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. . . . . . . . . . . . . . 71 9 Matrix Algebra 75 9.1 Sums of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 9.2 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 9.3 Matrix Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8... | {
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. . 93 11.3 Triangular Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 12 Properties of the Determinant 97 12.1 ERO and Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 12.2 Determinants and Invertibility of Matrices . . . . . . . . . . . . . . . . . . . 100 12.3 P... | {
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. . 109 14.2 Subspaces of Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 15 Linear Maps 117 15.1 Linear Maps on Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 117 15.2 Null space and Column space . . . . . . . . . . . . . . . . . . . . . . . . . . 121 16 Linear Independence... | {
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. . . . . . . . . . . . . . 137 18.2 Coordinate Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 18.3 Matrix Representation of a Linear Map . . . . . . . . . . . . . . . . . . . . . 142 19 Change of Basis 147 19.1 Review of Coordinate Mappings on Rn . . . . . . . . . . . . . . . . . . . . . 14... | {
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λ = 0 is an eigenvalue . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 22 The Characteristic Polynomial 169 22.1 The Characteristic Polynomial of a Matrix . . . . . . . . . . . . . . . . . . . 169 22.2 Eigenvalues and Similarity Transformations . . . . . . . . . . . . . . . . . . 176 23 Diagonalization 179 23... | {
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. . . . . . . . . . . . . . . . . . . . . . . 191 25.2 A Description of the PageRank Algorithm . . . . . . . . . . . . . . . . . . . 192 25.3 Computation of the PageRank Vector . . . . . . . . . . . . . . . . . . . . . . 195 26 Discrete Dynamical Systems 197 26.1 Discrete Dynamical Systems . . . . . . . . . . . . . . .... | {
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x2 + a23 x3 + · · · + a2nxn = b2 a31 x1 + a32 x2 + a33 x3 + · · · + a3nxn = b3 ... ... ... ... ... am1x1 + am2x2 + am3x3 + · · · + amn xn = bm (1.1) The numbers aij are called the coefficients of the linear system; because there are m equa-tions and n unknown variables there are thefore m × n coefficients. The main pro... | {
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and n = 6 unknowns: −5x1 + x3 − 44 x4 − 55 x6 = −1 πx 1 − 5x2 − x3 + 4 x4 − 5x5 + √5x6 = 0 63 x1 − √2x2 − 1 > 5 x3 + ln(3) x4 + 4 x5 − 1 > 33 x6 = 0 63 x1 − √2x2 − 1 5x3 − 1 > 8 x4 − 5x6 = 5 Example 1.2. Verify that (1 , 2, −4) is a solution to the system of equations 2x1 + 2 x2 + x3 = 2 x1 + 3 x2 − x3 = 11 . Is (1 , −... | {
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system will have either just one solution or infinitely many solutions. For example, a linear system cannot have just 4 or 5 solutions. If it has multiple solutions, then it will have infinitely many solutions. Example 1.3. Show that the linear system does not have a solution. −x1 + x2 = 3 x1 − x2 = 1 . Solution. If we... | {
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and to study the properties of the solution set of a linear system. Informally speaking, a matrix is an array or table consisting of rows and columns . For example, A = 1 −2 1 00 2 −8 8 −4 7 11 −5 is a matrix having m = 3 rows and n = 4 columns. In general, a matrix with m rows and n columns is a m × n matrix and... | {
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can write this as A ∈ Mm×n.If we are given an augmented matrix, we can write down the associated linear system in an obvious way. For example, the linear system associated to the augmented matrix 1 4 −2 8 12 0 1 −7 2 −40 0 5 −1 7 is x1 + 4 x2 − 2x3 + 8 x4 = 12 x2 − 7x3 + 2 x4 = −45x3 − x4 = 7 . > 4 Lecture 1 We c... | {
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. The goal with row reducing is to transform the original linear system into one having a triangular structure and then perform back substitution to solve the system. This is best explained via an example. Example 1.5. Use back substitution on the augmented matrix 1 0 −2 −40 1 −1 00 0 1 1 to solve the associated ... | {
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0 −2 −40 2 −2 02 −3 4 −3 −2R1+R3 −−−−−→ 1 0 −2 −40 2 −2 00 −3 8 5 Multiply R2 by 1 > 2 : 1 0 −2 −40 2 −2 00 −3 8 5 > 1 > 2R2 −−→ 1 0 −2 −40 1 −1 00 −3 8 5 Add 3 R2 to R3: 1 0 −2 −40 1 −1 00 −3 8 5 3R2+R3 −−−−→ 1 0 −2 −40 1 −1 00 0 5 5 Multiply R3 by 1 > 5 : 1 0 −2 −40 1 −1 00 0 5 5 ... | {
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1 0 −2 −40 1 −1 00 0 1 1 → x1 − 2x3 = −4 x2 − x3 = 0 x3 = 1 . Although the two augmented matrices M and N are clearly distinct, it is a fact that they have the same solution set. Example 1.7. Using elementary row operations, show that the linear system is inconsistent. x1 + 2 x3 = 1 x2 + x3 = 0 2x1 + 4 x3 = 1 Soluti... | {
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the equations of the system. The solution for this system is (3 , 2). The two lines intersect at the point ( x1, x 2) = (3 , 2), see Figure 1.1 . Figure 1.1: The intersection point of the two lines is the solution of the linear system ( 1.2 )Similarly, the solution of the linear system x1 − 2x2 + x3 = 0 2x2 − 8x3 = 8 −... | {
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10 Lecture 2 # Lecture 2 Row Reduction and Echelon Forms In this lecture, we will get more practice with row reduction and in the process introduce two important types of matrix forms. We will also discuss when a linear system has a unique solution, infinitely many solutions, or no solution. Lastly, we will introduce a... | {
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5 0 0 0 0 0 0 0 We can perform elementary row operations, or row reduction , to transform a matrix into REF. Example 2.1. Explain why the following matrices are not in REF. Use elementary row operations to put them in REF. M = 3 −1 0 30 0 0 00 1 3 0 N = 7 5 0 −30 3 −1 10 6 −5 2 Solution. Matrix M fai... | {
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the first equation we obtain that x1 = 1. # 2.2 Reduced row echelon form (RREF) Although REF simplifies the problem of solving a linear system, later on in the course we will need to completely row reduce matrices into what is called reduced row echelon form (RREF) . A matrix is in RREF if it is in REF (so it satisfies... | {
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the newly created leading 1: 1 −3 4 −3 2 53 −7 8 −5 8 90 3 −6 6 4 −5 −3R1+R2 −−−−−→ 1 −3 4 −3 2 50 2 −4 4 2 −60 3 −6 6 4 −5 Create a leading 1 in the second row: 1 −3 4 −3 2 50 2 −4 4 2 −60 3 −6 6 4 −5 > 1 > 2R2 −−→ 1 −3 4 −3 2 50 1 −2 2 1 −30 3 −6 6 4 −5 Create zeros under the newly created lea... | {
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the row reduction algorithm and the matrix is in RREF. Example 2.3. Use row reduction to solve the linear system. 2x1 + 4 x2 + 6 x3 = 8 x1 + 2 x2 + 4 x3 = 8 3x1 + 6 x2 + 9 x3 = 12 Solution. The augmented matrix is 2 4 6 81 2 4 83 6 9 12 > 14 Lecture 2 Create a leading 1 in the first row: 2 4 6 81 2 4 83 6 9 12... | {
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− 2tx2 = tx3 = 4 (2.1) where t can be any real number . If we had chosen x1 to be the parameter, say x1 = t,then the solution set can be written as x1 = tx2 = −4 − 1 > 2 tx3 = 4 (2.2) Although ( 2.1 ) and ( 2.2 ) are two different parameterizations, they both give the same solution set. > 15 Row Reduction and Echelon F... | {
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− x5 = 4. We choose x5 to be the first parameter so we set x5 = t. Therefore, x4 = 4 + t. The second equation of the augmented matrix is x2 − 3x3 + 3 x4 + x5 = −5and the unassigned variables are x2 and x3. We choose x3 to be the second parameter, say x3 = s. Then x2 = −5 + 3 x3 − 3x4 − x5 = −5 + 3 s − 3(4 + t) − t = −1... | {
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3. All the rows of the augmented matrix are consistent and there are d ≥ 1 variables that must be set to arbitrary parameters In Case 1., the linear system is inconsistent and thus has no solution. In Case 2., the linear system is consistent and has only one (and thus unique ) solution. This case occurs when r = rank( ... | {
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n × 1 matrix. From now on, we will drop the “column” descriptor and simply use the word vectors . It is important to emphasize that a vector in Rn is simply a list of n numbers; you are safe (and highly encouraged!) to forget the idea that a vector is an object with an arrow. Here is a vector in R2: v = [ 3 −1 ] . Here... | {
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0 1 −3 3 1 −50 0 0 1 −1 4 Solution. The number of unknowns is n = 5 and the associated coefficient matrix A has rank r = 3. Thus, the solution set is parametrized by d = n − r = 2 parameters. This system was considered in Example 2.4 and the general solution was found to be x1 = −89 − 31 t1 + 19 t2 x2 = −17 − 4t1 + ... | {
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v2. This motivates the following definition. Definition 3.2: Let v1, v2, . . . , vp be vectors in Rn. A vector b is said to be a linear combination of the vectors v1, v2, . . . , vp if there exists scalars x1, x 2, . . . , x p such that x1v1 + x2v2 + · · · + xpvp = b.The scalars in a linear combination are called the c... | {
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, 2, 0), and let b2 = ( −3, 0, 7). Are b1 and b2 linear combinations of v1, v2? Solution. For any scalars x1 and x2 x1v1 + x2v2 = x1 00 + 00 x2 = x1 0 x2 6 = 020 and thus no, b1 is not a linear combination of v1, v2, v3. On the other hand, by inspection we have that −3v1 + 7 v2 = −300 + ... | {
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= 01 −2 . Comparing component-by-component in the above relationship, we seek scalars x1, x 2, x 3 satisfying the equations x1 + x2 + 2 x3 = 0 2x1 + x2 + x3 = 1 x1 + 2 x3 = −2. (3.3) This is just a linear system consisting of m = 3 equations and n = 3 unknowns! Thus, the linear combination problem can be solved b... | {
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uniqueness Theorem 2.5 , the only three possibilities to the linear combination problem are: 1. If the linear system is inconsistent then b is not a linear combination of v1, v2, . . . , vp,i.e., there does not exist scalars x1, x 2, . . . , x p such that x1v1 + x2v2 + · · · + xpvp = b. 2. If the linear system is consi... | {
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linear combination of the vectors v1 = 102 , v2 = 010 , v3 = 214 ? > 24 Lecture 3 Solution. The augmented matrix of the corresponding linear system is 1 0 2 10 1 1 02 0 4 1 . After row reducing we obtain that 1 0 2 10 1 1 00 0 0 −1 . The last row is inconsistent, and therefore the linear sy... | {
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−2 we obtain x1 = −4, x2 = −2, and x3 = 4, and you can verify that −4v1 − 2v2 + 4 v3 = −4 213 − 2 426 + 4 649 = 8812 = b > 25 Vector Equations We make a few important observations on linear combinations of vectors. Given vectors v1, v2, . . . , vp, there are certain vectors b that can be written... | {
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vector b is a linear combination of {v1, v2, . . . , vp}. We now take another point of view and instead consider the idea of generating all vectors that are a linear combination of {v1, v2, . . . , vp}. So how do we generate a vector that is guaranteed to be a linear combination of {v1, v2, . . . , vp}? For example, if... | {
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· · · + xpvp = b. Even though span {v1, v2, . . . , vp} is an infinite set of vectors, it is not necessarily true that it is the whole space Rn.The set span {v1, v2, . . . , vp} is just a collection of infinitely many vectors but it has some geometric structure. In R2 and R3 we can visualize span {v1, v2, . . . , vp}. ... | {
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the vectors v1 = (1 , −2, −5) , v2 =(2 , 5, 6)? In other words, is b ∈ span {v1, v2}? 27 Vector Equations Solution. By definition, b is in the span of v1 and v2 if there exists scalars x1 and x2 such that x1v1 + x2v2 = b, that is, if b can be written as a linear combination of v1 and v2. From our previous discussion on... | {
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3.11. Answer the following with True or False, and explain your answer. (a) The vector b = (1 , 2, 3) is in the span of the set of vectors −130 , 2 −70 , 4 −50 . (b) The solution set of the linear system whose augmented matrix is [v1 v2 v3 b] is the same as the solution set of the vector equatio... | {
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We begin with the definition of matrix-vector multiplication. Definition 4.1: Given a matrix A ∈ Mm×n and a vector x ∈ Rn, A = a11 a12 a13 · · · a1n a21 a22 a23 · · · a2n ... ... ... ... ... am1 am2 am3 · · · amn , x = x1 x2 ... xn , we define the product of A and x as the vector Ax in Rm given ... | {
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[−3] (b) Ax = [3 3 −24 −4 −1 ] 10 −1 = [ (3)(1) + (3)(0) + ( −2)( −1) (4)(1) + ( −4)(0) + ( −1)( −1) ] = [55 ] > 32 Lecture 4 (c) Ax = −1 1 04 1 −23 −3 30 −2 −3 −12 −2 = (−1)( −1) + (1)(2) + (0)( −2) (4)( −1) + (1)(2) + ( −2)( −2) (3)( −1) + ( −3)(2) + (3)( −2) (0)( −1) + ( −2)(2) + ( −3)( −2)... | {
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. . . , vn denote the columns of A and consider the following linear combination: x1v1 + x2v2 + · · · + xnvn = x1a11 x1a21 ... x1am1 + x2a12 x2a22 ... x2am2 + · · · + xna1n xna2n ... xnamn = x1a11 + x2a12 + · · · + xna1n x1a21 + x2a22 + · · · + xna2n ... x1am1 + x2am2 + · · · +... | {
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b. (⋆)Equation ( ⋆) is a matrix equation where the unknown variable is x. If u is a vector such that Au = b, then we say that u is a solution to the equation Ax = b. For example, > 34 Lecture 4 suppose that A = [1 01 0 ] , b = [−37 ] . Does the equation Ax = b have a solution? Well, for any x = [x1 x2 ] we have that Ax... | {
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... ... am1x1 + am2x2 + am3x3 + · · · + amn xn = bm will be written in the compact form Ax = b where A is the coefficient matrix of the linear system, b is the output vector, and x is the unknown vector to be solved for. We summarize our findings with the following theorem. Theorem 4.6: Let A ∈ Mm×n and b ∈ Rm. The fol... | {
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= b In other words, b is a linear combination of the columns of A: −11 11 −3 + 3 357 + 0 −42 −6 = −2412 > 36 Lecture 4 Example 4.8. Solve, if possible, the matrix equation Ax = b if A = [1 22 4 ] , b = [ 3 −4 ] . Solution. Row reducing the augmented matrix [A b] we get [1 2 32 4 −4 ] −2R1+R2 −−−... | {
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3 − 2tx3 = t where t is an arbitrary number. Therefore, the matrix equation Ax = b has an infinite number of solutions and they can all be written as x = > 5 > 3 − 4t −1 > 3 − 2tt > 37 The Matrix Equation Ax = b where t is an arbitrary number. Equivalently, b can be written as a linear combination of the columns ... | {
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is consistent and has solution x = [2.51.5 ] Therefore, b is in span {v1, v2}, and b can be written in terms of v1 and v2 as 2.5v1 + 1 .5v2 = b > 38 Lecture 4 If v1, v2, . . . , vp are vectors in Rn and it happens to be true that span {v1, v2, . . . , vp} = Rn then we would say that the set of vectors {v1, v2, . . . , ... | {
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= R3. In other words, every vector b ∈ R3 can be written as a linear combination of v1, v2, v3. After this lecture you should know the following: • how to multiply a matrix A with a vector x • that the product Ax is a linear combination of the columns of A • how to solve the matrix equation Ax = b if A and b are known ... | {
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solution set is d = n − r, where r is the rank of the coefficient matrix A and n is the number of unknowns. Example 5.2. Does the linear homogeneous system have any nontrivial solutions? 3x1 + x2 − 9x3 = 0 x1 + x2 − 5x3 = 0 2x1 + x2 − 7x3 = 0 Solution. The linear system will have a nontrivial solution if the solution s... | {
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previous example, when solving a homogeneous system Ax = 0 using row reduction, the last column of the augmented matrix [A 0] remains unchanged (always 0) after every elementary row operation. Hence, to solve a homogeneous system, we can row reduce the coefficient matrix A only and then set all rows equal to zero when ... | {
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> v3 = t1v1 + t2v2 + t3v3 where t1, t 2, t 3 are arbitrary parameters. In other words, any solution x is in the span of v1, v2, v3: x ∈ span {v1, v2, v3}. The form of the general solution in Example 5.3 holds in general and is summarized in the following theorem. Theorem 5.4: Consider the homogenous linear system Ax = ... | {
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= b. And suppose that v is a solution to the homogeneous system Ax = 0, that is, Av = 0. Now let q = p + v. Then Aq = A(p + v)= Ap + Av = b + 0 = b. Therefore, Aq = b. In other words, q = p + v is also a solution of Ax = b. We have therefore proved the following theorem. Theorem 5.5: Suppose that the linear system Ax =... | {
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p + t1v1 + t2v2 + · · · + tpvd. Therefore, the solution set of Ax = b is a shift of the span {v1, v2, . . . , vd} by the vector p.This is illustrated in Figure 5.1 . > bbbbb 0pv tvp + tv span {v} p + span {v} Figure 5.1: The solution sets of a homogeneous and nonhomogeneous system. Example 5.6. Write the general soluti... | {
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6 3 −1 1 −2 −12 −2 4 2 . Solution. Write the general solution, in parametric vector form, of the linear system repre-sented by the augmented matrix 3 −3 6 3 −1 1 −2 −12 −2 4 2 The RREF of the augmented matrix is 3 −3 6 3 −1 1 −2 −12 −2 4 2 ∼ 1 −1 2 10 0 0 00 0 0 0 Here n = 3, r = 1 and therefore th... | {
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vd} where p satisfies Ap = b and Av i = 0. After this lecture you should know the following: • what a homogeneous/nonhomogenous linear system is • when a homogeneous linear system has nontrivial solutions • how to write the general solution set of a homogeneous system in parametric vector form Theorem 5.4 ) • how to wr... | {
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− 7v3 x = 0 v1 − 4v2 − 3v3. The fact that x can be written in more than one way in terms of v1, v2, v3 suggests that there might be a redundancy in the set {v1, v2, v3}. In fact, it is not hard to see that v3 = −v1 +v2,and thus v3 ∈ span {v1, v2}. The preceding discussion motivates the following definition. Definition ... | {
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vn} is linearly independent if and only if 0 can be written in only one way as a linear combination of {v1, v2, . . . , vn}. In other words, if t1v1 + t2v2 + · · · + tnvn = 0 then necessarily the coefficients t1, t 2, . . . , t n are all zero. Proof. If {v1, v2, . . . , vn} is linearly independent then every vector x ∈... | {
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zero. Therefore, we must have r1 − s1 = 0, r2 − s2 = 0 , . . . , r n − sn = 0, or equivalently that r1 = s1, r 2 = s2, . . . , r n = sn. Therefore, the linear combinations r1v1 + r2v2 + · · · + rnvn = x s1v1 + s2v2 + · · · + snvn = x are actually the same. Therefore, each x ∈ span {v1, v2, . . . , vn} can be written un... | {
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+ · · · + xnvn = 0 (6.1) has only the trivial solution. Now, the vector equation ( 6.1 ) is a homogeneous linear system of equations with coefficient matrix A = [v1 v2 · · · vn ] . Therefore, the set {v1, v2, . . . , vn} is linearly independent if and only if the the homogeneous system Ax = 0 has only the trivial solut... | {
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Clearly, r = rank( A) = 2, which is not equal to the number of vectors, n = 3. Therefore, {v1, v2, v3} is linearly dependent. We will find a nontrivial linear combination of the vectors v1, v2, v3 that gives the zero vector 0. The REF of A = [ v1 v2 v3] is A ∼ 1 4 20 −3 −30 0 0 Since r = 2, the solution set of... | {
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• Any set {v1, v2, . . . , vp} containing the zero vector, say that vp = 0, is linearly depen-dent. For example, the linear combination 0v1 + 0 v2 + · · · + 0 vp−1 + 2 vp = 0 is a non-trivial linear combination giving the zero vector 0. # 6.2 The maximum size of a linearly independent set The next theorem puts a constr... | {
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below linearly independent? v1 = 830 −2 , v2 = 411 −46 , v3 = 2011 , v4 = 3 −9 −53 , v5 = 0 −2 −77 . Solution. The vectors v1, v2, v3, v4, v5 are in R4. Therefore, by Theorem 6.7 , the set {v1, . . . , v5} is linearly dependent. To see this explicitly, let A = [v1 v2 v3... | {
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, vd} is also linearly independent. After this lecture you should know the following: • the definition of linear independence and be able to explain it to a colleague • how to test if a given set of vectors are linearly independent (Theorem 6.4 ) • the relationship between the linear independence of {v1, v2, . . . , vp... | {
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+ x22 + 1 ] . The vector b = (3 , −1) is not in the range of T because the second component of T(x) is positive. On the other hand, b = ( −1, 2) is in the range of T because T ([ 10 ]) = [12 sin(0) − cos(1 2 − 1) 12 + 0 2 + 1 ] = [−12 ] = b. Hence, a corresponding input for this particular b is x = (1 , 0). In Figure 7... | {
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= 3 T ([ 10 ]) = 3 [−12 ] = [−36 ] . However, T ([ 30 ]) = [32 sin(0) − cos(3 2 − 1) 32 + 0 2 + 1 ] = [− cos(8) 10 ] 6 = [−36 ] . Example 7.3. Is the vector mapping T : R2 → R3 linear? T ([ x1 x2 ]) = 2x1 − x2 x1 + x2 −x1 − 3x2 Solution. We must verify that the two conditions in Definition 7.2 hold. For the first... | {
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c(−u1 − 3u2) = c 2u1 − u2 u1 + u2 −u1 − 3u2 = cT(u)Therefore, both conditions of Definition 7.2 hold, and thus T is a linear map. Example 7.4. Let α ≥ 0 and define the mapping T : Rn → Rn by the formula T(x) = αx.If 0 ≤ α ≤ 1 then T is called a contraction and if α > 1 then T is called a dilation . In either c... | {
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will use the notation TA to indicate that TA is associated to A. We proved in Lecture 4 (Theorem 4.3 ), that for any u, v ∈ Rn, and scalar c, matrix-vector multiplication satisfies the properties: 1. A(u + v) = Au + Av 2. A(cu) = cAu .The following theorem is therefore immediate. Theorem 7.5: To a given matrix A ∈ Rm×n... | {
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is clear that finding such a vector x is equivalent to solving the matrix equation Ax = b. In summary, we have the following theorem. Theorem 7.7: Let T : Rn → Rm be a matrix mapping corresponding to A, that is, T(x) = Ax . Then b ∈ Rm is in the range of T if and only if the matrix equation Ax = b has a solution. Let T... | {
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−6 12 ∼ 1 3 −4 −20 1 3 30 0 −12 0 The system is consistent and the (unique) solution is x = ( −11 , 3, 0). Therefore, b is in the range of T. # 7.4 Examples If T : Rn → Rm is a linear mapping, then for any vectors v1, v2, . . . , vp and scalars c1, c 2, . . . , c p, it holds that T(c1v1 + c2v2 + · · · + cpvd) ... | {
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θ)sin( α + θ) ] . Then from the angle sum trigonometric identities: Tθ(v) = [cos( α + θ)sin( α + θ) ] = [cos( α) cos( θ) − sin( α) sin( θ)cos( α) sin( θ) + sin( α) cos( θ) ] But Tθ(v) = [cos( α) cos( θ) − sin( α) sin( θ)cos( α) sin( θ) + sin( α) cos( θ) ] = [cos( θ) − sin( θ)sin( θ) cos( θ) ] [ cos( α)sin( α) ]︸ ︷︷ ︸ >... | {
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x’s such that T(x) = b. > b > bb x = x1 x2 x3 T(x) = x1 x2 0 Figure 7.2: Projection onto the ( x1, x 2) plane > 64 Lecture 7 After this lecture you should know the following: • what a vector mapping is • what the range of a vector mapping is • that the co-domain and range of a vector mapping are generally n... | {
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and only if the columns of A span all of Rm.Combining Theorem 4.11 and Theorem 8.2 we have: Theorem 8.3: Let TA : Rn → Rm be the matrix mapping TA(x) = Ax , where A ∈ Rm×n.Then TA is onto if and only if r = rank( A) = m. Example 8.4. Let T : R3 → R3 be the matrix mapping with corresponding matrix A = 1 2 −1 −3 −4 25... | {
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Below is a theorem which places restrictions on the size of the domain of an onto mapping. Theorem 8.6: Suppose that TA : Rn → Rm is a matrix mapping corresponding to A ∈ Mm×n. If TA is onto then m ≤ n. Proof. If TA is onto then the rref (A) has r = m leading 1’s. Therefore, A has at least m columns. The number of colu... | {
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in the range of T is an existence question. Indeed, if b ∈ Range( T) then there exists a x ∈ Rm such that T(x) = b. We now want to look at the problem of whether x is unique . That is, does there exist a distinct y such that T(y) = b. Definition 8.9: A vector mapping T : Rn → Rm is said to be one-to-one if for each b ∈... | {
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−2 0 2 . Is TA one-to-one? Solution. By Theorem 8.11 , TA is one-to-one if and only if the columns of A are linearly independent. The columns of A lie in R3 and there are n = 4 columns. From Lecture 6, we know then that the columns are not linearly independent. Therefore, TA is not one-to-one. Alternatively, A will ... | {
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, e2 = 010...0 , e3 = 001...0 , · · · , en = 000...1 . Every x ∈ Rn is in span {e1, e2, . . . , en} because: x = x1 x2 ... xn = x1 10...0 + x2 01...0 + · · · + xn 00...1 = x1e1 + x2e2 + · · · + xnen With this notation we pro... | {
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that if T is a linear mapping, then to derive properties of T we need only know the standard matrix A corresponding to T. Example 8.15. Let T : R2 → R2 be the linear mapping that rotates every vector by an angle θ. Use the standard unit vectors e1 = [10 ] and e2 = [01 ] in R2 to write down the matrix A ∈ R2×2 correspon... | {
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a11 a12 · · · a1n a21 a22 · · · a2n ... ... ... ... am1 am2 · · · amn , B = b11 b12 · · · b1n b21 b22 · · · b2n ... ... ... ... bm1 bm2 · · · bmn , both of the same dimension m × n, the sum A + B is defined as A + B = a11 + b11 a12 + b12 · · · a1n + b1n a21 + b21 a22 + b22 · · · a2n + b2n ... | {
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α, β be scalars. Then (a) A + B = B + A (d) α(A + B) = αA + αB (b) ( A + B) + C = A + ( B + C) (e) ( α + β)A = αA + βA (c) A + 0 = A (f) α(βA) = ( αβ )A # 9.2 Matrix Multiplication Let TB : Rp → Rn and let TA : Rn → Rm be linear mappings. If x ∈ Rp then TB(x) ∈ Rn and thus we can apply TA to TB(x). The resulting vector... | {
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1 is Be 1 = [b1 b2 · · · bp ] e1 = b1. And similarly Be i = bi for all i = 1 , 2, . . . , p . Therefore, C = [Ab 1 Ab 2 · · · Ab p ] is the standard matrix of TA ◦ TB. This computation motivates the following definition. Definition 9.5: For A ∈ Rm×n and B ∈ Rn×p, with B = [b1 b2 · · · bp ], we define the product AB by ... | {
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and BA . A = −4 4 33 −3 −1 −2 −1 1 , B = −1 −1 0 −3 0 −2 −2 1 −2 Solution. First AB = [ Ab 1 Ab 2 Ab 3]: AB = −4 4 33 −3 −1 −2 −1 1 −1 −1 0 −3 0 −2 −2 1 −2 = −14 83= −14 78 −43 3= −14 7 −14 8 −4 83 3 0 > 78 Lecture 9 Next BA = [ Ba 1 Ba 2 Ba 3]: BA = −1 −1 0 −3 0 −2 −2 1 −2 −4... | {
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Rn×n is a square matrix, the kth power of A is Ak = AAA · · · A︸ ︷︷ ︸ > ktimes > 79 Matrix Algebra Example 9.9. Compute A3 if A = [ −2 31 0 ] . Solution. Compute A2: A2 = [ −2 31 0 ] [ −2 31 0 ] = [ 7 −6 −2 3 ] And then A3: A3 = A2A = [ 7 −6 −2 3 ] [ −2 31 0 ] = [ −20 21 7 −6 ] We could also do: A3 = AA 2 = [ −2 31 0 ]... | {
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−1 −2 31 −10 −3 = 3 −5 −4 5 −4 9 = ( AB )T The following theorem summarizes properties of the transpose. Theorem 9.12: Let A and B be matrices of appropriate sizes. The following hold: (1) (AT )T = A (2) (A + B)T = AT + BT (3) (αA)T = αAT (4) (AB )T = BT AT A consequence of property (4) is that (A1A2 . . ... | {
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other hand, the standard matrix corresponding to a contraction by a factor k = 1 > 3 is [1 > 3 00 1 > 3 ] Therefore, [cos( θ) − sin( θ)sin( θ) cos( θ) ]︸ ︷︷ ︸ > rotation [1 > 3 00 1 > 3 ]︸ ︷︷ ︸ > contraction = [1 > 3 cos( θ) −1 > 3 sin( θ) > 1 > 3 sin( θ) 1 > 3 cos( θ) ] = A After this lecture you should know the follo... | {
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if A is invertible, it can have only one inverse. This motivates the following definition. Definition 10.2: If A is invertible then we denote the inverse of A by A−1. Thus, AA −1 = A−1A = In. Example 10.3. Given A and C below, show that C is the inverse of A. A = 1 −3 0 −1 2 −2 −2 6 1 , C = −14 −3 −6 −5 −1 −22... | {
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1 −3 0 −1 2 −2 −2 6 1 , b = 1 −3 −1 . Solution. We showed in Example 10.3 that A−1 = −14 −3 −6 −5 −1 −22 0 1 . Therefore, the unique solution to the linear system Ax = b is A−1b = −14 −3 −6 −5 −1 −22 0 1 1 −3 −1 = 101 > 84 Lecture 10 Verify: 1 −3 0 −1 2 −2 −2 6 1 101 = 1... | {
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AA −1 = [e1 e2 · · · en ]︸ ︷︷ ︸ > In . To find ci we therefore need to solve the linear system Ax = ei. Here the image vector “ b”is ei. To find c1 we form the augmented matrix [A e1 ] and find its RREF: [A e1 ] ∼ [In c1 ] . > 85 Invertible Matrices We will need to do this for each c2, . . . , cn so we might as well fo... | {
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invertible. The inverse is A−1 = [−2 −31 1 ] Verify: AA −1 = [ 1 3 −1 −2 ] [ −2 −31 1 ] = [1 00 1 ] . Example 10.8. Find the inverse of A = 1 0 31 1 0 −2 0 −7 if it exists. > 86 Lecture 10 Solution. Form the augmented matrix [A I3 ] and row reduce: 1 0 3 1 0 01 1 0 0 1 0 −2 0 −7 0 0 1 −R1+R2, 2R1+R2 −−−−−−−... | {
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0 1 −R1+R2, 2R1+R2 −−−−−−−−−−→ 1 0 1 1 0 00 1 −3 −1 1 00 0 0 2 0 1 We need not go further since the rref (A) is not I3 (rank( A) = 2 ). Therefore, A is not invertible. # 10.3 Invertible Linear Mappings Let TA : Rn → Rn be a matrix mapping with standard matrix A and suppose that A is invertible. Let TA−1 : Rn →... | {
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matrices and matrix mappings specialized to the case of square matrices A ∈ Rn×n. Note that for non-square matrices, one-to-one does not imply ontoness, and conversely. Example 10.11. Without doing any arithmetic, write down the inverse of the dilation matrix A = [3 00 5 ] . Example 10.12. Without doing any arithmetic,... | {
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a22 ∣∣∣∣ . 89 Determinants Example 11.2. Compute the determinant of A.(i) A = [3 −18 2 ] (ii) A = [ 3 1 −6 −2 ] (iii) A = [−110 0568 0 ] Solution. For (i): det( A) = ∣∣∣∣3 −18 2 ∣∣∣∣ = (3)(2) − (8)( −1) = 14 For (ii): det( A) = ∣∣∣∣ 3 1 −6 −2 ∣∣∣∣ = (3)( −2) − (−6)(1) = 0 For (iii): det( A) = ∣∣∣∣−110 0568 0 ∣∣∣∣ = ( −... | {
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a23 a31 a32 a33 −→ A11 = [a22 a23 a32 a33 ] . 90 Lecture 11 Similarly, the matrix A12 = [a21 a23 a31 a33 ] is obtained from A by deleting the 1st row and the 2nd column: A = a11 a12 a13 a21 a22 a23 a31 a32 a33 −→ A12 = [a21 a23 a31 a33 ] . Finally, the matrix A13 = [a21 a22 a31 a32 ] is obtained from A by dele... | {
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is C23 = ( −1) 2+3 det A23 = − det A23 . A helpful way to remember the sign ( −1) j+k of a cofactor is to use the matrix + − + − + − + − + . This works not just for 3 × 3 matrices but for any square n × n matrix. Example 11.4. Compute the determinant of the matrix A = 4 −2 32 3 51 0 6 > 91 Determinants Solu... | {
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along either row j or column k because ajk Cjk = 0 and we need not compute Cjk . Example 11.5. Compute the determinant of the matrix A = 4 −2 32 3 51 0 6 Solution. In Example 11.4 , we computed det( A) = 77 by expanding along the 1st row. > 92 Lecture 11 Notice that a32 = 0. Expanding along the 3rd row: det A = (... | {
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0. Proof. If the jth row contains all zeros then aj1 = aj2 = · · · = ajn = 0: det A = aj1Cj1 + aj2Cj2 + · · · + ajn Cjn = 0 . > 93 Determinants Corollary 11.9: For any square matrix A it holds that det A = det AT . Sketch of the proof. Expanding along the jth row of A is equivalent to expanding along the jth column of ... | {
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3 −20 0 1 −1 −3 0 ∣∣∣∣∣∣ − ∣∣∣∣∣∣ 1 3 00 0 2 −1 −3 1 ∣∣∣∣∣∣ = −1( −3 − 12) + 2( −1 − 4) + 2(0) − (0) = 5 # 11.3 Triangular Matrices Below we introduce a class of matrices for which the determinant computation is trivial. Definition 11.12: A square matrix A ∈ Rn×n is called upper triangular if ajk = 0 whenever j > k . I... | {
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the ( j, k )-cofactor of A and aj = [aj1 aj2 · · · ajn ] denotes the jth row of A. Notice that det A = [aj1 aj2 · · · ajn ] Cj1 Cj2 ... Cjn . If we let cj = [Cj1 Cj2 · · · Cjn ] then det A = aj · cTj . In this lecture, we will establish properties of the determinant under elementary row opera-tions and some ... | {
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Then det B = β det A. Proof. Suppose that B is obtained from A by multiplying the jth row by β. The rows of A and B different from j are equal, and therefore Bjk = Ajk , for k = 1 , 2, . . . , n .In particular, the ( j, k ) cofactors of A and B are equal. The jth row of B is βaj . Then, expanding det B along the jth ro... | {
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−11 . Example 12.6. Suppose that A is a 4 × 4 matrix and suppose that det A = 11. Let a1, a2, a3, a4 denote the rows of A. If B is obtained from A by replacing row a3 by 3 a1 + a3,what is det B? Solution. This is a Type 3 elementary row operation, which preserves the value of the de-terminant. Therefore, det B = 11 . E... | {
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a2 · cT > 3 = 0 + 0 = 0 # 12.2 Determinants and Invertibility of Matrices The following theorem characterizes invertibility of matrices with the determinant. Theorem 12.9: A square matrix A is invertible if and only if det A 6 = 0. Proof. Beginning with the matrix A, perform elementary row operations and generate a seq... | {
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βa 12 βa 22 ∣∣∣∣ = βa 11 · βa 22 − βa 12 · βa 21 = β2(a11 a22 − a12 a21 )= β2 det A. Thus, the statement holds for 2 × 2 matrices. Consider a 3 × 3 matrix A. Then det( βA) = βa 11 |βA11 | − βa 12 |βA12 | + βa 13 |βA13 | = βa 11 β2|A11 | − βa 12 β2|A12 | + βa 13 β2|A13 | = β3 (a11 |A11 | − a12 |A12 | + a13 |A13 |)= β3 d... | {
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AC is invertible. (ii): We have det( AB ) = det A det B = 3 · 0 = 0. Thus, AB is not invertible. (iii): We have det( ACB ) = det A det C det B = 3 ·7·0 = 0. Thus, ACB is not invertible. After this lecture you should know the following: • how the determinant behaves under elementary row operations • that A is invertible... | {
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103 Applications of the Determinant Form the Cofactor matrix Cof( A) = C11 C12 · · · C1n C21 C22 · · · C2n ... ... · · · ... Cn1 Cn2 · · · Cnn = c1 c2 ... cn . Then, A(Cof( A)) T = a1 a2 ... an [cT > 1 cT > 2 · · · cTn ] = a1cT > 1 a1cT > 2 · · · a1cTn a2cT > 1 a2cT > 2 · · ... | {
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Rm×n is called an integer matrix if every entry of A is an integer. Suppose that A ∈ Rn×n is an invertible integer matrix. Then det( A) is a non-zero integer and (Cof( A)) T is an integer matrix. If A−1 is also an integer matrix then det( A−1) is also an integer. Now det( A) det( A−1) = 1 thus it must be the case that ... | {
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