Math 110, Spring 2015: Homework 5 Solutions. Exercise 2.4.4: Let A and B be n n invertible matrices. Prove that AB is invertible and (AB) 1 = B 1 A 1.

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1 Math, Spring 25: Homework 5 Solutions Section 2.4 Exercise 2.4.4: Let A and B be n n invertible matrices. Prove that AB is invertible and (AB) = B A. Proof. We first note that the matrices AB and B A are n n matrices. invertible n n matrix (on page of the text), we need only prove that By the definition of an (AB)(B A ) = (B A )(AB) = I, where I denotes the n n identity matrix. We can readily check this by using the associativity of matrix multiplication (Theorem 2.6) and the fact that IC = CI = C for all C M n n (R) (Theorem 2.2(c)): (AB)(B A ) = A(BB )A = AIA = AA = I. Similarly, (B A )(AB) = B (A A)B = B IB = B B = I. Exercise 2.4.6: Prove that if A is invertible and AB =, then B =. Proof. Multiplying both sides of the equation AB = on the left by the inverse A of A, we obtain A AB = A = IB = B = as desired; here I denotes the identity matrix of the same dimensions as A.

2 Exercise 2.4.9: Let A and B be n n matrices such that AB is invertible. Prove that A and B are invertible. Give an example to show that arbitrary matrices A and B need not be invertible if AB is invertible. Solution: We first prove that A and B are invertible under the original hypotheses: Proof. We ll prove this by converting it into a claim about linear transformations: By Corollary 2 on page 2 of the text, A and B are invertible matrices if and only if the left-multiplication transformations L A and L B are invertible linear transformations. Furthermore, by the same Corollary, our assumption that AB is invertible tells us that the composition L A L B = L AB is an invertible linear transformation. Note first that, because A and B are n n matrices, each of L A and L B maps R n to itself. So to show that each of L A and L B is invertible, we may show either that it is one-to-one or that it is onto because, by the Dimension Theorem, the conditions of being one-to-one and onto are equivalent here. We first show that L B is one-to-one. Note that if L B (x) =, then L A L B (x) = L A ( ) =. Thus, N(L B ) N(L A L B ) = { }, where the last equality follows from the assumption that L A L B is invertible and hence one-to-one. So N(L B ) = { }, and therefore L B is one-to-one. Next, we show that L A is onto. Indeed, for any y R(L A L B ), we can write y = L A ( LB (x) ) for some x R n. Thus, R n = R(L A L B ) R(L A ), where the first equality follows from the assumption that L A L B is invertible and hence onto. So we have R(L A ) = R n, and therefore L A is onto. Thus, L A and L B are invertible linear transformations; as noted above, this is equivalent to saying that A and B are invertible matrices. To wrap up this exercise, we construct an example to show that for arbitrary (i.e., not necessarily square) matrices A and B with AB invertible, A and B need not be invertible. The point here is that, by definition, an invertible matrix must be a square matrix. So if AB is invertible, it must be square, say of dimensions n n; then we must have A M n m (R) and B M m n (R) for some m N. So we should look for such examples with m n, since by definition such A and B cannot be invertible as they are not square matrices. Possibly the simplest example is the following: Take A = [ [ M 2 (R) and B = M 2 (R). Then A and B are not square matrices, so they can t be invertible; on the other hand, AB = [ = I, the identity matrix, which is invertible. Remark: If you want a principled way to come up with an example instead of just playing around until you find one, try using ideas similar to the proof for the first part of the exercise to prove the following: Suppose A is an n m matrix such that L A is onto (i.e., rank(a) = n) and B is an m n matrix such that L B is one-to-one. Suppose further that R(L B ) N(L A ) = { }. Then AB M n n (R) is invertible. (I don t know what exact language was used in Math 54, but note that R(L B ) is the column space of B and N(L A ) is the null space of A. You can easily arrange for the three conditions we just listed by writing down matrices in reduced row echelon form. ) 2

3 Exercise 2.4.5: Let V and W be n-dimensional vector spaces, and let T : V W be a linear transformation. Suppose that β is a basis for V. Prove that T is an isomorphism if and only if T(β) is a basis for W. Proof. We first prove the only if implication. So assume that T : V W is an isomorphism; we first claim that then T(β) must be a linearly independent set of n vectors in W. To that end, write β = {v,..., v n }. Because T is one-to-one, the vectors T(v k ) are distinct for k n, and thus T(β) = {T(v ),..., T(v n )} is a set of n vectors in W. Now suppose that some linear combination of these vectors is equal to the zero vector of W ; i.e., suppose a T(v ) a n T(v n ) = for some scalars a,..., a n R. It suffices to prove that a k = for all k. Now, again because T is a one-to-one linear transformation, we have N(T) = { }; thus, a T(v ) a n T(v n ) = T(a v a n v n ) = = a v a n v n =. Since β is assumed to be a basis, in particular the v k are mutually linearly independent; thus, we must have a k = for all k as desired. So T(β) is a linearly independent set of n vectors in W ; because dim(w ) = n, T(β) is a basis for W. Now let us prove the if implication. To that end, assume that T(β) = {T(v ),..., T(v n )} is a basis for W. Let y W be arbitrary. Using the assumed linearity of T, we can find scalars b,..., b n R such that y = b T(v ) b n T(v n ) = T(b v b n v n ) = T(x), where x = b v b n v n V. So we ve shown that for all y W, y R(T). In other words, T is onto. By the Dimension Theorem, because dim(v ) = dim(w ), T must also be one-to-one; thus, T is an isomorphism. Remarks: For the only if direction, we chose to show that T(β) must be linearly independent; of course one could also have chosen to show that T(β) spans W. Similarly, for the if direction, it would have been totally acceptable to show directly that T must be one-to-one and then to deduce that T must also be onto from the Dimension Theorem. Note also that the if implication of this exercise is not true in general if we don t assume dim(v ) = dim(w ); try to think of a counterexample. Exercise 2.4.6: Let B be an n n invertible matrix. Define Φ : M n n (R) M n n (R) by Φ(A) = B AB. Prove that Φ is an isomorphism. Proof. We first show that Φ is a linear transformation. Let A, A 2 M n n (R) and c R be arbitrary. Then Φ(cA + A 2 ) = B (ca + A 2 )B = (cb A + B A 2 )B = cb A B + B A 2 B = cφ(a ) + Φ(A 2 ), so Φ is linear. Next, let C M n n (R) be any n n matrix. Then Φ(BCB ) = B (BCB )B = (B B) C (B B) = I n C I n = C, where I n denotes the n n identity matrix. In particular, C R(Φ); as C was arbitrary, Φ is onto. Since Φ is a linear transformation from the finite-dimensional vector space M n n (R) to itself, by the Dimension Theorem Φ must also be one-to-one. Thus, Φ is an isomorphism. 3

4 Exercise 2.4.7: Let V and W be finite-dimensional vector spaces and T : V W be an isomorphism. Let V be a subspace of V. (a) Prove that T(V ) is a subspace of W. Proof. We need only use the fact that T is a linear transformation here; i.e., we do not need to use the full fact that T is an isomorphism. Because V is a subspace of V, the zero vector V of V is an element of V. So W = T( V ) T(V ). To see that T(V ) is closed under addition and scalar multiplication in W, let w, w 2 T(V ) and c R be arbitrary. Then there exist v, v 2 V such that T(v ) = w and T(v 2 ) = w 2. Furthermore, because V is a subspace of V, we have cv +v 2 V. Thus, cw + w 2 = ct(v ) + T(v 2 ) = T(cv + v 2 ) T(V ). As c, w, and w 2 were arbitrary, T(V ) is closed under addition and scalar multiplication in W. Thus, T(V ) is a subspace of W. (b) Prove that dim(v ) = dim ( T(V ) ). Proof. Here we need to use the full fact that T is an isomorphism. Consider the restriction T V of T to the subspace V. The restriction of a linear transformation to a subspace of its domain was defined in a previous homework; in this case we may view it as the linear transformation T V : V T(V ) defined by T V (x) := T(x) for all x V. By definition, R ( ) T V = T(V ), so T V is onto. Moreover, if x N(T V ), then W = T V (x) = T(x) = x = V, because T is an isomorphism, hence one-to-one. Thus, T V is one-to-one; since we also noted that it s onto, T V : V T(V ) is an isomorphism. In other words, V and T(V ) are isomorphic vector spaces; by Theorem 2.9, dim(v ) = dim ( T(V ) ). Exercise 2.4.2: Let V and W be finite-dimensional vector spaces with ordered bases β = {v, v 2,..., v n } and γ = {w, w 2,..., w m }, respectively. By Theorem 2.6 (p. 72), there exist linear transformations T ij : V W such that { wi if k = j T ij (v k ) = if k j. First prove that {T ij i m, j n} is a basis for L(V, W ). Remark: One could make this proof a bit shorter by using Theorem 2.2 and its Corollary; that would tell us that dim ( L(V, W ) ) = mn. Once we knew the dimension, we could get away with showing either linear independence or spanning and then immediately deduce the other. But since the spirit of this exercise seems to be to reprove those results from a slightly different perspective, we ll just do everything from scratch. Note also that the arguments below can be motivated by considering the matrix representations of linear transformations with respect to the bases β and γ. 4

5 Proof. We first show that the set {T ij i m, j n} is linearly independent. Let T : V W denote the zero transformation, which is the zero vector of the vector space L(V, W ). Assume there exist scalars a ij R such that m n a ij T ij = T ; i= j= we aim to show that we must have a ij = for all i m and j n. Our assumption is equivalent to the condition m n a ij T ij (v) = () i= j= for all v V. Fix k {,..., n}. Then plugging v = v k into equation () gives m n a ij T ij (v k ) = i= j= m a ik w i = a k w a mk w m =. i= Since γ = {w,..., w m } is a basis for W, the w i are linearly independent, and thus we must have a ik = for all i m. Since k was arbitrary, we have shown a ij = for all i m and j n, as desired. So the given set is linearly independent in L(V, W ). Next, we show that {T ij i m, j n} spans L(V, W ). Let S L(V, W ) be an arbitrary linear transformation. Again using the fact that γ = {w,..., w m } is a basis for W, for each k n we can find scalars b k, b 2k,..., b mk R such that S(v k ) = b k w b mk w m = m b ik w i. i= Consider the linear transformation m n b ij T ij span ( {T ij i m, j m} ). Note that, for each k n, i= j= m n b ij T ij (v k ) = i= j= m b ik w i = S(v k ). i= Since β = {v,..., v n } is a basis for v, by the Corollary to Theorem 2.6 on page 73 of the text we have S = m n b ij T ij span ( {T ij i m, j n} ). i= j= Since S L(V, W ) was arbitrary, the given set spans L(V, W ). Since we also showed it is linearly independent, it is a basis for L(V, W ). 5

6 Exercise (continued): Then let M ij be the m n matrix with in the ith row and jth column and elsewhere, and prove that [T ij γ β = M ij. Proof. We just compute to show that the corresponding entries of the two matrices are equal; i.e., we show that for all k m and all l n we have ( [Tij γ β) kl = ( M ij) kl. To do that, we just use the definition of the matrix representation of a linear transformation with respect to a pair of bases β and γ. Fix an arbitrary such k and l. Note that the lth column of [T ij γ β is the coordinate vector { { [wi [T ij (v l ) γ = γ if l = j ei R = m if l = j [ γ if l j R m if l j. (Here e i denotes the ith standard basis vector in R m.) So the kth entry of this vector is ( [Tij γ β)kl = { if l = j and k = i otherwise = ( M ij) kl, which is what we wanted to show. Exercise (continued): Again by Theorem 2.6, there exists a linear transformation Φ : L(V, W ) M m n (R) such that Φ(T ij ) = M ij. Prove that Φ is an isomorphism. Proof. Note that from the first part of this exercise, in which we showed that {T ij i m, j n} is a basis for L(V, W ) consisting of mn distinct linear transformations, we can deduce dim ( L(V, W ) ) = mn. Thus, because we also have dim ( M m n (R) ) = mn, we may apply Exercise 5 above to the linear transformation Φ : L(V, W ) M m n (R). Once we note that the set {M ij i m, j n} is just the standard basis for M m n (R), the if implication of Exercise 5 tells us that Φ is an isomorphism. Remark: If you think about it, Φ here is in fact the exact same isomorphism Φ appearing in Theorem 2.2 of the text. We ve basically given a somewhat lengthy elaboration of the proof of that theorem; comparing the proof we did here with that appearing in the book may be instructive. 6

7 Exercise : Let V denote the vector space defined in Example 5 of Section.2, and let W = P(R). Define n T : V W by T(σ) = σ(i)x i, where n is the largest integer such that σ(n). Prove that T is an isomorphism. Remark : To be utterly pedantic, the statement of the exercise does not define T( V ). (The zero vector V of V is the sequence σ defined by σ (m) = for all m, so it does not make sense to say n is the largest integer such that σ (n). ) So if we want to be completely precise, we should further define T( V ) = W, where W is the zero polynomial in P(R). This exercise is also confusing as stated because Example 5 in Section.2 defines V to be the space of functions from the positive integers to R; thus, σ() does not make sense for σ V under that definition. The natural thing to do is to modify the definition of V slightly, taking V to be the vector space of functions from the nonnegative integers to R. (N.B.: The two definitions yield isomorphic vector spaces. To check your understanding, find a reasonably obvious isomorphism from the new V to the old V. ) We ll use that modified definition of V for the following proof. Remark 2: Note that V and W here are infinite-dimensional vector spaces; accordingly, we cannot hope to make any use of the Dimension Theorem or other finite-dimensional techniques. Proof. We first show that T is a linear transformation. Let c R and σ, η V be arbitrary. Let N be the largest integer such that either σ(n) or η(n) ; if σ = η = V, then set N =. Then T(cσ + η) = i= i= N N (cσ + η)(i)x i = c σ(i)x i + i= N η(i)x i = ct(σ) + T(η), so T is linear. Next, we show T is one-to-one. Indeed, suppose T(σ) = W, the zero polynomial. Then by definition of the zero polynomial and the map T, we know that: ˆ σ(n) = for all integers n >, and ˆ σ() =. In other words, σ(n) = for all nonnegative integers n, and thus σ = V. So N(T) = { V }, and T is one-to-one. Finally, let f W = P(R) be an arbitrary polynomial, and write n f(x) = a i x i, so that deg(f) = n. Define a sequence σ V by { ak for k n σ(k) = for k > n. i= Then T(σ) = f; since f W was arbitrary, T is onto. Thus, we ve shown that T is a linear transformation that is both one-to-one and onto; in other words, T is an isomorphism. i= 7

8 Section 2.5 Exercise 2.5.2: For each of the following pairs of ordered bases β and β for R 2, find the change of coordinate matrix that changes β -coordinates into β-coordinates. (a) β = {e, e 2 } and β = {(a, a 2 ), (b, b 2 )}. Solution: For j =, 2, the j-th column of the change-of-coordinates matrix Q is given by the β-coordinate vector of the j-th vector in the basis β. We compute these coordinate vectors as follows: [ a (a, a 2 ) = a e + a 2 e 2 [(a, a 2 ) β = ; (b, b 2 ) = b e + b 2 e 2 [(b, b 2 ) β = So the change-of-coordinates matrix is [ Q = [I R 2 β a b β = a 2 b 2, a 2 [ b where I R 2 denotes the identity transformation on the vector space R 2. (d) β = {( 4, 3), (2, )} and β = {(2, ), ( 4, )}. Solution: With the same approach as in part (a), we compute the relevant coordinate vectors: { [ [ 4a + 2c = 2 a 2 (2, ) = a( 4, 3) + c(2, ) = [(2, ) 3a c = c β = ; 5 ( 4, ) = b( 4, 3) + d(2, ) { 4b + 2d = 4 3b d = So the change-of-coordinates matrix is [ a b Q = = c d [ b 2 [ b d. = [( 4, ) β = [ 4. 8

9 Exercise 2.5.3: For each of the following pairs of ordered bases β and β for P 2 (R), find the change of coordinate matrix that changes β -coordinates into β-coordinates. (b) β = {, x, x 2 } and β = {a 2 x 2 + a x + a, b 2 x 2 + b x + b, c 2 x 2 + c x + c }. Solution: We follow the same basic approach as in Exercise 2. The β-coordinate vectors of the basis vectors in β can easily be computed by inspection in this case: a [a 2 x 2 + a x + a β = a a 2, [b 2 x 2 + b x + b β =, [c 2 x 2 + c x + c β =. So the change-of-coordinates matrix is Q = [I P2 (R) β β = b b b 2 a b c a b c a 2 b 2 c 2 (d) β = {x 2 x +, x +, x 2 + } and β = {x 2 + x + 4, 4x 2 3x + 2, 2x 2 + 3}. Solution: For convenience, write β = {f, f 2, f 3 } and β = {g, g 2, g 3 } (as ordered bases). Then we compute the coordinate vectors: f = a g + a 2 g 2 + a 3 g 3 f 2 = b g + b 2 g 2 + b 3 g 3 f 3 = c g + c 2 g 2 + c 3 g 3 a + a 3 = a + a 2 = a + a 2 + a 3 = 4 b + b 3 = 4 b + b 2 = 3 b + b 2 + b 3 = 2 So the change-of-coordinates matrix is Q = [I P2 (R) β β = c + c 3 = 2 c + c 2 = c + c 2 + c 3 = 3 a b c a 2 b 2 c 2 a 3 b 3 c 3. [f β = [f 2 β = [f 2 β = a a 2 a 3 b b 2 b 3 c c 2 c c c c ;. ; 9

10 (f) β = {2x 2 x +, x 2 + 3x 2, x 2 + 2x + } and β = {9x 9, x 2 + 2x 2, 3x 2 + 5x + 2}. Solution: Again, write β = {f, f 2, f 3 } and β = {g, g 2, g 3 }. Compute the coordinate vectors (in the interest of saving space I will not show my work in solving the systems of equations, but you should): f = a g + a 2 g 2 + a 3 g 3 f 2 = b g + b 2 g 2 + b 3 g 3 f 3 = c g + c 2 g 2 + c 3 g 3 2a + a 2 a 3 = a + 3a 2 + 2a 3 = 9 a 2a 2 + a 3 = 9 So the change-of-coordinates matrix is Q = [I P2 (R) β β = 2b + b 2 b 3 = b + 3b 2 + 2b 3 = 2 b 2b 2 + b 3 = 2 2c + c 2 c 3 = 3 c + 3c 2 + 2c 3 = 5 c 2c 2 + c 3 = 2 a b c a 2 b 2 c 2 a 3 b 3 c 3 [f β = [f 2 β = [f 3 β = a a 2 a 3 b b 2 b 3 c c 2 c ;. ; Exercise 2.5.6: For each matrix A and ordered basis β, find [L A β. Also, find an invertible matrix Q such that [L A β = Q AQ. [ {[ [ } 2 (b) A = and β =,. 2 Solution: For convenience, write β = {v, v 2 } (as an ordered basis). Then the j-th column of [L A β is the β-coordinate vector [L A (v j ) β. So we compute [ [ 3 3 L A (v ) = = 3v 3 + v 2 = [L A (v ) β = ; [ [ L A (v 2 ) = = v + ( )v 2 = [L A (v 2 ) β =. So [L A β = [ 3 By the Corollary on page 5 of the text, the change-of-coordinates matrix Q that changes β- coordinates into standard-basis-coordinates, given by [ Q =, satisfies [L A β = Q AQ..

11 3 4 (d) A = 3 4 and β = 4 4 2,,. Solution: Again, for convenience write β = {v, v 2, v 3 }. Compute the relevant β-coordinate vectors: So we have L A (v ) = = 6v + v 2 + v 3 = [L A (v ) β = ; 2 L A (v 2 ) = 2 = v + 2v 2 + v 3 = [L A (v 2 ) β = 2 ; 8 L A (v 3 ) = 8 = v + v 2 + 8v 3 = [L A (v 3 ) β = [L A β = 2. 8 Again using the Corollary on page 5 of the text, the change-of-coordinates matrix Q = 2 satisfies [L A β = Q AQ. Remark: In both parts of this exercise, it was easy enough to solve (by inspection) the systems of linear equations required to obtain the relevant coordinate vectors to construct [L A β. On the other hand, if the vectors of the basis β are explicitly presented to you in standard coordinates, the matrix Q is always easy to construct: its columns are simply the standard coordinate vectors of the basis vectors from β, arranged in their proper order. So in general, you have a choice of which computation you d rather do: () solve the linear systems needed to compute the relevant coordinate vectors, or (2) invert the matrix Q, by whatever technique you desire (probably either by row operations or by Cramer s rule). Which option is the less tedious varies on a case-by-case basis, of course.

12 Exercise 2.5.7(b): In R 2, let L be the line y = mx, where m. Find an expression for T(x, y), where T is the projection on L along the line perpendicular to L. (See the definition of projection in the exercises of Section 2..) Solution: We follows the geometrically intuitive approach of Example 3 on pages 3 4 of the text. Let L denote the line in R 2 perpendicular to L; it has equation y = m x. Since L and L are lines through the origin in R 2, they are one-dimensional subspaces of R 2. Indeed, β = {(, m)} is a basis for L, and β = {( m, )} is a basis for L. We first verify that R 2 = L L. Geometrically, it is clear that the lines L and L intersect (only) at the origin; that is, L L = {(, )}. Moreover, since neither of the basis vectors (, m) and ( m, ) is a scalar multiple of the other, the set γ = {(, m), ( m, )} is linearly independent in R 2 ; since dim(r 2 ) = 2 the set must also span R 2, which implies R 2 = L + L. Note that γ is in fact a basis of R 2. By definition of the projection on L along L, we see that T(, m) = (, m) and T( m, ) = (, ). So we have [ [T γ =. Moreover, we can form the change-of-coordinates matrix Q that converts γ-coordinates into standardbasis-coordinates, and we can take its inverse using the familiar formula for the inverse of a 2 2 matrix: [ [ m Q =, Q m = m m 2. + m Then, as in Example 3, T = L A, where A is the matrix So we just compute A [ x = y A = Q [T γ Q = [ x + my m 2 + mx + m 2 y m 2 + = T(x, y) = [ m m m 2. ( x + my m 2 +, mx + ) m2 y m 2. + Remark: If you have learned elsewhere how to take orthogonal projections using dot products, you can verify that T(x, y) is indeed the same orthogonal projection of (x, y) to L as computed using the dot-product method. Exercise 2.5.: Prove that if A and B are similar n n matrices, then tr(a) = tr(b). Hint: Use Exercise 3 of Section 2.3. Proof. Exercise 3 of Section 2.3 gives the following commutativity property of the trace: For any C, D M n n (R), we have tr(cd) = tr(dc). Since A and B are similar matrices, there exists some invertible n n matrix Q such B = Q AQ, by definition. So we just compute, using the commutativity property: tr(b) = tr ( Q (AQ) ) = tr ( (AQ)Q ) = tr ( A(QQ ) ) = tr(ai) = tr(a), where I denotes the n n identity matrix, as desired. 2