equal to this term, A sub 1, 1, times this, but this is equal to this for the n by n case. Remember, we're doing the n plus 1 by n plus 1 case. But these sub-matrices are one dimension smaller in each direction, has one less row and one less column."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But these guys right here are n by n. So if we assume for the n by n case that the determinant of a matrix is equal to the determinant of a transpose, this is the determinant of a matrix, this is the determinant of its transpose, these two things have to be equal. So we can then say that the determinant of A transpose is equal to this term, A sub 1, 1, times this, but this is equal to this for the n by n case. Remember, we're doing the n plus 1 by n plus 1 case. But these sub-matrices are one dimension smaller in each direction, has one less row and one less column. So these two things are equal, so instead of writing this, I can just write this. So times the determinant of A sub 1, 1, and then you keep going, minus A sub 1, 2 times the determinant. Instead of writing this, I could write that because they're equal."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But these sub-matrices are one dimension smaller in each direction, has one less row and one less column. So these two things are equal, so instead of writing this, I can just write this. So times the determinant of A sub 1, 1, and then you keep going, minus A sub 1, 2 times the determinant. Instead of writing this, I could write that because they're equal. Determinant A sub 1, 2, all the way to plus minus 1 to the 1 plus m, times the determinant of this. These two things are equal. That is equal to that."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Instead of writing this, I could write that because they're equal. Determinant A sub 1, 2, all the way to plus minus 1 to the 1 plus m, times the determinant of this. These two things are equal. That is equal to that. That was our assumption in this inductive proof. A sub 1, m. And then you see that, of course, this line, this blue line here, is equivalent to this blue line there. So we get that the determinant of A, which is an n plus 1 by n plus 1, so this is the n plus 1 by n plus 1 case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That is equal to that. That was our assumption in this inductive proof. A sub 1, m. And then you see that, of course, this line, this blue line here, is equivalent to this blue line there. So we get that the determinant of A, which is an n plus 1 by n plus 1, so this is the n plus 1 by n plus 1 case. We get the determinant of A is equal to the determinant of A transpose. And we got this assuming that it is true. Let me write, assuming that it's true for n by n case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we get that the determinant of A, which is an n plus 1 by n plus 1, so this is the n plus 1 by n plus 1 case. We get the determinant of A is equal to the determinant of A transpose. And we got this assuming that it is true. Let me write, assuming that it's true for n by n case. And then we're done. We've now proved that this is true in general because we've proven the base case. We've proven it for the 2 by 2 case."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write, assuming that it's true for n by n case. And then we're done. We've now proved that this is true in general because we've proven the base case. We've proven it for the 2 by 2 case. And then we show that if it's true for the n case, it's true for the n plus 1 case. So if it's true for the 2 case, it's going to be true for the 3 by 3 case. If it's true for the 3 by 3 case, it's true for the 4 by 4 case, so on and so forth."}, {"video_title": "Determinant of transpose Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've proven it for the 2 by 2 case. And then we show that if it's true for the n case, it's true for the n plus 1 case. So if it's true for the 2 case, it's going to be true for the 3 by 3 case. If it's true for the 3 by 3 case, it's true for the 4 by 4 case, so on and so forth. But the takeaway is pretty neat. You can take the transpose. The determinant doesn't change."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If that's my vector a and that's my vector b right there, the angle between them is this angle. And we defined it in this way. And actually, if you ever want to solve, if you have two vectors and you want to solve for that angle, and I've never done this before explicitly, and I thought, well, I might as well do it right now, you could just solve for your theta. So it would be a dot b divided by the length of your two vectors multiplied by each other is equal to the cosine of theta. And then to solve for theta, you would have to take the inverse cosine of both sides, or the arc cosine of both sides, and you would get theta is equal to arc cosine of a dot b over the magnitudes or the lengths of the products, or the lengths of each of those vectors multiplied by each other. So if I give you two arbitrary vectors, and the neat thing about it is, this might be pretty straightforward, if I just drew something in two dimensions right here, you could just take your protractor out and measure this angle. But if a and b each have a hundred components, it becomes hard to visualize the idea of an angle between the two vectors."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it would be a dot b divided by the length of your two vectors multiplied by each other is equal to the cosine of theta. And then to solve for theta, you would have to take the inverse cosine of both sides, or the arc cosine of both sides, and you would get theta is equal to arc cosine of a dot b over the magnitudes or the lengths of the products, or the lengths of each of those vectors multiplied by each other. So if I give you two arbitrary vectors, and the neat thing about it is, this might be pretty straightforward, if I just drew something in two dimensions right here, you could just take your protractor out and measure this angle. But if a and b each have a hundred components, it becomes hard to visualize the idea of an angle between the two vectors. But you don't need to visualize them anymore, because you just have to calculate this thing right here, you just have to calculate this value right there, and then go to your calculator and type in arc cosine or the inverse cosine, they're the equivalent functions, and then it'll give you an angle. And that, by definition, is the angle between those two vectors, which is a very neat concept, and then you can start addressing issues of perpendicularity and whatever else. Well, this was a bit of a tangent."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But if a and b each have a hundred components, it becomes hard to visualize the idea of an angle between the two vectors. But you don't need to visualize them anymore, because you just have to calculate this thing right here, you just have to calculate this value right there, and then go to your calculator and type in arc cosine or the inverse cosine, they're the equivalent functions, and then it'll give you an angle. And that, by definition, is the angle between those two vectors, which is a very neat concept, and then you can start addressing issues of perpendicularity and whatever else. Well, this was a bit of a tangent. But the other outcome that I painstakingly proved to you in the previous video was that the length of the cross product of two vectors is equal to, it's a very similar expression, it's equal to the product of the two vectors' lengths, so the length of a times the length of b, times the sine of the angle between them. So it's the same angle. So what I want to do is take these two ideas, and this was a bit of a diversion there, just to kind of show you how to solve for theta, because I realize I've never done that for you before, but what I want to do is I want to take this expression up here, and this expression up here, and see if we can develop an intuition, at least in R3, because right now we've only defined our cross product, or the cross product of two vectors is only defined in R3."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this was a bit of a tangent. But the other outcome that I painstakingly proved to you in the previous video was that the length of the cross product of two vectors is equal to, it's a very similar expression, it's equal to the product of the two vectors' lengths, so the length of a times the length of b, times the sine of the angle between them. So it's the same angle. So what I want to do is take these two ideas, and this was a bit of a diversion there, just to kind of show you how to solve for theta, because I realize I've never done that for you before, but what I want to do is I want to take this expression up here, and this expression up here, and see if we can develop an intuition, at least in R3, because right now we've only defined our cross product, or the cross product of two vectors is only defined in R3. Let's take these two ideas in R3 and see if we can develop an intuition. And I've done a very similar video in the physics playlist, where I compared the dot product to the cross product. Now if I'm talking about, let me redraw my vectors."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what I want to do is take these two ideas, and this was a bit of a diversion there, just to kind of show you how to solve for theta, because I realize I've never done that for you before, but what I want to do is I want to take this expression up here, and this expression up here, and see if we can develop an intuition, at least in R3, because right now we've only defined our cross product, or the cross product of two vectors is only defined in R3. Let's take these two ideas in R3 and see if we can develop an intuition. And I've done a very similar video in the physics playlist, where I compared the dot product to the cross product. Now if I'm talking about, let me redraw my vectors. So the length of a, so let me draw a, b, I want to do it bigger than that, so let me do it like that. So that is my vector b, so this is b, that is a. What is the length of a times the length of b times the cosine of the angle?"}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now if I'm talking about, let me redraw my vectors. So the length of a, so let me draw a, b, I want to do it bigger than that, so let me do it like that. So that is my vector b, so this is b, that is a. What is the length of a times the length of b times the cosine of the angle? So let me do that right there. So this is the angle. So the length of a, if I were to draw these vectors, is this length right here."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What is the length of a times the length of b times the cosine of the angle? So let me do that right there. So this is the angle. So the length of a, if I were to draw these vectors, is this length right here. This is the length of a. It's this distance right here, the way I've drawn this vector. So this is literally the length of vector a, and I'm doing it in R3 or maybe a version of it that I can fit onto my little blackboard right here."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the length of a, if I were to draw these vectors, is this length right here. This is the length of a. It's this distance right here, the way I've drawn this vector. So this is literally the length of vector a, and I'm doing it in R3 or maybe a version of it that I can fit onto my little blackboard right here. So it'll just be the length of this line right there. And then the length of b is the length of that line right there. So that is the length of b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is literally the length of vector a, and I'm doing it in R3 or maybe a version of it that I can fit onto my little blackboard right here. So it'll just be the length of this line right there. And then the length of b is the length of that line right there. So that is the length of b. So what is, let me say, let me rewrite this thing up here. Let me write it as b, the length of b, times the length, I want to be careful, I don't want to do the dot there because you'll think it's a dot product, times a cosine of theta. All I did is I rearranged this thing here."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that is the length of b. So what is, let me say, let me rewrite this thing up here. Let me write it as b, the length of b, times the length, I want to be careful, I don't want to do the dot there because you'll think it's a dot product, times a cosine of theta. All I did is I rearranged this thing here. It's the same thing as a dot b. What is a times the cosine of theta? Let's get out our basic trigonometry tools."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "All I did is I rearranged this thing here. It's the same thing as a dot b. What is a times the cosine of theta? Let's get out our basic trigonometry tools. Soh-cah-toa. Cosine of theta is equal to adjacent over hypotenuse. So if I drop, if I create a right triangle here, and let me introduce some new colors just to ease the monotony."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's get out our basic trigonometry tools. Soh-cah-toa. Cosine of theta is equal to adjacent over hypotenuse. So if I drop, if I create a right triangle here, and let me introduce some new colors just to ease the monotony. If I drop a right triangle right here, and I create a right triangle right there, and this is theta, then what is the cosine of theta? It's equal to this, let me do it in another color, it's equal to this, the adjacent side, it's equal to this little magenta thing, not all of b, just this part that goes up to my right angle. That's my adjacent, I want to do it a little bit bigger."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if I drop, if I create a right triangle here, and let me introduce some new colors just to ease the monotony. If I drop a right triangle right here, and I create a right triangle right there, and this is theta, then what is the cosine of theta? It's equal to this, let me do it in another color, it's equal to this, the adjacent side, it's equal to this little magenta thing, not all of b, just this part that goes up to my right angle. That's my adjacent, I want to do it a little bit bigger. It's equal to the adjacent side over the hypotenuse. So let me write this down. So cosine of theta is equal to this little adjacent side, I'm just going to write it like that, is equal to this adjacent side, that's not a nice, adjacent side over the hypotenuse."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's my adjacent, I want to do it a little bit bigger. It's equal to the adjacent side over the hypotenuse. So let me write this down. So cosine of theta is equal to this little adjacent side, I'm just going to write it like that, is equal to this adjacent side, that's not a nice, adjacent side over the hypotenuse. But what is the hypotenuse? It is the length of vector a. It's this, that's my hypotenuse right there."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So cosine of theta is equal to this little adjacent side, I'm just going to write it like that, is equal to this adjacent side, that's not a nice, adjacent side over the hypotenuse. But what is the hypotenuse? It is the length of vector a. It's this, that's my hypotenuse right there. So my hypotenuse is the length of vector a. And so if I multiply both sides by the length of vector a, I get the length of vector a times the cosine of theta is equal to the adjacent side. I'll do that in magenta, is equal to the adjacent side."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's this, that's my hypotenuse right there. So my hypotenuse is the length of vector a. And so if I multiply both sides by the length of vector a, I get the length of vector a times the cosine of theta is equal to the adjacent side. I'll do that in magenta, is equal to the adjacent side. So this expression right here, which was just a dot b, can be rewritten as, I just told you that the length of vector a times cosine of theta is equal to this little magenta adjacent side. So this is equal to the adjacent side. So you can view a dot b as being equal to the length of vector b, that length, times that adjacent side."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'll do that in magenta, is equal to the adjacent side. So this expression right here, which was just a dot b, can be rewritten as, I just told you that the length of vector a times cosine of theta is equal to this little magenta adjacent side. So this is equal to the adjacent side. So you can view a dot b as being equal to the length of vector b, that length, times that adjacent side. And you're saying, Sal, what does that do for me? Well, what it tells you is you're multiplying, essentially, the length of vector b times the amount of vector a that's going in the same direction as vector b. You can kind of view this as the shadow of vector a."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you can view a dot b as being equal to the length of vector b, that length, times that adjacent side. And you're saying, Sal, what does that do for me? Well, what it tells you is you're multiplying, essentially, the length of vector b times the amount of vector a that's going in the same direction as vector b. You can kind of view this as the shadow of vector a. And I'll talk about projections in the future, and I'll more formally define them. But if the word projection helps you, just think of it that word. If you have a light that shines down from above, this adjacent side is kind of like the shadow of a onto vector b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You can kind of view this as the shadow of vector a. And I'll talk about projections in the future, and I'll more formally define them. But if the word projection helps you, just think of it that word. If you have a light that shines down from above, this adjacent side is kind of like the shadow of a onto vector b. So you can imagine if these two vectors looked more like this, if they were really going in the same direction, let's say that's vector a and that's vector b, then the adjacent side that I care about, they're going to have a lot more in common. The part of a that is going in the same direction of b will be a lot larger. So I have a larger dot product."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you have a light that shines down from above, this adjacent side is kind of like the shadow of a onto vector b. So you can imagine if these two vectors looked more like this, if they were really going in the same direction, let's say that's vector a and that's vector b, then the adjacent side that I care about, they're going to have a lot more in common. The part of a that is going in the same direction of b will be a lot larger. So I have a larger dot product. Because a dot product is essentially saying how much of those vectors are going in the same direction, but it's just a number. So it'll just be this adjacent side times the length of b. And what if I had vectors that are pretty perpendicular to each other?"}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I have a larger dot product. Because a dot product is essentially saying how much of those vectors are going in the same direction, but it's just a number. So it'll just be this adjacent side times the length of b. And what if I had vectors that are pretty perpendicular to each other? So what if I had two vectors that were like this? What if my vector a looked like that, and my vector b looked like that? Well now the adjacent, the way I define it here, if I had to make a right triangle like that, the adjacent side's very small."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And what if I had vectors that are pretty perpendicular to each other? So what if I had two vectors that were like this? What if my vector a looked like that, and my vector b looked like that? Well now the adjacent, the way I define it here, if I had to make a right triangle like that, the adjacent side's very small. So your dot product, even though a is still a reasonably large vector, is now much smaller, because a and b have very little commonality in the same direction. And you could do it the other way. You could draw this down like that, and you could do the adjacent the other way, but it doesn't matter because these a's and b's are arbitrary."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well now the adjacent, the way I define it here, if I had to make a right triangle like that, the adjacent side's very small. So your dot product, even though a is still a reasonably large vector, is now much smaller, because a and b have very little commonality in the same direction. And you could do it the other way. You could draw this down like that, and you could do the adjacent the other way, but it doesn't matter because these a's and b's are arbitrary. So the takeaway is the fact that a dot b is equal to the length of each of those times the cosine of theta, to me it says that the dot product tells me how much are my vectors moving together, or the product of the part of the vectors that are moving together. The product of the lengths of the vectors that are moving together, or in the same direction. You could view this adjacent side here as the part of a that's going in the direction of b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could draw this down like that, and you could do the adjacent the other way, but it doesn't matter because these a's and b's are arbitrary. So the takeaway is the fact that a dot b is equal to the length of each of those times the cosine of theta, to me it says that the dot product tells me how much are my vectors moving together, or the product of the part of the vectors that are moving together. The product of the lengths of the vectors that are moving together, or in the same direction. You could view this adjacent side here as the part of a that's going in the direction of b. You're multiplying that times b itself. So that's what the dot product is. How much are two things going in the same direction?"}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You could view this adjacent side here as the part of a that's going in the direction of b. You're multiplying that times b itself. So that's what the dot product is. How much are two things going in the same direction? And notice, when two things are orthogonal, or when they're perpendicular, when a dot b is equal to zero, we say they're perpendicular. And that makes complete sense based on this kind of intuition of what the dot product is doing, because that means that they're perfectly perpendicular. So that's b, and that's a."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "How much are two things going in the same direction? And notice, when two things are orthogonal, or when they're perpendicular, when a dot b is equal to zero, we say they're perpendicular. And that makes complete sense based on this kind of intuition of what the dot product is doing, because that means that they're perfectly perpendicular. So that's b, and that's a. And so the adjacent part of a, if I had to draw a right triangle, it would come straight down, and if I were to say the projection of a, and I haven't drawn that, or if I put a light shining down from above, and I'd say what's the shadow of a onto b, you'd get nothing, you'd get zero. This arrow has no width, even though I've drawn it to have some width. It has no width."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that's b, and that's a. And so the adjacent part of a, if I had to draw a right triangle, it would come straight down, and if I were to say the projection of a, and I haven't drawn that, or if I put a light shining down from above, and I'd say what's the shadow of a onto b, you'd get nothing, you'd get zero. This arrow has no width, even though I've drawn it to have some width. It has no width. So you'd have a zero down here, the part of a that goes in the same direction as b. No part of this vector goes in the same direction as this vector, so you're going to have this zero kind of adjacent side times b, so you're going to get something that's zero. So hopefully that makes a little sense."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It has no width. So you'd have a zero down here, the part of a that goes in the same direction as b. No part of this vector goes in the same direction as this vector, so you're going to have this zero kind of adjacent side times b, so you're going to get something that's zero. So hopefully that makes a little sense. Now let's think about the cross product. The cross product tells us, well, the length of a cross b, I painstakingly showed you, is equal to the length of a times the length of b times the sine of the angle between them. So let me do the same example."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So hopefully that makes a little sense. Now let's think about the cross product. The cross product tells us, well, the length of a cross b, I painstakingly showed you, is equal to the length of a times the length of b times the sine of the angle between them. So let me do the same example. Let me draw my two vectors. That's my vector a. This is my vector b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me do the same example. Let me draw my two vectors. That's my vector a. This is my vector b. Now sine, SOH CAH TOA, so sine of theta, let me write that, sine of theta, SOH CAH TOA is equal to opposite over the hypotenuse. So if I were to draw a little right triangle here, so if I were to draw a perpendicular right there, and this is theta, what is the sine of theta equal to in this context? The sine of theta is equal to what?"}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is my vector b. Now sine, SOH CAH TOA, so sine of theta, let me write that, sine of theta, SOH CAH TOA is equal to opposite over the hypotenuse. So if I were to draw a little right triangle here, so if I were to draw a perpendicular right there, and this is theta, what is the sine of theta equal to in this context? The sine of theta is equal to what? It's equal to this side over here. Let me call that just the opposite. It's equal to the opposite side over the hypotenuse."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The sine of theta is equal to what? It's equal to this side over here. Let me call that just the opposite. It's equal to the opposite side over the hypotenuse. So the hypotenuse is the length of this vector a right there. It's the length of this vector a. So the hypotenuse is the length over my vector a."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's equal to the opposite side over the hypotenuse. So the hypotenuse is the length of this vector a right there. It's the length of this vector a. So the hypotenuse is the length over my vector a. So if I multiply both sides of this by my length of vector a, I get the length of vector a times the sine of theta is equal to the opposite side. So if we rearrange this a little bit, I can rewrite this as equal to, I'm just going to swap them. I had to do the dot product as well."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the hypotenuse is the length over my vector a. So if I multiply both sides of this by my length of vector a, I get the length of vector a times the sine of theta is equal to the opposite side. So if we rearrange this a little bit, I can rewrite this as equal to, I'm just going to swap them. I had to do the dot product as well. This is equal to b, the length of vector b times the length of vector a sine of theta. Well, this thing is just the opposite side as I've defined it right here. So this right here is just the opposite side."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I had to do the dot product as well. This is equal to b, the length of vector b times the length of vector a sine of theta. Well, this thing is just the opposite side as I've defined it right here. So this right here is just the opposite side. This side right there. So when we're taking the cross product, we're essentially multiplying the length of vector b times the part of a that's going perpendicular to b. This opposite side is the part of a that's going perpendicular to b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this right here is just the opposite side. This side right there. So when we're taking the cross product, we're essentially multiplying the length of vector b times the part of a that's going perpendicular to b. This opposite side is the part of a that's going perpendicular to b. So they're kind of opposite ideas. The dot product, you're multiplying the part of a that's going in the same direction as b with b, while when you're taking the cross product, you're multiplying the part of a that's going in the perpendicular direction to b with the length of b. So you're saying, it's a measure, especially when you take the length of this, it's a measure of how perpendicular these two guys are."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This opposite side is the part of a that's going perpendicular to b. So they're kind of opposite ideas. The dot product, you're multiplying the part of a that's going in the same direction as b with b, while when you're taking the cross product, you're multiplying the part of a that's going in the perpendicular direction to b with the length of b. So you're saying, it's a measure, especially when you take the length of this, it's a measure of how perpendicular these two guys are. And this is, it's a measure of how much do they move in the same direction. And let's just look at a couple of examples. So if you take two right triangles, if you take two right triangles, so if that's a and that's, or if you take two vectors that are perpendicular to each other, the length of a cross b, a cross b, is it going to be equal to, if we just use this formula right there, the length of a times the length of b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you're saying, it's a measure, especially when you take the length of this, it's a measure of how perpendicular these two guys are. And this is, it's a measure of how much do they move in the same direction. And let's just look at a couple of examples. So if you take two right triangles, if you take two right triangles, so if that's a and that's, or if you take two vectors that are perpendicular to each other, the length of a cross b, a cross b, is it going to be equal to, if we just use this formula right there, the length of a times the length of b. And what's the sine of 90 degrees? It's one. So in this case, you kind of have maximized the length of your cross product."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if you take two right triangles, if you take two right triangles, so if that's a and that's, or if you take two vectors that are perpendicular to each other, the length of a cross b, a cross b, is it going to be equal to, if we just use this formula right there, the length of a times the length of b. And what's the sine of 90 degrees? It's one. So in this case, you kind of have maximized the length of your cross product. This is as high as it can go. Cosine of theta, it's a maximum value. Sine of theta is always less than or equal to one."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in this case, you kind of have maximized the length of your cross product. This is as high as it can go. Cosine of theta, it's a maximum value. Sine of theta is always less than or equal to one. So this is as good as you're ever going to get. This is the highest possible value when you have perfectly perpendicular vectors. Now, when is, actually just to kind of go back to make the same point here, when do you get the maximum value for your cosine of, for your dot product?"}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Sine of theta is always less than or equal to one. So this is as good as you're ever going to get. This is the highest possible value when you have perfectly perpendicular vectors. Now, when is, actually just to kind of go back to make the same point here, when do you get the maximum value for your cosine of, for your dot product? Well, it's when your two vectors are collinear. If I have one vector, if my vector a looks like that and my vector b is essentially another vector that's going in the same direction, then theta is zero, there's no angle between them, and then you have a dot b is equal to the magnitude or the length of vector a times the length of vector b times the cosine of the angle between them. The cosine of the angle between them, the cosine of that angle is zero, or the cosine of the angle is zero, so the cosine of that is one."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, when is, actually just to kind of go back to make the same point here, when do you get the maximum value for your cosine of, for your dot product? Well, it's when your two vectors are collinear. If I have one vector, if my vector a looks like that and my vector b is essentially another vector that's going in the same direction, then theta is zero, there's no angle between them, and then you have a dot b is equal to the magnitude or the length of vector a times the length of vector b times the cosine of the angle between them. The cosine of the angle between them, the cosine of that angle is zero, or the cosine of the angle is zero, so the cosine of that is one. So when you have two vectors that go exactly in the same direction or they're collinear, you kind of maximize your dot product. You maximize your cross product when they're perfectly perpendicular to each other. And just to make the analogy clear, when they're perpendicular to each other, you've minimized, or at least the magnitude of your dot product."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The cosine of the angle between them, the cosine of that angle is zero, or the cosine of the angle is zero, so the cosine of that is one. So when you have two vectors that go exactly in the same direction or they're collinear, you kind of maximize your dot product. You maximize your cross product when they're perfectly perpendicular to each other. And just to make the analogy clear, when they're perpendicular to each other, you've minimized, or at least the magnitude of your dot product. You can get negative dot products, but the absolute size of your dot product, the absolute value of your dot product is minimized when they're perpendicular to each other. Similarly, if you were to take two vectors that are collinear and they're moving in the same direction, so if that's vector a, and then I have vector b that just is another vector that, I don't want to draw them on top of each other, but I think you get the idea. Let's say vector b is like that, vector b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And just to make the analogy clear, when they're perpendicular to each other, you've minimized, or at least the magnitude of your dot product. You can get negative dot products, but the absolute size of your dot product, the absolute value of your dot product is minimized when they're perpendicular to each other. Similarly, if you were to take two vectors that are collinear and they're moving in the same direction, so if that's vector a, and then I have vector b that just is another vector that, I don't want to draw them on top of each other, but I think you get the idea. Let's say vector b is like that, vector b. Then theta is zero, you can't even see it. It's been squeezed out. I've just brought these two things on top of each other."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say vector b is like that, vector b. Then theta is zero, you can't even see it. It's been squeezed out. I've just brought these two things on top of each other. And then the cross product in this situation, a cross b is equal to, well, the length of both of these things times the sine of theta. Sine of zero is zero, so it's just zero. So two collinear vectors, the magnitude of their cross product is zero, but the magnitude of their dot product, the a dot b, is going to be maximized."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I've just brought these two things on top of each other. And then the cross product in this situation, a cross b is equal to, well, the length of both of these things times the sine of theta. Sine of zero is zero, so it's just zero. So two collinear vectors, the magnitude of their cross product is zero, but the magnitude of their dot product, the a dot b, is going to be maximized. It's going to be as high as you can get. It's going to be the length of a times the length of b. Now the opposite scenario is right here."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So two collinear vectors, the magnitude of their cross product is zero, but the magnitude of their dot product, the a dot b, is going to be maximized. It's going to be as high as you can get. It's going to be the length of a times the length of b. Now the opposite scenario is right here. When they're perpendicular to each other, the cross product is maximized because it's measuring on how much of the vectors, how much of the perpendicular part of a is multiplying that times the length of b. And then when you have two orthogonal vectors, your dot product is minimized or the absolute value of your dot product. So a dot b in this case is equal to zero."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now the opposite scenario is right here. When they're perpendicular to each other, the cross product is maximized because it's measuring on how much of the vectors, how much of the perpendicular part of a is multiplying that times the length of b. And then when you have two orthogonal vectors, your dot product is minimized or the absolute value of your dot product. So a dot b in this case is equal to zero. Anyway, I wanted to make all of this clear because sometimes you can get into the formulas and the definitions and you lose the intuition about what are all of these ideas really for. And actually, before I move on, let me just make another kind of idea about what the cross product can be interpreted as. Because a cross product tends to give people more trouble."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So a dot b in this case is equal to zero. Anyway, I wanted to make all of this clear because sometimes you can get into the formulas and the definitions and you lose the intuition about what are all of these ideas really for. And actually, before I move on, let me just make another kind of idea about what the cross product can be interpreted as. Because a cross product tends to give people more trouble. So if I have that's my a and that's my b, what if I wanted to figure out the area of this parallelogram? So if I were to shift a and have that there and if I were to shift b and draw a line parallel to b, and if I wanted to figure out the area of this parallelogram right there, how would I do it just using regular geometry? Well, I would drop a perpendicular right there."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Because a cross product tends to give people more trouble. So if I have that's my a and that's my b, what if I wanted to figure out the area of this parallelogram? So if I were to shift a and have that there and if I were to shift b and draw a line parallel to b, and if I wanted to figure out the area of this parallelogram right there, how would I do it just using regular geometry? Well, I would drop a perpendicular right there. If this is perpendicular and this length is h, let me use it for height, then the area of this, the area of the parallelogram is just equal to the length of my base, which is just the length of vector b, times my height. But what is my height? Well, let me just draw a little theta there."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I would drop a perpendicular right there. If this is perpendicular and this length is h, let me use it for height, then the area of this, the area of the parallelogram is just equal to the length of my base, which is just the length of vector b, times my height. But what is my height? Well, let me just draw a little theta there. Let me do a green theta, it's more visible. So theta. So we know already that the sine of this theta is equal to the opposite over the hypotenuse."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, let me just draw a little theta there. Let me do a green theta, it's more visible. So theta. So we know already that the sine of this theta is equal to the opposite over the hypotenuse. So it's equal to the height over the hypotenuse. The hypotenuse is just the length of vector a. Or we could just solve for height and we'd get the height is equal to the length of vector a times the sine of theta."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we know already that the sine of this theta is equal to the opposite over the hypotenuse. So it's equal to the height over the hypotenuse. The hypotenuse is just the length of vector a. Or we could just solve for height and we'd get the height is equal to the length of vector a times the sine of theta. So I can rewrite this here, I can replace it with that, and I get the area of this parallelogram is equal to the length of vector b times the length of vector a sine theta. Well, this is just the length of the cross product of the two vectors, a cross b. This is the same thing, I mean, you can rearrange the a and the b."}, {"video_title": "Dot and cross product comparison intuition Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Or we could just solve for height and we'd get the height is equal to the length of vector a times the sine of theta. So I can rewrite this here, I can replace it with that, and I get the area of this parallelogram is equal to the length of vector b times the length of vector a sine theta. Well, this is just the length of the cross product of the two vectors, a cross b. This is the same thing, I mean, you can rearrange the a and the b. So we now have another way of thinking about what the cross product is. The cross product of two vectors, or at least the magnitude, or the length of the cross product of two vectors, obviously the cross product, you're going to get a third vector, but the length of that third vector is equal to the area of the parallelogram that's defined, or that you can create from those two vectors. Anyway, hopefully you found this a little bit intuitive and it'll give you a little bit more of a sense of what the dot product and cross product are all about."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep it really general. So this is really any 2 by 2 matrix. What I want to do is use our technique for finding an inverse of this matrix to essentially find a formula for the inverse of a 2 by 2 matrix. So I want to essentially find A inverse. And I want to do it just using a formula that just applies to this matrix right here. So how can I do that? Well, we know a technique."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I want to essentially find A inverse. And I want to do it just using a formula that just applies to this matrix right here. So how can I do that? Well, we know a technique. We just create an augmented matrix. So let's just create an augmented matrix right here. So we have ABCD."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, we know a technique. We just create an augmented matrix. So let's just create an augmented matrix right here. So we have ABCD. And we augment it with the identity in R2. So 1, 0, 0, 1. And we know if we perform a series of row operations on this augmented matrix to get the left-hand side in reduced row echelon form, the right-hand side, if the reduced row echelon form here gets to the identity, then the right-hand side is going to be the inverse."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we have ABCD. And we augment it with the identity in R2. So 1, 0, 0, 1. And we know if we perform a series of row operations on this augmented matrix to get the left-hand side in reduced row echelon form, the right-hand side, if the reduced row echelon form here gets to the identity, then the right-hand side is going to be the inverse. So let's do it in this general case, not dealing with particular numbers here. So the first thing I want to do, or I would like to do, is I would like to zero this guy out. What we want to do is we want to zero that out, zero that out, and then these two terms have to become equal to 1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And we know if we perform a series of row operations on this augmented matrix to get the left-hand side in reduced row echelon form, the right-hand side, if the reduced row echelon form here gets to the identity, then the right-hand side is going to be the inverse. So let's do it in this general case, not dealing with particular numbers here. So the first thing I want to do, or I would like to do, is I would like to zero this guy out. What we want to do is we want to zero that out, zero that out, and then these two terms have to become equal to 1. So the best way to zero this out, let's perform a little transformation here. So if I perform the transformation on the columns, so those are the entries of a column. This would be one column right here."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "What we want to do is we want to zero that out, zero that out, and then these two terms have to become equal to 1. So the best way to zero this out, let's perform a little transformation here. So if I perform the transformation on the columns, so those are the entries of a column. This would be one column right here. That would be another column right there. That's the third column. That's the fourth column."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This would be one column right here. That would be another column right there. That's the third column. That's the fourth column. But the transformation I'm going to perform on each of these columns, and we know this is equivalent to a row operation, is going to be equal to, since I want to zero this one out, I'm going to keep my first row the same. So it's going to be C1. And I'm going to replace my second row with A times my second row minus C times my first row."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's the fourth column. But the transformation I'm going to perform on each of these columns, and we know this is equivalent to a row operation, is going to be equal to, since I want to zero this one out, I'm going to keep my first row the same. So it's going to be C1. And I'm going to replace my second row with A times my second row minus C times my first row. Now why am I doing that? Because A times C minus C times A is going to be 0. So this guy's going to be 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I'm going to replace my second row with A times my second row minus C times my first row. Now why am I doing that? Because A times C minus C times A is going to be 0. So this guy's going to be 0. So that's the row operation I'm going to perform. And I'm doing this so we can kind of keep track, account for what we're doing, because the algebra is going to get hairy in a little bit. So let me perform this operation."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this guy's going to be 0. So that's the row operation I'm going to perform. And I'm doing this so we can kind of keep track, account for what we're doing, because the algebra is going to get hairy in a little bit. So let me perform this operation. So if I perform that operation on our matrix, what do we have? So our first row is going to be the same. Let me start with our second row, because that's a little bit more complicated."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me perform this operation. So if I perform that operation on our matrix, what do we have? So our first row is going to be the same. Let me start with our second row, because that's a little bit more complicated. So I'm going to replace C with A times C minus C times A. That's AC. So let me put it this way."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me start with our second row, because that's a little bit more complicated. So I'm going to replace C with A times C minus C times A. That's AC. So let me put it this way. So that's going to be 0 right there. I'm going to replace D with D times A, or A times D. Let me write that, A times D minus C times C1 in this column vector, so minus C times B. Let me write this as BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me put it this way. So that's going to be 0 right there. I'm going to replace D with D times A, or A times D. Let me write that, A times D minus C times C1 in this column vector, so minus C times B. Let me write this as BC. And then let me augment it. And then this guy is going to be A times 0, because he's C2, minus C times C1. So it's going to be minus C. And then finally, this guy right here is going to be A times 1, A times this 1 right here, minus C times 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me write this as BC. And then let me augment it. And then this guy is going to be A times 0, because he's C2, minus C times C1. So it's going to be minus C. And then finally, this guy right here is going to be A times 1, A times this 1 right here, minus C times 0. So that's just going to be an A. And then the first row is pretty straightforward. We know that the first row, or the first entries in our column vectors, just stay the same through this transformation."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be minus C. And then finally, this guy right here is going to be A times 1, A times this 1 right here, minus C times 0. So that's just going to be an A. And then the first row is pretty straightforward. We know that the first row, or the first entries in our column vectors, just stay the same through this transformation. So it's A, B, 1, 0. And just to make sure it's clear what we're doing, when you perform this transformation on this column vector right here, you got this column vector right there. When you perform the transformation on this column vector right there, you get this column vector right there."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We know that the first row, or the first entries in our column vectors, just stay the same through this transformation. So it's A, B, 1, 0. And just to make sure it's clear what we're doing, when you perform this transformation on this column vector right here, you got this column vector right there. When you perform the transformation on this column vector right there, you get this column vector right there. And I just want to make that clear, because I did all of the second entries of all of the column vectors at once, because we all were essentially performing the same row operation. So that just helped me simplify at least my thinking a little bit, let me stay in this mode. So let's continue to get this in reduced row echelon form."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When you perform the transformation on this column vector right there, you get this column vector right there. And I just want to make that clear, because I did all of the second entries of all of the column vectors at once, because we all were essentially performing the same row operation. So that just helped me simplify at least my thinking a little bit, let me stay in this mode. So let's continue to get this in reduced row echelon form. The next thing we want to do is, let's make another transformation. We'll call this T1. That was our first transformation."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let's continue to get this in reduced row echelon form. The next thing we want to do is, let's make another transformation. We'll call this T1. That was our first transformation. Let's do another transformation, T2, or another set of row operations. So if I start with the column vector C1, C2, what I want to do now is, I want to keep my second row the same, and I want to zero out this character right here. I want to zero him out."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That was our first transformation. Let's do another transformation, T2, or another set of row operations. So if I start with the column vector C1, C2, what I want to do now is, I want to keep my second row the same, and I want to zero out this character right here. I want to zero him out. So I know I'm going to keep my second row the same, so C2 is just going to still be C2. But in order to zero this out, what I can do is, I can replace the first row with this scaling factor times the first row minus this scaling factor times the second row. So it'll be AD minus BC times your first entry in your column vector minus B times your second entry."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I want to zero him out. So I know I'm going to keep my second row the same, so C2 is just going to still be C2. But in order to zero this out, what I can do is, I can replace the first row with this scaling factor times the first row minus this scaling factor times the second row. So it'll be AD minus BC times your first entry in your column vector minus B times your second entry. And the whole reason why I'm doing that is so that this guy zeros out. So if we apply that to this matrix up here, let's do the first row first. So this first entry right here is going to be AD minus BC times A, because that's C1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it'll be AD minus BC times your first entry in your column vector minus B times your second entry. And the whole reason why I'm doing that is so that this guy zeros out. So if we apply that to this matrix up here, let's do the first row first. So this first entry right here is going to be AD minus BC times A, because that's C1. Let me write that down. So it's AD minus BC times A minus B minus B times C2 minus 0. So it's just going to be that."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this first entry right here is going to be AD minus BC times A, because that's C1. Let me write that down. So it's AD minus BC times A minus B minus B times C2 minus 0. So it's just going to be that. That second term just becomes 0. Fair enough. Now what is this guy going to be?"}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's just going to be that. That second term just becomes 0. Fair enough. Now what is this guy going to be? He's going to be, I'll write it out. It doesn't hurt to write it out. It's going to be AD minus BC times B minus B times your C2 in this column vector minus B times AD minus BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now what is this guy going to be? He's going to be, I'll write it out. It doesn't hurt to write it out. It's going to be AD minus BC times B minus B times your C2 in this column vector minus B times AD minus BC. And you can see immediately that these two guys are going to cancel out. And you're going to get a 0 there. And then we've got to augment it."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's going to be AD minus BC times B minus B times your C2 in this column vector minus B times AD minus BC. And you can see immediately that these two guys are going to cancel out. And you're going to get a 0 there. And then we've got to augment it. I want to make sure I don't run out of space. I should have started to the left a little bit more. So what's this guy going to be?"}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we've got to augment it. I want to make sure I don't run out of space. I should have started to the left a little bit more. So what's this guy going to be? Well, I'm going to have this guy times AD minus BC. Do it in pink. So you're going to have AD minus BC times 1, which is just AD minus BC, minus B times C2."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what's this guy going to be? Well, I'm going to have this guy times AD minus BC. Do it in pink. So you're going to have AD minus BC times 1, which is just AD minus BC, minus B times C2. So minus B times minus C. So that's plus BC. So 1 times AD minus BC minus B times minus C is equal to that. And you can immediately see that these two guys will cancel out."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you're going to have AD minus BC times 1, which is just AD minus BC, minus B times C2. So minus B times minus C. So that's plus BC. So 1 times AD minus BC minus B times minus C is equal to that. And you can immediately see that these two guys will cancel out. You're just going to have AD. And then this guy over here, you're going to have 0 times AD minus BC, which is just a 0, minus B times A. So you have minus AB."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And you can immediately see that these two guys will cancel out. You're just going to have AD. And then this guy over here, you're going to have 0 times AD minus BC, which is just a 0, minus B times A. So you have minus AB. Just squeezed it in there. And we know that our second row just stays the same. Our second row just stays the same in this transformation."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you have minus AB. Just squeezed it in there. And we know that our second row just stays the same. Our second row just stays the same in this transformation. So we had a 0 here. We're still going to have a 0. We had an AD minus BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our second row just stays the same in this transformation. So we had a 0 here. We're still going to have a 0. We had an AD minus BC. We'll still have an AD minus BC. We had a minus C. And we had an A, just like that. Now, let me rewrite this matrix, just so it gets cleaned up a little bit."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We had an AD minus BC. We'll still have an AD minus BC. We had a minus C. And we had an A, just like that. Now, let me rewrite this matrix, just so it gets cleaned up a little bit. Let me rewrite it right here. I'll do it in my orange. Well, let me do it in this yellow."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, let me rewrite this matrix, just so it gets cleaned up a little bit. Let me rewrite it right here. I'll do it in my orange. Well, let me do it in this yellow. OK. So I have AD minus BC times A. And then this term right here just became a 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, let me do it in this yellow. OK. So I have AD minus BC times A. And then this term right here just became a 0. This term right here is a 0. This term right here is an AD minus BC. And then in our augmented part, this part was just an AD."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then this term right here just became a 0. This term right here is a 0. This term right here is an AD minus BC. And then in our augmented part, this part was just an AD. This was a minus AB. This is a minus C. And then this is an A. Now, we're almost at reduced row echelon form right here."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then in our augmented part, this part was just an AD. This was a minus AB. This is a minus C. And then this is an A. Now, we're almost at reduced row echelon form right here. These two things just have to be equal to 1 in order to get reduced row echelon form. So let's define a transformation that will make both of these equal to 1. So if this was T2, let me define a transformation T3."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, we're almost at reduced row echelon form right here. These two things just have to be equal to 1 in order to get reduced row echelon form. So let's define a transformation that will make both of these equal to 1. So if this was T2, let me define a transformation T3. You give it a column vector, C1, C2. And it's just going to scale each of the column vectors. So what I want to do is I want to divide my first entries by this scaling factor right here so that this becomes a 1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if this was T2, let me define a transformation T3. You give it a column vector, C1, C2. And it's just going to scale each of the column vectors. So what I want to do is I want to divide my first entries by this scaling factor right here so that this becomes a 1. So I'm essentially going to multiply 1 over AD minus BC times A. So 1 over AD BC A times my first entry in each of my column vectors. And then my second one, I want to divide by this so that this guy becomes a 1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what I want to do is I want to divide my first entries by this scaling factor right here so that this becomes a 1. So I'm essentially going to multiply 1 over AD minus BC times A. So 1 over AD BC A times my first entry in each of my column vectors. And then my second one, I want to divide by this so that this guy becomes a 1. So I'm doing two scalar divisions in kind of one transformation. So this one's going to be 1 over AD minus BC times C2. So I'm just scaling everything by these two scaling factors."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then my second one, I want to divide by this so that this guy becomes a 1. So I'm doing two scalar divisions in kind of one transformation. So this one's going to be 1 over AD minus BC times C2. So I'm just scaling everything by these two scaling factors. So if you apply this transformation to that right there, what do we get? We get a matrix. And this guy, I'm going to divide him by AD minus BC times A."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So I'm just scaling everything by these two scaling factors. So if you apply this transformation to that right there, what do we get? We get a matrix. And this guy, I'm going to divide him by AD minus BC times A. So I'm dividing it by itself. So that guy's going to be 1. I'm going to divide 0 by this, but 0 divided by anything is just 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this guy, I'm going to divide him by AD minus BC times A. So I'm dividing it by itself. So that guy's going to be 1. I'm going to divide 0 by this, but 0 divided by anything is just 0. And then we're in our augmented part. AD divided by, so let me write it like this. So AD, I'm going to divide by this."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to divide 0 by this, but 0 divided by anything is just 0. And then we're in our augmented part. AD divided by, so let me write it like this. So AD, I'm going to divide by this. So it's going to be AD minus BC times A. You immediately see that the A's cancel out. This is going to be minus AB divided by AD minus BC times A."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So AD, I'm going to divide by this. So it's going to be AD minus BC times A. You immediately see that the A's cancel out. This is going to be minus AB divided by AD minus BC times A. Once again, the A's cancel out. And then in my second row, or my second entries in my column vectors, 0 divided by anything is 0. So 0 divided by this thing is going to be 0, assuming we can divide by that."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This is going to be minus AB divided by AD minus BC times A. Once again, the A's cancel out. And then in my second row, or my second entries in my column vectors, 0 divided by anything is 0. So 0 divided by this thing is going to be 0, assuming we can divide by that. And we're going to talk about that in a second. This guy divided by this guy, we're just dividing by himself. So it's going to be equal to 1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So 0 divided by this thing is going to be 0, assuming we can divide by that. And we're going to talk about that in a second. This guy divided by this guy, we're just dividing by himself. So it's going to be equal to 1. Now we have minus C divided by this. So our AD minus BC. And then we have an A divided by AD minus BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's going to be equal to 1. Now we have minus C divided by this. So our AD minus BC. And then we have an A divided by AD minus BC. And we're done. We put the left-hand side of our augmented matrix into reduced row echelon form. And now this is going to be our inverse."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we have an A divided by AD minus BC. And we're done. We put the left-hand side of our augmented matrix into reduced row echelon form. And now this is going to be our inverse. And let me clean it up a little bit. So so far we said we started off with a matrix. I'll do it in purple."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And now this is going to be our inverse. And let me clean it up a little bit. So so far we said we started off with a matrix. I'll do it in purple. We started off with a matrix A is equal to ABCD. And now just using our technique, we figured out that A inverse is equal to this thing right here. And just to simplify it, well let me just write it the way I have it there."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I'll do it in purple. We started off with a matrix A is equal to ABCD. And now just using our technique, we figured out that A inverse is equal to this thing right here. And just to simplify it, well let me just write it the way I have it there. I don't want to skip any steps. This is equal to D over AD minus BC. Right?"}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And just to simplify it, well let me just write it the way I have it there. I don't want to skip any steps. This is equal to D over AD minus BC. Right? This guy and that guy canceled out. And then we have a minus B over AD minus BC, because that guy and that guy canceled out. Then you have a minus C over AD minus BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Right? This guy and that guy canceled out. And then we have a minus B over AD minus BC, because that guy and that guy canceled out. Then you have a minus C over AD minus BC. And then finally you have an A over AD minus BC, which is our inverse. But one thing I might point out, just pop out at you immediately, is that everything in our inverse is being divided by this. So maybe an easier way to write our inverse."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then you have a minus C over AD minus BC. And then finally you have an A over AD minus BC, which is our inverse. But one thing I might point out, just pop out at you immediately, is that everything in our inverse is being divided by this. So maybe an easier way to write our inverse. We could also write our inverse like this. We could just write it as 1 over AD minus BC times the matrix D minus B minus C and A. And just like that, we have come up with a formula for the inverse of a 2 by 2 matrix."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So maybe an easier way to write our inverse. We could also write our inverse like this. We could just write it as 1 over AD minus BC times the matrix D minus B minus C and A. And just like that, we have come up with a formula for the inverse of a 2 by 2 matrix. You give me any real numbers here, and I'm going to give you its inverse. That's straightforward. Now one thing you might be saying, hey, but not all 2 by 2 matrices are invertible."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And just like that, we have come up with a formula for the inverse of a 2 by 2 matrix. You give me any real numbers here, and I'm going to give you its inverse. That's straightforward. Now one thing you might be saying, hey, but not all 2 by 2 matrices are invertible. How can this be the case for all of them? And then I'll give you a question. When will this thing right here not be defined?"}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now one thing you might be saying, hey, but not all 2 by 2 matrices are invertible. How can this be the case for all of them? And then I'll give you a question. When will this thing right here not be defined? When is this thing not defined? Every operation I did, I can do with any real numbers. And this applies to any real numbers."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "When will this thing right here not be defined? When is this thing not defined? Every operation I did, I can do with any real numbers. And this applies to any real numbers. But when is this thing not defined? Well, it's not defined when I divide by 0. And when would I divide by 0?"}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this applies to any real numbers. But when is this thing not defined? Well, it's not defined when I divide by 0. And when would I divide by 0? Everything else you can multiply and subtract and add 0 to anything. But you just can't divide by 0. We've never defined what it means when you divide something by 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And when would I divide by 0? Everything else you can multiply and subtract and add 0 to anything. But you just can't divide by 0. We've never defined what it means when you divide something by 0. So it's not defined if AD minus BC is equal to 0. So this is an interesting thing. I can always find the inverse of a 2 by 2 matrix as long as AD minus BC is not equal to 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We've never defined what it means when you divide something by 0. So it's not defined if AD minus BC is equal to 0. So this is an interesting thing. I can always find the inverse of a 2 by 2 matrix as long as AD minus BC is not equal to 0. I mean, we came up with all of these fancy things for invertibility. You've got to put it into reduced row echelon form. And before that, we talked about it being on 2 and 1 to 1."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "I can always find the inverse of a 2 by 2 matrix as long as AD minus BC is not equal to 0. I mean, we came up with all of these fancy things for invertibility. You've got to put it into reduced row echelon form. And before that, we talked about it being on 2 and 1 to 1. But for at least a 2 by 2 matrix, we've really simplified things, as long as AD minus BC does not equal 0, we can use this formula. And then we know that A, and it goes both ways, A is invertible. And not only is it invertible, but we can just apply this formula to it."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And before that, we talked about it being on 2 and 1 to 1. But for at least a 2 by 2 matrix, we've really simplified things, as long as AD minus BC does not equal 0, we can use this formula. And then we know that A, and it goes both ways, A is invertible. And not only is it invertible, but we can just apply this formula to it. So immediately, something interesting might, you know, you might say, hey, this is an interesting number. We should come up with some name for it. And lucky for us, we have come up with a name for it."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And not only is it invertible, but we can just apply this formula to it. So immediately, something interesting might, you know, you might say, hey, this is an interesting number. We should come up with some name for it. And lucky for us, we have come up with a name for it. This is called the determinant. Let me write it in pink. So the determinant of A, and it's also written like this with these little straight lines around A."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And lucky for us, we have come up with a name for it. This is called the determinant. Let me write it in pink. So the determinant of A, and it's also written like this with these little straight lines around A. And you could also write it like this. ABCD. But most people kind of think this is redundant to have brackets and these lines."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of A, and it's also written like this with these little straight lines around A. And you could also write it like this. ABCD. But most people kind of think this is redundant to have brackets and these lines. So then they just write it like this. This is equal to just, they just write the lines, ABCD. And I want to make this very clear."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But most people kind of think this is redundant to have brackets and these lines. So then they just write it like this. This is equal to just, they just write the lines, ABCD. And I want to make this very clear. If you have the brackets, you're dealing with a matrix. If you have these straight lines, you're talking about the determinant of the matrix. But this is defined for the 2 by 2 case to be equal to AD minus BC."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I want to make this very clear. If you have the brackets, you're dealing with a matrix. If you have these straight lines, you're talking about the determinant of the matrix. But this is defined for the 2 by 2 case to be equal to AD minus BC. This is a definition. This is a definition. Definition of the determinant."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But this is defined for the 2 by 2 case to be equal to AD minus BC. This is a definition. This is a definition. Definition of the determinant. So we can rewrite if we have some matrix here. We have some matrix A, which is equal to ABCD. We can now write its inverse."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Definition of the determinant. So we can rewrite if we have some matrix here. We have some matrix A, which is equal to ABCD. We can now write its inverse. A inverse is equal to 1 over this thing, which we've defined as the determinant of A. The determinant of A times, and let's just see a good way of kind of memorizing this. We're swapping these two guys."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can now write its inverse. A inverse is equal to 1 over this thing, which we've defined as the determinant of A. The determinant of A times, and let's just see a good way of kind of memorizing this. We're swapping these two guys. The A and the D get swapped. So you get a D and an A. And then these two guys stay the same, they just become negative."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We're swapping these two guys. The A and the D get swapped. So you get a D and an A. And then these two guys stay the same, they just become negative. So minus B and minus C. So that's the general formula for the determinant of a 2 by 2 matrix. Let's try to do a couple. Let's try to find the determinant of the matrix 1, 2, 3, 4."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then these two guys stay the same, they just become negative. So minus B and minus C. So that's the general formula for the determinant of a 2 by 2 matrix. Let's try to do a couple. Let's try to find the determinant of the matrix 1, 2, 3, 4. Easy enough. So the determinant of, let's say this is the matrix B. So the determinant of B, or we could write it like that, that's equal to the determinant of B."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's try to find the determinant of the matrix 1, 2, 3, 4. Easy enough. So the determinant of, let's say this is the matrix B. So the determinant of B, or we could write it like that, that's equal to the determinant of B. That is just equal to, that's this thing right here, 1 times 4 minus 3 times 2, which is equal to 4 minus 6, which is equal to minus 2. So the determinant is minus 2. So this is invertible."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the determinant of B, or we could write it like that, that's equal to the determinant of B. That is just equal to, that's this thing right here, 1 times 4 minus 3 times 2, which is equal to 4 minus 6, which is equal to minus 2. So the determinant is minus 2. So this is invertible. Not only is it invertible, but it's very easy to find its inverse now. We can just apply this formula. The inverse of B in this case, let me do it in this color, B inverse is equal to 1 over the determinant, so it's 1 over minus 2, times the matrix where we swap, well this is the determinant of B. I want to be careful."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this is invertible. Not only is it invertible, but it's very easy to find its inverse now. We can just apply this formula. The inverse of B in this case, let me do it in this color, B inverse is equal to 1 over the determinant, so it's 1 over minus 2, times the matrix where we swap, well this is the determinant of B. I want to be careful. B is the same thing, but with brackets. 1, 2, 3, 4. So B inverse is going to be 1 over the determinant of B, which is equal to minus 2."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "The inverse of B in this case, let me do it in this color, B inverse is equal to 1 over the determinant, so it's 1 over minus 2, times the matrix where we swap, well this is the determinant of B. I want to be careful. B is the same thing, but with brackets. 1, 2, 3, 4. So B inverse is going to be 1 over the determinant of B, which is equal to minus 2. So 1 over minus 2. We swap these two guys, so we get a 4 and a 1, and then these two guys become negative. Minus 2 and then minus 3."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So B inverse is going to be 1 over the determinant of B, which is equal to minus 2. So 1 over minus 2. We swap these two guys, so we get a 4 and a 1, and then these two guys become negative. Minus 2 and then minus 3. And then if we were to multiply this out, it would be equal to minus 1 half times 4 is minus 2. Minus 1 half times minus 2 is 1. Minus 1 half times minus 3 is 3 halves."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 2 and then minus 3. And then if we were to multiply this out, it would be equal to minus 1 half times 4 is minus 2. Minus 1 half times minus 2 is 1. Minus 1 half times minus 3 is 3 halves. Minus 1 half times 1 is minus 1 half. So that there is the inverse of B. Let's say we have another matrix."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Minus 1 half times minus 3 is 3 halves. Minus 1 half times 1 is minus 1 half. So that there is the inverse of B. Let's say we have another matrix. Let's say we have the matrix C. And C is equal to 1, 2, 3, 6. What is the determinant of C? We could write it this way."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have another matrix. Let's say we have the matrix C. And C is equal to 1, 2, 3, 6. What is the determinant of C? We could write it this way. 1, 2, 3, 6. And it is equal to 1 times 6 minus 3 times 2, which is equal to 6 minus 6, which is equal to 0. And there you see it's equal to 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We could write it this way. 1, 2, 3, 6. And it is equal to 1 times 6 minus 3 times 2, which is equal to 6 minus 6, which is equal to 0. And there you see it's equal to 0. And so you cannot find, so this is not invertible. Not invertible. So we can't find its inverse because if we were to try to apply this formula right here, you'd have a 1 over 0."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And there you see it's equal to 0. And so you cannot find, so this is not invertible. Not invertible. So we can't find its inverse because if we were to try to apply this formula right here, you'd have a 1 over 0. But we know this formula just comes out of that attempt to put it into reduced row echelon form. And in that last step, we just had to essentially divide everything by these terms. So these terms would be 0 in this matrix C that I just constructed for you."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can't find its inverse because if we were to try to apply this formula right here, you'd have a 1 over 0. But we know this formula just comes out of that attempt to put it into reduced row echelon form. And in that last step, we just had to essentially divide everything by these terms. So these terms would be 0 in this matrix C that I just constructed for you. And the reason why I knew, I just pulled this out of my brain. I knew this wasn't going to be invertible because I constructed a situation where I have columns that are linear combinations of each other. I have 1, 3."}, {"video_title": "Formula for 2x2 inverse Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So these terms would be 0 in this matrix C that I just constructed for you. And the reason why I knew, I just pulled this out of my brain. I knew this wasn't going to be invertible because I constructed a situation where I have columns that are linear combinations of each other. I have 1, 3. You multiply that by 2, you get 2 and 6. So I knew that these aren't linearly independent columns. So you know that its rank wasn't going to be equal to."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I have here three linear equations of four unknowns. And like the first video where I talked about reduced row echelon form and solving systems of linear equations using augmented matrices, at least my gut feeling says, look, I have fewer equations than variables, so I probably won't be able to constrain this enough. Or maybe I'll have an infinite number of solutions. But let's see if I'm right. So let's construct the augmented matrix for this system of equations. So my coefficients on the x1 terms are 1, 1, and 2. Coefficients on the x2 are 2, 2, and 4."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's see if I'm right. So let's construct the augmented matrix for this system of equations. So my coefficients on the x1 terms are 1, 1, and 2. Coefficients on the x2 are 2, 2, and 4. Coefficients on the x3 are 1, 2, and 0. There's, of course, no x3 term, so we can view it as a 0 coefficient. Coefficients on the x4 are 1, minus 1, and 6."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Coefficients on the x2 are 2, 2, and 4. Coefficients on the x3 are 1, 2, and 0. There's, of course, no x3 term, so we can view it as a 0 coefficient. Coefficients on the x4 are 1, minus 1, and 6. And then on the right-hand side of the equal sign, I have 8, 12, and 4. There's my augmented matrix. Now let's put this guy into reduced row echelon form."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Coefficients on the x4 are 1, minus 1, and 6. And then on the right-hand side of the equal sign, I have 8, 12, and 4. There's my augmented matrix. Now let's put this guy into reduced row echelon form. So the first thing I want to do is I want to zero out these two rows right here. So what can we do? So I'm going to keep my first row the same for now."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now let's put this guy into reduced row echelon form. So the first thing I want to do is I want to zero out these two rows right here. So what can we do? So I'm going to keep my first row the same for now. So it's 1, 2, 1, 1, 8. And that line essentially represents my equal sign. And if I, let's see, what I can do is let me replace the second row with the second row minus the first row."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I'm going to keep my first row the same for now. So it's 1, 2, 1, 1, 8. And that line essentially represents my equal sign. And if I, let's see, what I can do is let me replace the second row with the second row minus the first row. So 1 minus 1 is 0. 2 minus 2 is 0. 2 minus 1 is 1."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if I, let's see, what I can do is let me replace the second row with the second row minus the first row. So 1 minus 1 is 0. 2 minus 2 is 0. 2 minus 1 is 1. Negative 1 minus 1 is minus 2. And then 12 minus 8 is 4. There you go."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "2 minus 1 is 1. Negative 1 minus 1 is minus 2. And then 12 minus 8 is 4. There you go. That looks good so far. It looks like column, or x2, which is represented by column 2, looks like it might be a free variable, but we're not 100% sure yet. Let's do all of our rows."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "There you go. That looks good so far. It looks like column, or x2, which is represented by column 2, looks like it might be a free variable, but we're not 100% sure yet. Let's do all of our rows. So let's take, let's, to get rid of this guy right here, let's replace our third equation with our third equation minus 2 times our first equation. So we get 2 minus 2 times 1 is 0. 4 minus 2 times 2, well that's 0."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's do all of our rows. So let's take, let's, to get rid of this guy right here, let's replace our third equation with our third equation minus 2 times our first equation. So we get 2 minus 2 times 1 is 0. 4 minus 2 times 2, well that's 0. 0 minus 2 times 1, that's minus 2. 6 minus 2 times 1, well that's 4, right? 6 minus 2."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4 minus 2 times 2, well that's 0. 0 minus 2 times 1, that's minus 2. 6 minus 2 times 1, well that's 4, right? 6 minus 2. And then 4 minus 2 times 8. Minus 2 times 8 is minus 16, 4 minus 16 is minus 12. Now what can we do?"}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "6 minus 2. And then 4 minus 2 times 8. Minus 2 times 8 is minus 16, 4 minus 16 is minus 12. Now what can we do? Well, let's see if we can get rid of this minus 2 term right there. So let me rewrite my augmented matrix. I'm going to keep row 2 the same this time."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now what can we do? Well, let's see if we can get rid of this minus 2 term right there. So let me rewrite my augmented matrix. I'm going to keep row 2 the same this time. So I get a 0, 0, 1, minus 2. And essentially my equal sign, or the augmented part of the matrix. And now, let's see what I can do."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm going to keep row 2 the same this time. So I get a 0, 0, 1, minus 2. And essentially my equal sign, or the augmented part of the matrix. And now, let's see what I can do. Well, actually let me get rid of this 0 up here, because I want to get it in a reduced row echelon form. So any of my pivot entries, which are always going to have the coefficient 1 or the entry 1, it should be the only non-zero term in my row. So how do I get rid of this 1 here?"}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And now, let's see what I can do. Well, actually let me get rid of this 0 up here, because I want to get it in a reduced row echelon form. So any of my pivot entries, which are always going to have the coefficient 1 or the entry 1, it should be the only non-zero term in my row. So how do I get rid of this 1 here? Well, I can subtract, I can replace row 1 with row 1 minus row 2. So 1 minus 0 is 1, 2 minus 0 is 2, 1 minus 1 is 0, 1 minus minus 2, that's 1 plus 2, which is 3. And then 8 minus 4 is 4."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So how do I get rid of this 1 here? Well, I can subtract, I can replace row 1 with row 1 minus row 2. So 1 minus 0 is 1, 2 minus 0 is 2, 1 minus 1 is 0, 1 minus minus 2, that's 1 plus 2, which is 3. And then 8 minus 4 is 4. There you go. And now how can I get rid of this guy? Well, let me replace row 3 with row 3 plus 2 times row 1."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then 8 minus 4 is 4. There you go. And now how can I get rid of this guy? Well, let me replace row 3 with row 3 plus 2 times row 1. Sorry, with row 3 plus 2 times row 2, right? Because I need to have minus 2 plus 2 times this, and they'd cancel out. So let's see, the 0's, 0 plus 2 times 0, that's 0."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, let me replace row 3 with row 3 plus 2 times row 1. Sorry, with row 3 plus 2 times row 2, right? Because I need to have minus 2 plus 2 times this, and they'd cancel out. So let's see, the 0's, 0 plus 2 times 0, that's 0. 0 plus 2 times 0, that's 0. Minus 2 plus 2 times 1 is 0. 4 plus 2 times minus 2, that's 4 minus 4, that's 0."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's see, the 0's, 0 plus 2 times 0, that's 0. 0 plus 2 times 0, that's 0. Minus 2 plus 2 times 1 is 0. 4 plus 2 times minus 2, that's 4 minus 4, that's 0. And then I have minus 12 plus 2 times 4. That's minus 12 plus 8, that's minus 4. Now this is interesting right now."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "4 plus 2 times minus 2, that's 4 minus 4, that's 0. And then I have minus 12 plus 2 times 4. That's minus 12 plus 8, that's minus 4. Now this is interesting right now. I've essentially put this in reduced row echelon form. I have two pivot entries. That's a pivot entry right there."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now this is interesting right now. I've essentially put this in reduced row echelon form. I have two pivot entries. That's a pivot entry right there. They're the only non-zero term in their respective columns. And this is just kind of a style issue, but this pivot entry is in a lower row than that one. So it's to the right."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That's a pivot entry right there. They're the only non-zero term in their respective columns. And this is just kind of a style issue, but this pivot entry is in a lower row than that one. So it's to the right. It's in a column to the right of this one right there. And when I just inspect it, this looks like a free variable, there's no pivot entry there. But let's see, let's map this back to our system of equations."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it's to the right. It's in a column to the right of this one right there. And when I just inspect it, this looks like a free variable, there's no pivot entry there. But let's see, let's map this back to our system of equations. These are just numbers to me, and I just kind of mechanically, almost like a computer, put this in reduced row echelon form. Actually, almost exactly like a computer. But let me put it back to my system of linear equations to see what our result is."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's see, let's map this back to our system of equations. These are just numbers to me, and I just kind of mechanically, almost like a computer, put this in reduced row echelon form. Actually, almost exactly like a computer. But let me put it back to my system of linear equations to see what our result is. So we get 1 times x1. Let me write it in yellow. So I get 1 times x1 plus 2 times x2 plus 0 times x3 plus 3 times x4 is equal to 4."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let me put it back to my system of linear equations to see what our result is. So we get 1 times x1. Let me write it in yellow. So I get 1 times x1 plus 2 times x2 plus 0 times x3 plus 3 times x4 is equal to 4. Obviously, I could ignore this term right there. I didn't even have to write it. Then, actually, I'm not going to write that."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So I get 1 times x1 plus 2 times x2 plus 0 times x3 plus 3 times x4 is equal to 4. Obviously, I could ignore this term right there. I didn't even have to write it. Then, actually, I'm not going to write that. Then I get 0 times x1 plus 0 times x2 plus 1 times x3. So I could just write that. Well, I'll just write the 1."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then, actually, I'm not going to write that. Then I get 0 times x1 plus 0 times x2 plus 1 times x3. So I could just write that. Well, I'll just write the 1. 1 times x3 minus 2 times x4 is equal to 4. And then this last term, what do I get? I get 0 x1 plus 0 x2 plus 0 x3 plus 0 x4."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I'll just write the 1. 1 times x3 minus 2 times x4 is equal to 4. And then this last term, what do I get? I get 0 x1 plus 0 x2 plus 0 x3 plus 0 x4. Well, all of that's equal to 0, and I've got to write something on the left-hand side. So let me just write a 0, and that's got to be equal to minus 4. Well, this doesn't make any sense."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I get 0 x1 plus 0 x2 plus 0 x3 plus 0 x4. Well, all of that's equal to 0, and I've got to write something on the left-hand side. So let me just write a 0, and that's got to be equal to minus 4. Well, this doesn't make any sense. Whatsoever. 0 equals minus 4. This is a nonsensical constraint."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this doesn't make any sense. Whatsoever. 0 equals minus 4. This is a nonsensical constraint. This is impossible. 0 can never equal minus 4. This is impossible."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a nonsensical constraint. This is impossible. 0 can never equal minus 4. This is impossible. Which means that it is essentially impossible to find an intersection of these three systems of equations, or a solution set that satisfies all of them. So when we looked at this initially, at the beginning of the video, we said, oh, there's only three equations that are four unknowns. Maybe there's going to be an infinite set of solutions."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is impossible. Which means that it is essentially impossible to find an intersection of these three systems of equations, or a solution set that satisfies all of them. So when we looked at this initially, at the beginning of the video, we said, oh, there's only three equations that are four unknowns. Maybe there's going to be an infinite set of solutions. But it turns out that these three, I guess you can call them, surfaces don't intersect in R4. These are all four-dimensional. We're dealing in R4 right here because I guess each vector has four components, or we have four variables, I guess, is the way you could think about it."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe there's going to be an infinite set of solutions. But it turns out that these three, I guess you can call them, surfaces don't intersect in R4. These are all four-dimensional. We're dealing in R4 right here because I guess each vector has four components, or we have four variables, I guess, is the way you could think about it. And it's always hard to visualize things in R4, but if we were doing things in R3, we can imagine a situation where, you know, let's say we had two planes in R3. So that's one plane right there. And then I had another completely parallel plane to that one."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We're dealing in R4 right here because I guess each vector has four components, or we have four variables, I guess, is the way you could think about it. And it's always hard to visualize things in R4, but if we were doing things in R3, we can imagine a situation where, you know, let's say we had two planes in R3. So that's one plane right there. And then I had another completely parallel plane to that one. So I had another completely parallel plane to that first one. Even though these would be two planes in R3, so let me give an example. So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then I had another completely parallel plane to that one. So I had another completely parallel plane to that first one. Even though these would be two planes in R3, so let me give an example. So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5. And the second plane was represented by the equation 3x plus 6y plus 9z is equal to 2. These two planes in R3, this is the case of R3, so this is R3 right here. These two planes clearly, they'll never intersect because obviously this one has the same coefficients adding up to 5."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5. And the second plane was represented by the equation 3x plus 6y plus 9z is equal to 2. These two planes in R3, this is the case of R3, so this is R3 right here. These two planes clearly, they'll never intersect because obviously this one has the same coefficients adding up to 5. This one has the same coefficients adding up to 2. And if we just looked at this initially, if it wasn't so obvious, we said, oh, we have only two equations with three unknowns. Maybe this has an infinite set of solutions."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These two planes clearly, they'll never intersect because obviously this one has the same coefficients adding up to 5. This one has the same coefficients adding up to 2. And if we just looked at this initially, if it wasn't so obvious, we said, oh, we have only two equations with three unknowns. Maybe this has an infinite set of solutions. But it won't be the case because you can actually just subtract this equation from the bottom equation from the top equation, and what do you get? You would get a very familiar, so if you just subtract the bottom equation from the top equation, you get 3x minus 3x, 6y minus 6y, 9z minus 9z. Actually, let me do it right here."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Maybe this has an infinite set of solutions. But it won't be the case because you can actually just subtract this equation from the bottom equation from the top equation, and what do you get? You would get a very familiar, so if you just subtract the bottom equation from the top equation, you get 3x minus 3x, 6y minus 6y, 9z minus 9z. Actually, let me do it right here. So that minus that, you get 0 is equal to 5 minus 2, which is 3, which is a very similar result that we got up there. So when you have two parallel planes, in this case in R3, or really do any kind of two parallel equations or a set of parallel equations, they won't intersect and you're going to get, when you either put in reduced row echelon form or you just do basic elimination or you solve the system, you're going to get a statement that 0 is equal to something, and that means that there are no solutions. So the general takeaway, if you have 0 equals something, no solutions."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Actually, let me do it right here. So that minus that, you get 0 is equal to 5 minus 2, which is 3, which is a very similar result that we got up there. So when you have two parallel planes, in this case in R3, or really do any kind of two parallel equations or a set of parallel equations, they won't intersect and you're going to get, when you either put in reduced row echelon form or you just do basic elimination or you solve the system, you're going to get a statement that 0 is equal to something, and that means that there are no solutions. So the general takeaway, if you have 0 equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations, let me write this down. This is good to know. If you have 0 is equal to anything, then that means no solution."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the general takeaway, if you have 0 equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations, let me write this down. This is good to know. If you have 0 is equal to anything, then that means no solution. If you're dealing in R3, you're probably dealing with parallel planes. In R2, you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, and this is a case of R4 maybe right there."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you have 0 is equal to anything, then that means no solution. If you're dealing in R3, you're probably dealing with parallel planes. In R2, you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, and this is a case of R4 maybe right there. I think you get the idea. That equals A, B, C, D, then you have a unique solution. Now, if you have any free variables, so free variables look like this."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, and this is a case of R4 maybe right there. I think you get the idea. That equals A, B, C, D, then you have a unique solution. Now, if you have any free variables, so free variables look like this. Let's say I have 1, 0, 1, 0, and then I have the entry 1, let me be careful, 0, let me do it like this, 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of 0's over here, and then this has to equal 0. Remember, if this was a bunch of 0's equaling some variable, then I would have no solution or equaling some constant. Let's say this is equal to 5, this is equal to 2."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, if you have any free variables, so free variables look like this. Let's say I have 1, 0, 1, 0, and then I have the entry 1, let me be careful, 0, let me do it like this, 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of 0's over here, and then this has to equal 0. Remember, if this was a bunch of 0's equaling some variable, then I would have no solution or equaling some constant. Let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free column to some degree. This one would be 2, because it has no pivot entries."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free column to some degree. This one would be 2, because it has no pivot entries. These are the pivot entries. So this is, let's say, variable x2, and that's variable x4. Then these would be free."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This one would be 2, because it has no pivot entries. These are the pivot entries. So this is, let's say, variable x2, and that's variable x4. Then these would be free. We can set them equal to anything. So then here we have unlimited solutions or no unique solutions, and that was actually the first example we saw, and these are really the three cases that you're going to see every time. And it's good to get familiar with them, so you're never going to get stumped up when you have something like 0 equals minus 4 or 0 equals 3."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Then these would be free. We can set them equal to anything. So then here we have unlimited solutions or no unique solutions, and that was actually the first example we saw, and these are really the three cases that you're going to see every time. And it's good to get familiar with them, so you're never going to get stumped up when you have something like 0 equals minus 4 or 0 equals 3. Or if you have just a bunch of 0's and a bunch of rows. But I want to make that very clear. Sometimes you see a bunch of 0's here on the left-hand side of kind of the augmented divide, and you might say, oh, maybe I have no unique solution."}, {"video_title": "Matrices Reduced row echelon form 3 Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And it's good to get familiar with them, so you're never going to get stumped up when you have something like 0 equals minus 4 or 0 equals 3. Or if you have just a bunch of 0's and a bunch of rows. But I want to make that very clear. Sometimes you see a bunch of 0's here on the left-hand side of kind of the augmented divide, and you might say, oh, maybe I have no unique solution. I have an infinite number of solutions. But you've got to look at this entry right here. Only if this whole thing is 0 and you have free variables, then you have an infinite number of solutions."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "When you get into higher mathematics, you might see a professor write something like this on a board, where it's this R with this extra backbone right over here. And maybe they write R2. Or if you're looking at it in a book, it might just be a bolded capital R with a 2 superscript like this. And if you see this, they're referring to the two-dimensional real coordinate space, which sounds very fancy, but one way to think about it, it's really just the two-dimensional space that you're used to dealing with in your coordinate plane. To go a little bit more abstract, this isn't necessarily this visual representation. This visual representation is one way to think about this real coordinate space. If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space, let me write this down, and the 2, the two-dimensional real coordinate space."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if you see this, they're referring to the two-dimensional real coordinate space, which sounds very fancy, but one way to think about it, it's really just the two-dimensional space that you're used to dealing with in your coordinate plane. To go a little bit more abstract, this isn't necessarily this visual representation. This visual representation is one way to think about this real coordinate space. If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space, let me write this down, and the 2, the two-dimensional real coordinate space. And just to break down the notation, the 2 tells us how many dimensions we're dealing with, and the R tells us this is a real coordinate space. The two-dimensional real coordinate space is all the possible real-valued 2-tuples. Let me write that down."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space, let me write this down, and the 2, the two-dimensional real coordinate space. And just to break down the notation, the 2 tells us how many dimensions we're dealing with, and the R tells us this is a real coordinate space. The two-dimensional real coordinate space is all the possible real-valued 2-tuples. Let me write that down. This is all possible real-valued 2-tuples. So what is a 2-tuple? Well, a tuple is an ordered list of numbers."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write that down. This is all possible real-valued 2-tuples. So what is a 2-tuple? Well, a tuple is an ordered list of numbers. And since we're talking about real values, it's going to be an ordered list of real numbers. And a 2-tuple just says it's an ordered list of 2 numbers. So this is an ordered list of 2 real-valued numbers."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, a tuple is an ordered list of numbers. And since we're talking about real values, it's going to be an ordered list of real numbers. And a 2-tuple just says it's an ordered list of 2 numbers. So this is an ordered list of 2 real-valued numbers. Well, that's exactly what we did here when we thought about a two-dimensional vector. This right over here is a 2-tuple. And this is a real-valued 2-tuple."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this is an ordered list of 2 real-valued numbers. Well, that's exactly what we did here when we thought about a two-dimensional vector. This right over here is a 2-tuple. And this is a real-valued 2-tuple. Neither of these have any imaginary parts. So you have a 3 and a 4. Order matters."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And this is a real-valued 2-tuple. Neither of these have any imaginary parts. So you have a 3 and a 4. Order matters. We view this as a different 2-tuple than, say, 4, 3. And even if we were to try to represent them in our axis right over here, this vector, 4, 3, would be 4 along the horizontal axis and then 3 along the vertical axis. And so it would look something like this."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Order matters. We view this as a different 2-tuple than, say, 4, 3. And even if we were to try to represent them in our axis right over here, this vector, 4, 3, would be 4 along the horizontal axis and then 3 along the vertical axis. And so it would look something like this. And remember, we don't have to draw it just over there. We just care about its magnitude and direction. We could draw it right over here."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And so it would look something like this. And remember, we don't have to draw it just over there. We just care about its magnitude and direction. We could draw it right over here. That this would also be 4, 3, the column vector 4, 3. So when we're talking about R2, we're talking about all of the possible real-valued 2-tuples. So all of the possible vectors that you can have where each of its components, and the components are these numbers right over here, where each of its components are a real number."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We could draw it right over here. That this would also be 4, 3, the column vector 4, 3. So when we're talking about R2, we're talking about all of the possible real-valued 2-tuples. So all of the possible vectors that you can have where each of its components, and the components are these numbers right over here, where each of its components are a real number. So you might have 3, 4. You could have negative 3, negative 4. So that would be 1, 2, 3."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So all of the possible vectors that you can have where each of its components, and the components are these numbers right over here, where each of its components are a real number. So you might have 3, 4. You could have negative 3, negative 4. So that would be 1, 2, 3. 1, 2, 3, 4. Might look something like, actually, I should make the scale a little bit bigger, that so it looks the same. 1, 2, 3, 4."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So that would be 1, 2, 3. 1, 2, 3, 4. Might look something like, actually, I should make the scale a little bit bigger, that so it looks the same. 1, 2, 3, 4. So it might look something like that. So that would be the vector negative 3. Let me write a little bit better than that."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "1, 2, 3, 4. So it might look something like that. So that would be the vector negative 3. Let me write a little bit better than that. Negative 3 and negative 4. So if you were to take all of the possible 2-tuples, including the vector 0, 0, so it has no magnitude and you could debate what its direction is right over there. You take all of those combined, and then you have created your two-dimensional real coordinate space."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me write a little bit better than that. Negative 3 and negative 4. So if you were to take all of the possible 2-tuples, including the vector 0, 0, so it has no magnitude and you could debate what its direction is right over there. You take all of those combined, and then you have created your two-dimensional real coordinate space. And that is referred to as R2. Now, as you can imagine, the fact that we wrote this 2 here, we had to specify. You're just like, hey, well, can I put a 3 there?"}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You take all of those combined, and then you have created your two-dimensional real coordinate space. And that is referred to as R2. Now, as you can imagine, the fact that we wrote this 2 here, we had to specify. You're just like, hey, well, can I put a 3 there? And I would say, absolutely, you could put a 3 there. So R3 would be the three-dimensional real coordinate space. So 3D real coordinate space."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You're just like, hey, well, can I put a 3 there? And I would say, absolutely, you could put a 3 there. So R3 would be the three-dimensional real coordinate space. So 3D real coordinate space. And so you would view this as all the possible real-valued 3-tuples. So real-valued 3-tuples. So for example, that would be a member of R3."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So 3D real coordinate space. And so you would view this as all the possible real-valued 3-tuples. So real-valued 3-tuples. So for example, that would be a member of R3. And let me actually label these vectors just so we get in the habit of it. So let's say this vector, we call this vector x. Let's say we have a vector b that looks like this."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So for example, that would be a member of R3. And let me actually label these vectors just so we get in the habit of it. So let's say this vector, we call this vector x. Let's say we have a vector b that looks like this. Negative 1, 5, 3. Both of these would be members of R3. And if you want to see some fancy notation, a member of a set."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say we have a vector b that looks like this. Negative 1, 5, 3. Both of these would be members of R3. And if you want to see some fancy notation, a member of a set. So this is a member of R3. It is a real-valued 3-tuple. Now, you say, well, what would not be a member of R3?"}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And if you want to see some fancy notation, a member of a set. So this is a member of R3. It is a real-valued 3-tuple. Now, you say, well, what would not be a member of R3? Well, this right over here isn't a 3-tuple. This right over here is a member of R2. Now, you might be able to extend it in some way at a 0 or something."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, you say, well, what would not be a member of R3? Well, this right over here isn't a 3-tuple. This right over here is a member of R2. Now, you might be able to extend it in some way at a 0 or something. But formally, this is not a 3-tuple. Another thing that would not be a member of R3, let's say someone wanted to make some type of vector that had some imaginary parts in it. So let's say it had i, 0, 1."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Now, you might be able to extend it in some way at a 0 or something. But formally, this is not a 3-tuple. Another thing that would not be a member of R3, let's say someone wanted to make some type of vector that had some imaginary parts in it. So let's say it had i, 0, 1. This is no longer real-valued. We have put an imaginary. This number up here has an imaginary part."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say it had i, 0, 1. This is no longer real-valued. We have put an imaginary. This number up here has an imaginary part. So this is no longer a real-valued 3-tuple. And what's neat about linear algebra is we don't have to stop there. R3, we can visualize."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This number up here has an imaginary part. So this is no longer a real-valued 3-tuple. And what's neat about linear algebra is we don't have to stop there. R3, we can visualize. We can plot these things. We've already, probably in your previous mathematical career, especially if you have some type of a hologram or something, it's not hard to visualize things in three dimensions. But what's neat is that we can keep extending this."}, {"video_title": "Real coordinate spaces Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "R3, we can visualize. We can plot these things. We've already, probably in your previous mathematical career, especially if you have some type of a hologram or something, it's not hard to visualize things in three dimensions. But what's neat is that we can keep extending this. We can go into 4, 5, 6, 7, 20, 100 dimensions. And obviously, there it becomes much harder, if not impossible, to visualize it. But then we can at least represent it mathematically with a n-tuple of vectors."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's review our notions of subspaces again, and then let's see if we can define some interesting subspaces dealing with matrices and vectors. So a subspace, let me just, subspace, let's say that I have some subspace, oh let me just call it some subspace S. This is a subspace if the following are true, and this is all a review. That the zero vector, I'll just do it like that, the zero vector is a member of S, it contains the zero vector. That if V1 and V2 are both members of my subspace, then V1 plus V2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. You can add any of their two members and you'll get another member of the subspace. And then the last requirement, if you remember, is that subspaces are closed under multiplication."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That if V1 and V2 are both members of my subspace, then V1 plus V2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. You can add any of their two members and you'll get another member of the subspace. And then the last requirement, if you remember, is that subspaces are closed under multiplication. So that if C is a real number, and it's just a scalar, and if I multiply and V1 is a member of my subspace, that if I multiply that arbitrary real number times my member of my subspace, V1, I'm going to get another member of the subspace. So it's closed under multiplication. These are all of what a subspace is, this is our definition of a subspace."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then the last requirement, if you remember, is that subspaces are closed under multiplication. So that if C is a real number, and it's just a scalar, and if I multiply and V1 is a member of my subspace, that if I multiply that arbitrary real number times my member of my subspace, V1, I'm going to get another member of the subspace. So it's closed under multiplication. These are all of what a subspace is, this is our definition of a subspace. If you call something a subspace, these need to be true. Now let's see if we can do something interesting with what we understand about matrix vector multiplication. Let's say I have the matrix A, I'll make it nice and bold, and it's an m by n matrix."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "These are all of what a subspace is, this is our definition of a subspace. If you call something a subspace, these need to be true. Now let's see if we can do something interesting with what we understand about matrix vector multiplication. Let's say I have the matrix A, I'll make it nice and bold, and it's an m by n matrix. And I'm interested in the following situation. I want to set up the homogeneous equation, and we'll talk about why it's homogeneous. Well, I'll tell you in a second."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's say I have the matrix A, I'll make it nice and bold, and it's an m by n matrix. And I'm interested in the following situation. I want to set up the homogeneous equation, and we'll talk about why it's homogeneous. Well, I'll tell you in a second. So let's say we set up the equation, my matrix A times vector X is equal to the zero vector. This is a homogeneous equation because we have a zero there. And I want to ask the question, I talked about subspaces, if I take all of the X's, if I take the world, the universe, the set of all of the X's that satisfy this equation, do I have a valid subspace?"}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, I'll tell you in a second. So let's say we set up the equation, my matrix A times vector X is equal to the zero vector. This is a homogeneous equation because we have a zero there. And I want to ask the question, I talked about subspaces, if I take all of the X's, if I take the world, the universe, the set of all of the X's that satisfy this equation, do I have a valid subspace? So if I take, let's think about this, I want to take all of the X's that are a member of Rn. Remember, if our matrix A has n columns, then I've only defined this matrix vector multiplication. If X is a member of Rn, if X has to have exactly n components, only then is it defined."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And I want to ask the question, I talked about subspaces, if I take all of the X's, if I take the world, the universe, the set of all of the X's that satisfy this equation, do I have a valid subspace? So if I take, let's think about this, I want to take all of the X's that are a member of Rn. Remember, if our matrix A has n columns, then I've only defined this matrix vector multiplication. If X is a member of Rn, if X has to have exactly n components, only then is it defined. So let me define a set of all of the vectors that are a member of Rn, where they satisfy the equation A times my vector X is equal to the zero vector. So my question is, is this a subspace? So the first question is, does it contain the zero vector?"}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If X is a member of Rn, if X has to have exactly n components, only then is it defined. So let me define a set of all of the vectors that are a member of Rn, where they satisfy the equation A times my vector X is equal to the zero vector. So my question is, is this a subspace? So the first question is, does it contain the zero vector? Well, in order for this to contain the zero vector, the zero vector must satisfy this equation. So what is any m by n matrix A times the zero vector? Let's write out my matrix A."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the first question is, does it contain the zero vector? Well, in order for this to contain the zero vector, the zero vector must satisfy this equation. So what is any m by n matrix A times the zero vector? Let's write out my matrix A. My matrix A, A11, A12, all the way to A1n. And then this, as we go down a column, we go all the way down to AM1. And then as we go all the way to the bottom right, we go to AMn."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's write out my matrix A. My matrix A, A11, A12, all the way to A1n. And then this, as we go down a column, we go all the way down to AM1. And then as we go all the way to the bottom right, we go to AMn. And I'm going to multiply that times a unit vector, I'm sorry, I'm going to multiply that times a zero vector that has exactly n components. So the zero vector with n components is 0, 0, and you're going to have n of these. The number of components here has to be the exact same number of the number of columns you have."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And then as we go all the way to the bottom right, we go to AMn. And I'm going to multiply that times a unit vector, I'm sorry, I'm going to multiply that times a zero vector that has exactly n components. So the zero vector with n components is 0, 0, and you're going to have n of these. The number of components here has to be the exact same number of the number of columns you have. But when you take this product, this matrix vector product, what do you get? What do we get? Well, this first term up here is going to be A11 times 0 plus A12 times 0 plus each of these terms times 0."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "The number of components here has to be the exact same number of the number of columns you have. But when you take this product, this matrix vector product, what do you get? What do we get? Well, this first term up here is going to be A11 times 0 plus A12 times 0 plus each of these terms times 0. And you add them all up. A11 times 0 plus A12 times 0 all the way to A1n times this 0. So you get 0."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, this first term up here is going to be A11 times 0 plus A12 times 0 plus each of these terms times 0. And you add them all up. A11 times 0 plus A12 times 0 all the way to A1n times this 0. So you get 0. Now this term is going to be A21 times 0 plus A22 times 0 plus A23 times 0 all the way to A2n times 0. Well, that's obviously going to be 0. And you're going to keep doing that because all of these are essentially, you can kind of view it as the dot product of, well, I haven't defined dot products with row vectors and column vectors, but I think you get the idea."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So you get 0. Now this term is going to be A21 times 0 plus A22 times 0 plus A23 times 0 all the way to A2n times 0. Well, that's obviously going to be 0. And you're going to keep doing that because all of these are essentially, you can kind of view it as the dot product of, well, I haven't defined dot products with row vectors and column vectors, but I think you get the idea. The sum of each of these elements multiplied with the corresponding component in this vector. And of course, you're just always multiplying by 0 and then adding up, so you're going to get nothing but a bunch of 0's. So the zero vector does satisfy the equation."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "And you're going to keep doing that because all of these are essentially, you can kind of view it as the dot product of, well, I haven't defined dot products with row vectors and column vectors, but I think you get the idea. The sum of each of these elements multiplied with the corresponding component in this vector. And of course, you're just always multiplying by 0 and then adding up, so you're going to get nothing but a bunch of 0's. So the zero vector does satisfy the equation. A times the zero vector is equal to the zero vector. And this is a very unconventional notation. I'm just writing it like that because I don't feel like bolding out my 0's all the time to make you realize that that's a vector."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So the zero vector does satisfy the equation. A times the zero vector is equal to the zero vector. And this is a very unconventional notation. I'm just writing it like that because I don't feel like bolding out my 0's all the time to make you realize that that's a vector. So we meet our first requirement. The zero vector is a member of this set. So it does contain, so let me define my set here."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just writing it like that because I don't feel like bolding out my 0's all the time to make you realize that that's a vector. So we meet our first requirement. The zero vector is a member of this set. So it does contain, so let me define my set here. Let me define it n. And I'll tell you in a second why I'm calling it n. So we now know that the zero vector is a member of my set n. Now let's say I have two vectors, v1 and v2, that are members. Let me write this. So let's say I have two vectors, v1 and v2."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it does contain, so let me define my set here. Let me define it n. And I'll tell you in a second why I'm calling it n. So we now know that the zero vector is a member of my set n. Now let's say I have two vectors, v1 and v2, that are members. Let me write this. So let's say I have two vectors, v1 and v2. This was a v2 over here. v1 and v2 that are both members of our set. What does that mean?"}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let's say I have two vectors, v1 and v2. This was a v2 over here. v1 and v2 that are both members of our set. What does that mean? That means that they both satisfy this equation. So that means that my matrix A times vector 1 is equal to 0. This is by definition."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "What does that mean? That means that they both satisfy this equation. So that means that my matrix A times vector 1 is equal to 0. This is by definition. I'm saying that they're a member of this set, which means they must satisfy this. And that also means that A times vector 2 is equal to our zero vector. So in order for this to be closed under addition, A times vector 1 plus vector 2, the sum of these two vectors should also be a member of n. But let's figure out what this is."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is by definition. I'm saying that they're a member of this set, which means they must satisfy this. And that also means that A times vector 2 is equal to our zero vector. So in order for this to be closed under addition, A times vector 1 plus vector 2, the sum of these two vectors should also be a member of n. But let's figure out what this is. The sum of these two vectors is this vector right here. This is equal to, and I haven't proven this to you yet. I haven't made a video where I prove this, but it's very easy to prove just using the definition of matrix-vector multiplication, that matrix-vector multiplication does display the distributive property."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So in order for this to be closed under addition, A times vector 1 plus vector 2, the sum of these two vectors should also be a member of n. But let's figure out what this is. The sum of these two vectors is this vector right here. This is equal to, and I haven't proven this to you yet. I haven't made a video where I prove this, but it's very easy to prove just using the definition of matrix-vector multiplication, that matrix-vector multiplication does display the distributive property. Maybe I'll make a video on that. Literally, you just have to go through the mechanics of each of the terms. This is equal to A v1 plus A v2."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I haven't made a video where I prove this, but it's very easy to prove just using the definition of matrix-vector multiplication, that matrix-vector multiplication does display the distributive property. Maybe I'll make a video on that. Literally, you just have to go through the mechanics of each of the terms. This is equal to A v1 plus A v2. We know that this is equal to the zero vector, and this is equal to the zero vector. If you add the zero vector to itself, this whole thing is going to be equal to the zero vector. So if v1 is a member of n and v2 is a member of n, which means they both satisfy this equation, then v1 plus v2 is definitely still a member of n because when I multiply A times that, I get the zero vector again."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is equal to A v1 plus A v2. We know that this is equal to the zero vector, and this is equal to the zero vector. If you add the zero vector to itself, this whole thing is going to be equal to the zero vector. So if v1 is a member of n and v2 is a member of n, which means they both satisfy this equation, then v1 plus v2 is definitely still a member of n because when I multiply A times that, I get the zero vector again. Let me write that result as well. We are also... Let me write this right here. We also know that v1 plus v2 is also a member of n. The last thing we have to show is that it's closed under multiplication."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So if v1 is a member of n and v2 is a member of n, which means they both satisfy this equation, then v1 plus v2 is definitely still a member of n because when I multiply A times that, I get the zero vector again. Let me write that result as well. We are also... Let me write this right here. We also know that v1 plus v2 is also a member of n. The last thing we have to show is that it's closed under multiplication. Let's say that v1 is a member of our space that I defined here where they satisfy this equation. What about c times v1? Is that a member of n?"}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "We also know that v1 plus v2 is also a member of n. The last thing we have to show is that it's closed under multiplication. Let's say that v1 is a member of our space that I defined here where they satisfy this equation. What about c times v1? Is that a member of n? Let's think about it. What's our matrix A times the vector? I'm just multiplying this times the scalar."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Is that a member of n? Let's think about it. What's our matrix A times the vector? I'm just multiplying this times the scalar. I'm just going to get another vector. I don't want to write a capital V there. Lowercase v, so it's a vector."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "I'm just multiplying this times the scalar. I'm just going to get another vector. I don't want to write a capital V there. Lowercase v, so it's a vector. What's this equal to? Once again, I haven't proven it to you yet, but it's actually a very straightforward thing to do to show that when you're dealing with scalars, if you have a scalar here, it doesn't matter if you multiply the scalar times the vector before multiplying it times the matrix or multiplying the matrix times the vector and then doing the scalar. It's fairly straightforward to prove that this is equal to c times our matrix A, I'll make that nice and bold, times our vector v, that these two things are equivalent."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Lowercase v, so it's a vector. What's this equal to? Once again, I haven't proven it to you yet, but it's actually a very straightforward thing to do to show that when you're dealing with scalars, if you have a scalar here, it doesn't matter if you multiply the scalar times the vector before multiplying it times the matrix or multiplying the matrix times the vector and then doing the scalar. It's fairly straightforward to prove that this is equal to c times our matrix A, I'll make that nice and bold, times our vector v, that these two things are equivalent. Maybe I should just churn out the video that does this, but I'll leave it to you. You literally just go through the mechanics by component. You show this."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's fairly straightforward to prove that this is equal to c times our matrix A, I'll make that nice and bold, times our vector v, that these two things are equivalent. Maybe I should just churn out the video that does this, but I'll leave it to you. You literally just go through the mechanics by component. You show this. Clearly, if this is true, we already know that v1 is a member of our set, which means that A times v1 is equal to the zero vector. That means this will reduce to c times the zero vector, which is still the zero vector. So cv1 is definitely a member of N. It's closed under multiplication."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "You show this. Clearly, if this is true, we already know that v1 is a member of our set, which means that A times v1 is equal to the zero vector. That means this will reduce to c times the zero vector, which is still the zero vector. So cv1 is definitely a member of N. It's closed under multiplication. I kind of assumed this right here, but maybe I'll prove that in a different video. But I want to do all this to show that this set N is a valid subspace. This is a valid subspace."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So cv1 is definitely a member of N. It's closed under multiplication. I kind of assumed this right here, but maybe I'll prove that in a different video. But I want to do all this to show that this set N is a valid subspace. This is a valid subspace. It contains the zero vector. It's closed under addition. It's closed under multiplication."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This is a valid subspace. It contains the zero vector. It's closed under addition. It's closed under multiplication. We actually have a special name for this. We call this right here, we call N the null space of A. Or we could write N is equal to, maybe I shouldn't have written an N, maybe I'll write an orange N there."}, {"video_title": "Introduction to the null space of a matrix Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's closed under multiplication. We actually have a special name for this. We call this right here, we call N the null space of A. Or we could write N is equal to, maybe I shouldn't have written an N, maybe I'll write an orange N there. Our orange N is equal to, the notation is just the null space of A. Or we could write the null space is equal to the orange notation of N. And so literally, if I just give you some arbitrary matrix, A, and I say, hey, find me N of A, what is that? Literally, your goal is to find the set of all X's that satisfy the equation A times X is equal to zero."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So if we have one matrix, A, and it's an m by n matrix, and then we have some other matrix, B, let's say that's an n by k matrix, then we've defined the product of A and B to be equal to. And actually, before I define the product, let me just write B out as just a collection of column vectors. So we know that B can be written as a column vector B1, another column vector B2, and all the way it's going to have k of them, because it has exactly k columns. So Bk. So in the last video, we defined the product of A and B, and they have to have the columns of A have to be the same as the rows of B in order for this to be well defined. But we defined this product to be equal to A times each of the column vectors of B. So it's equal to, let me switch back to that color, it's equal to A times, let me do it in the color that I did it, A times B1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So Bk. So in the last video, we defined the product of A and B, and they have to have the columns of A have to be the same as the rows of B in order for this to be well defined. But we defined this product to be equal to A times each of the column vectors of B. So it's equal to, let me switch back to that color, it's equal to A times, let me do it in the color that I did it, A times B1. And then the second column of our product is going to be A times B2. A times B2. Our third product is going to be A times B3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it's equal to, let me switch back to that color, it's equal to A times, let me do it in the color that I did it, A times B1. And then the second column of our product is going to be A times B2. A times B2. Our third product is going to be A times B3. All the way to A times Bk. Right there. And the whole motivation for this, you've probably seen this before, maybe in your Algebra 2 class."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Our third product is going to be A times B3. All the way to A times Bk. Right there. And the whole motivation for this, you've probably seen this before, maybe in your Algebra 2 class. It might have not been defined exactly this way, but this is equivalent to what you probably saw in your Algebra 2 class. But the neat thing about this definition is that the motivation came from the composition of two linear transformations whose transformation matrices were the matrices A and B. And I showed you that in the last video."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And the whole motivation for this, you've probably seen this before, maybe in your Algebra 2 class. It might have not been defined exactly this way, but this is equivalent to what you probably saw in your Algebra 2 class. But the neat thing about this definition is that the motivation came from the composition of two linear transformations whose transformation matrices were the matrices A and B. And I showed you that in the last video. With that said, let's actually compute some matrix-matrix products, just so you get the hang of it. So let's say that I have the matrix A. Let's say that A is equal to the matrix 1, minus 1, 2, 0, minus 2, and 1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And I showed you that in the last video. With that said, let's actually compute some matrix-matrix products, just so you get the hang of it. So let's say that I have the matrix A. Let's say that A is equal to the matrix 1, minus 1, 2, 0, minus 2, and 1. I keep the numbers low to keep our arithmetic fairly straightforward. And let's say that I have the matrix B. And let's say that it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and then 3, 1, 0, 2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let's say that A is equal to the matrix 1, minus 1, 2, 0, minus 2, and 1. I keep the numbers low to keep our arithmetic fairly straightforward. And let's say that I have the matrix B. And let's say that it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and then 3, 1, 0, 2. So A is a 2 by 3 matrix. Two rows, three columns. And B is a 3 by 4 matrix."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And let's say that it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and then 3, 1, 0, 2. So A is a 2 by 3 matrix. Two rows, three columns. And B is a 3 by 4 matrix. So by our definition, what is the product AB going to be equal to? Well, we know it's well-defined because the number of columns here is equal to the number of rows. So we can actually take these matrix-vector products."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And B is a 3 by 4 matrix. So by our definition, what is the product AB going to be equal to? Well, we know it's well-defined because the number of columns here is equal to the number of rows. So we can actually take these matrix-vector products. You'll see that in a second. So AB is equal to the matrix A times the column vector 1, 2, 3. That's going to be the first column in our product matrix."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we can actually take these matrix-vector products. You'll see that in a second. So AB is equal to the matrix A times the column vector 1, 2, 3. That's going to be the first column in our product matrix. And the second one is going to be the matrix A times the column 0, 0, 1. The third column is going to be the matrix A times the column vector 1, 1, 0. And then the fourth column in our product vector is going to be the matrix A times the column vector 1, minus 1, 2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's going to be the first column in our product matrix. And the second one is going to be the matrix A times the column 0, 0, 1. The third column is going to be the matrix A times the column vector 1, 1, 0. And then the fourth column in our product vector is going to be the matrix A times the column vector 1, minus 1, 2. And when we write it like this, it should be clear why this has to be why the number of columns in A have to be the number of rows in B. Because the column vectors in B are going to have the same number of components as the number of rows in B. So all of the column vectors in B, so if we call this B1, B2, B3, B4, all of my Bi's, let me write it this way, all of my Bi's, where this i could be 1, 2, 3, or 4, are all members of R3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then the fourth column in our product vector is going to be the matrix A times the column vector 1, minus 1, 2. And when we write it like this, it should be clear why this has to be why the number of columns in A have to be the number of rows in B. Because the column vectors in B are going to have the same number of components as the number of rows in B. So all of the column vectors in B, so if we call this B1, B2, B3, B4, all of my Bi's, let me write it this way, all of my Bi's, where this i could be 1, 2, 3, or 4, are all members of R3. So we only have matrix vector products well defined when the number of columns in your matrix are equivalent to essentially the dimensionality of your vectors. That's why that number and that number has to be the same. Well, now we've reduced our matrix matrix product problem to just a bunch of four different matrix vector product problems, so we can just multiply these."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So all of the column vectors in B, so if we call this B1, B2, B3, B4, all of my Bi's, let me write it this way, all of my Bi's, where this i could be 1, 2, 3, or 4, are all members of R3. So we only have matrix vector products well defined when the number of columns in your matrix are equivalent to essentially the dimensionality of your vectors. That's why that number and that number has to be the same. Well, now we've reduced our matrix matrix product problem to just a bunch of four different matrix vector product problems, so we can just multiply these. This is nothing new to us. So let's do it. And so what is this equal to?"}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Well, now we've reduced our matrix matrix product problem to just a bunch of four different matrix vector product problems, so we can just multiply these. This is nothing new to us. So let's do it. And so what is this equal to? So AB, let me rewrite it, AB, my product vector, is going to be equal to, so this first column is the matrix A times the column vector 1, 2, 3. And how did we define that? Remember, one way to think about it is that this is equal to, you can kind of think of it as each of the rows of A dotted with the column here of B."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And so what is this equal to? So AB, let me rewrite it, AB, my product vector, is going to be equal to, so this first column is the matrix A times the column vector 1, 2, 3. And how did we define that? Remember, one way to think about it is that this is equal to, you can kind of think of it as each of the rows of A dotted with the column here of B. Or even better, this is the transpose of some matrix. Like let me write this this way. If A is equal to, sorry, the transpose of some vector."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, one way to think about it is that this is equal to, you can kind of think of it as each of the rows of A dotted with the column here of B. Or even better, this is the transpose of some matrix. Like let me write this this way. If A is equal to, sorry, the transpose of some vector. Let's say that A is equal to the column vector 0, minus 1, 2, then A transpose, and I haven't talked about transposes a lot yet, but I think you get the idea. You just change all of the columns into rows. So A transpose will just be equal to 0, minus 1, 2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If A is equal to, sorry, the transpose of some vector. Let's say that A is equal to the column vector 0, minus 1, 2, then A transpose, and I haven't talked about transposes a lot yet, but I think you get the idea. You just change all of the columns into rows. So A transpose will just be equal to 0, minus 1, 2. You just go from a column vector to a row vector. So if we call this thing here A transpose, then when we take the product of our matrix A times this vector, we essentially are just taking A and dotting with this guy for our first row and our first column. So let me do it that way."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So A transpose will just be equal to 0, minus 1, 2. You just go from a column vector to a row vector. So if we call this thing here A transpose, then when we take the product of our matrix A times this vector, we essentially are just taking A and dotting with this guy for our first row and our first column. So let me do it that way. So let me write in that notation. So this is going to be the vector 1, minus 1, 2. That's essentially that row right there represented as a column dotted with 1, 2, 3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me do it that way. So let me write in that notation. So this is going to be the vector 1, minus 1, 2. That's essentially that row right there represented as a column dotted with 1, 2, 3. Actually, let me do it in that color. Just so I can later switch to one color to make things simple, but dotted with 1, 2, 3. So we just took that row, or I guess the column equivalent of that row, and dotted with this."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "That's essentially that row right there represented as a column dotted with 1, 2, 3. Actually, let me do it in that color. Just so I can later switch to one color to make things simple, but dotted with 1, 2, 3. So we just took that row, or I guess the column equivalent of that row, and dotted with this. And I wrote it like this because we've only defined dot products for column vectors. I could do it maybe for row vectors, but who would need to make a new definition? So that's going to be the first entry in this matrix vector product."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we just took that row, or I guess the column equivalent of that row, and dotted with this. And I wrote it like this because we've only defined dot products for column vectors. I could do it maybe for row vectors, but who would need to make a new definition? So that's going to be the first entry in this matrix vector product. The second entry is going to be the second row of A, essentially dotted with this vector right there. So it's going to be equal to 0, minus 2, and 1, dotted with 1, 2, 3. Dotted with 1, 2, 3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So that's going to be the first entry in this matrix vector product. The second entry is going to be the second row of A, essentially dotted with this vector right there. So it's going to be equal to 0, minus 2, and 1, dotted with 1, 2, 3. Dotted with 1, 2, 3. And we just keep doing that. And I'll just switch maybe to one neutral color now. So then A times 0, 0, 1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Dotted with 1, 2, 3. And we just keep doing that. And I'll just switch maybe to one neutral color now. So then A times 0, 0, 1. That's going to be the first row of A expressed as a column vector. So we can write it like this. 1, minus 1, 2, dot 0, 0, 1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So then A times 0, 0, 1. That's going to be the first row of A expressed as a column vector. So we can write it like this. 1, minus 1, 2, dot 0, 0, 1. And then, actually, and then we have our, and then the second row of A dotted with this column vector. So we have 0, minus 2, 1, dotted with 0, 0, 1. Two more rows left."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1, minus 1, 2, dot 0, 0, 1. And then, actually, and then we have our, and then the second row of A dotted with this column vector. So we have 0, minus 2, 1, dotted with 0, 0, 1. Two more rows left. This can get a little tedious. And it's inevitable that I'll probably make a careless mistake, but as long as you understand the process, that's the important thing. So the next one, this row of A expressed as a column vector."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Two more rows left. This can get a little tedious. And it's inevitable that I'll probably make a careless mistake, but as long as you understand the process, that's the important thing. So the next one, this row of A expressed as a column vector. 1, minus 1, 2, we're going to dot it with this vector right there. 1, 1, 0. And then this row of A, I could just look over here as well, 0, minus 2, 1, dotted with 1, 1, 0."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So the next one, this row of A expressed as a column vector. 1, minus 1, 2, we're going to dot it with this vector right there. 1, 1, 0. And then this row of A, I could just look over here as well, 0, minus 2, 1, dotted with 1, 1, 0. And then finally, the last two entries are going to be the top row of A. 1, minus 1, 2, dotted with this column vector, 1, minus 1, 2, put a little dot there. Remember, we're taking the dot product."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then this row of A, I could just look over here as well, 0, minus 2, 1, dotted with 1, 1, 0. And then finally, the last two entries are going to be the top row of A. 1, minus 1, 2, dotted with this column vector, 1, minus 1, 2, put a little dot there. Remember, we're taking the dot product. And then finally, this second row of A. So 0, minus 2, 1, dotted with this column vector. 1, minus 1, 2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, we're taking the dot product. And then finally, this second row of A. So 0, minus 2, 1, dotted with this column vector. 1, minus 1, 2. And that is going to be our product matrix. And this looks very complicated right now, but now we just have to compute in dot products, tend to simplify things a good bit. So what is our matrix, our product, going to simplify to?"}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "1, minus 1, 2. And that is going to be our product matrix. And this looks very complicated right now, but now we just have to compute in dot products, tend to simplify things a good bit. So what is our matrix, our product, going to simplify to? I'll do it in pink. AB is equal to, let me draw the matrix right there. So what's the dot product of these two things?"}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what is our matrix, our product, going to simplify to? I'll do it in pink. AB is equal to, let me draw the matrix right there. So what's the dot product of these two things? It's 1 times 1. I'll just write it out. It's 1 times 1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So what's the dot product of these two things? It's 1 times 1. I'll just write it out. It's 1 times 1. I'll just write 1 times 1 is 1, plus minus 1 times 2, so minus 2, plus 2 times 3, plus 6. Now we'll do this term right here. 0 times 1 is 0, plus minus 2 times 2, so that's minus 4, plus 1 times 3, plus 3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's 1 times 1. I'll just write 1 times 1 is 1, plus minus 1 times 2, so minus 2, plus 2 times 3, plus 6. Now we'll do this term right here. 0 times 1 is 0, plus minus 2 times 2, so that's minus 4, plus 1 times 3, plus 3. Now we're on to this term. 1 times 0 is 0, plus minus 1 times 0, plus 0, plus 2 times 1, is equal to plus 2. This term, 0 times 0 is 0, plus minus 2 times 0, let me write it as 0, plus minus 2 times 0 is 0, plus 1 times 1."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "0 times 1 is 0, plus minus 2 times 2, so that's minus 4, plus 1 times 3, plus 3. Now we're on to this term. 1 times 0 is 0, plus minus 1 times 0, plus 0, plus 2 times 1, is equal to plus 2. This term, 0 times 0 is 0, plus minus 2 times 0, let me write it as 0, plus minus 2 times 0 is 0, plus 1 times 1. So plus 1. Then here you have 1 times 1 is 1, plus minus 1 times 1 is minus 1, plus 2 times 0, so plus 0. Here, 0 times 1 is 0, minus 2 times 1 is minus 2, and then 1 times 0 is plus 0."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "This term, 0 times 0 is 0, plus minus 2 times 0, let me write it as 0, plus minus 2 times 0 is 0, plus 1 times 1. So plus 1. Then here you have 1 times 1 is 1, plus minus 1 times 1 is minus 1, plus 2 times 0, so plus 0. Here, 0 times 1 is 0, minus 2 times 1 is minus 2, and then 1 times 0 is plus 0. Almost done. 1 times 1 is 1, minus 1 times minus 1 is 1, 2 times 2 is 4. Finally, 0 times 1 is 0, minus 2 times minus 1 is 2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Here, 0 times 1 is 0, minus 2 times 1 is minus 2, and then 1 times 0 is plus 0. Almost done. 1 times 1 is 1, minus 1 times minus 1 is 1, 2 times 2 is 4. Finally, 0 times 1 is 0, minus 2 times minus 1 is 2. 1 times 2 is also 2. And we're in the home stretch. Now we just have to add up these values."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Finally, 0 times 1 is 0, minus 2 times minus 1 is 2. 1 times 2 is also 2. And we're in the home stretch. Now we just have to add up these values. So our dot product of the two matrices is equal to the 2 by 4 matrix, 1 minus 2 plus 6, that's equal to 5, minus 4 plus 3 is minus 1. This is just 2, this is just 1. Then we have 1 minus 1 plus 0 is just 0, minus 2, we just have a minus 2 there, 1 plus 1 plus 4 is 6, and then 2 plus 2 is 4, and we are done."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now we just have to add up these values. So our dot product of the two matrices is equal to the 2 by 4 matrix, 1 minus 2 plus 6, that's equal to 5, minus 4 plus 3 is minus 1. This is just 2, this is just 1. Then we have 1 minus 1 plus 0 is just 0, minus 2, we just have a minus 2 there, 1 plus 1 plus 4 is 6, and then 2 plus 2 is 4, and we are done. The product of AB is equal to this matrix right here. Let me get my A and B back. We can talk a little bit more about what this product actually represented."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Then we have 1 minus 1 plus 0 is just 0, minus 2, we just have a minus 2 there, 1 plus 1 plus 4 is 6, and then 2 plus 2 is 4, and we are done. The product of AB is equal to this matrix right here. Let me get my A and B back. We can talk a little bit more about what this product actually represented. So let me copy and paste this. And then I'll paste it. Let me scroll down a little bit."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We can talk a little bit more about what this product actually represented. So let me copy and paste this. And then I'll paste it. Let me scroll down a little bit. Go down here, paste. There you go. So this was our A and our B."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Let me scroll down a little bit. Go down here, paste. There you go. So this was our A and our B. And when we took the product, we got this matrix here. Now there are a couple of interesting things to notice. Remember, I said that this product is only well defined when the number of columns in A is equal to the number of rows in B."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So this was our A and our B. And when we took the product, we got this matrix here. Now there are a couple of interesting things to notice. Remember, I said that this product is only well defined when the number of columns in A is equal to the number of rows in B. So that was the case in this situation. And then notice, we got a 2 by 4 matrix, which is the number of rows in A times the number of columns in B. So we got a 2 by 4 matrix."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Remember, I said that this product is only well defined when the number of columns in A is equal to the number of rows in B. So that was the case in this situation. And then notice, we got a 2 by 4 matrix, which is the number of rows in A times the number of columns in B. So we got a 2 by 4 matrix. So another natural question is, could we have found, or is it even equal, if we were to take the product BA? So if we tried to apply our definition there, what would it be equal to? It would be equal to the matrix B times the column 1, 0."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we got a 2 by 4 matrix. So another natural question is, could we have found, or is it even equal, if we were to take the product BA? So if we tried to apply our definition there, what would it be equal to? It would be equal to the matrix B times the column 1, 0. Then the matrix B times the column minus 1, minus 2. And then it would be the matrix B times the column 2, 1. Now, can we take this matrix vector product?"}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It would be equal to the matrix B times the column 1, 0. Then the matrix B times the column minus 1, minus 2. And then it would be the matrix B times the column 2, 1. Now, can we take this matrix vector product? We have a 3 by 4. This right here is a 3 by 4 matrix. And this guy right here is a member of R2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Now, can we take this matrix vector product? We have a 3 by 4. This right here is a 3 by 4 matrix. And this guy right here is a member of R2. So this is not well defined. We have more columns here than entries here. So we have never defined a matrix vector product like this."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And this guy right here is a member of R2. So this is not well defined. We have more columns here than entries here. So we have never defined a matrix vector product like this. So not only is this not equal to this, it's not even defined. So it's not defined when you take a 3 by 4 matrix and you take the product of that with a 2 by 4 matrix. It's not defined because that number and that number is not equal."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So we have never defined a matrix vector product like this. So not only is this not equal to this, it's not even defined. So it's not defined when you take a 3 by 4 matrix and you take the product of that with a 2 by 4 matrix. It's not defined because that number and that number is not equal. And so obviously, since this is defined and this isn't defined, you know that AB is not always equal to BA. In fact, it's not usually equal to BA. And sometimes it's not even defined."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "It's not defined because that number and that number is not equal. And so obviously, since this is defined and this isn't defined, you know that AB is not always equal to BA. In fact, it's not usually equal to BA. And sometimes it's not even defined. And the last point I want to make is you probably learned to do matrix-matrix products in Algebra 2, but you didn't have any motivation for what you were doing. But now we do have a motivation. Because when you're taking the product of A and B, we learned in the last video that if we have two transformations, we have two transformations."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And sometimes it's not even defined. And the last point I want to make is you probably learned to do matrix-matrix products in Algebra 2, but you didn't have any motivation for what you were doing. But now we do have a motivation. Because when you're taking the product of A and B, we learned in the last video that if we have two transformations, we have two transformations. Let's say we have the transformation S is a transformation from R3 to R2. And that S is represented by the matrix. So S, given some matrix in R3, if you apply the transformation S to it, it's equivalent to multiplying that, or given any vector in R3, applying the transformation S is equivalent to multiplying that vector times A."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "Because when you're taking the product of A and B, we learned in the last video that if we have two transformations, we have two transformations. Let's say we have the transformation S is a transformation from R3 to R2. And that S is represented by the matrix. So S, given some matrix in R3, if you apply the transformation S to it, it's equivalent to multiplying that, or given any vector in R3, applying the transformation S is equivalent to multiplying that vector times A. We can say that. And I used R3 and R2 because the number of columns in A is 3. So it can apply to a three-dimensional vector."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So S, given some matrix in R3, if you apply the transformation S to it, it's equivalent to multiplying that, or given any vector in R3, applying the transformation S is equivalent to multiplying that vector times A. We can say that. And I used R3 and R2 because the number of columns in A is 3. So it can apply to a three-dimensional vector. And similarly, we can imagine B as being the matrix transformation of some transformation T that is a mapping from R4 to R3, where if you give it some vector x in R4, it will produce, you take the product of that with B, and you're going to get some vector in R3. Now, if we think of the composition of the two, so let's think about it a little bit. If we have R4 here, let me switch colors."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So it can apply to a three-dimensional vector. And similarly, we can imagine B as being the matrix transformation of some transformation T that is a mapping from R4 to R3, where if you give it some vector x in R4, it will produce, you take the product of that with B, and you're going to get some vector in R3. Now, if we think of the composition of the two, so let's think about it a little bit. If we have R4 here, let me switch colors. We have R4 here. We have R3 here. And then we have R2 here."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "If we have R4 here, let me switch colors. We have R4 here. We have R3 here. And then we have R2 here. T is a transformation from R4 to R3. So T would look like that. T is a transformation."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And then we have R2 here. T is a transformation from R4 to R3. So T would look like that. T is a transformation. It's B times x. That's what T is equal to. So T is this transformation."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "T is a transformation. It's B times x. That's what T is equal to. So T is this transformation. And then S is a transformation from R3 to R2. So S looks like that. And S is equivalent to A times any vector in R3."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So T is this transformation. And then S is a transformation from R3 to R2. So S looks like that. And S is equivalent to A times any vector in R3. So that is S. So now we know how to visualize or how to think about what the product of A and B are. The product of A and B is essentially, you apply the transformation B first. So let me think of the composition of S. Let me write it this way."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "And S is equivalent to A times any vector in R3. So that is S. So now we know how to visualize or how to think about what the product of A and B are. The product of A and B is essentially, you apply the transformation B first. So let me think of the composition of S. Let me write it this way. So what is the composition of S with T? This is equal to S of T of x. So you take a transformation from R4 to R3, and then you take the S transformation from R3 to R2."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So let me think of the composition of S. Let me write it this way. So what is the composition of S with T? This is equal to S of T of x. So you take a transformation from R4 to R3, and then you take the S transformation from R3 to R2. So this is S of T. S of T is a transformation from R4 all the way to R2. And then the neat thing about this, if you were to just write this out in its matrix representations, we did this in the last video, this would be equal to the S matrix, A, times this vector right here, which is Bx. But now we know that the matrix, by our definition of matrix vector products, that this guy right here is going to have a transformation."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "So you take a transformation from R4 to R3, and then you take the S transformation from R3 to R2. So this is S of T. S of T is a transformation from R4 all the way to R2. And then the neat thing about this, if you were to just write this out in its matrix representations, we did this in the last video, this would be equal to the S matrix, A, times this vector right here, which is Bx. But now we know that the matrix, by our definition of matrix vector products, that this guy right here is going to have a transformation. It's going to be equal to, so the composition S of T of x is going to be equal to the matrix AB, based on our definition, so the transformation AB times some vector x. So the reason why I'm going all this is because we just did a matrix matrix product up here. We took the pain of multiplying the matrix A times the matrix B, and we got this value here."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "But now we know that the matrix, by our definition of matrix vector products, that this guy right here is going to have a transformation. It's going to be equal to, so the composition S of T of x is going to be equal to the matrix AB, based on our definition, so the transformation AB times some vector x. So the reason why I'm going all this is because we just did a matrix matrix product up here. We took the pain of multiplying the matrix A times the matrix B, and we got this value here. And hopefully I didn't make any careless mistakes. But the big idea here, the idea that you probably weren't exposed to in your Algebra 2 class, is that this is the matrix of the composition of the transformations S and T. So right here. It's the matrix of the composition of S and T. So you're not just blindly doing some matrix matrix products."}, {"video_title": "Matrix product examples Matrix transformations Linear Algebra Khan Academy.mp3", "Sentence": "We took the pain of multiplying the matrix A times the matrix B, and we got this value here. And hopefully I didn't make any careless mistakes. But the big idea here, the idea that you probably weren't exposed to in your Algebra 2 class, is that this is the matrix of the composition of the transformations S and T. So right here. It's the matrix of the composition of S and T. So you're not just blindly doing some matrix matrix products. It can be pretty tedious. But now you know what they're for. They're actually for the composition of two transformations, where each of A and B are the transformation matrices for each of the individual linear transformations."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we could draw it right over here. So it's equal to two, one. So the horizontal, if we were to start at the origin, we would move two in the horizontal direction and one in the vertical direction. So we would end up right over here. Now what I wanna do is think about how we can define multiplying this vector by a scalar. So for example, if I were to say, if I were to say three times, three times the vector a, which is the same thing as saying three times two, one. So three is just a number."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So we would end up right over here. Now what I wanna do is think about how we can define multiplying this vector by a scalar. So for example, if I were to say, if I were to say three times, three times the vector a, which is the same thing as saying three times two, one. So three is just a number. One way to think about a scalar quantity, it is just a number versus a vector. This is giving you, it's giving you how much you're moving in the various directions right over here. It's giving you both a magnitude, magnitude and a direction, while this is just a plain number right over here."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So three is just a number. One way to think about a scalar quantity, it is just a number versus a vector. This is giving you, it's giving you how much you're moving in the various directions right over here. It's giving you both a magnitude, magnitude and a direction, while this is just a plain number right over here. But how would we define multiplying three times this vector right over here? Well, one reasonable thing that might jump out at you is well, why don't we just multiply the three times each of these components? So this could be equal to, so we have two and one, and we're gonna multiply each of these times three."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It's giving you both a magnitude, magnitude and a direction, while this is just a plain number right over here. But how would we define multiplying three times this vector right over here? Well, one reasonable thing that might jump out at you is well, why don't we just multiply the three times each of these components? So this could be equal to, so we have two and one, and we're gonna multiply each of these times three. So three times two and three times one, and then the resulting vector is still going to be a two-dimensional vector, and it's going to be the two-dimensional vector six, three. Now I encourage you to get some graph paper out and to actually plot this vector and think about how it relates to this vector right over, this vector right over here. So let me do that."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So this could be equal to, so we have two and one, and we're gonna multiply each of these times three. So three times two and three times one, and then the resulting vector is still going to be a two-dimensional vector, and it's going to be the two-dimensional vector six, three. Now I encourage you to get some graph paper out and to actually plot this vector and think about how it relates to this vector right over, this vector right over here. So let me do that. So the vector six, three, if we started at the origin, we would move six in the horizontal direction, one, two, three, four, five, six, and three in the vertical, one, two, three. So it gets us right over there. So it would look like this."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So let me do that. So the vector six, three, if we started at the origin, we would move six in the horizontal direction, one, two, three, four, five, six, and three in the vertical, one, two, three. So it gets us right over there. So it would look like this. So what just happened to this vector? Well, notice, one way to think about it is what's changed and what has not changed about this vector? Well, what's not changed is still pointing in the same direction."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it would look like this. So what just happened to this vector? Well, notice, one way to think about it is what's changed and what has not changed about this vector? Well, what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it, did not change the direction that my vector is going in, or at least in this case it didn't, but it did change its magnitude. Its magnitude is now three times longer, which makes sense, because we multiplied it by three."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Well, what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it, did not change the direction that my vector is going in, or at least in this case it didn't, but it did change its magnitude. Its magnitude is now three times longer, which makes sense, because we multiplied it by three. One way to think about it is we scaled it up by three. The scalar scaled up the vector. That might make sense, or it might give an intuition of where that word scalar came from."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Its magnitude is now three times longer, which makes sense, because we multiplied it by three. One way to think about it is we scaled it up by three. The scalar scaled up the vector. That might make sense, or it might give an intuition of where that word scalar came from. The scalars, when you multiply it, it scales up a vector. It increased its magnitude by three without changing its direction. But let's do something interesting."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "That might make sense, or it might give an intuition of where that word scalar came from. The scalars, when you multiply it, it scales up a vector. It increased its magnitude by three without changing its direction. But let's do something interesting. Let's multiply our vector A. Let's now multiply it by a negative number. Let's actually just multiply it by negative one, just for simplicity."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "But let's do something interesting. Let's multiply our vector A. Let's now multiply it by a negative number. Let's actually just multiply it by negative one, just for simplicity. So let's just multiply negative one times A. Well, using the convention that we just came up with, we would multiply each of the components by negative one. So two times negative one is negative two, and one times negative one is negative one."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let's actually just multiply it by negative one, just for simplicity. So let's just multiply negative one times A. Well, using the convention that we just came up with, we would multiply each of the components by negative one. So two times negative one is negative two, and one times negative one is negative one. So now negative one times A is going to be negative two, negative one. So if we started at the origin, we would move in the horizontal direction, negative two, and in the vertical direction, negative one. So now what happened to the vector?"}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So two times negative one is negative two, and one times negative one is negative one. So now negative one times A is going to be negative two, negative one. So if we started at the origin, we would move in the horizontal direction, negative two, and in the vertical direction, negative one. So now what happened to the vector? Now what happened to the vector? When I did that, well, now it flipped its direction. Multiplying it by this negative one, it flipped its direction."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So now what happened to the vector? Now what happened to the vector? When I did that, well, now it flipped its direction. Multiplying it by this negative one, it flipped its direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction, which makes sense that multiplying by a negative number would do that. In fact, when we just dealt with the traditional number line, that's what happened. If you took five times negative one, well, now you're going in the other direction."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Multiplying it by this negative one, it flipped its direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction, which makes sense that multiplying by a negative number would do that. In fact, when we just dealt with the traditional number line, that's what happened. If you took five times negative one, well, now you're going in the other direction. You're at negative five. You're five to the left of zero. So it makes sense that this would flip its direction."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "If you took five times negative one, well, now you're going in the other direction. You're at negative five. You're five to the left of zero. So it makes sense that this would flip its direction. So you could imagine if you were to take something like negative two times your vector A, negative two times your vector A, and I encourage you to pause this video and try this on your own, what would this give and what would be the resulting visualization of the vector? Well, let's see. This would be equal to negative two times two is negative four negative two times one is negative two."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So it makes sense that this would flip its direction. So you could imagine if you were to take something like negative two times your vector A, negative two times your vector A, and I encourage you to pause this video and try this on your own, what would this give and what would be the resulting visualization of the vector? Well, let's see. This would be equal to negative two times two is negative four negative two times one is negative two. So this vector, if you were to start at the origin, remember, you don't have to start at the origin, but if you were, it would be, so you'd go zero, one, two, three, four, one, two. It looks just like this. It looks just like this."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "This would be equal to negative two times two is negative four negative two times one is negative two. So this vector, if you were to start at the origin, remember, you don't have to start at the origin, but if you were, it would be, so you'd go zero, one, two, three, four, one, two. It looks just like this. It looks just like this. And so just to remind ourselves, our original vector A looked like this. Our original vector A looked like this. Two, one looks like this."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "It looks just like this. And so just to remind ourselves, our original vector A looked like this. Our original vector A looked like this. Two, one looks like this. And then when you multiply it by negative two, you get a vector that looks like this. You get a vector that looks like this. Let me draw it like this."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Two, one looks like this. And then when you multiply it by negative two, you get a vector that looks like this. You get a vector that looks like this. Let me draw it like this. And purposely not having them all start at the origin because they don't have to all start at the origin, but you get a vector that looks like this. That looks like this. So what's the difference between A and negative two times A?"}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "Let me draw it like this. And purposely not having them all start at the origin because they don't have to all start at the origin, but you get a vector that looks like this. That looks like this. So what's the difference between A and negative two times A? Well, the negative flipped it over and then the two flipped it over and now it has twice the magnitude, but because of the negative, it has twice the magnitude in the other direction."}, {"video_title": "Multiplying a vector by a scalar Vectors and spaces Linear Algebra Khan Academy.mp3", "Sentence": "So what's the difference between A and negative two times A? Well, the negative flipped it over and then the two flipped it over and now it has twice the magnitude, but because of the negative, it has twice the magnitude in the other direction."}]