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, then it follows that Bob and Brenda have the same hair color, so the transitive property holds too and we are done. 374 G.2 Systems of Linear Equations 375 Hint for Review Question 5 The first part for Review Question 5 is simple--just write out the associated linear system and you will find the equation 0 = 6 which ...
. Hint for Review Question 5 Lets worry about the last part of the problem. The problem can be solved by considering a non-zero simple polynomial, such as a degree 0 polynomial, and multiplying by i ∈ C. That is to say we take a vector p ∈ P R 3 and then considering i·p. This will violate one of the vector space rules ...
ch says that (AB)−1 = B−1A−1. Lets check to see if this works A−1 = 5 −3 1 −1 1 2 and B−1 = 1 −2 0 1 and B−1A−1 = 1 0 −2 1 5 −3 1 −1 = 2 0 1 2 0 2 Hint for Review Problem 3 Firstnote that (b) implies (a) is the easy direction: just think about what it means for M to be non-singular and for a linear function to be well-...
or maybe, to make this a square matrix M we could write this as The matrix for this would be M = 1 0 determinant, this row of all zeros gets multiplied in every term. If instead we were given redundant equations and det(M ) = 0. When we compute the 1 0 x + y = 1 2x + 2y = 2 The matrix for this would be M = 1 2 1 2 and...
of possible vectors in the span of two vectors. 406 G.10 Basis and Dimension 407 G.10 Basis and Dimension Proof Explanation through the proof of theorem 11.0.1. We want to show that for Lets walk S = {v1, . . . , vn} a basis for a vector space V , then every vector w ∈ V can be written uniquely as a linear combination ...
such that v(0) = v(1) = v(2) . . ., we must look for a vector that satisfies v = M v. What eigenvalue does this correspond to? If you found a v(0) with this property would cv(0) for a scalar c also work? Remember that eigenvectors have to be nonzero, so what if c = 0? For part (c) if we tried an eigenvector would we ha...
Q =    1√ 2 0 − 1√ 2 Overview − 1√ 2 2 3 1√ 3 1√ 3 3 − 1√ 3 2 3 This video depicts the ideas of a subspace sum, a direct sum and an orthogonal complement in R3. Firstly, lets start with the subspace sum. Remember that even if U and V are subspaces, their union U ∪ V is usually not a subspace. However, the span of t...
is one-one. If we found x and y with L(x) = L(y), then we know x = y. But if L(v) = 0 then L(x) + L(v) = L(y). Does this cause a problem? 431 432 Movie Scripts G.16 Least Squares and Singular Values Least Squares: Hint for Review Problem 1 Lets work through this problem. Let L : U → V be a linear transformation. Suppo...
8 2y + 5z = 19. This system can be written as an augmented matrix as follows: 1 3 6 25 2 7 14 58 0 2 5 19   .   A matrix is just a 2-dimensional array of numbers. An augmented matrix has two parts separated by a vertical line. Notice that the augmented matrix notation has exactly the same information as the origina...
y, substituting y = 1 and z = 3 into the ﬁrst equation, we have x + 4(1) + 3(3) = 11, which we can solve to get x = 2. Therefore we have the solution (x, y, z) = ( 2, 1, 3).   At this point, it is a good idea to double-check this solution using the original equations − − x + 4y + 3z = 11 2x + 10y + 7z = 27 x + y + 2z...
ue     ∗ ∗ ∗ ∗     denotes an arbitrary number and Exercise 1.4.4 Consider the following augmented matrix in which denotes a non-zero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique ∗         26 Systems of linear equations Exercise 1.4.5 Suppose ...
atrix with only one constant vector, we create an augmented matrix containing both constant vectors at the same time     Then we row-reduce the coefﬁcient matrix to reduced echelon form as usual. (We do not need to bother reducing the right-hand side to reduced echelon form).    R2 R3 ← ← 2R1 3R1 − − R2 R3 ≃  ...
k 2. (d) 3 equations in 4 variables, rank 2. Exercise 1.6.3 My system of equations has a solution (x, y, z) = (1, 2, 4). The associated homogeneous system has basic solutions (x, y, z) = (1, 0, 1) and (x, y, z) = (0, 1, 1). What is the general solution of my system of equations? − 1.7 Uniqueness of the reduced echelon ...
on: 1a = a; (M4) Multiplicative inverse: for each non-zero a aa− 1 = 1; (D) Distributivelaw: a(b + c) = ab + ac. 1 K, there exists an element a− ∈ K such that ∈ Properties (A1)–(A4) are about addition, properties (M1)–(M4) are about multiplication, and property (D) is about both addition and multiplication. Here are so...
erefore the general solution is z = t, y = 2 − i.e., t is any complex number. In vector form, the general solution is i)t, x = 1 + i (1 − − R1 ← R1+R2 ≃ )t, where t − ∈ C is a parameter, ♠ Exercises Exercise 1.8.1 Solve each of the following systems of equations with scalars in Z2. If there is more than one solution, w...
ts, to get a formula for the lift that would be reasonable. You could also consider more variables and more complicated situations in the same way. Exercises Exercise 1.10.1 In this section, we observed that ρV 2AB has the units of force. Describe a systematic way to obtain such combinations of the variables that will ...
is similar. Here, a vector is described by three components, namely, its . The same idea x-component, y-component, and z-component. The three components are written as generalizes to n-dimensional vectors when n is greater than 3. x y z h i Deﬁnition 2.1: Column vectors and Rn A n-dimensionalcolumnvector,oftensimplyca...
Exercise 2.3.2 Find .     Exercise 2.3.3 Use the properties of scalar multiplication from Proposition 2.11 and the properties of vector addition from Proposition 2.8 to prove the following equalities. Justify every step. (a) (k + ℓ)(u + v) = ku + kv + ℓu + ℓv. (b) 0u = 0. u. 1)u = − − (ku) = k( (c) ( (d) − u) = ( −...
two vectors, the result is a scalar. For this reason, the dot product is also called the scalar product. Sometimes it is also called the inner product. The deﬁnition is as follows. Deﬁnition 2.23: Dot product u1 u2 ... un Let =      v1 v2 ... vn  betwovectorsin Rn. Wedeﬁnetheirdotproduct as     u · v = u1v1 ...
2.6.4 Find cosθ where θ is the angle between the vectors     Exercise 2.6.5 Find cosθ where θ is the angle between the vectors   −   Exercise 2.6.6 Use the formula given in Proposition 2.30 to verify the Cauchy Schwarz inequality and to show that equality occurs if and only if one of the vectors is a scalar mul...
k k u v w cosθ = (u v) w. Consider what happens if you interchange v with w or u with w. Geometrically, we can see that this merely introduces a minus sign. We ﬁnd that the box product of three vectors equals the volume of the 96 Vectors in Rn parallelepiped determined by the three vectors if the three vectors form a r...
n the line L if and only if there exists some such that ∈ Subtracting [1, 2, 1]T from both sides of the equation, this is equivalent to 3.1. Lines 105 2 3 1   We can write this as a set of parametric equations:   t . 4 6 3   =   2t = 4, 3t = 6, t = 3. This is a system of three linear equations in one variable, ...
tion If p and q are the position vectors of P and Q, respectively, we have −→PQ = q of the plane can be written as − 3.2. Planes 113 or equivalently, n (q p) = 0, − · n · q = n p. · This is called the normal equation of the plane. Note that in this equation, n and p are given and ﬁxed, whereas q is a variable ranging o...
rcise 3.2.14 Find the shortest distance from the point P = (1, 1, 6, and ﬁnd the point Q on the plane that is closest to P. 1) to the plane given by x+2y+2z = − Exercise 3.2.15 Use Exercise 2.7.15 to ﬁnd an equation of a plane containing the two vectors p and q and the point 0. Hint: If (x, y, z) is a point in this pla...
ensional × a12 a11 a22 a21 ... ... am1 am2 a1n a2n ... amn · · · · · · ... · · · x1 x2 ... xn           Ax =           = x1      a11 a21 ... am1      + x2      a12 a22 ... am2      + . . . + xn      a1n a2n ... amn  .     In other words, we can think of the vector x as ...
tiplying by the identity matrix Calculate AI,where I isthe 2 2 identitymatrixand × Solution. We need to calculate a11 a12 a21 a22 a31 a32 .   A =   AI =  a11 a12 a21 a22 a31 a32 1 0 0 1 .  Using the column method, we ﬁnd that the ﬁrst column of AI is   a11 a12 a21 a22 a31 a32 = 1 1 0 a11 a21 a31   + 0    ...
n-matrixoftheform × B] . [I \| Ifthiscanbedone,then A isinvertibleand A− formof A haslessthan n pivotentries),then A isnotinvertible. 1 = B. Ifitisnotpossible(i.e.,ifthereducedechelon This algorithm shows how to ﬁnd the inverse if it exists. It also tells us if A does not have an inverse. Example 4.42: Finding the inver...
a32 a21 a22 · · · · · · · · · a1n a3n a2n .   154 Matrices So the effect of multiplying A by E on the left is exactly the same as switching rows 2 and 3. We say that E is the elementary matrix for switching rows 2 and 3. ♠ Example 4.50: Elementary matrix for multiplying a row by a non-zero number Let   . ×  Whatis...
vot column, and it follows that R = I. Hence, UA = I, and therefore A is left invertible. Moreover, we have B = IB = (UA)B = U (AB) = U I = U , and therefore B = U . It follows that BA = UA = I, as claimed. To prove the last claim, note that we just proved that for square matrices, AB = I implies BA = I. Therefore, eve...
er, the diffusion property means that changing one symbol of a plaintext block potentially affects every symbol of the ciphertext block. The following is an example of a block cipher. Deﬁnition 4.73: Hill cipher n-matrix A with scalars from Z29. The Hill cipher of block size n has as its key an invertible n Each cipher...
iphers like this actually used in the real world? The answer is yes. While the cipher of Deﬁnition 4.77 is greatly simpliﬁed, it has the same basic structure as modern real-world block ciphers (such as AES, the Advanced Encryption Standard). Naturally, these real-world ciphers differ in some details, such as the alphab...
endence In Example 5.4, we encountered three vectors u, v, and w such that span happens, then the vector w does not contribute anything to the span of redundant. The following deﬁnition generalizes this notion. u, v, w { u, v, w { = span u, v . If this } , and we say that w is } { } 180 Spans, linear independence, and ...
)w = 0. − (5.2) Since u, v, and w are linearly independent, we know, again by Theorem 5.12, that equation (5.2) only has the trivial solution. Therefore, a + 2b = 0, a + c = 0, 5c = 0. b − We can solve this system of three equations in three variables, and we ﬁnd that it has the unique solution a = b = c = 0. Therefore...
eneralize this concept of subspace to arbitrary dimensions. Deﬁnition 5.21: Subspace AsubsetV of Rn iscalledasubspaceof Rn if 1. V containsthezerovectorof Rn,i.e., 0 ∈ 2. V isclosedunderaddition,i.e.,forall u, w V; V,wehave u + w V; ∈ ∈ 3. V isclosedunderscalarmultiplication,i.e.,forall u V andscalars k,wehave k u V. ∈...
and will still span S. In other from words, the remaining vectors will be a basis for S. We use the casting-out algorithm to identity the redundant vectors 10 ≃  .  ≃    Since columns 2, 4, and 5 are the non-pivot columns, it follows that the vectors u2, u4, and u5 are redundant. Therefore, the desired basis is . ...
ep the rest. The resulting basis is u1, u2, e2, e3 {  ,         ,         ,        .        } =        ♠ Example 5.46: Extending a linearly independent set to a basis Let Notethat u1, u2 V,where ∈ V =        u1 =     Extend u1, u2 { } toabasisofV.  , u2 =  ...
space of R5. Exercise 5.4.15 If you have 6 vectors in R5, is it possible they are linearly independent? Explain. Exercise 5.4.16 Suppose V and W both have dimension equal to 7 and they are subspaces of R10. What W ? Hint: Remember that a linear independent set can be are the possibilities for the dimension of V extende...
formation, or simply linear, if it satisﬁes 1. T preservesaddition,i.e.,forall v, w Rn,wehave T (v + w) = T (v) + T (w); ∈ 2. T preservesscalarmultiplication,i.e,forall v Rn andscalars k,wehave T (kv) = kT (v). ∈ 217 218 Linear transformations in Rn Example 6.2: Linear and non-linear transformations Whichofthevectorfun...
the pink letter “F”) by 90 degrees counterclockwise. The picture also illustrates that when we apply the rotation T to the ﬁrst and second standard basis vectors e1 and e2, we obtain the vectors T (e1) = 0 1 and T (e2) = 1 − 0 . The matrix of T has these vectors as its columns. Therefore, the matrix of T is A = 0 1 1 ...
x + 4y . Solution. The easiest way to do this is to ﬁnd the matrix of T . We have 1 0 T = 2 7 and T 0 1 = 1 4 . 234 Linear transformations in Rn Therefore, the matrix of T is The inverse of A is 4 7 − 1 is the linear transformation deﬁned by 1 = A− Therefore = Exercises y 4x − 7x + 2y − . ♠ Exercise 6.4.1 Find the matr...
is also often denoted by enclosing the matrix with two vertical lines. Thus ♠ det a b c d = 3-matrix = ad bc. − a b c d Deﬁnition 7.3: Determinant of a 3 × Let A =  a11 a12 a13 a21 a22 a23 a31 a32 a33 . Then   det(A) = a11a22a33 + a12a23a31 + a13a21a32  a31a22a13 − − a32a23a11 − a33a21a12. 241 242 Determinants The ...
ainedfrom A byaddingamultipleofonerowtoanotherrow,then det(B) = det(A). Notice that the second part of this theorem is true when we multiply one row of the matrix by k. If we were to multiply two rows of A by k to obtain B, we would have det(B) = k2 det(A). Example 7.17: Using row operations to calculate a determinant ...
Exercise 7.5.9 Let A and B be two n exists an invertible matrix P such that A = P− n-matrices. We say that A is similar to B, in symbols A B, then det(A) = det(B). 1BP. Show that if A × ∼ B, if there ∼ Exercise 7.5.10 Find the determinant of A =  1 1 1 1 a a2 1 b b2   .  For which values of a and b is this matrix ...
when A is not invertible. Theorem 7.34: Cramer’s rule Suppose A is an invertible n [x1, . . . , xn]. Then xi canbecomputedbytherule × n-matrix and we wish to solve the system Ax = b, where x = where Ai isthematrixobtainedbyreplacingthe i th columnof A with b. xi = det(Ai) det(A) , Proof. Since A is invertible, the sol...
1λ + 6 λ). − − λ)(2 − 8.2. Finding eigenvalues 275 − 2( 4)(2 − λ) − The eigenvalues are the roots of this polynomial, i.e., the solutions of the equation (λ 0. Therefore, the eigenvalues of A are λ = 1, λ = 2, and λ = 3. 3)(λ 1)(2 − − − λ) = ♠ Example 8.9: No real eigenvalue Findtheeigenvaluesofthematrix A = 0 1 1 − 0 ...
ansformation that reﬂects all vectors in R3 about the xy-plane. Find a matrix for T and then obtain its eigenvalues and eigenvectors. 8.4 Diagonalization Outcomes A. Computesums,products,andpowersofdiagonalmatrices. B. Determinewhetherasquarematrixisdiagonalizable,anddiagonalizeitifpossible. A square matrix D is called...
− 2 0 ±   It follows that A has 8 different square roots as well. We leave it as an exercise to compute them all. ♠ The same method can also be used to compute other powers of a matrix, for example a cube root. Exercises Exercise 8.5.1 Let A = Exercise 8.5.2 Let A = Exercise 8.5.3 Let A = Exercise 8.5.4 Let . Find A...
= a + bx, y = ae√q x + be− √q x, where a and b are parameters. The fact that each of these solutions is indeed the most general one is proved in a calculus course. ♠ 8.7.2. Systems of linear differential equations In the same way that a system of linear equations consists of several linear equations about several varia...
λ = 4. − • Solutions for λ = 0: The last eigenvalue is zero. By Theorem 8.32, the general solution of w′′ = 0 is w = a + bt, where a and b are arbitrary constants. This translates into the following solution for w: w = (a + bt) and by the change of variables v = Pw, we get ,   Writing this in terms of the coordinates...
envalue Algebraic multiplicity Geometric multiplicity ♠ In the example, the geometric multiplicity is either smaller than or equal to the algebraic multiplicity. The following proposition states that this is always the case. Proposition 8.42: Algebraic and geometric multiplicity Let ˆλ be an eigenvalue of a matrix A, w...
p(λ) = ( 3 − − λ)(1 − By trial and error, we ﬁnd that λ = λ)(3 λ) 16 + 8(1 λ) + 4(3 λ) = − − 1 is one of the roots. We factor out (λ + 1): − − − λ3 +λ2 − p(λ) = (λ + 1)( λ2 + 2λ − 5). − We then use the quadratic formula to ﬁnd the other two eigenvalues, i.e., the roots of are: Eigenvectors: 2 λ = − 16 − √ ± 2 − = 1 2i...
+ b0) = (a2 + b2)x2 + (a1 + b1)x + (a0 + b0) 9.1. Deﬁnition of vector spaces 327 = (b2 + a2)x2 + (b1 + a1)x + (b0 + a0) = (b2x2 + b1x + b0) + (a2x2 + a1x + a0) = q(x) + p(x). (A2) We prove the associative law of addition. (p(x) + q(x)) + r(x) = ((a2x2 + a1x + a0) + (b2x2 + b1x + b0)) + (c2x2 + c1x + c0) = ((a2 + b2)x2 ...
s deﬁned on a set X that have values in a vector space W . Is this a vector space? Explain. Exercise 9.1.8 Consider the set R2 with the following non-standard operations of addition and scalar multiplication: (a, b) k (c, d) = (a + c − (c, d) = (kc + (1 1, b + d 1), k), kd + (1 − − k)). − ⊕ ⊙ Show that R2 is a vector s...
M3 = 0. Notice that this equation has non-trivial solutions, for example a = 2, b = 3 and c = matrices are linearly dependent. 1. Therefore the − ♠ Example 9.21: Linearly independent functions In the vector space FuncR,R of real-valued functions on the real numbers, show that the functions f (x) = sin x and g(x) = cos ...
ns that are solutions of the differentialequation f ′′ = f . ThenW isasubspaceof V. − Proof. 1. The zero function f (x) = 0 is a solution of the differential equation, and therefore an element of W . 2. To show that W is closed under addition, let f , g g, and therefore h′′ = f ′′ + g′′ = f + ( W and consider h = f + g...
x3, x4, . . . . 0 and a0, . . . , an p(x) = anxn + an 1xn 1 + . . . + a1x + a0, − − K. But then p(x) is a linear combination of 1, . . . , xn, i.e., it is in the span ∈ ♠ Example 9.39: Not a basis Consider the vector space SeqK of inﬁnite sequences. As before, let ek be the sequence whose k th elementis 1 andthatis 0 ...
ralinearcode. D. Usethesyndromemethodtocorrecterrorsincodeblocks. E. ConstructanduseHammingcodes. Binary codes When transmitting or storing information on digital media, the information is usually encoded as a sequence of bits, i.e., 0s and 1s. For example, in the ASCII code, each symbol is encoded as a sequence of 8 b...
5.5, we know that every subspace of Zn 2, and therefore every linear code, is the column space of some matrix G, and also the null space of some matrix H. Such matrices are called a generator matrix and a check matrix for the code, respectively. } Deﬁnition 9.61: Generator matrix and check matrix Consider a linear cod...
1 0001111 1110000 1010 1001 0001 1110 Thus, the decoded message is 1010 1001 0001 1110. ♠ Example 9.70: Hamming code for r = 2 WhatistheHammingcodefor r = 2? Haveyouseenthiscodebefore? Solution. The check matrix and generator matrix for the Hamming code with r = 2 are H = 1 1 0 1 0 1 and G =  . 1 1 1    The code ha...
he equation T (kv) = kT (v). Finally, the third property is a direct consequence of the fact that T preserves addition and scalar multiplication: − T (a1v1 + a2v2 + . . . + akvk) = T (a1v1) + T (a2v2) + . . . + T (akvk) = a1T (v1) + a2T (v2) + . . . + akT (vk). ♠ Exercises Exercise 10.1.1 Consider the following functio...
ular, on a basis), then we know how it acts on the entire space. There is also a kind of converse to this: given a basis of V , we can map the basis vectors to any elements of W we like, and this will always determine a unique linear transformations. This is the content of the following theorem. Theorem 10.20: Linear t...
nsformation such that T → 3 3 3    be the standard basis of R3, and let   −   Let E = e1, e2, e3 { } be another basis. B = { v1, v2, v3 = } , a) Find the matrix of T with respect to E, i.e., ﬁnd [T ]E,E. 386 Linear transformation of vector spaces (b) Find [T ]B,B. Exercise 10.4.4 Let T : R2 → R2 be a linear tran...
rmula product on R2. h u, v (a) A = 1 0 0 2 , (b) A = 3 1 1 3 , (c) A = 1 1 0 2 , (d) A = i 1 0 = uT Av determines an inner 0 1 − , (e) A = 1 1 1 1 . Exercise 11.1.2 Consider the inner product space C[0, 1] as in Example 11.4. Compute the following inner products: (a) , 1, x i h (b) h x, x2 , i (c) h 1 + x, 2 + x2 . i ...
A word of warning is in order: Fourier coefﬁcients do not work when the basis is not orthogonal. Consider the following picture, where u1 and u2 are not orthogonal. a2 u2 v u1 a1 The coordinates of v with respect to u1, u2 are (2, 1), because v = 2u1 + 1u2, as indicated by the shaded parallelogram. On the other hand, ...
vT Aw. Let u1, u2, u3 { } for R3 with Solution. As usual, we start with u1 = v1 = . 1 0 0     u1, v2 h u1, u1 h i i = uT = uT 1 Av2 = 2, 1 Au1 = 1. u2 = v2 u1, v2 h u1, u1 h i i − u1 =   u1, v3 h u2, v3 h u2, u2 h i i i 2, = uT = uT = uT 1 Av3 = − 2 Av3 = 3, 2 Au2 = 2. Next, we calculate Therefore, Finally, we ca...
lations in this example are. All we had to do is calculate a few inner products. It is not even necessary to solve a system of equations. Such is the power of orthogonal bases. The next example shows that we can use exactly the same method to ﬁnd approximations of functions. 416 Inner product spaces Example 11.40: Appr...
( if x < 0, 0. if x x − Find the closest approximation to f by a polynomial of degree at most 0, 1, 2, 3, and 4. Graph both f and the approximating polynomials. ≥ Exercise 11.4.4 In the inner product space C[ ample 11.21: Fourier series of f . 1, sin x, cos x, sin 2x, cos 2x, sin 3x, cos 3x, . . . } − { π,π], consider ...
for each data point. x Actual y Best ﬁt parabola Error 0.0 3 +1.5 4 5 3.0 − +1.5 6 1.5 7 − 8 +3.0 1.5 9 23.5 13.5 12.5 5.5 9.0 8.0 19.0 23.5 15.0 9.5 7.0 7.5 11.0 17.5 − The sum of the squares of the errors is 02 + 1.52 + 32 + 1.52 + 1.52 + 32 + 1.52 = 27. ♠ Exercises Exercise 11.5.1 Find the least squares approximati...
it follows that λ = λ, i.e., λ is real. (b) Suppose v is an eigenvector for eigenvalue λ, w is an eigenvector for eigenvalue µ, and λ evaluate vT Aw in two different ways. On the on hand, we have = µ. We vT Aw = vT (Aw) = vT (µw) = µvT w. On the other hand, we have vT Aw = (vT A)w = (AT v)T w = (Av)T w = (λv)T w = λvT ...
e following proposition gives us a more practical method for determining whether a matrix is positive (semi)deﬁnite. ≥ Proposition 11.69: Characterization of positive (semi)deﬁnite matrices using eigenvalues Let A beasymmetricn initeifandonlyifλ1, . . . ,λn n-matrix,andletλ1, . . . ,λn beitseigenvalues. Then A ispositi...
f how to simplify quadratic forms. The primary tool we have for doing so is a change of variables. This means replacing the variables x1, . . . , xn by new variables y1, . . . , yn that are linear combinations of x1, . . . , xn. Example 11.79: Change of variables Applythechangeofvariables tothequadraticform 3x2 + y2 + ...
the transpose as usual. As a matter of fact, when working with complex of a vector), and ( vectors and matrices, it turns out that we should almost always take the complex conjugate at the same time as taking the transpose. For this reason, we introduce a special name and notation for the conjugate transpose of a vect...
the norm of each of the following vectors in Cn: (a) 1 i , (b) 1 0 1 + 2i     , (c) . 1 + i i 1 3i  −    u = 1, v Cn and u ∈ · · v = 2, and u · v = 2i. Then compute: Exercise 11.10.3 Suppose u, v (a) (2u + v) (u (b) (iu v) − v). − 2iv). − · (u · Exercise 11.10.4 Compute the adjoint of the following matrices. i...
sis. Imagine we are given a collection of data points such as the following: (11.6) Although these points are spread out in two dimensions, they seem to be located pretty close to a 1dimensional subspace. Probably the best way to interpret this particular data set is to think of the points as being “essentially” on a l...
Note that this is the equation of a line passing through the centroid v, and with direction vector u1. The points v1, . . . , v9, their centroid, and the afﬁne subspace v +W are shown in the following illustration: 10 centroid 10 Application to U.S. Senate voting data ♠ The United States Senate votes on a lot of things...
e, it 1, 2, 3, . . . } { became a problem that certain equations, such as x + 5 = 3, do not have a solution in the natural numbers. • To solve this problem, zero and negative numbers were invented, resulting in the set of integers . In the integers, the equation x + 5 = 3 has a solution, namely . . . , { 2. But some ot...
Example A.10: Roots of a quadratic polynomial Findtherootsofthepolynomial p(z) = z2 2z + 2. − Solution. The quadratic formula gives 2 ± z = Of course, in the real numbers, the square root of numbers. However, in the complex numbers, the square root of roots of p(z) are: − − Indeed, we can double-check that 1 + i and 1 ...
the distributive law, properties of scalars, the property of part (b), and the additive unit law, respectively. 1)u + (u + ( − u) = 0 + ( − u)) = (( u) = − u) = 0u + ( 1)u + u) + ( u) = (( − − − − − − − −     = a1 + a2 3 1 1   −   + a3 .4.2 2.4.3 The system has no solution. 2.5.1 √17. 2.5.2 √27. 501 2.5.3 The ...
the reduced . C− 1 does not exist because the reduced eche =      , which has a row of zeros. E− 1 does not exist because E is not a square 4.5. ad bc − − c ad bc ad ad bc . − a bc # " − − .5.8 (a)  x y z    =  −   12 1 5   .  3c 1 3b a 2a b.5.9 Multiply both sides of AX = B on the left by A−   1. 4.5...
wi { null(A) ∩ ∈ ∈ This implies and so it is of the form It follows that Bx = c1w1 + . . . + ckwk = B(c1z1 + . . . + ckzk). (c1z1 + . . . + ckzk) x − null(B) ∈ (c1z1 + . . . + ckzk) = d1u1 + . . . + drur. x − span x ∈ { z1, . . . , zk, u1, . . . , ur . } Since we have shown that every element of null(AB) is in the spa...
has eigenvalue 1 (with all vectors being eigenvectors). − 8.3.3 The matrix of T is 1 0 0 1 0 1 − . The eigenvectors and eigenvalues are: for eigenvalue 1, − 1 0 for eigenvalue 1. 8.3.4 The matrix of T is 0 1 1 0 . The eigenvectors and eigenvalues are: 1 1 8.3.5 The matrix of T is  1 0 0 1 0 0 0 0 1   − for eigenval...
rected code block is 011000010010000, and the decoding is 01100001001. The error syndrome for 010000100000001 is 0110, which corresponds to an error in bit 3. The corrected code block is 011000100000001 and the decoding is 01100010000. However, if you have used a different check matrix and/or generator matrix, your ans...
agonal, 11.8.1 (a) Positive semideﬁnite. (b) Positive deﬁnite. (c) Not symmetric (therefore neither). (d) Positive semideﬁnite. (e) Neither. 11.8.2 (a) Eigenvalues: Eigenvalues: 0, 2, 3 1, 6 . Positive deﬁnite. (b) Eigenvalues: } . Positive semideﬁnite. (d) Eigenvalues: } 0, 13 . Positive semideﬁnite. (c) { } . Neither...
on of complex numbers, 487 dot product, 80 complex, 457 from matrix product, 163 properties, 80 echelon form, 18 algorithm, 19 reduced, 29 eigenfrequency, 305 eigenspace, 271 eigenvalue, 269 algebraic multiplicity, 312, 321 calculating, 276 complex, 318 geometric multiplicity, 312, 321 of triangular matrix, 279 eigenve...
ions, 111 standard equation, 114 vector equation, 111 point, 61 polynomial, 275, 326, 330, 491 addition, 326, 330 characteristic, see characteristic polynomial coefﬁcient, 275, 491 constant, 491 degree, 491 Legendre polynomial, 408, 409 root, 276, 491 scalar multiplication, 326, 330 zero, 327 position vector, 63 positi...
um. Then im α = U + W and ker α = {(u, −u) : u ∈ U ∩ W } ∼= dim(U ∩ W ). Then we have dim U + dim W = dim(U ⊕ W ) = r(α) + n(α) = dim(U + W ) + dim(U ∩ W ). This is a result we’ve previously obtained through ﬁddling with basis and horrible stuﬀ. Corollary. Suppose α : U → V is a linear map between vector spaces over F ...
If V is a ﬁnite-dimensional vector space over f with basis (e1, · · · , en), then there is a basis (ε1, · · · , εn) for V ∗ (called the dual basis to (e1, · · · , en)) such that εi(ej) = δij. Proof. Since linear maps are characterized by their values on a basis, there exists unique choices for ε1, · · · , εn ∈ V ∗. No...
(θ) = 0 for all θ ∈ V ∗. So dimv0 = dim V ∗ = dim V . So dimv = 0. So v = 0. So ev is injective. Since V and V ∗∗ have the same dimension, this is also surjective. So done. From now on, we will just pretend that V and V ∗∗ are the same thing, at least when V is ﬁnite dimensional. Note that this lemma does not just say ...
ﬁnition (Volume form). A volume form on Fn is a function d : Fn×· · ·×Fn → F that is (i) Multilinear, i.e. for all i and all v1, · · · , vi−1, vi+1, · · · , vn ∈ Fn, we have d(v1, · · · , vi−1, · , vi+1, · · · , vn) ∈ (Fn)∗. (ii) Alternating, i.e. if vi = vj for some i = j, then d(v1, · · · , vn) = 0. We should think o...
vector space over F. An endomorphism of V is a linear map α : V → V . We write End(V ) for the F-vector space of all such linear maps, and I for the identity map V → V . When we think about matrices representing an endomorphism of V , we’ll use the same basis for the domain and the range. We are going to study some pro...
en for any v, we get p(α)(v) = k i=1 p(α)(vi) = k i=1 p(λi)vi = 0. So p(α) = 0. By construction, p has distinct linear factors. Conversely, suppose we have our polynomial k p(t) = (t − λi), i=1 with λ1, · · · , λk ∈ F distinct, and p(α) = 0 (we can wlog assume p is monic, i.e. the leading coeﬃcient is 1). We will show ...
e get n (α − λiι)(V ) ⊆ V0 = 0. i=1 So χα(α) = 0 as required. Note that if our ﬁeld F is not C but just a subﬁeld of C, say R, we can just pretend it is a complex matrix, do the same proof. 60 6 Endomorphisms IB Linear Algebra We can see this proof more “visually” as follows: for simplicity of expression, we suppose n ...
mutation of blocks. 67 6 Endomorphisms IB Linear Algebra Proof. We work blockwise for  Jn1 (λ1) A =     Jn2 (λ2) . . .      . Jnk (λk) We have previously computed n((Jm(λ) − λIm)r . Hence we know n((Jm(λ) − λIm)r) − n((Jm(λ) − λIm)r−1) = 1 0 r ≤ m otherwise. It is also easy to see that for µ = λ, n((Jm(µ) − λ...
ut we can for bilinear forms over arbitrary ﬁelds. Proof. We induct over n = dim V . The cases n = 0 and n = 1 are trivial, since all matrices are diagonal. Suppose we have proven the result for all spaces of dimension less than n. First consider the case where φ(v, v) = 0 for all v ∈ V . We want to show that we must h...
nalogous fact for the bilinear form on ¯V × W → C. Let v = λivi and W = µjwj. Then we have φ(v, w) = i,j λiµjφ(vi, wj) = λ†Aµ. 80 7 Bilinear forms II IB Linear Algebra We now want the right deﬁnition of symmetric sesquilinear form. We cannot just require φ(v, w) = φ(w, v), since φ is linear in the second variable and c...
that the orthogonal complement is a complement and orthogonal, i.e. V is the orthogonal direct sum of W and W ⊥. Proposition. Let V be a ﬁnite-dimensional inner product space, and W ≤ V . Then V = W ⊥ W ⊥. Proof. There are three things to prove, and we know (iii) implies (ii). Also, (iii) is obvious by deﬁnition of W ⊥...
is of U , then α is represented by a real symmetric matrix, say a b b a But then χα\|U (t) = (t − a)2 − b2, which has real roots, namely a ± b. This is a contradiction, since Mα\|U = f , but f is irreducible. (ii) Now suppose αv = λv, αw = µw and λ = µ. We need to show (v, w) = 0. We know (αv, w) = (v, αw) by deﬁnition. ...
E ⊆ X is nowhere dense if ¯E has empty interior. Usually, we will pick E to be closed so that the deﬁnition just says that E has an empty interior. Deﬁnition (First/second category, meagre and residual). Let X be a topological space. We say that Z ⊆ X is of ﬁrst category or meagre if it is a countable union of nowhere ...
ar Analysis Note that in this proof, we required both V and W to be Banach spaces. However, we used the completeness in diﬀerent ways. We used the completeness of V to extract a limit, but we just used the completeness of W to say it is of second category. In particular, it suﬃces to assume the image of T is of second ...
e objective of this chapter is to show that if we have a continuous function deﬁned on a closed subset of a normal space, then we can extend this to the whole of the space. We start a special case of this theorem. Lemma (Urysohn’s lemma). Let X be normal and C0, C1 be disjoint closed subsets of X. Then there is a f ∈ C...

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