Solutions to Axler, Linear Algebra Done Right 2nd Ed.Edvard Fagerholm edvard.fagerholm@¦helsinki.fi[gmail.com¦ Beware of errors. I read the book and solved the exercises during spring break (one week), so the problems were solved in a hurry. However, if you do find better or interesting solutions to the problems, I’d still like to hear about them. Also please don’t put this on the Internet to encourage copying homework solutions... 1 Vector Spaces 1. Assuming that C is a field, write z = a + bi. Then we have 1/z = z/zz = z/[z[ 2 . Plugging in the numbers we get 1/(a + bi) = a/(a 2 + b 2 ) − bi/(a 2 + b 2 ) = c + di. A straightforward calculation of (c+di)(a+bi) = 1 shows that this is indeed an inverse. 2. Just calculate ((1 + √ 3)/2) 3 . 3. We have v + (−v) = 0, so by the uniqueness of the additive inverse (prop. 1.3) −(−v) = v. 4. Choose a = 0 and v = 0. Then assuming av = 0 we get v = a −1 av = a −1 0 = 0. Contradiction. 5. Denote the set in question by A in each part. (a) Let v, w ∈ A, v = (x 1 , x 2 , x 3 ), w = (y 1 , y 2 , y 3 ). Then x 1 + 2x 2 + 3x 3 = 0 and y 1 + 2y 2 + 3y 3 = 0, so that 0 = x 1 + 2x 2 + 3x 3 + y 1 + 2y 2 + 3y 3 = (x 1 + y 1 ) + 2(x 2 +y 2 ) +3(x 3 +y 3 ), so v +w ∈ A. Similarly 0 = a0 = ax 1 +2ax 2 +3ay 3 , so av ∈ A. Thus A is a subspace. (b) This is not a subspace as 0 ∈ A. (c) We have that (1, 1, 0) ∈ A and (0, 0, 1) ∈ A, but (1, 1, 0)+(0, 0, 1) = (1, 1, 1) ∈ A, so A is not a subspace. (d) Let (x 1 , x 2 , x 3 ), (y 1 , y 2 , y 3 ) ∈ A. If x 1 = 5x 3 and y 1 = 5y 3 , then ax 1 = 5ax 3 , so a(x 1 , x 2 , x 3 ) ∈ A. Similarly x 1 + y 1 = 5(x 3 + y 3 ), so that (x 1 , x 2 , x 3 ) + (y 1 , y 2 , y 3 ) ∈ A. Thus A is a subspace. 1 6. Set U = Z 2 . 7. The set ¦(x, x) ∈ R 2 [ x ∈ R¦ ∪¦(x, −x) ∈ R 2 [ x ∈ R¦ is closed under multiplication but is trivially not a subspace ((x, x) + (x, −x) = (2x, 0) doesn’t belong to it unless x = 0). 8. Let ¦V i ¦ be a collection of subspaces of V . Set U = ∩ i V i . Then if u, v ∈ U. We have that u, v ∈ V i for all i because V i is a subspace. Thus au ∈ V i for all i, so that av ∈ U. Similarly u + v ∈ V i for all i, so u + v ∈ U. 9. Let U, W ⊂ V be subspaces. Clearly if U ⊂ W or W ⊂ U, then U ∪ W is clearly a subspace. Assume then that U ⊂ W and W ⊂ U. Then we can choose u ∈ U ` W and w ∈ W ` U. Assuming that U ∪ W is a subspace we have u + w ∈ U ∪ W. Assuming that u + w ⊂ U we get w = u + w − u ⊂ U. Contradiction. Similarly for u + w ∈ W. Thus U ∪ W is not a subspace. 10. Clearly U = U + U as U is closed under addition. 11. Yes and yes. Follows directly from commutativity and associativity of vector addition. 12. The zero subspace, ¦0¦, is clearly an additive identity. Assuming that we have inverses, then the whole space V should have an inverse U such that U + V = ¦0¦. Since V + U = V this is clearly impossible unless V is the trivial vector space. 13. Let W = R 2 . Then for any two subspaces U 1 , U 2 of W we have U 1 + W = U 2 + W, so the statement is clearly false in general. 14. Let W = ¦p ∈ {(F) [ p = ¸ n i=0 a i x i , a 2 = a 5 = 0¦. 15. Let V = R 2 . Let W = ¦(x, 0) ∈ R 2 [ x ∈ R¦. Set U 1 = ¦(x, x) ∈ R 2 [ x ∈ R¦, U 2 = ¦(x, −x) ∈ R 2 [ x ∈ R¦. Then it’s easy to see that U 1 + W = U 2 + W = R 2 , but U 1 = U 2 , so the statement is false. 2 Finite-Dimensional Vector Spaces 1. Let u n = v n and u i = v i +v i+1 , i = 1, . . . , n−1. Now we see that v i = ¸ n j=i u i . Thus v i ∈ span(u 1 , . . . , u n ), so V = span(v 1 , . . . , v n ) ⊂ span(u 1 , . . . , u n ). 2. From the previous exercise we know that the span is V . As (v 1 , . . . , v n ) is a linearly independent spanning list of vectors we know that dimV = n. The claim now follows from proposition 2.16. 2 3. If (v 1 + w, . . . , v n + w) is linearly dependent, then we can write 0 = a 1 (v 1 + w) + . . . + a n (v n + w) = n ¸ i=1 a i v i + w n ¸ i=1 a i , where a i = 0 for some i. Now ¸ n i=1 a i = 0 because otherwise we would get 0 = n ¸ i=1 a i v i + w n ¸ i=1 a i = n ¸ i=1 a i v i and by the linear independence of (v i ) we would get a i = 0, ∀i contradicting our assumption. Thus w = − n ¸ i=1 a i −1 n ¸ i=1 a i v i , so w ∈ span(v 1 , . . . , v n ). 4. Yes, multiplying with a nonzero constant doesn’t change the degree of a polynomial and adding two polynomials either keeps the degree constant or makes it the zero polynomial. 5. (1, 0, . . .), (0, 1, 0, . . .), (0, 0, 1, 0, . . .), . . . is trivially linearly independent. Thus F ∞ isn’t finite-dimensional. 6. Clearly {(F) is a subspace which is infinite-dimensional. 7. Choose v 1 ∈ V . Assuming that V is infinite-dimensional span(v 1 ) = V . Thus we can choose v 2 ∈ V ` span(v 1 ). Now continue inductively. 8. (3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1) is clearly linearly independent. Let (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ U. Then (x 1 , x 2 , x 3 , x 4 , x 5 ) = (3x 2 , x 2 , 7x 4 , x 4 , x 5 ) = x 2 (3, 1, 0, 0, 0)+x 4 (0, 0, 7, 1, 0)+ x 5 (0, 0, 0, 0, 1), so the vectors span U. Hence, they form a basis. 9. Choose the polynomials (1, x, x 2 −x 3 , x 3 ). They clearly span { 3 (F) and form a basis by proposition 2.16. 10. Choose a basis (v 1 , . . . , v n ). Let U i = span(v i ). Now clearly V = U 1 ⊕ . . . ⊕ U n by proposition 2.19. 11. Let dimU = n = dimV . Choose a basis (u 1 , . . . , u n ) for U and extend it to a basis (u 1 , . . . , u n , v 1 , . . . , v k ) for V . By assumption a basis for V has length n, so (u 1 , . . . , u n ) spans V . 3 12. Let U = span(u 0 , . . . , u m ). As U = { m (F) we have by the previous exercise that dimU < dim{ m (F) = m + 1. As (p 0 , . . . , p m ) is a spanning list of vectors having length m + 1 it is not linearly independent. 13. By theorem 2.18 we have 8 = dimR 8 = dimU + dimW − dim(U ∩ W) = 4 + 4 − dim(U ∩W). Since U +W = R 8 , we have that dim(U ∩W) = 0 and the claim follows 14. Assuming that U ∩ W = ¦0¦ we have by theorem 2.18 that 9 = dimR 9 = dimU + dimW −dim(U ∩ W) = 5 + 5 −0 = 10. Contradiction. 15. Let U 1 = ¦(x, 0) ∈ R 2 [ x ∈ R¦, U 2 = ¦(0, x) ∈ R 2 [ x ∈ R¦, U 3 = ¦(x, x) ∈ R 2 [ x ∈ R¦. Then U 1 ∩ U 2 = ¦0¦, U 1 ∩ U 3 = ¦0¦ and U 2 ∩ U 3 = ¦0¦. Thus the left-hand side of the equation in the problem is 2 while the right-hand side is 3. Thus we have a counterexample. 16. Choose a basis (u i 1 , . . . , u i n i ) for each U i . Then the list of vectors (u 1 1 , . . . , u m nm ) has length dimU 1 + . . . + dimU m and clearly spans U 1 + . . . + U m proving the claim. 17. By assumption the list of vectors (u 1 1 , . . . , u m nm ) from the proof of the previous exercise is linearly independent. Since V = U 1 +. . .+U n = span(u 1 1 , . . . , u m nm ) the claim follows. 3 Linear Maps 1. Any v = 0 spans V . Thus, if w ∈ V , we have w = av for some a ∈ F. Now T(v) = av for some a ∈ F, so for w = bv ∈ V we have T(w) = T(bv) = bT(v) = bav = aw. Thus T is multiplication by a scalar. 2. Define f by e.g. f(x, y) = x, x = y 0, x = y , then clearly f satisfies the condition, but is non-linear. 3. To define a linear map it’s enough to define the image of the elements of a basis. Choose a basis (u 1 , . . . , u m ) for U and extend it to a basis (u 1 , . . . , u m , v 1 , . . . , v n ) of V . If S ∈ L(U, W). Choose some vectors w 1 , . . . , w n ∈ W. Now define T ∈ L(V, W) by T(u i ) = S(u i ), i = 1, . . . , m and T(v i ) = w i , i = 1, . . . , n. These relations define the linear map and clearly T |U = S. 4. By theorem 3.4 dimV = dimnull T +dimrange T. If u ∈ null T, then dimrange T > 0, but as dimrange T ≤ dimF = 1 we have dimrange T = 1 and it follows that 4 dimnull T = dimV −1. Now choose a basis (u 1 , . . . , u n ) for null T. By our assumption (u 1 , . . . , u n , u) is linearly independent and has length dimV . Hence, it’s a basis for V . This implies that V = span(u 1 , . . . , u n , u) = span(u 1 , . . . , u n ) ⊕span(u) = null T ⊕¦au [ a ∈ F¦. 5. By linearity 0 = ¸ n i=1 a i T(v i ) = ¸ n i=1 T(a i v i ) = T( ¸ n i=1 a i v i ). By injectivity we have ¸ n i=1 a i v i = 0, so that a i = 0 for all i. Hence (T(v 1 ), . . . , T(v n )) is linearly independent. 6. If n = 1 the claim is trivially true. Assume that the claim is true for n = k. If S 1 , . . . , S k+1 satisfy the assumptions, then S 1 S k is injective by the induction hypothesis. Let T = S 1 S k . If u = 0, then by injectivity of S k+1 we S k+1 u = 0 and by injectivity of T we have TS k+1 u = 0. Hence TS k+1 is injective and the claim follows. 7. Let w ∈ W. By surjectivity of T we can find a vector v ∈ V such that T(v) = w. Writing v = a 1 v 1 + . . . + a n v n we get w = T(v) = T(a 1 v 1 + . . . + a n v n ) = a 1 T(v 1 ) + . . . + a n T(v n ) proving the claim. 8. Let (u 1 , . . . , u n ) be a basis for null T and extend it to a basis (u 1 , . . . , u n , v 1 , . . . , v k ) of V . Let U := span(v 1 , . . . , v k ). Then by construction null T ∩ U = ¦0¦ and for an arbitrary v ∈ V we have that v = a 1 u 1 + . . . + a n u n + b 1 v n + . . . + b k v k , so that T(v) = b 1 T(v 1 ) + . . . + b k T(v k ). Hence range T = T(U). 9. It’s easy to see that (5, 1, 0, 0), (0, 0, 7, 1) is a basis for null T (see exercise 2.8). Hence dimrange T = dimF 4 − dimnull T = 4 − 2 = 2. Thus range T = F 2 , so that T is surjective. 10. It’s again easy to see that (3, 1, 0, 0, 0), (0, 0, 1, 1, 1) is a basis of null T. Hence, we get dimrange T = dimF 5 −dimnull T = 5 −2 = 3 which is impossible. 11. This follows trivially from dimV = dimnull T + dimrange T. 12. From dimV = dimnull T +dimrange T it follows trivially that if we have a surjective linear map T ∈ L(V, W), then dimV ≥ dimW. Assume then that dimV ≥ dimW. Choose a basis (v 1 , . . . , v n ) for V and a basis (w 1 , . . . , w m ) for W. We can define a linear map T ∈ L(V, W) by letting T(v i ) = w i , i = 1, . . . , m, T(v i ) = 0, i = m + 1, . . . , n. Clearly T is surjective. 5 13. We have that dimrange T ≤ dimW. Thus we get dimnull T = dimV −dimrange T ≥ dimV −dimW. Choose an arbitrary subspace U of V of such that dimU ≥ dimV − dimW. Let (u 1 , . . . , u n ) be a basis of U and extend it to a basis (u 1 , . . . , u n , v 1 , . . . , v k ) of V . Now we know that k ≤ dimW, so let (w 1 , . . . , w m ) be a basis for W. Define a linear map T ∈ L(V, W) by T(u i ) = 0, i = 1, . . . , n and T(v i ) = w i , i = 1, . . . , k. Clearly null T = U. 14. Clearly if we can find such a S, then T is injective. Assume then that T is injective. Let (v 1 , . . . , v n ) be a basis of V . By exercise 5 (T(v 1 ), . . . , T(v n )) is linearly inde- pendent so we can extend it to a basis (T(v 1 ), . . . , T(v n ), w 1 , . . . , w m ) of W. Define S ∈ L(W, V ) by S(T(v i )) = v i , i = 1, . . . , n and S(w i ) = 0, i = 1, . . . , m. Clearly ST = I V . 15. Clearly if we can find such a S, then T is surjective. Otherwise let (v 1 , . . . , v n ) be a basis of V . By assumption (T(w 1 ), . . . , T(w n )) spans W. By the linear dependence lemma we can make the list of vectors (T(w 1 ), . . . , T(w n )) a basis by removing some vectors. Without loss of generality we can assume that the first m vectors form the basis (just permute the indices). Thus T(w 1 ), . . . , T(w m )) is a basis of W. Define the map S ∈ L(W, V ) by S(T(w i )) = w i , i = 1, . . . , m. Now clearly TS = I W . 16. We have that dimU = dimnull T + dimrange T ≤ dimnull T + dimV . Substituting dimV = dimnull S + dimrange S and dimU = dimnull ST + dimrange ST we get dimnull ST + dimrange ST ≤ dimnull T + dimnull S + dimrange S. Clearly dimrange ST ≤ dimrange S, so our claim follows. 17. This is nothing but pencil pushing. Just take arbitrary matrices satisfying the re- quired dimensions and calculate each expression and the equalities easily fall out. 18. Ditto. 19. Let V = (x 1 , . . . , n ). From proposition 3.14 we have that ´(Tv) = ´(T)´(v) = a 1,1 a 1,n . . . . . . . . . a m,1 a m,n ¸ ¸ ¸ x 1 . . . x n ¸ ¸ ¸ = a 1,1 x 1 + . . . + a 1,n x n . . . a m,1 x 1 + . . . + a m,n x n ¸ ¸ ¸ which shows that Tv = (a 1,1 x 1 + . . . + a 1,n x n , . . . , a m,1 x 1 + . . . + a m,n x n ). 20. Clearly dimMat(n, 1, F) = n = dimV . We have that Tv = 0 if and only if v = 0v 1 + . . . , 0v n = 0, so null T = ¦0¦. Thus T is injective and hence invertible. 6 21. Let e i denote the n 1 matrix with a 1 in the ith row and 0 everywhere else. Let T be the linear map. Define a matrix A := [T(e 1 ), . . . , T(e n )]. Now it’s trivial to verify that Ae i = T(e i ). By distributivity of matrix multiplication (exercise 17) we get for an arbitrary v = a 1 e 1 + . . . + a n e n that Av = A(a 1 e 1 + . . . + a n e n ) = a 1 Ae 1 + . . . + a n Ae n = a 1 T(e 1 ) + . . . + a n T(e n ) = T(a 1 e 1 + . . . + a n e n ), so the claim follows. 22. From theorem 3.21 we have ST invertible ⇔ ST bijective ⇔ S and T bijective ⇔ S and T invertible. 23. By symmetry it’s sufficient to prove this in one direction only. Thus if TS = I. Then ST(Su) = S(TSu) = Su for all u. As S is bijective Su goes through the whole space V as u varies, so ST = I. 24. Clearly TS = ST if T is a scalar multiple of the identity. For the other direction assume that TS = ST for every linear map S ∈ L(V ). 25. Let T ∈ L(F 2 ) be the operator T(a, b) = (a, 0) and S ∈ L(F 2 ) the operator S(a, b) = (0, b). Then neither one is injective and hence invertible by 3.21. However, T + S is the identity operator which is trivially invertible. This can be trivially generalized to spaces arbitrary spaces of dimension ≥ 2. 26. Write A := a 1,1 a 1,n . . . . . . . . . a n,1 a n,n ¸ ¸ ¸, x = x 1 . . . x n ¸ ¸ ¸. Then the system of equations in a) reduces to Ax = 0. Now A defines a linear map from Mat(n, 1, F) to Mat(n, 1, F). What a) states is now that the map is injective while b) states that it is surjective. By theorem 3.21 these are equivalent. 4 Polynomials 1. Let λ 1 , . . . , λ m be m distinct numbers and k 1 , . . . , k m > 0 such that their sum is n. Then ¸ m i=1 (x −λ i ) k i is clearly such a polynomial. 7 2. Let p i (x) = ¸ j=i (x − z j ), so that deg p i = m. Now we have that p i (z j ) = 0 if and only if i = j. Let c i = p i (z i ). Define p(x) = m+1 ¸ i=1 w i c −1 i p i (x). Now deg p = m and p clearly p(z i ) = w i . 3. We only need to prove uniqueness as existence is theorem 4.5. Let s , r be other such polynomials. Then we get that 0 = (s −s )p + (r −r ) ⇔ r −r = (s −s )p We know that for any polynomials p = 0 = q we have deg pq = deg p + deg q. Assuming that s = s we have deg(r − r) < deg(s − s )p which is impossible. Thus s = s implying r = r . 4. Let λ be a root of p. Thus we can write p(x) = (x −λ)q(x). Now we get p (x) = q(x) + (x −λ)q (x). Thus λ is a root of p if and only if λ is a root of q i.e. λ is a multiple root. The statement follows. 5. Let p(x) = ¸ n i=0 a i x i , where a i ∈ R. By the fundamental theorem of calculus we have a complex root z. By proposition 4.10 z is also a root of p. Then z and z are roots of equal multiplicity (divide by (x − z)(x − z) which is real). Thus if we have no real roots we have an even number of roots counting multiplicity, but the number of roots counting multiplicity is deg p hence odd. Thus p has a real root. 5 Eigenvalues and Eigenvectors 1. An arbitrary element of U 1 +. . . +U n is of the form u 1 +. . . +u n where u i ∈ U i . Thus we get T(u 1 + . . . + u n ) = T(u 1 ) + . . . + T(u n ) ∈ U 1 + . . . + U n by the assumption that T(u i ) ∈ U i for all i. 2. Let V = ∩ i U i where U i is invariant under T for all i. Let v ∈ V , so that v ∈ U i for all i. Now T(v) ∈ U i for all i by assumption, so T(v) ∈ ∩U i = V . Thus V is invariant under T. 3. The clame is clearly true for U = ¦0¦ or U = V . Assume that ¦0¦ = U = V . Let (u 1 , . . . , u n ) be a basis for U and extend it to a basis (u 1 , . . . , u n , v 1 , . . . , v m ). By our assumption m ≥ 1. Define a linear operator by T(u i ) = v 1 , i = 1, . . . , n and T(v i ) = v 1 , i = 1, . . . , m. Then clearly U is not invariant under T. 8 4. Let u ∈ null(T −λI), so that T(u) −λu = 0. ST = TS gives us 0 = S(Tu −λu) = STu −λSu = TSu −λSu = (T −λI)Su, so that Su ∈ null(T −λI). 5. Clearly T(1, 1) = (1, 1), so that 1 is an eigenvalue. Also T(−1, 1) = (1, −1) = −(−1, 1) so −1 is another eigenvalue. By corollary 5.9 these are all eigenvalues. 6. We easily see that T(0,0,1)=(0,0,5), so that 5 is an eigenvalue. Also T(1, 0, 0) = (0, 0, 0) so 0 is an eigenvalue. Assume that λ = 0 and T(z 1 , z 2 , z 3 ) = (2z 2 , 0, 5z 3 ) = λ(z 1 , z 2 , z 3 ). From the assumption λ = 0 we get z 2 = 0, so the equation is of the form (0, 0, 5z 3 ) = λ(z 1 , 0, z 3 ). Again we see that z 1 = 0 so we get the equation (0, 0, 5z 3 ) = λ(0, 0, z 3 ). Thus 5 is the only non-zero eigenvalue. 7. Notice that the range of T is the subspace ¦(x, . . . , x) ∈ F n [ x ∈ F¦ and has dimension 1. Thus dimrange T = 1, so dimnull T = n − 1. Assume that T has two distinct eigenvalues λ 1 , λ 2 and assume that λ 1 = 0 = λ 2 . Let v 1 , v 2 be the corresponding eigenvectors, so by theorem 5.6 they are linearly independent. Then v 1 , v 2 ∈ null T, but dimnull T = n − 1, so this is impossible. Hence T has at most one non-zero eigenvalue hence at most two eigenvalues. Because T is not injective we know that 0 is an eigenvalue. We also see that T(1, . . . , 1) = n(1, . . . , 1), so n is another eigenvalue. By the previous paragraph, these are all eigenvalues of T. 8. Let a ∈ F. Now we have T(a, a 2 , a 3 , . . .) = (a 2 , a 3 , . . .) = a(a, a 2 , . . .), so every a ∈ F is an eigenvalue. 9. Assume that T has k +2 distinct eigenvalues λ 1 , . . . , λ k+2 with corresponding eiven- vectors v 1 , . . . , v k+2 . By theorem 5.6 these eigenvectors are linearly independent. Now Tv i = λ i v i and dimspan(Tv 1 , . . . , Tv k+2 ) = dimspan(λ 1 v 1 , . . . , λ k+2 v k+2 ) ≥ k + 1 (it’s k + 2 if all λ i are non-zero, otherwise k + 1). This is a contradiction as dimrange T = k and span(λ 1 v 1 , . . . , λ k+2 v k+2 ) ⊂ range T. 10. As T = (T −1 ) −1 we only need to show this in one direction. If T is invertible, then 0 is not an eigenvalue. Now let λ be an eigenvalue of T and v the corresponding eigenvector. From Tv = λv we get that T −1 λv = λT −1 v = v, so that T −1 v = λ −1 v. 11. Let λ be an eigenvalue of TS and v the corresponding eigenvector. Then we get STSv = Sλv = λSv, so if Sv = 0, then it is is an eigenvector for the eigenvalue λ. If Sv = 0, then TSv = 0, so λ = 0. As Sv = 0 we know that S is not injective, so ST is not injective and it has eigenvalue 0. Thus if λ is an eigenvalue of TS, then it’s an eigenvalue of ST. The other implication follows by symmetry. 9 12. Let (v 1 , . . . , v n ) be a basis of V . By assumption (Tv 1 , . . . , Tv n ) = (λ 1 v, . . . , λ n v n ). We need to show that λ i = λ j for all i, j. Choose i = j, then by our assumption v i + v j is an eigenvector, so T(v i + v j ) = λ i v i + λ j v j = λ(v i + v j ) = λv i + λv j . This means that (λ i − λ)v i + (λ j − λ)v j = 0. Because (v i , v j ) is linearly independent we get that λ i −λ = 0 = λ j −λ i.e. λ i = λ j . 13. Let v 1 ∈ V and extend it to a basis (v 1 , . . . , v n ) of V . Now Tv 1 = a 1 v 1 + . . . + a n v n . Let U i be the subspace generated by the vectors v j , j = i. By our assumption each U i is an invariant subspace. Let Tv 1 = a 1 v 1 + . . . + a n v n . Now v 1 ∈ U i for i > 1. So let j > 1, then Tv ∈ U j imples a j = 0. Thus Tv 1 = a 1 v 1 . We see that v 1 was an eigenvector. The result now follows from the previous exercise. 14. Clearly (STS −1 ) n = ST n S −1 , so n ¸ i=0 a i (STS −1 ) i = n ¸ i=0 a i ST i S −1 = S n ¸ i=0 a i T i S −1 . 15. Let λ be an eigenvalue of T and v ∈ V a corresponding eigenvector. Let p(x) = ¸ n i=0 a i x i . Then we have p(T)v = n ¸ i=0 a i T i v = n ¸ i=0 a i λ i v = p(λ)v. Thus p(λ) is an eigenvalue of p(T). Then let a be an eigenvalue of p(T) and v ∈ V the corresponding eigenvector. Let q(x) = p(x) −a. By the fundamental theorem of algebra we can write q(x) = c ¸ n i=1 (x −λ i ). Now q(T)v = 0 and q(T) = c ¸ n i=1 (T − λ i I). As q(T) is non-injective we have that T −λ i I is non-injective for some i. Hence λ i is an eigenvalue of T. Thus we get 0 = q(λ i ) = p(λ i ) −a, so that a = p(λ i ). 16. Let T ∈ L(R 2 ) be the map T(x, y) = (−y, x). On page 78 it was shown that it has no eigenvalue. However, T 2 (x, y) = T(−y, x) = (−x, −y), so −1 it has an eigenvalue. 17. By theorem 5.13 T has an upper-triangular matrix with respect to some basis (v 1 , . . . , v n ). The claim now follows from proposition 5.12. 18. Let T ∈ L(F 2 ) be the operator T(a, b) = (b, a) with respect to the standard basis. Now T 2 = I, so T is clearly invertible. However, T has the matrix ¸ 0 1 1 0 . 10 19. Take the operator T in exercise 7. It is clearly not invertible, but has the matrix 1 1 . . . . . . . . . 1 1 ¸ ¸ ¸. 20. By theorem 5.6 we can choose basis (v 1 , . . . , v n ) for V where v i is an eigenvector of T corresponding to the eigenvalue λ i . Let λ be the eigenvalue of S corresponding to v i . Then we get STv i = Sλ i v i = λ i Sv i = λ i λv i = λλ i v i = λTv i = Tλv i = TSv i , so ST and TS agree on a basis of V . Hence they are equal. 21. Clearly 0 is an eigenvalue and the corresponding eigenvectors are null T = W ` ¦0¦. Now assume λ = 0 is an eigenvalue and v = u + w is a corresponding eigenvector. Then P U,W (u + w) = u = λu + λw ⇔ (1 −λ)u = λw. Thus λw ∈ U, so λw = 0, but λ = 0, so we have w = 0 which implies (1 −λ)u = 0. Because v is an eigenvector we have v = u + w = u = 0, so that 1 − λ = 0. Hence λ = 1. As we can choose freely our u ∈ U, so that it’s non-zero we see that the eigenvectors corresponding to 1 are u = U ` ¦0¦. 22. As dimV = dimnull P + dimrange P, it’s clearly sufficient to prove that null P ∩ range P = ¦0¦. Let v ∈ null P ∩ range P. As v ∈ range P, we can find a u ∈ V such that Pu = v. Thus we have that v = Pu = P 2 u = Pv, but v ∈ null P, so Pv = 0. Thus v = 0. 23. Let T(a, b, c, d) = (−b, a, −d, c), then T is clearly injective, so 0 is not an eigenvalue. If λ = 0 is an eigenvalue and λ(a, b, c, d) = (−b, a, −d, c). We clearly see that a = 0 ⇔ b = 0 and similarly with c, d. By symmetry we can assume that a = 0 (an eigenvector is non-zero). Then we have λa = −b and λb = a. Substituting we get λ 2 b = −b ⇔ (λ 2 + 1)b = 0. As b = 0 we have λ 2 + 1 = 0, but this equation has no solutions in R. Hence T has no eigenvalue. 24. If U is an invariant subspace of odd degree, then by theorem 5.26 T |U has an eigenvalue λ with eigenvector v. Then λ is an eigenvalue of T with eigenvector v against our assumption. Thus T has no subspace of odd degree. 6 Inner-Product Spaces 1. From the law of cosines we get |x − y| 2 = |x| 2 + |y| 2 − 2|x||y| cos θ. Solving we get |x||y| cos θ = |x| 2 +|y| 2 −|x −y| 2 2 . 11 Now let x = (x 1 , x2), y = (y 1 , y 2 ). A straight calculation shows that |x| 2 +|y| 2 −|x−y| 2 = x 2 1 +x 2 2 +y 2 1 +y 2 2 −(x 1 −y 1 ) 2 −(x 2 −y 2 ) 2 = 2(x 1 y 1 +x 2 y 2 ), so that |x||y| cos θ = |x| 2 +|y| 2 −|x −y| 2 2 = 2(x 1 y 1 + x 2 y 2 ) 2 = 'x, y` . 2. If 'u, v` = 0, then 'u, av` = 0, so by the Pythagorean theorem |u| ≤ |u| + |av| = |u+av|. Then assume that |u| ≤ |u+av| for all a ∈ F. We have |u| 2 ≤ |u+av| 2 ⇔ 'u, u` ≤ 'u + av, u + av`. Thus we get 'u, u` ≤ 'u, u` +'u, av` +'av, u` +'av, av` , so that −2Re a 'u, v` ≤ [a[ 2 |v| 2 . Choose a = −t 'u, v` with t > 0, so that 2t 'u, v` 2 ≤ t 2 'u, v` 2 |v| 2 ⇔ 2 'u, v` 2 ≤ t 'u, v` 2 |v| 2 . If v = 0, then clearly 'u, v` = 0. If not choose t = 1/|v| 2 , so that we get 2 'u, v` 2 ≤ 'u, v` 2 . Thus 'u, v` = 0. 3. Let a = (a 1 , √ 2a 2 , . . . , √ na n ) ∈ R n and b = (b 1 , b 2 / √ 2, . . . , b n / √ n) ∈ R n . This equality is then simply 'a, b` 2 ≤ |a| 2 |b| 2 which follows directly from the Cauchy- Schwarz inequality. 4. From the parallelogram equality we have |u + v| 2 + |u − v| 2 = 2(|u| 2 + |v| 2 ). Solving for |v| we get |v| = √ 17. 5. Set e.g. u = (1, 0), v = (0, 1). Then |u| = 1, |v| = 1, |u + v| = 2, |u − v| = 2. Assuming that the norm is induced by an inner-product, we would have by the parallelogram inequality 8 = 2 2 + 2 2 = 2(1 2 + 1 2 ) = 4, which is clearly false. 6. Just use |u| = 'u, u` and simplify. 7. See previous exercise. 12 8. This exercise is a lot trickier than it might seem. I’ll prove it for R, the proof for C is almost identical except for the calculations that are longer and more tedious. All norms on a finite-dimensional real vector space are equivalent. This gives us lim n→∞ |r n x + y| = |λx + y| when r n → λ. This probably doesn’t make any sense, so check a book on topology or functional analysis or just assume the result. Define 'u, v` by 'u, v` = |u + v| 2 −|u −v| 2 4 . Trivially we have |u| 2 = 'u, u`, so positivity definiteness follows. Now 4('u + v, w` −'u, w` −'v, w`) = |u + v + w| 2 −|u + v −w| 2 −(|u + w| 2 −|u −w| 2 ) −(|v + w| 2 −|v −w| 2 ) = |u + v + w| 2 −|u + v −w| 2 +(|u −w| 2 +|v −w| 2 ) −(|u + w| 2 +|v + w| 2 ) We can apply the parallelogram equality to the two parenthesis getting 4('u + v, w` −'u, w` −'v, w`) = |u + v + w| 2 −|u + v −w| 2 + 1 2 (|u + w −2w| 2 +[u −v[ 2 ) − 1 2 (|u + v + 2w| 2 +|u −v| 2 ) = |u + v + w| 2 −|u + v −w| 2 + 1 2 |u + w −2w| 2 − 1 2 |u + v + 2w| 2 = (|u + v + w| 2 +|w| 2 ) + 1 2 |u + v −2w| 2 − (|u + v −w| 2 +|w| 2 ) − 1 2 |u + v + 2w| 2 Applying the parallelogram equality to the two parenthesis we and simplifying we are left with 0. Hence we get additivity in the first slot. It’s easy to see that '−u, v` = −'u, v`, so we get 'nu, v` = n'u, v` for all n ∈ Z. Now we get 'u, v` = 'nu/n, v` = n'u/n, v` , 13 so that 'u/n, v` = 1/n'u, v`. Thus we have 'nu/m, v` = n/m'u, v`. It follows that we have homogeneity in the first slot when the scalar is rational. Now let λ ∈ R and choose a sequence (r n ) of rational numbers such that r n → λ. This gives us λ'u, v` = lim n→∞ r n 'u, v` = lim n→∞ 'r n u, v` = lim n→∞ 1 4 (|r n u + v| 2 −|r n u −v| 2 ) = 1 4 (|λu + v| 2 −|λu −v| 2 ) = 'λu, v` Thus we have homogeneity in the first slot. We trivially also have symmetry, so we have an inner-product. The proof for F = C can be done by defining the inner-product from the complex polarizing identity. And by using the identities 'u, v` 0 = 1 4 (|u + v| 2 −|u −v| 2 ) ⇒ 'u, v` = 'u, v` 0 +'u, iv` 0 i. and using the properties just proved for ', ` 0 . 9. This is really an exercise in calculus. We have integrals of the types 'sin nx, sin mx` = π −π sin nxsin mxdx 'cos nx, cos mx` = π −π cos nxcos mxdx 'sin nx, cos mx` = π −π sin nxcos mxdx and they can be evaluated using the trigonometric identities sin nxsin mx = cos((n −m)x) −cos((n + m)x) 2 cos nxcos mx = cos((n −m)x) + cos((n + m)x) 2 cos nxsin mx = sin((n −m)x) −sin((n + m)x) 2 . 10. e 1 = 1, then e 2 = x−x,11 x−x,11 . Where 'x, 1` = 1 0 x = 1/2, 14 so that |x −1/2| = 'x −1/2, x −1/2` = 1 0 (x −1/2) 2 dx = 1/12, which gives e 2 = √ 12(x − 1/2) = √ 3(2x − 1). Then to continue pencil pushing we have x 2 , 1 = 1/3, x 2 , √ 3(2x −1) = √ 3/6, so that x 2 − x 2 , 1 1 − x 2 , √ 3(2x −1) √ 3(2x −1) = x 2 −1/3 −1/2(2x −1). Now |x 2 −1/3 −1/2(2x −1)| = 1/(6 √ 5) giving e 3 = √ 5(6x 2 −6x+1). The orthonormal basis is thus (1, √ 3(2x−1), √ 5(6x 2 − 6x + 1). 11. Let (v 1 , . . . , v n ) be a linearly dependent list. We can assume that (v 1 , . . . , v n−1 ) is linearly independent and v n ∈ span(v 1 , . . . , v n ). Let P denote the projection on the subspace spanned by (v 1 , . . . , v n−1 ). Then calculating e n with the Gram-Schmidt algorithm gives us e n = v n −P(v n ) |v n −Pv n | , but v n = Pv n by our assumption, so e n = 0. Thus the resulting list will simply contain zero vectors. It’s easy to see that we can extend the algorithm to work for linearly dependent lists by tossing away resulting zero vectors. 12. Let (e 1 , . . . , e n−1 ) be an orthogonal list of vectors. Assume that (e 1 , . . . , e n ) and (e 1 , . . . , e n−1 , e n ) are orthogonal and span the same subspace. Then we can write e n = a 1 e 1 + . . . + a n e n . Now we have 'e n , e i ` = 0 for all i < n, so that a i = 0 for i < n. Thus we have e n = a n e n and from |e n | = 1 we have a n = ±1. Let (e 1 , . . . , e n ) be the orthonormal base produced from (v 1 , . . . , v n ) by Gram-Schmidt. Then if (e 1 , . . . , e n ) satisfies the hypothesis from the problem we have by the previous paragraph that e i = ±e i . Thus we have 2 n possible such orthonormal lists. 13. Extend (e 1 , . . . , e m ) to a basis (e 1 , . . . , e n ) of V . By theorem 6.17 v = 'v, e 1 ` e 1 + . . . +'v, e n ` e n , so that |v| = | 'v, e 1 ` e 1 + . . . +'v, e m ` e n | = | 'v, e 1 ` e 1 | + . . . +| 'v, e m ` e n | = [ 'v, e 1 ` [ + . . . +[ 'v, e n ` [. Thus we have |v| = [ 'v, e 1 ` [ + . . . +[ 'v, e m ` [ if and only if 'v, e i ` = 0 for i > m i.e. v ∈ span(e 1 , . . . , e m ). 15 14. It’s easy to see that the differentiation operator has a upper-triangular matrix in the orthonormal basis calculated in exercise 10. 15. This follows directly from V = U ⊕U ⊥ . 16. This follows directly from the previous exercise. 17. From exercise 5.21 we have that V = range P ⊕ null P. Let U := range P, then by assumption null P ⊂ U ⊥ . From exercise 15, we have dimU = dimV − dimU = dimnull P, so that null P = U ⊥ . An arbitrary v ∈ V can be written as v = Pv +(v − Pv). From P 2 = P we have that P(v − Pv) = 0, so v − Pv ∈ null P = U ⊥ . Hence the decomposition v = Pv + (v −Pv) is the unique decomposition in U ⊕U ⊥ . Now Pv = P(Pv +(v −Pv)) = Pv, so that P is the identity on U. By definition P = P U . 18. Let u ∈ range P, then u = Pv for some v ∈ V hence Pu = P 2 v = Pv = u, so P is the identity on range P. Let w ∈ null P. Then for a ∈ F |u| 2 = |P(u + aw)| 2 ≤ |u + aw| 2 . By exercise 2 we have 'u, w` = 0. Thus null P ⊂ (range P) ⊥ and from the dimension equality null P = (range P) ⊥ . Hence the claim follows. 19. If TP U = P U TP U clearly U is invariant. Now assume that U is invariant. Then for u ∈ U we have P U TP U u = P U Tu = Tu = TP U u. 20. Let u ∈ U, then we have Tu = TP U u = P U Tu ∈ U. Thus U is invariant. Then let w ∈ U ⊥ . Now we can write Tw = u+u where u ∈ U and u ∈ U ⊥ . Now P U Tw = u, but u = P U Tw = TP U w = T0 = 0, so Tw ∈ U ⊥ . Thus U ⊥ is also invariant. 21. First we need to find an orthonormal basis for U. With Gram-Schmidt we get e 1 = (1/ √ 2, 1/ √ 2, 0, 0), e 2 = (0, 0, 1/ √ 5, 2/ √ 5). Let U = span(e 1 , e 2 ). Then we have u = P U (1, 2, 3, 4) = '(1, 2, 3, 4), e 1 ` e 1 +'(1, 2, 3, 4), e 2 ` e 2 = (3/2, 3/2, 11/5, 22/5). 22. If p(0) = 0 and p (0) = 0, then p(x) = ax 2 + bx 3 . Thus we want to find the projection of 2 + 3x to the subspace U := span(x 2 , x 3 ). With Gram-Schmidt we get the orthonormal basis ( √ 3x 2 , 420/11(x 3 − 1 2 x 2 )). We get P U (2+3x) = 2 + 3x, √ 3x 2 √ 3x 2 + 2 + 3x, 420/11(x 3 − 1 2 x 2 ) 420/11(x 3 − 1 2 x 2 ). Here 2 + 3x, √ 3x 2 = 17 12 √ 3, 2 + 3x, 420/11(x 3 − 1 2 x 2 ) = 47 660 √ 1155, so that P U (2 + 3x) = − 71 22 x 2 + 329 22 x 3 . 16 23. There’s is nothing special with this exercise, compared to the previous two, except that it takes ages to calculate. 24. We see that the map T : { 2 (R) → R defined by p → p(1/2) is linear. From exercise 10 we get that (1, √ 3(2x − 1), √ 5(6x 2 − 6x + 1) is an orthonormal basis for { 2 (R). From theorem 6.45 we see that q(x) = 1 + √ 3(2 1/2 −1) √ 3(2x −1) + √ 5(6 (1/2) 2 −6 1/2 + 1) √ 5(6x 2 −6x + 1) = −3/2 + 15x −15x 2 . 25. The map T : { 2 (R) → R defined by p → 1 0 p(x) cos(πx)dx is clearly linear. Again we have the orthonormal basis (1, √ 3(2x −1), √ 5(6x 2 −6x + 1), so that q(x) = 1 0 cos(πx)dx + 1 0 √ 3(2x −1) cos(πx)dx √ 3(2x −1) + 1 0 √ 5(6x 2 −6x + 1) cos(πx)dx √ 5(6x 2 −6x + 1) = 0 − 4 √ 3 π 2 √ 3(2x −1) + 0 = − 12 π 2 (2x −1). 26. Choose an orthonormal basis (e 1 , . . . , e n ) for V and let F have the usual basis. Then by proposition 6.47 we have ´(T, (e 1 , . . . , e n )) = ['e 1 , v` 'e n , v`] , so that ´(T ∗ , (e 1 , . . . , e n )) = 'e 1 , v` . . . 'e n , v` ¸ ¸ ¸ and finally T ∗ a = (a'e 1 , v`, . . . , a'e n , v`). 27. with the usual basis for F n we get ´(T) = 0 1 0 . . . . . . . . . 1 0 0 ¸ ¸ ¸ ¸ ¸ ¸ so by proposition 6.47 ´(T ∗ ) = 0 0 1 . . . . . . . . . 0 1 0 ¸ ¸ ¸ ¸ ¸ ¸ . 17 Clearly T ∗ (z 1 , . . . , z n ) = (z 2 , . . . , z n , 0). 28. By additivity and conjugate homogeneity we have (T −λI) ∗ = T ∗ −λI. From exercise 31 we see that T ∗ −λI is non-injective if and only if T −λI is non-injective. That is λ is an eigenvalue of T ∗ if and only if λ is an eigenvalue of T. 29. As (U ⊥ ) ⊥ = U and (T ∗ ) ∗ = T it’s enough to show only one of the implications. Let U be invariant under T and choose w ∈ U ⊥ . Now we can write T ∗ w = u + v, where u ∈ U and v ∈ U ⊥ , so that 0 = 'Tu, w` = 'u, T ∗ w` = 'u, u + v` = 'u, u` = |u| 2 . Thus u = 0 which completes the proof. 30. As (T ∗ ) ∗ = T it’s sufficient to prove the first part. Assume that T is injective, then by exercise 31 we have dimV = dimrange T = dimrange T ∗ , but range T is a subspace of V , so that range T = V . 31. From proposition 6.46 and exercise 15 we get dimnull T = dim(range T) ⊥ = dimW −dimrange T = dimnull T + dimW −dimV. For the second part we have dimrangeT ∗ = dimW −dimnullT ∗ = dimV −dimnull T = dimrange T, where the second equality follows from the first part. 32. Let T be the operator induced by A. The columns are the images of the basis vectors under T, so the generate range T. Hence the dimension of the span of the column vectors equal dimrange T. By proposition 6.47 the span of the row vectors equals range T ∗ . By the previous exercise dimrange T = dimrange T ∗ , so the claim follows. 7 Operators on Inner-Product Spaces 1. For the first part, let p(x) = x and q(x) = 1. Then clearly 1/2 = 'Tp, q` = 'p, Tq` = 'p, 0` = 0. For the second part it’s not a contradiction since the basis is not orthog- onal. 18 2. Choose the standard basis for R 2 . Let T and S the the operators defined by the matrices ¸ 0 1 1 0 , ¸ 0 0 0 1 . Then T and S as clearly self-adjoint, but TS has the matrix ¸ 0 1 0 0 . so TS is not self-adjoint. 3. If T and S are self-adjoint, then (aT +bS) ∗ = (aT) ∗ +(bS) ∗ = aT +bS for any a, b ∈ R, so self-adjoint operators form a subspace. The identity operator I is self-adjoint, but clearly (iI) ∗ = −iI, so iI is not self-adjoint. 4. Assume that P is self-adjoint. Then from proposition 6.46 we have null P = null P ∗ = (range P) ⊥ . Now P is clearly a projection to range P. Then assume that P is a projection. Let (u 1 , . . . , u n ) an orthogonal basis for range P and extend it to a basis of V . Then clearly P has a diagonal matrix with respect to this basis, so P is self-adjoint. 5. Choose the operators corresponding to the matrices A = ¸ 0 −1 1 0 , B = ¸ 1 1 1 0 . Then we easily see that A ∗ A = AA ∗ and B ∗ B = BB ∗ . However, A + B doesn’t define a normal operator which is easy to check. 6. From proposition 7.6 we get null T = null T ∗ . It follows from proposition 6.46 that range T = (null T ∗ ) ⊥ = (null T) ⊥ = range T ∗ . 7. Clearly null T ⊂ null T k . Let v ∈ null T k . Then we have T ∗ T k−1 v, T ∗ T k−1 v = T ∗ T k v, T k−1 v = 0, so that T ∗ T k−1 v = 0. Now T k−1 v, T k−1 v = T ∗ T k−1 v, T k−2 v = 0, so that v ∈ null T k−1 . Thus null T k ⊂ null T k−1 and continuing we get null T k ⊂ null T. Now let u ∈ range T k , then we can find a v ∈ V such that u = T k v = T(T k−1 v), so that range T k ⊂ range T. From the first part we get dimrange T k = dimrange T, so range T k = range T. 19 8. The vectors u = (1, 2, 3) and v = (2, 5, 7) are both eigenvectors corresponding to different eigenvalues. A self-adjoint operator is normal, so by corollary 7.8 if T is normal that would imply orthogonality of u and v. Clearly 'u, v` = 0, so T can’t be even normal much less self-adjoint. 9. If T is normal, we can choose a basis of V consisting of eigenvectors of T. Let A be the matrix of T corresponding to the basis. Now T is self-adjoint if and only if the conjugate transpose of A equals A that is the eigenvalues of T are real. 10. For any v ∈ V we get 0 = T 9 v −T 8 v = T 8 (Tv −v). Thus Tv −v ∈ null T 8 = null T by the normality of T. Hence T(Tv −v) = T 2 v −Tv = 0, so T 2 = T. By the spectral theorem we can choose a basis of eigenvectors for T such that T has a diagonal matrix with λ 1 , . . . , λ n on the diagonal. Now T 2 has the diagonal matrix with λ 2 1 , . . . , λ 2 n on the diagonal and from T 2 = T we must have λ 2 i = λ i for all i. Hence λ i = 0 or λ i = 1, so the matrix of T equals its conjugate transpose. Hence T is self-adjoint. 11. By the spectral theorem we can choose a basis of T such that the matrix of T corresponding to the basis is a diagonal matrix. Let λ 1 , . . . , λ n be the diagonal elements. Let S be the operator corresponding to the diagonal matrix having the elements √ λ 1 , . . . , √ λ n on the diagonal. Clearly S 2 = T. 12. Let T the operator corresponding to the matrix ¸ 0 −1 1 0 . Then T 2 + I = 0. 13. By the spectral theorem we can choose a basis consisting of eigenvectors of T. Then T has a diagonal matrix with respect to the basis. Let λ 1 , . . . , λ n be the diago- nal elements. Let S be the operator corresponding to the diagonal matrix having 3 √ λ 1 , . . . , 3 √ λ n on the diagonal. Then clearly S 3 = T. 14. By the spectral theorem we can choose a basis (v 1 , . . . , v n ) consisting of eigenvectors of T. Let v ∈ V be such that |v| = 1. If v = a 1 v 1 + . . . + a n v n this implies that ¸ n i=1 [a i [ = 1. Assume that |Tv −λv| < ε. If [λ i −λ[ ≥ ε for all i, then |Tv −λv| = | n ¸ i=1 (a i Tv i −λv i )| = | n ¸ i=1 (a i λ i v i −λv i )| = | n ¸ i=1 a i (λ i −λ)v i | = n ¸ i=1 [a i [[λ i −λ[|v 1 | = n ¸ i=1 [a i [[λ i −λ[ ≥ ε n ¸ i=1 [a i [ = ε, which is a contradiction. Thus we can find an eigenvalue λ i such that [λ −λ i [ < ε. 20 15. If such an inner-product exists we have a basis consisting of eigenvectors by the spectral theorem. Assume then that (v 1 , . . . , v n ) is basis such that the matrix of T is a matrix consisting of eigenvectors of T. Define an inner-product by 'v i , v j ` = 0, i = j 1, i = j . Extend this to an inner-product by bilinearity and homogeneity. Then we have an inner-product and clearly T is self-adjoint as (v 1 , . . . , v n ) is an orthonormal basis of U. 16. Let T ∈ L(R 2 ) be the operator T(a, b) = (a, a + b). Then span((1, 0)) is invariant under T, but it’s orthogonal complement span((0, 1)) is clearly not. 17. Let T, S be two positive operators. Then they are self-adjoint and (S + T) ∗ = S ∗ +T ∗ = S +T, so S +T is self-adjoint. Also '(S + T)v, v` = 'Sv, v` +'Tv, v` ≥ 0. Thus, S + T is positive. 18. Clearly T k is self-adjoint for every positive integer. For k = 2 we have T 2 v, v = 'Tv, Tv` ≥ 0. Assume that the result is true for all positive k < n and n ≥ 2. Then 'T n v, v` = T n−2 Tv, Tv ≥ 0, by hypothesis and self-adjointness. Hence the result follows by induction. 19. Clearly 'Tv, v` > 0 for all v ∈ V ` ¦0¦ implies that T is injective hence invertible. So assume that T is invertible. Since T is self-adjoint we can choose an orthonormal basis (v 1 , . . . , v n ) of eigenvectors such that T has a diagonal matrix with respect to the basis. Let the elements on the diagonal be λ 1 , . . . , λ n and by assumption λ i > 0 for all i. Let v = a 1 v 1 + . . . + a n v n ∈ V ` ¦0¦, so that 'Tv, v` = [a 1 [ 2 λ 1 + . . . +[a n [ 2 λ n > 0. 20. ¸ sin θ cos θ cos θ −sin θ is a square root for every θ ∈ R, which shows that it has infinitely many square roots. 21. Let S ∈ L(R 2 ) be defined by S(a, b) = (a + b, 0). Then |S(1, 0)| = |(1, 0)| = 1 and |S(0, 1)| = 1, but S is clearly not an isometry. 22. R 3 is an odd-dimensional real vector space. Hence S has an eigenvalue λ and a corresponding eigenvector v. Since |Sv| = [λ[|v| = |v| we have λ 2 = 1. Hence S 2 v = λ 2 v = v. 21 23. The matrices corresponding to T and T ∗ are ´(T) = 0 0 1 2 0 0 0 3 0 ¸ ¸ , ´(T ∗ ) = 0 2 0 0 0 3 1 0 0 ¸ ¸ . Thus we get ´(T ∗ T) = 4 0 0 0 9 0 0 0 1 ¸ ¸ , so that T ∗ T is the operator (z 1 , z 2 , z 3 ) → (4z 1 , 9z 2 , z 3 ). Hence √ T ∗ T(z 1 , z 2 , z 3 ) = (2z 1 , 3z 2 , z 3 ). Now we just need to permute the indices which corresponds to the isometry S(z 1 , z 2 , z 3 ) = (z 3 , z 1 , z 2 ). 24. Clearly T ∗ T is positive, so by proposition 7.26 it has a unique positive square root. Thus it’s sufficient to show that R 2 = T ∗ T. Now we have T ∗ = (SR) ∗ = R ∗ S ∗ = RS ∗ by self-adjointness of R. Thus R 2 = RIR = RS ∗ SR = T ∗ T. 25. Assume that T is invertible. Then √ T ∗ T must be invertible (polar decomposition). Hence S = T √ T ∗ T −1 , so S is uniquely determined. Now assume that T is not invertible. Then range √ T ∗ T is not invertible (hence not surjective). Now assume that T is not invertible, so that √ T ∗ T is not invertible. By the spectral theorem we can choose a basis (v 1 , . . . , v n ) of eigenvectors of √ T ∗ T and we can assume that v 1 corresponds to the eigenvalue 0. Now let T = S √ T ∗ T. Define U by Uv 1 = −Sv 1 , Uv i = Sv i , i > 1. Clearly U = S and T = U √ T ∗ T. Now choose u, v ∈ V . Then it’s easy to verify that 'Uu, Uv` = 'Su, Sv` = 'u, v`, so U is an isometry. 26. Choose a basis of eigenvectors for T such that T has a diagonal matrix with eigenval- ues λ 1 , . . . , λ n on the diagonal. Now T ∗ T = T 2 corresponds to the diagonal matrix with λ 2 1 , . . . , λ 2 n on the diagonal and the square root √ T ∗ T corresponds to the diag- onal matrix having [λ 1 [, . . . , [λ n [ on the diagonal. These are the singular values, so the claim follows. 27. Let T ∈ L(R 2 ) be the map T(a, b) = (0, a). Then from the matrix representation we see that T ∗ T(a, 0) = (a, 0), hence √ T ∗ T = T ∗ T and 1 is clearly a singular value. However, T 2 = 0, so (T 2 ) ∗ T 2 = 0, so 1 2 = 1 is not a singular value of T 2 . 28. The composition of two bijections is a bijection. If T=S √ T ∗ T, then since S is an isometry we have that T is bijective hence invertible if and only if √ T ∗ T is bijective. But √ T ∗ T is injective hence bijective if and only if 0 is not an eigenvalue. Thus T is bijective if and only if 0 is not a singular value. 22 29. From the polar decomposition theorem we see that dimrange T = dimrange √ T ∗ T. Since √ T ∗ T is self-adjoint we can choose a basis of eigenvectors of √ T ∗ T such that the matrix of √ T ∗ T is a diagonal matrix. Clearly dimrange √ T ∗ T equals the number of non-zero elements on the diagonal i.e. the number of non-zero singular values. 30. If S is an isometry, then clearly √ S ∗ S = I, so all singular values are 1. Now assume that all singular values are 1. Then √ S ∗ S is a self-adjoint (positive) operator with all eigenvalues equal to 1. By the spectral theorem we can choose a basis of V such that √ S ∗ S has a diagonal matrix. As all eigenvalues are one, this means that the matrix of √ S ∗ S is the identity matrix. Thus √ S ∗ S = I, so S is an isometry. 31. Let T 1 = S 1 T ∗ 1 T 1 and T 2 = S 2 T ∗ 2 T 2 . Assume that T 1 and T 2 have the same singular values s 1 , . . . , s n . Then we can choose bases of eigenvectors of T ∗ 1 T 1 and T ∗ 2 T 2 such that they have the same matrix. Let these bases be (v 1 , . . . , v n ) and (w 1 , . . . , w n ). Let S the operator defined by S(w i ) = v i , then clearly S is an isometry and T ∗ 2 T 2 = S −1 T ∗ 1 T 1 S. Thus we get T 2 = S 2 T ∗ 2 T 2 = S 2 S −1 T ∗ 1 T 1 S. Writing S 3 = S 2 S −1 S −1 1 , which is clearly an isometry, we get T 2 = S 3 S 1 T ∗ 1 T 1 S = S 3 T 1 S. Now let T 2 = S 1 T 1 S 2 . Then we have T ∗ 2 T 2 = (S 1 T 1 S 2 ) ∗ (S 1 T 1 S 2 ) = S ∗ 1 T ∗ 1 T 1 S 1 = S −1 1 T ∗ 1 T 1 S 1 . Let λ be an eigenvalue of T ∗ 1 T 1 and v a corresponding eigenvector. Because S 1 is bijective we have a u ∈ V such that v = S 1 u. This gives us T ∗ 2 T 2 u = S −1 1 T ∗ 1 T 1 S 1 u = S −1 1 T ∗ 1 T 1 v = S −1 1 λv = λu, so that λ is an eigenvalue of T ∗ 2 T 2 . Let (v 1 , . . . , v n ) be an orthonormal basis of eigen- vectors of T ∗ 1 T 1 , then (S −1 1 v 1 , . . . , S −1 v n ) is an orthonormal basis for V of eigenvectors of T ∗ 2 T 2 . Hence T ∗ 2 T 2 and T ∗ 1 T 1 have the same eigenvalues. The singular values are simply the positive square roots of these eigenvalues, so the claim follows. 32. For the first part, denote by S the map defined by the formula. Set A := ´(S, (f 1 , . . . , f n ), (e 1 , . . . , e n )), B := ´(T, (e 1 , . . . , e n ), (f 1 , . . . , f n )). Clearly A = B and is a diagonal matrix with s 1 , . . . , s n on the diagonal. All the s i are positive real numberse, so that B ∗ = B = A. It follows from proposition 6.47 that S = T ∗ . For the second part observe that a linear map S defined by the formula maps f i to s −1 i e i which is mapped by T to f i . Thus TS = I. The claim now follows from exercise 3.23. 23 33. By definition √ T ∗ T is positive. Thus all singular values of T are positive. From the singular value decomposition theorem we can choose bases of V such that |Tv| = | n ¸ i=1 s i 'v, e i ` f i | = n ¸ i=1 s i [ 'v, e i ` [ since all s i are positive. Now clearly ˆ s|v| = ˆ s n ¸ i=1 [ 'v, e i ` [ ≤ n ¸ i=1 s i [ 'v, e i ` [ = |Tv| and similarly for s|v|. 34. From the triangle equality and the previous exercise we get |(T + T )v| ≤ |T v| +|T v| ≤ (s + s )|v|. Now let v be the eigenvector of (T + T ) ∗ (T + T ) corresponding to the singular value s. Let T + T = S (T + T ) ∗ (T + T ), so that |(T + T )v| = |S (T + T ) ∗ (T + T )v| = |Ssv| = s|v| ≤ (s + s )|v|, so that s ≤ s + s . 8 Operators on Complex Vector Spaces 1. T is not injective, so 0 is an eigenvalue of T. We see that T 2 = 0, so that any v ∈ V is a generalized eigenvector corresponding to the eigenvalue 0. 2. On page 78 it’s shown that ±i are the eigenvalues of T. We have dimnull(T −iI) 2 + dimnull(T + iI) 2 = dimC 2 = 2, so that dimnull(T − iI) 2 = dimnull(T + iI) 2 = 1. Because (1, −i) is an eigenvector corresponding to the eigenvalue i and (1, i) is an eigenvector corresponding to the eigenvalue −i. The set of all generalized eigenvectors are simply the spans of these corresponding eigenvectors. 3. Let ¸ m−1 i=0 a i T i v = 0, then 0 = T m−1 ( ¸ m−1 i=0 a i T i v = a 0 T m−1 v, so that a 0 = 0. Ap- plying repeatedly T m−i for i = 2, . . . we see that a i = 0 for all i. Hence (v, . . . , T m−1 v) is linearly independent. 4. We see that T 3 = 0, but T 2 = 0. Assume that S is a square root of T, then S is nilpotent, so by corollary 8.8 we have S dimV = S 3 = 0, so that 0 = S 4 = T 2 . Contradiction. 5. If ST is nilpotent, then ∃n ∈ N such that (ST) n = 0, so (TS) n+1 = T(ST) n S = 0. 24 6. Assume that λ = 0 is an eigenvalue of N with corresponding eigenvector v. Then N dimV v = λ dimV v = 0. This contradicts corollary 8.8. 7. By lemma 8.26 we can find a basis of V such that the matrix of N is upper-triangular with zeroes on the diagonal. Now the matrix of N ∗ is the conjugate transpose. Since N ∗ = N, the matrix of N must be zero, so the claim follows. 8. If N dimV −1 = N dimV , then dimnull N i−1 < dimnull N i for all i ≤ dimV by propo- sition 8.5. By corollary dimnull N dimV = dimV , so the claim clearly follows. 9. range T m = range T m+1 implies dimnull T m = dimnull T m+1 i.e. null T m = null T m+1 . From proposition 8.5 we get dimnull T m = dimnull T m+k for all k ≥ 1. Again this implies that dimrange T m = dimrange T m+k for all k ≥ 1, so the claim follows. 10. Let T be the operator defined in exercise 1. Clearly null T ∩ range T = ¦0¦, so the claim is false. 11. We have dimV = dimnull T n +dimrange T n , so it’s sufficient to prove that null T n ∩ range T n = ¦0¦. Let v ∈ null T n ∩ range T n . Then we can find a u ∈ V such that T n u = v. From 0 = Tv = T n+1 u we see that u ∈ null T n+1 = null T n which implies that v = T n u = 0. 12. From theorem 8.23 we have V = null T dimV , so that T dimV = 0. Then let T ∈ L(R 3 ) be the operator T(a, b, c) = (−b, a, 0). Then clearly 0 is the only eigenvalue, but T is not nilpotent. 13. From null T n−2 = null T n−1 and from proposition 8.5 we see that dimnull T n ≥ n−1. Assume that T has three different eigenvalues 0, λ 1 , λ 2 . Then dimnull (T −λ i I) n ≥ 1, so from theorem 8.23 n ≥ dimnull T n +dimnull (T −λ 1 I) n +dimnull (T −λ 2 I) n ≥ n −1 +1 +1 = n +1, which is impossible, so T has at most two different eigenvalues. 14. Let T ∈ L(C 4 ) be defined by T(a, b, c, d) = (7a, 7b, 8c, 8d) from the matrix of T it’s easy to see that the characteristic polynomial is (z −7) 2 (z −8) 2 . 15. Let d 1 be the multiplicity of the eigenvalue 5 and d 2 the multiplicity of the eigenvalue 6. Then d 1 , d 2 ≥ 1 and d 1 + d 2 = n. It follows that d 1 , d 2 ≤ n − 1, so that (z − 5) d 1 (z − 6) d 2 divides (z − 5) n−1 (z − 6) n−1 . By the Cayley-Hamilton theorem (T −5I) d 1 (T −6I) d 2 = 0, so that (T −5I) n−1 (T −6I) n−1 = 0. 16. If every generalized eigenvector is an eigenvector, then by theorem 8.23 T has a basis of eigenvectors. If there’s a generalized eigenvector that is not an eigenvector, then we have an eigenvalue λ such that dimnull(T −λI) dimV = dimnull(T −λI). Thus if 25 λ 1 , . . . , λ m are the eigenvalues of T, then ¸ m i=1 dimnull(T − λ i I) < dimV , so there doesn’t exists a basis of eigenvectors. 17. By lemma 8.26, choose a basis (v 1 , . . . , v n ) such that N has an upper-triangular matrix. Apply Gram-Schmidt orthogonalization to the basis. Then Ne 1 = 0 and assume that Ne i ∈ span(e 1 , . . . , e i ), so that Ne i+1 = N v i+1 −'v i+1 , e 1 ` e 1 −. . . −'v i+1 , e i ` e i |v i+1 −'v i+1 , e 1 ` e 1 −. . . −'v i+1 , e i ` e i | ∈ span(e 1 , . . . , e i+1 ). Thus N has an upper-triangular matrix in the basis (e 1 , . . . , e n ). 18. Continuing in the proof of lemma 8.30 up to j = 4 we see that a 4 = −5/128. So that √ I + N = I + 1 2 N − 1 8 N 2 + 1 16 N 3 − 5 128 N 4 19. Just replace the Taylor-polynomial in lemma 8.30 with the Taylor polynomial of 3 √ 1 + x and copy the proof of the lemma and theorem 8.32. 20. Let p(x) = ¸ m i=0 a i x i be the minimal polynomial of T. If a 0 = 0, then p(x) = x ¸ m i=1 a i x i−1 . Since T is invertible we must have ¸ m i=1 a i T i−1 = 0 which contradicts the minimality of p. Hence a 0 = 0, so solving for I in p(T) we get −a −1 0 (a 1 + . . . + a m T m−1 )T = I. Hence setting q(x) = −a −1 0 (a 1 + . . . + a m x m−1 ) we have q(T) = T −1 . 21. The operator defined by the matrix 0 0 1 0 0 0 0 0 0 ¸ ¸ is clearly an example. 22. Choose the matrix A = 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 ¸ ¸ ¸ ¸ . It’s easy to see that A(A−I) 2 = 0. Thus, the minimal polynomial divides z(z −1) 2 . However, the whole polynomial is the only factor that annihilates A, so z(z −1) 2 is the minimal polynomial. 26 23. Let λ 1 , . . . , λ m be the eigenvalues of T with multiplicity d 1 , . . . , d m . Assume that V has a basis of eigenvectors (v 1 , . . . , v n ). Let p(z) = ¸ m i=1 (z −λ i ). Then we get p(T)v i = ¸ ¸ j=i (T −λ j I) ¸ (T −λ i I)v i = 0, so that the minimal polynomial divides p and hence has no double root. Now assume that the minimal polynomial has no double roots. Let the minimal polynomial be p and let (v 1 , . . . , v n ) be a Jordan-basis of T. Let A be the largest Jordan-block of T. Now clearly p(A) = 0. However, by exercise 29, the minimal polynomial of (A − λI) is z m+1 where m is the length of the longest consecutive string of 1’s that appear just above the diagonal, so the minimal polynomial of A is (z − λ) m+1 . Hence m = 0, so that A is a 1 1 matrix and the basis vector corresponding to A is an eigenvalue. Since A was the largest Jordan-block it follows that all basis vectors are eigenvalues (corresponding to a 1 1 matrix). 24. If V is a complex vector space, then V has a basis of eigenvectors. Now let λ 1 , . . . , λ m be the distinct eigenvalues. Now clearly p = ¸ m i=1 (z −λ i ) annihilates T, so that the minimal polynomial being a factor of p doesn’t have a double root. Now assume that V is a real vector space. By theorem 7.25 we can find a basis of T such that T has a block diagonal matrix where each block is a 1 1 or 2 2 matrix. Let λ 1 , . . . , λ m be the distinct eigenvalues corresponding to the 1 1 blocks. Then p(z) = ¸ m i=1 (z − λ i ) annihilates all but the 2 2 blocks of the matrix. Now it’s sufficient to show that each 2 2 block is annihilated by a polynomial which doesn’t have real roots. By theorem 7.25 we can choose the 2 2 blocks to be of the form ¸ a −b b a where b > 0 and it’s easy to see that ¸ a −b b a −2a ¸ a −b b a + (a 2 + b 2 ) ¸ 1 0 0 1 = 0. Clearly this polynomial has a negative discriminant, so the claim follows. 25. Let q be the minimal polynomial. Write q = sp + r, where deg r < deg p, so that 0 = q(T)v = s(T)p(T)v +r(T)v = r(T)v. Assuming r = 0, we can multiply the both sides with the inverse of the highest coefficient of r yielding a monic polynomial r 2 of degree less than p such that r 2 (T)v = 0 contradicting the minimality of p. Hence r = 0 and p divides q. 27 26. It’s easy to see that no proper factor of z(z −1) 2 (z −3) annihilates the matrix A = 3 1 1 1 0 1 1 0 0 0 1 0 0 0 0 0 ¸ ¸ ¸ ¸ , so the it’s minimal polynomial is z(z − 1) 2 (z − 3) which by definition is also the characteristic polynomial. 27. It’s easy to see that no proper factor of z(z −1)(z −3) annihilates the matrix A = 3 1 1 1 0 1 1 0 0 0 1 0 0 0 0 0 ¸ ¸ ¸ ¸ , but A(A−I)(A−3I) = 0, so the claim follows. 28. We see that T(e i ) = e i+1 for i < n. Hence (e 1 , Te 1 , . . . , T n−1 e 1 ) = (e 1 , . . . , e n ) is linearly independent. Thus for any non-zero polynomial p of degree less than n we have p(e i ) = 0. Hence the minimal polynomial has degree n, so that the minimal polynomial equals the characteristic polynomial. Now from the matrix of T we see that T n (e i ) = −a 0 −a 1 T(e 1 ) −. . . −a n−1 T n−1 . Set p(z) = a 0 +a 1 z +. . . +a n−1 z n−1 +z n . Now p(T)(e 1 ) = 0, so by exercise 25 p divides the minimal polynomial. However, the minimal polynomial is monic and has degree n, so p is the minimal polynomial. 29. The biggest Jordan-block of N is of the form 0 1 0 1 . . . . . . 0 1 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ . Now clearly N m+1 = 0, so the minimal polynomial divides z m+1 . Let v i , . . . , v i+m be the basis elements corresponding to the biggest Jordan-block. Then N m e i+m = e i = 0, so the minimal polynomial is z m+1 . 30. Assume that V can’t be decomposed into two proper subspaces. Then T has only one eigenvalue. If there is more than one Jordan-block, then Let (v 1 , . . . , v m ) be the vectors corresponding to all but the last Jordan-block and (v m+1 , . . . , v n ) the 28 vectors corresponding to the last Jordan-block. Then clearly V = span(v 1 , . . . , v m ) ⊕ span(v m+1 , . . . , v n ). Thus, we have only one Jordan-block and T − λI is nilpotent and has minimal polynomial z dimV by the previous exercise. Hence T has minimal polynomial (z −λ) dimV . Now assume that T has minimal polynomial (z −λ) dimV . If V = U ⊕W, let p 1 be the minimal polynomial of T |U and p 2 the minimal polynomial of T |W . Now (p 1 p 2 )(T) = 0, so that (z − λ) dimV divides p 1 p 2 . Hence deg p 1 p 2 = deg p 1 + deg p 2 ≥ dimV , but deg p 1 +deg p 2 ≤ dimU +dimW = dimV . Thus deg p 1 p 2 = dimV . This means that p 1 (z) = (z − λ) dimU and p 2 (z) = (z − λ) dimW . Now if n = max¦dimU, dimW¦ we have (T |U − λI) n = (T |W − λI) n = 0. This means that (T − λI) n = 0 contradicting the fact that (z −λ) dimV is the minimal polynomial. 31. Reversing the Jordan-basis simply reverses the order of the Jordan-blocks and each block needs to be replaced by its transpose. 9 Operators on Real Vector Spaces 1. Let a be the value in the upper-left corner. Then the matrix must have the form ¸ a 1 −a 1 −a a and it’s easy to see that ¸ 1 1 is an eigenvector corresponding to the eigenvalue 1. 2. The characteristic polynomial of the matrix is p(x) = (x − a)(x − d) − bc. Now the discriminant equals (a + d) 2 −4(ad −bc) = (a −d) 2 + 4bc. Hence the characteristic polynomial has a root if and only if (a − d) 2 + 4bc ≥ 0. If p(x) = (x − λ 1 )(x − λ 2 ), then p(T) = 0 implies that either A − λ 1 I or A − λ 2 I is not invertible hence A has an eigenvalue. If p doesn’t have a root then assuming that Av = λv we get 0 = p(A)v = p(λ)v, so that v = 0. Hence A has no eigenvalues. The claim follows. 3. See the proof of the next problem, though in this special case the proof is trivial. 4. First let λ be an eigenvalue of A. Assume that λ is not an eigenvalue of A 1 , . . . , A m , so that A i −λI is invertible for each i. Now let B be the matrix (A 1 −λI) −1 ∗ . . . 0 (A m −λI) −1 ¸ ¸ ¸. 29 It’s now easy to verify that B(A−λI) is a diagonal matrix with all 1s on the diagonal, hence it’s invertible. However, this is impossible, since A −λI is not. Thus, λ is an eigenvalue of some A i . Now let λ be an eigenvalue of A i . Let T be the operator corresponding to A − λI in L(F n ). Let A i correspond to the columns j, . . . , j + k and let (e 1 , . . . , e n ) be the standard basis. Let (a 1 , . . . , a k ) be the eigenvector of A i corresponding to λ. Set v = a 1 e j +. . . +a k e j+k . Then it’s easy to see that T maps the space span(e 1 , . . . , e j−1 , v) to the space span(e 1 , . . . , e j−1 ). Hence T is not injective, so that λ is an eigenvalue of A. 5. The proof is identical to the argument used in the solution of exercise 2. 6. Let (v 1 , . . . , v n ) be the basis with the respect to which T has the given matrix. Applying Gram-Schmidt to the basis the matrix trivially has the same form. 7. Choose a basis (v 1 , . . . , v n ) such that T has a matrix of the form in theorem 9.10. Now if v j corresponds to the second vector in a pair corresponding to a 2 2 matrix or corresponds to a 1 1 matrix, then span(v 1 , . . . , v j ) is an invariant subscape. The second posibility means that v j corresponds to the first vector in a pair corresponding to a 2 2 matrix, so that span(v 1 , . . . , v j+1 ) is an invariant subspace. 8. Assuming that such an operator existed we would have basis (v 1 , . . . , v 7 ) such that the matrix of T would have a matrix with eigenpair (1, 1) dimnull(T 2 + T + I) dimV 2 = 7/2 times on the diagonal. This would contradict theorem 9.9 as 7/2 is not a whole number. It follows that such an operator doesn’t exist. 9. The equation x 2 + x + 1 = 0 has a solution in C. Let λ be a solution. Then the operator corresponding to the matrix λI clearly is an example. 10. Let T be the operator in L(R 2k ) corresponding to the block diagonal matrix where each block has characteristic polynomial x 2 +αx +β. Then by theorem 9.9 we have k = dimnull(T 2 + αT + βI) 2k 2 , so that dimnull(T 2 + αT + βI) 2k is even. However, the minimal polynomial of T divides (T 2 +αT +βI) 2k and has degree less than 2k, so that (T 2 +αT +βI) k = 0. It follows that dimnull(T 2 + αT + βI) 2k = dimnull(T 2 + αT + βI) k and the claim follows. 30 11. We see that (α, β) is an eigenpair and from the nilpotency of T 2 + αT + βI and theorem 9.9 we have dimnull(T 2 + αT + βI) dimV 2 = (dimV )/2 it follows that dimV is even. For the second part we know from the Cayley-Hamilton theorem that the minimal polynomial p(x) of T has degree less than dimV . Thus we have p(x) = (x 2 + αx + β) k and from deg p ≤ dimV we get k ≤ (dimV )/2. Hence (T + αT + βI) (dimV )/2 = 0. 12. By theorem 9.9 we can choose a basis such that the matrix of T has the form of 9.10. Now we have the matrices [5] and [7] at least once on the diagonal. Assuming that T has an eigenpair we would have at least one 2 2 matrix on the diagonal. This is impossible as the diagonal has only length 3. 13. By proposition 8.5 dimnull T n−1 ≥ n−1. Like in the previous exercise we know that the matrix [0] is at least n−1 times on the diagonal and it’s impossible to fit a 2 2 matrix in the only place left. 14. We see that A satisfies the polynomial p(z) = (z − a)(z − d) − bc no matter if F is R or C. If F = C, then because p is monic, has degree 2 and annihilates A it follows that p is the characteristic polynomial. Assume then that F = R. Now if A has no eigenvalues, then p is the characteristic polynomial by definition. Assume then that p(z) = (z − λ 1 )(z − λ 2 ) which implies that p(T) = (T −λ 1 I)(T −λ 2 I) = 0. If neither T −λ 1 I or T −λ 2 I is injective, then by definition p is the minimal polynomial. Assume then that T − λ 2 I is invertible. Then dimnull(T − λ 1 I) = 2, so that T − λ 1 I = 0. Hence we get c = b = 0 and a = d = λ 1 . Hence p(z) = (z − λ 1 ) 2 , so that p is the characteristic polynomial by definition. 15. S is normal, so there’s an orthonormal basis of V such that S has a block diagonal matrix with respect to the basis and each block is a 1 1 or 2 2 matrix and the 22 blocks have no eigenvalue (theorem 7.25). From S ∗ S = I we see that each block A i satisfy A ∗ i A i = I, so that they are isometries. Hence a 2 2 block is of the form ¸ cos θ −sin θ sin θ cos θ . We see that T 2 +αT +βI is not injective if and only if A 2 i +αA i +βI is not injective for some 22 matrix A i . Hence x 2 +αx+β must equal the characteristic polynomial of some A i , so by the previous exercise it’s of the form (x −cos θ)(z −cos θ) + sin 2 θ = x 2 −2xcos θ + cos 2 θ + sin 2 θ so that β = cos 2 θ + sin 2 θ = 1. 31 10 Trace and Determinant 1. The map T → ´(T, (v 1 , . . . , v n )) is bijective and satisfies ST → ´(ST, (v 1 , . . . , v n )) = ´(S, (v 1 , . . . , v n ))´(T, (v 1 , . . . , v n )). Hence ST = I ⇔ ´(ST, (v 1 , . . . , v n )) = ´(S, (v 1 , . . . , v n ))´(T, (v 1 , . . . , v n )) = I. The claim now follows trivially. 2. Both matrices represent an operator in L(F n ). The claim now follows from exercise 3.23. 3. Choose a basis (v 1 , . . . , v n ) and let A = (a ij ) be the matrix corresponding to the basis. Then Tv 1 = a 11 v 1 + . . . + a n1 v n . Now (v 1 , 2v 2 , . . . , 2v n ) is also a basis and we have by our assumption Tv = a 11 v 1 + 2(a 21 v 2 + . . . + a n1 v n ). We thus get a 21 v 2 + . . . + a n1 v n = 2(a 21 v 2 + . . . + a n1 v n ), which implies (a 21 v 2 + . . . + a n1 v n = 0. By linear independence we get a 21 = . . . = a n1 = 0, so that Tv 1 = a 11 v. The claim clearly follows. 4. Follows directly from the definition of ´(T, (u 1 , . . . , u n ), (v 1 , . . . , v n )). 5. Let (e 1 , . . . , e n ) be the standard basis for C n . Then we can find an operator T ∈ L(C n ) such that ´(T, (e 1 , . . . , e n )) = B. Now T has an upper-triangular matrix corresponding to some basis (v 1 , . . . , v n ) of V . Let A = ´((v 1 , . . . , v n ), (e 1 , . . . , e n )). Then A −1 BA = ´(T, (v 1 , . . . , v n )) which is upper-triangular. Clearly A is an invertible square matrix. 6. Let T be the operator corresponding to the matrix ¸ 0 −1 1 0 , ¸ 0 −1 1 0 2 = ¸ −1 0 0 −1 . By theorem 10.11 we have trace(T 2 ) = −2 < 0. 7. Let (v 1 , . . . , v n ) be a basis of eigenvectors of T and λ 1 , . . . , λ n be the corresponding eigenvalues. Then T has the matrix λ 1 0 . . . 0 λ n ¸ ¸ ¸. Clearly trace(T 2 ) = λ 2 1 + . . . + λ 2 n ≥ 0 by theorem 10.11. 32 8. Extend v/|v| to an orthonormal basis (v/|v|, e 1 , . . . , e n ) of V and let A = ´(T, (v/|v|, e 1 , . . . , e n )). Let a, a 1 , . . . , a n denote the diagonal elements of A. Then we have a i = 'Te i , e i ` = ''e i , v` w, e i ` = '0, e i ` = 0, so that trace(T) = a = 'Tv/|v|, v/|v|` = ''v/|v|, v` w, v/|v|` = 'w, v` . 9. From exercise 5.21 we have V = null P ⊕range P. Now let v ∈ range P. Then v = Pu for some u ∈ V . Hence Pv = P 2 w = Pw = v, so that P is the identity on range P. Now choose a basis (v 1 , . . . , v m ) for range P and extend it with a basis (u 1 , . . . , u n ) of null P to get a basis for V . Then clearly the matrix for P in this basis consists of a diagonal matrix with 1s on part of the diagonal corresponding to the vectors (v 1 , . . . , v m ) and 0s on the rest of the diagonal. Hence trace P = dimrange P ≥ 0. 10. Let (v 1 , . . . , v n ) be some basis of V and let A = ´(T, (v 1 , . . . , v n )). Let a 1 , . . . , a n be the diagonal elements of A. By proposition 6.47 T ∗ has the matrix A ∗ , so that the diagonal elements of A ∗ are a 1 , . . . , a n . By theorem 10.11 we have trace(T ∗ ) = a 1 + . . . + a n = a 1 + . . . + a n = trace(T). 11. A positive operator is self-adjoint. By the spectral theorem we can find a basis (v 1 , . . . , v n ) of eigenvectors of T. Then A = ´(T, (v 1 , . . . , v n )) is a diagonal matrix and by positivity of T all the diagonal elements a 1 , . . . , a n are positive. Hence a 1 + . . . + a n = 0 implies a 1 = . . . = a n = 0, so that A = 0 and hence T = 0. 12. The trace of T is the sum of the eigenvalues. Hence λ − 48 + 24 = 51 − 40 + 1, so that λ = 36. 13. Choose a basis of (v 1 , . . . , v n ) of V and let A = ´(T, (v 1 , . . . , v n )). Then the matrix of cT is cA. Let a 1 , . . . , a n be the diagonal elements of A. Then we have trace(cT) = ca 1 + . . . + ca n = c(a 1 + . . . + a n ) = ctrace(T). 14. The example in exercise 6 shows that this is false. For another example take S = T = I. 15. Choose a basis (v 1 , . . . , v n ) for V and let A = ´(T, (v 1 , . . . , v n )). It’s sufficient to prove that A = 0. Now let a i,j be the element in row i, column j of A. Let B be the matrix with 1 in row j column i and 0 elsewhere. Then it’s easy to see that in BA the only non-zero diagonal element is a i,j . Thus trace(BA) = a i,j . Let S be the operator corresponding to B. It follows that a i,j = trace(ST) = 0, so that A = 0. 33 16. Let T ∗ Te i = a 1 e 1 + . . . + a n e n . Then we have that a i = 'T ∗ Te i , e i ` = |Te i | 2 by the ortogonality of (e 1 , . . . , e n ). Clearly a i is the ith diagonal element in the matrix ´(T ∗ T, (e 1 , . . . , e n )), so that trace(T ∗ T) = |Te 1 | 2 + . . . +|Te n | 2 . The second assertion follows immediately. 17. Choose a basis (v 1 , . . . , v n ) such that T has an upper triangular matrix B = (b ij ) with eigenvalues λ 1 , . . . , λ n on the diagonal. Now the ith element on the diagonal of B ∗ B is ¸ n j=1 b ji b ji = ¸ n j=1 [b ji [ 2 , where [b ii [ 2 = [λ i [ 2 . Hence n ¸ i=1 [λ i [ 2 ≤ n ¸ i=1 n ¸ j=1 [b ji [ 2 = trace(B ∗ B) = trace(A ∗ A) = n ¸ k=1 n ¸ j=1 [a jk [ 2 . 18. Positivity and definiteness follows from exercise 16 and additivity in the first slot from corollary 10.12. Homogeneity in the first slot is exercise 13 and by exercise 10 'S, T` = trace(ST ∗ ) = trace((ST ∗ ) ∗ ) = trace(TS ∗ ) = 'T, S`. It follows that the formula defines an inner-product. 19. We have 0 ≤ 'Tv, Tv` − 'T ∗ v, T ∗ v` = '(T ∗ T −TT ∗ )v, v`. Hence T ∗ T − TT ∗ is a positive operator (it’s clearly self-adjoint), but trace(T ∗ T − TT ∗ ) = 0, so all the eigenvalues are 0. Hence T ∗ T −TT ∗ = 0 by the spectral theorem, so T is normal. 20. Let dimV = n and write A = ´(T). Then we have det cT = det cA = ¸ σ ca σ(1),1 ca σ(n),n = c n det A = c n det T 21. Let S = I and T = −I. Then we have 0 = det(S + T) = det S + det T = 1 + 1 = 2. 22. We can use the result of the next exercise which means that we only need to prove this for the complex case. Let T ∈ L(C n ) be the operator corresponding to the matrix A. Now each block A j on the diagonal corresponds to some basis vectors (e i , . . . , e i+k ). These can be replaced with another set of vectors, (v i , . . . , v i+k ), span- ning the same subspace such that A j in this new basis is upper-triangular. We have T(span(v i , . . . , v i+k )) ⊂ span(v i , . . . , v i+k ), so after doing this for all blocks we can assume that A is upper-triangular. The claim follows immediately. 23. This follows trivially from the formula of trace and determinant for a matrix, because the formulas only depends on the elements of the matrix. 34 24. Let dimV = n and let A = ´(T), so that B = ´(T) equals the conjugate transpose of A i.e. b ij = a ji . Then we have det T = det A = ¸ σ a σ(1),1 a σ(n),n = ¸ σ a 1,σ(1) a n,σ(n) = ¸ σ a 1,σ(1) a n,σ(n) = ¸ σ b σ(1),1 b σ(n),n = det B = det T ∗ . The first equality on the second line follows easily that every term in the upper sum is represented by a term in the lower sum and vice versa. The second claim follows immediately from theorem 10.31. 25. Let Ω = ¦(x, y, z) ∈ R 3 [ x 2 + y 2 + z 2 < 1¦ and let T be the operator T(x, y, z) = (ax, by, cz). It’s easy to see that T(Ω) is the ellipsoid. Now Ω is a circle with radius 1. Hence [ det T[volume(Ω) = abc 4 3 π = 4 3 πabc, which is the volume of an ellipsoid. 35 6. Set U = Z2 . 7. The set {(x, x) ∈ R2 | x ∈ R} ∪ {(x, −x) ∈ R2 | x ∈ R} is closed under multiplication but is trivially not a subspace ((x, x) + (x, −x) = (2x, 0) doesn’t belong to it unless x = 0). 8. Let {Vi } be a collection of subspaces of V . Set U = ∩i Vi . Then if u, v ∈ U . We have that u, v ∈ Vi for all i because Vi is a subspace. Thus au ∈ Vi for all i, so that av ∈ U . Similarly u + v ∈ Vi for all i, so u + v ∈ U . 9. Let U, W ⊂ V be subspaces. Clearly if U ⊂ W or W ⊂ U , then U ∪ W is clearly a subspace. Assume then that U ⊂ W and W ⊂ U . Then we can choose u ∈ U \ W and w ∈ W \ U . Assuming that U ∪ W is a subspace we have u + w ∈ U ∪ W . Assuming that u + w ⊂ U we get w = u + w − u ⊂ U . Contradiction. Similarly for u + w ∈ W . Thus U ∪ W is not a subspace. 10. Clearly U = U + U as U is closed under addition. 11. Yes and yes. Follows directly from commutativity and associativity of vector addition. 12. The zero subspace, {0}, is clearly an additive identity. Assuming that we have inverses, then the whole space V should have an inverse U such that U + V = {0}. Since V + U = V this is clearly impossible unless V is the trivial vector space. 13. Let W = R2 . Then for any two subspaces U1 , U2 of W we have U1 + W = U2 + W , so the statement is clearly false in general. 14. Let W = {p ∈ P(F) | p = n i i=0 ai x , a2 = a5 = 0}. 15. Let V = R2 . Let W = {(x, 0) ∈ R2 | x ∈ R}. Set U1 = {(x, x) ∈ R2 | x ∈ R}, U2 = {(x, −x) ∈ R2 | x ∈ R}. Then it’s easy to see that U1 + W = U2 + W = R2 , but U1 = U2 , so the statement is false. 2 Finite-Dimensional Vector Spaces 1. Let un = vn and ui = vi + vi+1 , i = 1, . . . , n − 1. Now we see that vi = vi ∈ span(u1 , . . . , un ), so V = span(v1 , . . . , vn ) ⊂ span(u1 , . . . , un ). n j=i ui . Thus 2. From the previous exercise we know that the span is V . As (v1 , . . . , vn ) is a linearly independent spanning list of vectors we know that dim V = n. The claim now follows from proposition 2.16. 2 3. If (v1 + w, . . . , vn + w) is linearly dependent, then we can write n n 0 = a1 (v1 + w) + . . . + an (vn + w) = i=1 ai vi + w i=1 ai , where ai = 0 for some i. Now n n i=1 ai = 0 because otherwise we would get n n 0= i=1 ai vi + w i=1 ai = i=1 ai vi and by the linear independence of (vi ) we would get ai = 0, ∀i contradicting our assumption. Thus n −1 n w=− i=1 ai i=1 ai vi , so w ∈ span(v1 , . . . , vn ). 4. Yes, multiplying with a nonzero constant doesn’t change the degree of a polynomial and adding two polynomials either keeps the degree constant or makes it the zero polynomial. 5. (1, 0, . . .), (0, 1, 0, . . .), (0, 0, 1, 0, . . .), . . . is trivially linearly independent. Thus F∞ isn’t finite-dimensional. 6. Clearly P(F) is a subspace which is infinite-dimensional. 7. Choose v1 ∈ V . Assuming that V is infinite-dimensional span(v1 ) = V . Thus we can choose v2 ∈ V \ span(v1 ). Now continue inductively. 8. (3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1) is clearly linearly independent. Let (x1 , x2 , x3 , x4 , x5 ) ∈ U . Then (x1 , x2 , x3 , x4 , x5 ) = (3x2 , x2 , 7x4 , x4 , x5 ) = x2 (3, 1, 0, 0, 0)+x4 (0, 0, 7, 1, 0)+ x5 (0, 0, 0, 0, 1), so the vectors span U . Hence, they form a basis. 9. Choose the polynomials (1, x, x2 − x3 , x3 ). They clearly span P3 (F) and form a basis by proposition 2.16. 10. Choose a basis (v1 , . . . , vn ). Let Ui = span(vi ). Now clearly V = U1 ⊕ . . . ⊕ Un by proposition 2.19. 11. Let dim U = n = dim V . Choose a basis (u1 , . . . , un ) for U and extend it to a basis (u1 , . . . , un , v1 , . . . , vk ) for V . By assumption a basis for V has length n, so (u1 , . . . , un ) spans V . 3 . . we have that dim(U ∩ W ) = 0 and the claim follows 14. These relations define the linear map and clearly T|U = S. 13. If S ∈ L(U. . Define f by e. . . Now define T ∈ L(V. 15. . x) ∈ R2 | x ∈ R}. . x = y then clearly f satisfies the condition. .18 that 9 = dim R9 = dim U + dim W − dim(U ∩ W ) = 5 + 5 − 0 = 10. As (p0 . Since U + W = R8 . f (x. . . then dim range T > 0. Thus we have a counterexample. . U2 = {(0. . W ) by T (ui ) = S(ui ). x = y . . we have w = av for some a ∈ F. . so for w = bv ∈ V we have T (w) = T (bv) = bT (v) = bav = aw. U1 ∩ U3 = {0} and U2 ∩ U3 = {0}. . Choose a basis (u1 . Choose a basis (ui . . . By theorem 3. . . W ). . . By assumption the list of vectors (u1 . . 16. . 17. i = 1. m and T (vi ) = wi . Any v = 0 spans V . If u ∈ null T . . Choose some vectors w1 . Then U1 ∩ U2 = {0}. .+Un = span(u1 . . if w ∈ V . . . but as dim range T ≤ dim F = 1 we have dim range T = 1 and it follows that 4 . . Then the list of vectors (u1 . um ) has n nm 1 1 length dim U1 + . + dim Um and clearly spans U1 + . ui i ) for each Ui . n. y) = x. v1 . but is non-linear. x) ∈ R2 | x ∈ R}. Thus T is multiplication by a scalar. Thus the left-hand side of the equation in the problem is 2 while the right-hand side is 3. Since V = U1 +. um . Contradiction. U3 = {(x. . .18 we have 8 = dim R8 = dim U + dim W − dim(U ∩ W ) = 4 + 4 − dim(U ∩ W ). As U = Pm (F) we have by the previous exercise that dim U < dim Pm (F) = m + 1. . 4. um ). . . 2. . .12. Let U = span(u0 . Assuming that U ∩ W = {0} we have by theorem 2. . . .4 dim V = dim null T + dim range T . + Um proving the claim. pm ) is a spanning list of vectors having length m + 1 it is not linearly independent. 3. um ) from the proof of the previous exercise nm 1 is linearly independent. i = 1. . To define a linear map it’s enough to define the image of the elements of a basis. . Let U1 = {(x. vn ) of V . . um ) for U and extend it to a basis (u1 . . . um ) the claim follows. nm 1 3 Linear Maps 1.g. 0. . wn ∈ W . Now T (v) = av for some a ∈ F. . . . . . By theorem 2. 0) ∈ R2 | x ∈ R}. . . . Thus. . . . . Clearly T is surjective. . . . vk ). 5. then by injectivity of Sk+1 we Sk+1 u = 0 and by injectivity of T we have T Sk+1 u = 0. . 0. so that ai = 0 for all i. . . . 8. + an T (vn ) proving the claim. + an un + b1 vn + . . . Sk+1 satisfy the assumptions. . . From dim V = dim null T + dim range T it follows trivially that if we have a surjective linear map T ∈ L(V. + an vn ) = a1 T (v1 ) + . . n. . . By linearity 0 = n ai T (vi ) = n T (ai vi ) = T ( n ai vi ). Then by construction null T ∩ U = {0} and for an arbitrary v ∈ V we have that v = a1 u1 + . 7. It’s easy to see that (5. . . W ) by letting T (vi ) = wi . so that T (v) = b1 T (v1 ) + . un ) be a basis for null T and extend it to a basis (u1 . un ) ⊕ span(u) = null T ⊕ {au | a ∈ F}. we get dim range T = dim F5 − dim null T = 5 − 2 = 3 which is impossible.8). We can define a linear map T ∈ L(V. . 0. 10. Now choose a basis (u1 . 0. . Hence dim range T = dim F4 − dim null T = 4 − 2 = 2. . Assume then that dim V ≥ dim W . It’s again easy to see that (3. + bk T (vk ). . . 1. . . Choose a basis (v1 . . un . 11. 0. wm ) for W . i = m + 1. it’s a basis for V . . u) is linearly independent and has length dim V . . . . By surjectivity of T we can find a vector v ∈ V such that T (v) = w. . . + an vn we get w = T (v) = T (a1 v1 + . . . . By our assumption (u1 . Hence (T (v1 ). Let w ∈ W . (0. 1. . v1 . If u = 0. . If S1 . . . un ) for null T . . . Assume that the claim is true for n = k. 12. Let T = S1 · · · Sk . 1) is a basis for null T (see exercise 2. Hence. then dim V ≥ dim W . 0). . . . m. Hence T Sk+1 is injective and the claim follows. By injectivity we i=1 i=1 i=1 have n ai vi = 0. so that T is surjective. . W ). . 1) is a basis of null T . Hence range T = T (U ). 9. . . u) = span(u1 . T (vn )) is linearly i=1 independent. 1. Hence. i = 1. . . then S1 · · · Sk is injective by the induction hypothesis. . . 7. . This implies that V = span(u1 . . un . 5 . .dim null T = dim V −1. . vk ) of V . If n = 1 the claim is trivially true. 6. . . un . Let U := span(v1 . 0). Thus range T = F2 . (0. + bk vk . . . 1. This follows trivially from dim V = dim null T + dim range T . . vn ) for V and a basis (w1 . . Let (u1 . T (vi ) = 0. . . . Writing v = a1 v1 + . 0. . . . Just take arbitrary matrices satisfying the required dimensions and calculate each expression and the equalities easily fall out. W ) by T (ui ) = 0. . Thus T is injective and hence invertible. 17. Thus we get dim null T = dim V −dim range T ≥ dim V − dim W .n xn . . then T is injective. . . i = 1. . . . We have that dim range T ≤ dim W . . . Define a linear map T ∈ L(V. Define S ∈ L(W. .n xn ).1 · · · am. .1 · · · a1.n x1 a1. . Thus T (w1 ). . = . . w1 . 19. 0vn = 0. i = 1. 16. Substituting dim V = dim null S + dim range S and dim U = dim null ST + dim range ST we get dim null ST + dim range ST ≤ dim null T + dim null S + dim range S.n xn . Let (u1 . + a1. . . . . . Clearly dim Mat(n. n and T (vi ) = wi . M(T v) = M(T )M(v) = . . . . . . 20. i = 1. . . Define the map S ∈ L(W. Clearly if we can find such a S. vn ) be a basis of V . . . . . . n and S(wi ) = 0. . . . . Without loss of generality we can assume that the first m vectors form the basis (just permute the indices). . T (wn )) spans W . Clearly if we can find such a S. . . . . . . . . 18. Let V = (x1 . Assume then that T is injective. . . . From proposition 3. + am. .. . . T (wn )) a basis by removing some vectors. .n ). . We have that T v = 0 if and only if v = 0v1 + . F) = n = dim V . . v1 . . . un . T (wm )) is a basis of W . We have that dim U = dim null T + dim range T ≤ dim null T + dim V . Ditto. . .14 we have that a1. . . .1 x1 + . . 1. . wm ) of W . . By exercise 5 (T (v1 ). 6 . .n xn which shows that T v = (a1. un ) be a basis of U and extend it to a basis (u1 . . . . Clearly ST = IV . vn ) be a basis of V . 15. . This is nothing but pencil pushing. Now clearly T S = IW . . . so null T = {0}. so our claim follows. . Otherwise let (v1 . . T (vn ). i = 1. . By assumption (T (w1 ). . .13. . . k. . By the linear dependence lemma we can make the list of vectors (T (w1 ). T (vn )) is linearly independent so we can extend it to a basis (T (v1 ). .1 x1 + . i = 1. V ) by S(T (vi )) = vi . .1 x1 + . so let (w1 . . am. . .n xn am. Choose an arbitrary subspace U of V of such that dim U ≥ dim V − dim W . . m. . . . Clearly null T = U . . .1 x1 + . . Clearly dim range ST ≤ dim range S. Now we know that k ≤ dim W . + a1. Let (v1 . vk ) of V . . m. . am. . . wm ) be a basis for W . V ) by S(T (wi )) = wi . + am. then T is surjective. 14. . . . Now A defines a linear map from Mat(n. Then m (x − λi )ki is clearly such a polynomial. b) = (0. 22.21 these are equivalent. From theorem 3. This can be trivially generalized to spaces arbitrary spaces of dimension ≥ 2. Let λ1 . 1. . As S is bijective Su goes through the whole space V as u varies. 0) and S ∈ L(F2 ) the operator S(a. . xn Then the system of equations in a) reduces to Ax = 0. . . Clearly T S = ST if T is a scalar multiple of the identity. However. T + S is the identity operator which is trivially invertible. b) = (a. . 23. so ST = I. Let ei denote the n × 1 matrix with a 1 in the ith row and 0 everywhere else. . A := . i=1 7 . For the other direction assume that T S = ST for every linear map S ∈ L(V ). . + an en that Av = A(a1 e1 + . Then neither one is injective and hence invertible by 3. . . . F) to Mat(n. . . . . . . 26. 4 Polynomials 1. Let T ∈ L(F2 ) be the operator T (a. .21 we have ST invertible ⇔ ST bijective ⇔ S and T bijective ⇔ S and T invertible. 25. . . . T (en )].. λm be m distinct numbers and k1 .21. Thus if T S = I. . . By distributivity of matrix multiplication (exercise 17) we get for an arbitrary v = a1 e1 + . Write a1.n . . What a) states is now that the map is injective while b) states that it is surjective. x = .1 · · · a1.n x1 . so the claim follows. . + an en ) = a1 Ae1 + . Let T be the linear map. an. Define a matrix A := [T (e1 ). Now it’s trivial to verify that Aei = T (ei ). F). Then ST (Su) = S(T Su) = Su for all u. 24. . + an Aen = a1 T (e1 ) + . .21. + an T (en ) = T (a1 e1 + .1 · · · . . km > 0 such that their sum is n. b). . an. . By theorem 3. 1. . By symmetry it’s sufficient to prove this in one direction only. + an en ). i Now deg p = m and p clearly p(zi ) = wi . . . Then clearly U is not invariant under T . Thus λ is a root of p if and only if λ is a root of q i. . vm ). . where ai ∈ R. Let v ∈ V . .5. . . Define a linear operator by T (ui ) = v1 . The statement follows. Define m+1 p(x) = i=1 wi c−1 pi (x). Thus we get T (u1 + . Thus V is invariant under T . The clame is clearly true for U = {0} or U = V . 8 . We only need to prove uniqueness as existence is theorem 4. Thus p has a real root. so that deg pi = m. . By our assumption m ≥ 1. r be other such polynomials. . Let pi (x) = j=i (x − zj ). Assuming that s = s we have deg(r − r) < deg(s − s )p which is impossible. Let ci = pi (zi ). un . Let λ be a root of p. Then we get that 0 = (s − s )p + (r − r ) ⇔ r − r = (s − s )p We know that for any polynomials p = 0 = q we have deg pq = deg p + deg q. 4. . By the fundamental theorem of calculus we i=0 have a complex root z. . Assume that {0} = U = V . Thus s = s implying r = r . i = 1. so T (v) ∈ ∩Ui = V . . 5. Now we get p (x) = q(x) + (x − λ)q (x). 5 Eigenvalues and Eigenvectors 1. An arbitrary element of U1 + . but the number of roots counting multiplicity is deg p hence odd. Now we have that pi (zj ) = 0 if and only if i = j. . . λ is a multiple root. i = 1. so that v ∈ Ui for all i. + un where ui ∈ Ui . By proposition 4. Let (u1 . . v1 . 3. + Un by the assumption that T (ui ) ∈ Ui for all i. . Thus we can write p(x) = (x − λ)q(x). . un ) be a basis for U and extend it to a basis (u1 . + un ) = T (u1 ) + . . Now T (v) ∈ Ui for all i by assumption.2. . . Then z and z are roots of equal multiplicity (divide by (x − z)(x − z) which is real). Let p(x) = n ai xi . . . . . n and T (vi ) = v1 . Let V = ∩i Ui where Ui is invariant under T for all i. . . + Un is of the form u1 + . . m. + T (un ) ∈ U1 + . Thus if we have no real roots we have an even number of roots counting multiplicity.10 z is also a root of p. . . . Let s . 3. 2.e. . . a2 . By corollary 5. . As Sv = 0 we know that S is not injective. so ST is not injective and it has eigenvalue 0. so that T (u) − λu = 0.0. then it is is an eigenvector for the eigenvalue λ. so that Su ∈ null(T − λI). . 0) so 0 is an eigenvalue. 8. . .5). so dim null T = n − 1. Let λ be an eigenvalue of T S and v the corresponding eigenvector. so the equation is of the form (0.). . . 1) = (1. λk+2 with corresponding eivenvectors v1 . −1) = −(−1. Let u ∈ null(T − λI). . Now we have T (a. . . 10. 0. so if Sv = 0. . a3 . Now let λ be an eigenvalue of T and v the corresponding eigenvector. . Now T vi = λi vi and dim span(T v1 . Thus dim range T = 1.9 these are all eigenvalues. From T v = λv we get that T −1 λv = λT −1 v = v. Clearly T (1. Then v1 . If T is invertible. The other implication follows by symmetry. . .) = (a2 . λk+2 vk+2 ) ⊂ range T . . 0. By the previous paragraph. Again we see that z1 = 0 so we get the equation (0. From the assumption λ = 0 we get z2 = 0. z2 . 0) = (0. . z3 ). z3 ). T vk+2 ) = dim span(λ1 v1 . so that 1 is an eigenvalue. . . so that 5 is an eigenvalue. 5z3 ) = λ(0. x) ∈ Fn | x ∈ F} and has dimension 1. so λ = 0. 1). By theorem 5. . so n is another eigenvalue.4. 6. 0. 5z3 ) = λ(z1 . 11. vk+2 . z2 . λ2 and assume that λ1 = 0 = λ2 . 5z3 ) = λ(z1 . Also T (1. Hence T has at most one non-zero eigenvalue hence at most two eigenvalues. so that T −1 v = λ−1 v.6 they are linearly independent.0. Then we get ST Sv = Sλv = λSv. Assume that T has k + 2 distinct eigenvalues λ1 . . 9. Thus if λ is an eigenvalue of T S. λk+2 vk+2 ) ≥ k + 1 (it’s k + 2 if all λi are non-zero. . v2 be the corresponding eigenvectors. Notice that the range of T is the subspace {(x. 1) = n(1. . 7. 1). This is a contradiction as dim range T = k and span(λ1 v1 . so every a ∈ F is an eigenvalue. . . . but dim null T = n − 1. As T = (T −1 )−1 we only need to show this in one direction. 0. . We also see that T (1. We easily see that T(0. a3 . then T Sv = 0. . 0. . 9 . ST = T S gives us 0 = S(T u − λu) = ST u − λSu = T Su − λSu = (T − λI)Su. If Sv = 0. z3 ). Assume that λ = 0 and T (z1 . Thus 5 is the only non-zero eigenvalue. . then it’s an eigenvalue of ST . these are all eigenvalues of T . a2 . 1) = (1. . . . 0. 5. then 0 is not an eigenvalue. . . Also T (−1. . .6 these eigenvectors are linearly independent.) = a(a. Let v1 . so by theorem 5. Because T is not injective we know that 0 is an eigenvalue. Assume that T has two distinct eigenvalues λ1 . 0. .1)=(0. Let a ∈ F. otherwise k + 1). 1) so −1 is another eigenvalue. v2 ∈ null T . . z3 ) = (2z2 . . . so this is impossible. We see that v1 was an eigenvector. Let T v1 = a1 v1 + . T 2 (x. vn ). j = i. We need to show that λi = λj for all i. 15. so that a = p(λi ). The claim now follows from proposition 5. Now T 2 = I. .12. 13. Let (v1 . 18. 10 . However. . . a) with respect to the standard basis. λi = λj . . . . . . . Let Ui be the subspace generated by the vectors vj . . so n n n ai (ST S i=0 −1 i ) = i=0 ai ST S i −1 =S i=0 ai T i S −1 .12. By theorem 5. . Let q(x) = p(x) − a. On page 78 it was shown that it has no eigenvalue. 14. . vn ) of V . so T (vi + vj ) = λi vi + λj vj = λ(vi + vj ) = λvi + λvj . So let j > 1. Thus T v1 = a1 v1 . . By our assumption each Ui is an invariant subspace. . vn ) be a basis of V . Now T v1 = a1 v1 + . . . Thus p(λ) is an eigenvalue of p(T ). 17. Then we have n n p(T )v = i=0 ai T i v = i=0 ai λi v = p(λ)v. . By assumption (T v1 . Let T ∈ L(R2 ) be the map T (x. Let p(x) = n i i=0 ai x . y) = T (−y. b) = (b. so −1 it has an eigenvalue. Because (vi . . 16. T has the matrix 0 1 1 0 . Choose i = j. j. y) = (−y. Thus we get 0 = q(λi ) = p(λi ) − a. vj ) is linearly independent we get that λi − λ = 0 = λj − λ i. . . then by our assumption vi + vj is an eigenvector. By the fundamental theorem of algebra we can write q(x) = c n (x − λi ). + an vn . This means that (λi − λ)vi + (λj − λ)vj = 0. λn vn ). Let v1 ∈ V and extend it to a basis (v1 . + an vn . Hence λi is an eigenvalue of T . then T v ∈ Uj imples aj = 0. .e. Clearly (ST S −1 )n = ST n S −1 . x) = (−x. T vn ) = (λ1 v. However. . Let λ be an eigenvalue of T and v ∈ V a corresponding eigenvector. Let T ∈ L(F2 ) be the operator T (a.13 T has an upper-triangular matrix with respect to some basis (v1 . −y). . Now v1 ∈ Ui for i > 1. As q(T ) is non-injective we have that T − λi I is non-injective for some i. x). so T is clearly invertible. Then let a be an eigenvalue of p(T ) and v ∈ V the corresponding eigenvector. Now q(T )v = 0 and q(T ) = c n (T − i=1 i=1 λi I). . The result now follows from the previous exercise. clearly not invertible. If λ = 0 is an eigenvalue and λ(a. . b. . but λ = 0. d) = (−b. Because v is an eigenvector we have v = u + w = u = 0. Then we get ST vi = Sλi vi = λi Svi = λi λvi = λλi vi = λT vi = T λvi = T Svi . Then PU. 1 1 ··· 20. so λw = 0. Let v ∈ null P ∩ range P . it’s clearly sufficient to prove that null P ∩ range P = {0}. Hence λ = 1. b. we can find a u ∈ V such that P u = v. Solving we x + y − x−y 2 11 . Then λ is an eigenvalue of T with eigenvector v against our assumption. . If U is an invariant subspace of odd degree. Thus T has no subspace of odd degree. By symmetry we can assume that a = 0 (an eigenvector is non-zero). then T is clearly injective. 22. By theorem 5. We clearly see that a = 0 ⇔ b = 0 and similarly with c. so that 1 − λ = 0.19. . Clearly 0 is an eigenvalue and the corresponding eigenvectors are null T = W \ {0}. Then we have λa = −b and λb = a. . . . c. From the law of cosines we get x − y get x y cos θ = 2 = x 2 2 + y 2 2 −2 x 2 y cos θ. Now assume λ = 0 is an eigenvalue and v = u + w is a corresponding eigenvector. .W (u + w) = u = λu + λw ⇔ (1 − λ)u = λw. −d. Let T (a. d) = (−b. 23. . a. but this equation has no solutions in R. c). As v ∈ range P . . d. Hence they are equal. .. a. c. 6 Inner-Product Spaces 1. . Take the operator T in exercise 7. It is 1 . c). so ST and T S agree on a basis of V . As b = 0 we have λ2 + 1 = 0. As we can choose freely our u ∈ U . Thus we have that v = P u = P 2 u = P v. 24. vn ) for V where vi is an eigenvector of T corresponding to the eigenvalue λi . Thus v = 0. but has the matrix ··· 1 . so we have w = 0 which implies (1 − λ)u = 0. Hence T has no eigenvalue. −d. so 0 is not an eigenvalue. Thus λw ∈ U .26 T|U has an eigenvalue λ with eigenvector v. so that it’s non-zero we see that the eigenvectors corresponding to 1 are u = U \ {0}. As dim V = dim null P + dim range P . Let λ be the eigenvalue of S corresponding to vi . 21.6 we can choose basis (v1 . but v ∈ null P . then by theorem 5. so P v = 0. Substituting we get λ2 b = −b ⇔ (λ2 + 1)b = 0. u ≤ u. 6. Thus we get u. v 2 ≤ u. u + v = 2.g. If not choose t = 1/ v 2 . so that −2Re a u. v = 1. av = 0. y2 ). See previous exercise. 2a2 . so that 2t u. u = (1. Assuming that the norm is induced by an inner-product.Now let x = (x1 . v 2 ≤ t2 u. v = 0. b2 / 2. y = (y1 . Set e. v = (0. 2 2. . u + av. Then u = 1. av . u ≤ u + av. u and simplify. v = 0. x2). Choose a = −t u. u + av . 5. then u. v = 0. This equality is then simply a. which is clearly false. 2 + u−v 2 = 2( u 2 + v 2 ). 12 . v 2 . so by the Pythagorean theorem u ≤ u + av = u+av . u + u. If u. √ √ √ √ 3. If v = 0. v ≤ |a|2 v 2 . v 2 v 2. Just use u = u. 4. so that we get 2 u. We have u 2 ≤ u+av 2 ⇔ u. . v with t > 0. . Thus u. . 7. From the parallelogram equality we have u + v √ Solving for v we get v = 17. b 2 ≤ a 2 b 2 which follows directly from the CauchySchwarz inequality. 0). u − v = 2. Let a = (a1 . 1). . v 2 v 2 ⇔ 2 u. av + av. Then assume that u ≤ u+av for all a ∈ F. . . we would have by the parallelogram inequality 8 = 22 + 22 = 2(12 + 12 ) = 4. then clearly u. nan ) ∈ Rn and b = (b1 . v 2 ≤ t u. 1 2 so that x y cos θ = x 2 + y 2 − x−y 2 2 = 2(x1 y1 + x2 y2 ) = x. A straight calculation shows that x 2 + y 2 − x−y 2 2 2 = x2 + x2 + y1 + y2 − (x1 − y1 )2 − (x2 − y2 )2 = 2(x1 y1 + x2 y2 ). y . bn / n) ∈ Rn . . v for all n ∈ Z. v = Trivially we have u 2 u+v 2 − u−v 4 2 . Define u. v = n u. w − v. v = n u/n. v . This gives us n→∞ lim rn x + y = λx + y when rn → λ. so positivity definiteness follows. so check a book on topology or functional analysis or just assume the result. It’s easy to see that −u. This probably doesn’t make any sense. Hence we get additivity in the first slot. v = nu/n. u . w − u. w ) = + − = + = − u+v+w 2− u+v−w 2 1 ( u + w − 2w 2 + |u − v|2 ) 2 1 ( u + v + 2w 2 + u − v 2 ) 2 u+v+w 2− u+v−w 2 1 1 u + w − 2w 2 − u + v + 2w 2 2 2 1 2 ( u + v + w + w 2 ) + u + v − 2w 2 1 ( u + v − w 2 + w 2 ) − u + v + 2w 2 2 2 Applying the parallelogram equality to the two parenthesis we and simplifying we are left with 0. = u. the proof for C is almost identical except for the calculations that are longer and more tedious. This exercise is a lot trickier than it might seem. so we get nu. I’ll prove it for R. 13 .8. Now u+v+w −( u + w −( v + w = +( u − w −( u + w 2 2 2 4( u + v. v . All norms on a finite-dimensional real vector space are equivalent. Now we get u. v = − u. v by u. w − v. w ) = − u+v−w 2 − u − w 2) − v − w 2) 2 u+v+w 2 2 − u+v−w 2 + v − w 2) + v + w 2) We can apply the parallelogram equality to the two parenthesis getting 4( u + v. w − u. e1 = 1. Where 1 x. so we have an inner-product. v 0 1 = ( u+v 4 2 − u − v 2 ) ⇒ u.1 1 x− x. 9. v . sin mx cos nx.so that u/n. cos mx sin nx. cos mx = −π π sin nx sin mxdx cos nx cos mxdx −π π = = −π sin nx cos mxdx and they can be evaluated using the trigonometric identities sin nx sin mx = cos nx cos mx = cos nx sin mx = 10. Now let λ ∈ R and choose a sequence (rn ) of rational numbers such that rn → λ. This is really an exercise in calculus. and using the properties just proved for ·. v 0 + u. The proof for F = C can be done by defining the inner-product from the complex polarizing identity. · 0 . It follows that we have homogeneity in the first slot when the scalar is rational. And by using the identities u. v . We trivially also have symmetry. We have integrals of the types π sin nx. v = n/m u. 2 . iv 0 i. then e2 = x− x. v = n→∞ lim rn u. v n→∞ 1 = lim ( rn u + v 2 − rn u − v 2 ) n→∞ 4 1 = ( λu + v 2 − λu − v 2 ) 4 = λu. v = 1/n u. v Thus we have homogeneity in the first slot. Thus we have nu/m. v = u. This gives us λ u. v = lim rn u. 1 = 0 x = 1/2. 14 .1 1 cos((n − m)x) − cos((n + m)x) 2 cos((n − m)x) + cos((n + m)x) 2 sin((n − m)x) − sin((n + m)x) . em en = | v. . The orthonormal basis is thus (1. Now we have en . . . which gives e2 = have √ 12(x − 1/2) = √ 3(2x − 1). . . . + v. en ) are orthogonal and span the same subspace. . . . 3(2x − 1). . . so that ai = 0 for i < n. Then if (e1 . . vn ) be a linearly dependent list. vn−1 ). . en |. . . . . Now √ x2 − 1/3 − 1/2(2x − 1) = 1/(6 5) √ √ √ giving e3 = 5(6x2 − 6x + 1). . en−1 ) be an orthogonal list of vectors. en ) of V . . ei = 0 for i > m i.so that 1 x − 1/2 = x − 1/2. en ) and (e1 . vn ). . Thus we have v = | v. . . x2 . + v. √ √ so that x2 − x2 . Then we can write en = a1 e1 + . . . x − 1/2 = 0 (x − 1/2)2 dx = 1/12. .17 v = v. . . . so that v = = v. . e1 e1 + . e1 e1 + . . We can assume that (v1 . Then to continue pencil pushing we √ √ x2 . Assume that (e1 . . en ) be the orthonormal base produced from (v1 . . . ei = 0 for all i < n. v ∈ span(e1 . + | v. It’s easy to see that we can extend the algorithm to work for linearly dependent lists by tossing away resulting zero vectors. . . vn−1 ) is linearly independent and vn ∈ span(v1 . 1 = 1/3. + an en . en ) satisfies the hypothesis from the problem we have by the previous paragraph that ei = ±ei . Extend (e1 . 12. em ). 5(6x2 − 6x + 1). so en = 0. . . . . . vn ) by Gram-Schmidt. e1 | + . . . . + | v. . . en = vn − P vn but vn = P vn by our assumption. By theorem 6. . Let (e1 . em | if and only if v. . . Let (e1 . Let (v1 . Then calculating en with the Gram-Schmidt algorithm gives us vn − P (vn ) . 15 . . Thus we have en = an en and from en = 1 we have an = ±1. e1 e1 + . . . 1 1 − x2 . . en−1 . 13. em en v. . en en . . e1 | + . . . . 3(2x − 1) = 3/6. . . . . 3(2x − 1) 3(2x − 1) = x2 − 1/3 − 1/2(2x − 1). em ) to a basis (e1 . . Thus the resulting list will simply contain zero vectors. Thus we have 2n possible such orthonormal lists. 11. . Let P denote the projection on the subspace spanned by (v1 . + v. .e. 420/11(x3 − 2 x2 )). Hence the claim follows. If T PU = PU T PU clearly U is invariant. Thus null P ⊂ (range P )⊥ and from the dimension equality null P = (range P )⊥ . 2/ 5). Then for a ∈ F u 2 = P (u + aw) 2 ≤ u + aw 2 . Now PU T w = u. By exercise 2 we have u. Let u ∈ range P . Then for u ∈ U we have PU T PU u = PU T u = T u = T PU u. 1 420/11(x3 − x2 ) 2 71 2 329 3 x + x . 20. then we have T u = T PU u = PU T u ∈ U . so T w ∈ U ⊥ . 3. 3x2 3x + 2 + 3x. 12 so that PU (2 + 3x) = − 16 2 + 3x. w = 0.21 we have that V = range P ⊕ null P . so that null P = U ⊥ . Then we have u = PU (1. Now we can write T w = u + u where u ∈ U and u ∈ U ⊥ . 22/5). Let u ∈ U . 1/ 2. 0. First we need to find an orthonormal √ basis for U . Thus U ⊥ is also invariant. 15. e2 ). Now assume that U is invariant. 19. e1 e1 + (1. 2. 4). From P 2 = P we have that P (v − P v) = 0. so v − P v ∈ null P = U ⊥ . Let U := range P . x3 ). 4) = (1. 2. 22. 660 1 420/11(x3 − x2 ) 2 1 420/11(x3 − x2 ). 0. 3. 18. 16. but u = PU T w = T PU w = T 0 = 0. Thus U is invariant. 1/ 5. so P is the identity on range P . This follows directly from the previous exercise. 2. Thus we want to find the projection of 2 + 3x to the subspace U := span(x2 . so that P is the identity on U . then p(x) = ax2 + bx3 . If p(0) = 0 and p (0) = 0. then by assumption null P ⊂ U ⊥ .14. e2 e2 = (3/2. Hence the decomposition v = P v + (v − P v) is the unique decomposition in U ⊕ U ⊥ . 17. Let w ∈ null P . we have dim U = dim V − dim U = dim null P . 3/2. 21. Here √ 17 √ 2 + 3x. 4). 3. then u = P v for some v ∈ V hence P u = P 2 v = P v = u. With Gram-Schmidt we get e1 = √ √ √ (1/ 2. An arbitrary v ∈ V can be written as v = P v + (v − P v). Then let w ∈ U ⊥ . This follows directly from V = U ⊕ U ⊥ . Now P v = P (P v + (v − P v)) = P v. e2 = (0. From exercise 15. From exercise 5. We get √ √ 2 PU (2+3x) = 2 + 3x. 22 22 = 47 √ 1155. 2 . 11/5. 0). By definition P = PU . With Gram-Schmidt we get √ 1 the orthonormal basis ( 3x2 . It’s easy to see that the differentiation operator has a upper-triangular matrix in the orthonormal basis calculated in exercise 10. 3x2 = 3. Let U = span(e1 . . v ). . . . . Choose an orthonormal basis (e1 . . 3(2x − 1). v ] . 25. compared to the previous two. v . . 5(6x2 − 6x + 1) is an orthonormal basis for P2 (R). 24. . . . . We see that the map T : P2 (R) → R defined by p → p(1/2) is linear.. v · · · en . . There’s is nothing special with this exercise. π π 26. so that e1 . From theorem 6. 0 17 . so that 1 1 q(x) = 0 cos(πx)dx + 0 1√ 1√ 3(2x − 1) cos(πx)dx × √ 3(2x − 1) + √ 5(6x2 − 6x + 1) cos(πx)dx × 5(6x2 − 6x + 1) 0 √ 4 3√ 12 = 0 − 2 3(2x − 1) + 0 = − 2 (2x − 1). (e1 . en )) = . . a en . 0 . . v . . Then by proposition 6. en . . 27. en ) for V and let F have the usual basis..47 ∗ 0 1 . . . The map T : P2 (R) → R defined by p → 0 p(x) cos(πx)dx is clearly linear. v and finally T ∗ a = (a e1 . M(T ∗ . except that it takes ages to calculate.. 3(2x − 1)..45 we see that √ √ √ √ q(x) = 1 + 3(2 · 1/2 − 1) 3(2x − 1) + 5(6 · (1/2)2 − 6 · 1/2 + 1) 5(6x2 − 6x + 1) = −3/2 + 15x − 15x2 . . . 1 .. 0 0 1 . From exercise √ √ 10 we get that (1. (e1 .23.47 we have M(T.. Again √ √ we have the orthonormal basis (1. 1 0 0 . .. with the usual basis for Fn we get M(T ) = 0 so by proposition 6. . . 5(6x2 − 6x + 1). M(T ) = . en )) = [ e1 . zn . 0). 7 Operators on Inner-Product Spaces 1.Clearly T ∗ (z1 . Assume that T is injective. . That is λ is an eigenvalue of T ∗ if and only if λ is an eigenvalue of T . By the previous exercise dim range T = dim range T ∗ . By proposition 6.47 the span of the row vectors equals range T ∗ . so the claim follows. From exercise 31 we see that T ∗ − λI is non-injective if and only if T − λI is non-injective.46 and exercise 15 we get dim null T = dim(range T )⊥ = dim W − dim range T = dim null T + dim W − dim V. 32. . let p(x) = x and q(x) = 1. so that 0 = T u. . . u = u 2 . Now we can write T ∗ w = u + v. As (U ⊥ )⊥ = U and (T ∗ )∗ = T it’s enough to show only one of the implications. Let T be the operator induced by A. Then clearly 1/2 = T p. By additivity and conjugate homogeneity we have (T −λI)∗ = T ∗ −λI. w = u. T q = p. 30. 28. From proposition 6. As (T ∗ )∗ = T it’s sufficient to prove the first part. 18 . . so that range T = V . but range T is a subspace of V . T ∗ w = u. Thus u = 0 which completes the proof. . 29. Hence the dimension of the span of the column vectors equal dim range T . 31. so the generate range T . zn ) = (z2 . 0 = 0. For the second part we have dim rangeT ∗ = dim W − dim nullT ∗ = dim V − dim null T = dim range T. . q = p. Let U be invariant under T and choose w ∈ U ⊥ . For the first part. For the second part it’s not a contradiction since the basis is not orthogonal. then by exercise 31 we have dim V = dim range T = dim range T ∗ . u + v = u. where the second equality follows from the first part. The columns are the images of the basis vectors under T . where u ∈ U and v ∈ U ⊥ . . 4. If T and S are self-adjoint. The identity operator I is self-adjoint.6 we get null T = null T ∗ . Then assume that P is a projection. so that range T k ⊂ range T . but T S has the matrix 0 1 0 0 so T S is not self-adjoint. so that v ∈ null T k−1 . Choose the operators corresponding to the matrices A= 0 −1 1 0 . so iI is not self-adjoint. Let v ∈ null T k . then we can find a v ∈ V such that u = T k v = T (T k−1 v). so range T k = range T . Now P is clearly a projection to range P . Then we have T ∗ T k−1 v. T k−1 v = 0. Clearly null T ⊂ null T k . From the first part we get dim range T k = dim range T . A + B doesn’t define a normal operator which is easy to check. but clearly (iI)∗ = −iI. .2. 1 0 0 1 Then T and S as clearly self-adjoint. Now T k−1 v. T k−2 v = 0. b ∈ R. un ) an orthogonal basis for range P and extend it to a basis of V . However.46 we have null P = null P ∗ = (range P )⊥ . .46 that range T = (null T ∗ )⊥ = (null T )⊥ = range T ∗ . T k−1 v = T ∗ T k−1 v. 6. 5. It follows from proposition 6. Now let u ∈ range T k . so self-adjoint operators form a subspace. then (aT +bS)∗ = (aT )∗ +(bS)∗ = aT +bS for any a. From proposition 7. Assume that P is self-adjoint. Thus null T k ⊂ null T k−1 and continuing we get null T k ⊂ null T . . B= 1 1 1 0 . . Then we easily see that A∗ A = AA∗ and B ∗ B = BB ∗ . Let T and S the the operators defined by the matrices 0 1 0 0 . 19 . T ∗ T k−1 v = T ∗ T k v. Then clearly P has a diagonal matrix with respect to this basis. . 7. so P is self-adjoint. Let (u1 . Choose the standard basis for R2 . so that T ∗ T k−1 v = 0. Then from proposition 6. . 3. 8 if T is normal that would imply orthogonality of u and v. λn on the diagonal. . . Hence T (T v − v) = T 2 v − T v = 0. . . Let λ1 . . i so the matrix of T equals its conjugate transpose. . For any v ∈ V we get 0 = T 9 v − T 8 v = T 8 (T v − v). . A self-adjoint operator is normal. . we can choose a basis of V consisting of eigenvectors of T . . + an vn this implies that n i=1 |ai | = 1. . . 12. . so by corollary 7. 11. If T is normal. Then T has a diagonal matrix with respect to the basis. 7) are both eigenvectors corresponding to different eigenvalues. 10. Clearly u. If v = a1 v1 + . then n n 0 −1 . Let v ∈ V be such that v = 1. .8. Thus we can find an eigenvalue λi such that |λ − λi | < ε. By the spectral theorem we can choose a basis of eigenvectors for T such that T has a diagonal matrix with λ1 . If |λi − λ| ≥ ε for all i. vn ) consisting of eigenvectors of T . By the spectral theorem we can choose a basis of T such that the matrix of T corresponding to the basis is a diagonal matrix. 13. so T 2 = T . so T can’t be even normal much less self-adjoint. . Let S be the operator corresponding to the diagonal matrix having √ √ 3 λ1 . Hence T is self-adjoint. 14. Thus T v − v ∈ null T 8 = null T by the normality of T . Now T is self-adjoint if and only if the conjugate transpose of A equals A that is the eigenvalues of T are real. . . . 2. √Let S be the operator corresponding to the diagonal matrix having the √ elements λ1 . . . . Clearly S 2 = T . By the spectral theorem we can choose a basis consisting of eigenvectors of T . . Then clearly S 3 = T . Hence λi = 0 or λi = 1. Let λ1 . Now T 2 has the diagonal matrix with λ2 . . λ2 on n 1 the diagonal and from T 2 = T we must have λ2 = λi for all i. . 9. . By the spectral theorem we can choose a basis (v1 . 3 λn on the diagonal. 3) and v = (2. v = 0. Assume that T v − λv < ε. . Then 1 0 T v − λv = i=1 n (ai T vi − λvi ) = i=1 n (ai λi vi − λvi ) |ai ||λi − λ| v1 = i=1 n ai (λi − λ)vi = i=1 n = i=1 |ai ||λi − λ| ≥ ε i=1 |ai | = ε. λn on the diagonal. 5. λn be the diagonal elements. . Let T the operator corresponding to the matrix T 2 + I = 0. λn be the diagonal elements. 20 . Let A be the matrix of T corresponding to the basis. . . The vectors u = (1. which is a contradiction. . . . Hence S 2 v = λ2 v = v. 21 . 22. . by hypothesis and self-adjointness. So assume that T is invertible. . b) = (a. T v ≥ 0. v + T v. For k = 2 we have T 2 v. . Assume that the result is true for all positive k < n and n ≥ 2. Also (S + T )v. . v = T v. 20. 0)) is invariant under T . Let T. v > 0 for all v ∈ V \ {0} implies that T is injective hence invertible. Then T n v. Let the elements on the diagonal be λ1 . a + b). v = Sv. . . 1. vj = 0. 17. . v = |a1 |2 λ1 + . If such an inner-product exists we have a basis consisting of eigenvectors by the spectral theorem. . Then S(1. λn and by assumption λi > 0 for all i. which shows that it has infinitely cos θ − sin θ many square roots. . i = j Extend this to an inner-product by bilinearity and homogeneity. Define an inner-product by vi . Let S ∈ L(R2 ) be defined by S(a. 16. but S is clearly not an isometry. Assume then that (v1 . R3 is an odd-dimensional real vector space.15. Clearly T v. Since T is self-adjoint we can choose an orthonormal basis (v1 . Then they are self-adjoint and (S + T )∗ = S ∗ + T ∗ = S + T . + |an |2 λn > 0. S be two positive operators. so that T v. S + T is positive. v = T n−2 T v. Then we have an inner-product and clearly T is self-adjoint as (v1 . Hence S has an eigenvalue λ and a corresponding eigenvector v. Let v = a1 v1 + . . 19. Since Sv = |λ| v = v we have λ2 = 1. . 21. T v ≥ 0. Let T ∈ L(R2 ) be the operator T (a. 0) = (1. Then span((1. vn ) is an orthonormal basis of U. . Thus. 18. but it’s orthogonal complement span((0. vn ) of eigenvectors such that T has a diagonal matrix with respect to the basis. + an vn ∈ V \ {0}. i = j . Clearly T k is self-adjoint for every positive integer. . 1) = 1. 0) = 1 and S(0. sin θ cos θ is a square root for every θ ∈ R. . . . 0). . v ≥ 0. 1)) is clearly not. . vn ) is basis such that the matrix of T is a matrix consisting of eigenvectors of T . b) = (a + b. Hence the result follows by induction. . so S + T is self-adjoint. Choose a basis of eigenvectors for T such that T has a diagonal matrix with eigenvalues λ1 . z3 ) = (z3 . Now assume that T is not invertible. √ 25. √ 28. Thus it’s sufficient to show that R2 = T ∗ T . so U is an isometry. . . The matrices corresponding to T and T ∗ are 0 0 1 0 2 0 M(T ) = 2 0 0 . Thus R2 = RIR = RS ∗ SR = T ∗ T . so 12 = 1 is not a singular value of T 2 . . z2 . z3 ). v ∈ V . U vi = Svi . z2 ). If T=S T ∗ T . 22 . Then from the matrix representation √ we see that T ∗ T (a. Now choose u. Define U S by U v1 = −Sv1 . vn ) of eigenvectors of √∗ T and we can T assume that v1 corresponds to the eigenvalue 0. Let T ∈ L(R2 ) be the map T (a. √ −1 . 0 3 0 1 0 0 Thus we get 4 0 0 M(T ∗ T ) = 0 9 0 . so that T ∗ T is not invertible. |λn | on the diagonal. 9z2 . These are the singular values. b) = (0. so the claim follows. so by proposition 7. . i > 1. Now we have T ∗ = (SR)∗ = R∗ S ∗ = RS ∗ by self-adjointness of R. 26. Now T ∗ T = T 2 corresponds to the diagonal matrix √ with λ2 . . Thus T is bijective if and only if 0 is not a singular value. then since S is an √ isometry we have that T is bijective hence invertible if and only if T ∗ T is bijective. 0). Hence T ∗ T (z1 . 3z2 . By the spectral √ theorem we can choose a basis (v1 . Now let T =√ T ∗ T . z2 . The composition of two bijections is a bijection. a). Hence S = T T ∗ T √so S is uniquely determined. hence T ∗ T = T ∗ T and 1 is clearly a singular value. Then range T ∗ T is not invertible (hence not surjective).23. 0) = (a. Sv = u. . Clearly T ∗ T is positive. 24. However. U v = Su. z1 . z3 ). . 0 0 1 √ so that T ∗ T is the operator (z1 . so (T 2 )∗ T 2 = 0. z2 . . Then it’s easy to verify that U u. . . . λn on the diagonal. v . . z3 ) → (4z1 . . Then T ∗ T must be invertible (polar decomposition). λ2 on the diagonal and the square root T ∗ T corresponds to the diagn 1 onal matrix having |λ1 |. Now we just need to permute the indices which corresponds to the isometry S(z1 . T 2 = 0. . √ But T ∗ T is injective hence bijective if and only if 0 is not an eigenvalue. 27. Clearly U = S and T = U T ∗ T . √ Now assume that T is not invertible. . . z3 ) = (2z1 . M(T ∗ ) = 0 0 3 . Assume that T is invertible.26 it has a unique positive square root. . ∗ so that λ is an eigenvalue of T2 T2 . . . then clearly S ∗ S = I. . . . Writing −1 S −1 . √ 30. wn ). . this means that the √ √ matrix of S ∗ S is the identity matrix. the √ Since T ∗ T is self-adjoint we can choose a basis of eigenvectors of T ∗ T such that √ √ the matrix of T ∗ T is a diagonal matrix. . so that B ∗ = B = A. we get S3 = S2 S 1 T2 = S3 S1 ∗ T1 T1 S = S3 T1 S. . Thus T S = I. S −1 v ) is an orthonormal basis for V of eigenvectors vectors of T1 1 1 n 1 ∗ ∗ ∗ of T2 T2 . so the claim follows. (e1 . . It follows from proposition 6. . . . Clearly dim range T ∗ T equals the number of non-zero elements on the diagonal i. As all eigenvalues are one. Let T1 = S1 T1 T1 and T2 = S2 T2 T2 . . vn ) and (w1 . sn . . . . . .e. Thus we get T2 = S2 T2 T2 = S2 S −1 T1 T1 S. (f1 . Then we can choose bases of eigenvectors of T1 T1 and ∗ T2 T2 such that they have the same matrix. B := M(T. ∗ ∗ 31. The singular values are simply the positive square roots of these eigenvalues. Clearly A = B and is a diagonal matrix with s1 . . The claim now follows from i exercise 3. 23 .√ 29. . . so all singular values are 1. . Now assume √ that all singular values are 1. the number of non-zero singular values. ∗ ∗ ∗ Now let T2 = S1 T1 S2 . Then we have T2 T2 = (S1 T1 S2 )∗ (S1 T1 S2 ) = S1 T1 T1 S1 = −1 ∗ ∗ T and v a corresponding eigenvector. . From √ polar decomposition theorem we see that dim range T = dim range T ∗ T . 32. so S is an isometry. fn )). which is clearly an isometry. . S1 T1 T1 S1 . Let λ be an eigenvalue of T1 1 Because S1 is bijective we have a u ∈ V such that v = S1 u. For the second part observe that a linear map S defined by the formula maps fi to s−1 ei which is mapped by T to fi . . sn on the diagonal. . then (S −1 v . en ). Let these bases be (v1 . . By the spectral theorem we can choose a basis of V such √ that S ∗ S has a diagonal matrix. en )). (e1 . Set A := M(S.23. Assume that T1 and T2 have the same ∗ singular values s1 . fn ). This gives us −1 −1 ∗ −1 ∗ ∗ T2 T2 u = S1 T1 T1 S1 u = S1 T1 T1 v = S1 λv = λu. . . For the first part. . .47 that S = T ∗ . . . . (f1 . then clearly S is an isometry ∗ ∗ ∗ ∗ and T2 T2 = S −1 T1 T1 S. denote by S the map defined by the formula. . . Let S the operator defined by S(wi ) = vi . Hence T2 T2 and T1 T1 have the same eigenvalues. . . vn ) be an orthonormal basis of eigen∗ T . All the si are positive real numberse. Then S ∗ S is a self-adjoint (positive) operator with all eigenvalues equal to 1. Let (v1 . If S is an isometry. . Thus S ∗ S = I. T is not injective. but T 2 = 0. so that a0 = 0. We see that T 3 = 0. By definition T ∗ T is positive. ei | ≤ i=1 si | v. We see that T 2 = 0. If ST is nilpotent. so that any v ∈ V is a generalized eigenvector corresponding to the eigenvalue 0. i) is an eigenvector corresponding to the eigenvalue −i. then S is nilpotent. 34. so that 0 = S 4 = T 2 . From the triangle equality and the previous exercise we get (T + T )v ≤ T v + T v ≤ (s + s ) v . T m−1 v) is linearly independent. −i) is an eigenvector corresponding to the eigenvalue i and (1. Let T + T = S (T + T )∗ (T + T ). On page 78 it’s shown that ±i are the eigenvalues of T . ei | = T v and similarly for s v . ei | since all si are positive. so by corollary 8. Contradiction. Thus all singular values of T are positive. (T + T )∗ (T + T )v = Ssv = s v ≤ (s + s ) v . then ∃n ∈ N such that (ST )n = 0. The set of all generalized eigenvectors are simply the spans of these corresponding eigenvectors. Assume that S is a square root of T . 4. . Let m−1 ai T i v = 0. . Api=0 i=0 plying repeatedly T m−i for i = 2. 3. so (T S)n+1 = T (ST )n S = 0. 5. ei fi = i=1 si | v. Because (1. . . . we see that ai = 0 for all i. We have dim null(T − iI)2 + dim null(T + iI)2 = dim C2 = 2. 2. Now clearly n n s v =s ˆ ˆ i=1 | v. so 0 is an eigenvalue of T . so that dim null(T − iI)2 = dim null(T + iI)2 = 1. Hence (v. . 8 Operators on Complex Vector Spaces 1. then 0 = T m−1 ( m−1 ai T i v = a0 T m−1 v.√ 33. 24 . so that (T + T )v = S so that s ≤ s + s . From the singular value decomposition theorem we can choose bases of V such that n n Tv = i=1 si v. .8 we have S dim V = S 3 = 0. Now let v be the eigenvector of (T + T )∗ (T + T ) corresponding to the singular value s. so that (z − 5)d1 (z − 6)d2 divides (z − 5)n−1 (z − 6)n−1 . so the claim clearly follows. b.23 T has a basis of eigenvectors. d2 ≤ n − 1. then dim null N i−1 < dim null N i for all i ≤ dim V by proposition 8. b.e. Then we can find a u ∈ V such that T n u = v. so from theorem 8. c. Then d1 .5.5 we get dim null T m = dim null T m+k for all k ≥ 1. 16. 14. Assume that T has three different eigenvalues 0. Let d1 be the multiplicity of the eigenvalue 5 and d2 the multiplicity of the eigenvalue 6. λ1 . so it’s sufficient to prove that null T n ∩ range T n = {0}. Now the matrix of N ∗ is the conjugate transpose. so the claim follows. Then dim null (T −λi I)n ≥ 1. range T m = range T m+1 implies dim null T m = dim null T m+1 i. Since N ∗ = N . From null T n−2 = null T n−1 and from proposition 8. so the claim is false. the matrix of N must be zero. d) = (7a. We have dim V = dim null T n + dim range T n . If N dim V −1 = N dim V . Then N dim V v = λdim V v = 0. 7b. 9. By the Cayley-Hamilton theorem (T − 5I)d1 (T − 6I)d2 = 0. which is impossible.6. 15. Let T be the operator defined in exercise 1.26 we can find a basis of V such that the matrix of N is upper-triangular with zeroes on the diagonal.8. Then clearly 0 is the only eigenvalue. It follows that d1 . From proposition 8. so that (T − 5I)n−1 (T − 6I)n−1 = 0. If there’s a generalized eigenvector that is not an eigenvector. λ2 .5 we see that dim null T n ≥ n−1. Assume that λ = 0 is an eigenvalue of N with corresponding eigenvector v. then by theorem 8. Let v ∈ null T n ∩ range T n . This contradicts corollary 8. From 0 = T v = T n+1 u we see that u ∈ null T n+1 = null T n which implies that v = T n u = 0. If every generalized eigenvector is an eigenvector. 8c.23 n ≥ dim null T n + dim null (T − λ1 I)n + dim null (T − λ2 I)n ≥ n − 1 + 1 + 1 = n + 1. Then let T ∈ L(R3 ) be the operator T (a. so the claim follows. 0). c) = (−b. Let T ∈ L(C4 ) be defined by T (a. Again this implies that dim range T m = dim range T m+k for all k ≥ 1. 7.23 we have V = null T dim V . 13. so that T dim V = 0. From theorem 8. null T m = null T m+1 . then we have an eigenvalue λ such that dim null(T − λI)dim V = dim null(T − λI). but T is not nilpotent. Thus if 25 . a. 12. d2 ≥ 1 and d1 + d2 = n. 11. 10. By corollary dim null N dim V = dim V . Clearly null T ∩ range T = {0}. 8d) from the matrix of T it’s easy to see that the characteristic polynomial is (z − 7)2 (z − 8)2 . By lemma 8. 8. so T has at most two different eigenvalues. . The operator defined by the matrix 0 0 1 0 0 0 0 0 0 is clearly an example. choose a basis (v1 . . Then N e1 = 0 and assume that N ei ∈ span(e1 .30 up to j = 4 we see that a4 = −5/128.26. . the minimal polynomial divides z(z − 1)2 . + am xm−1 ) we have q(T ) = T −1 . then doesn’t exists a basis of eigenvectors. Since T is invertible we must have x i=1 ai x i=1 ai T the minimality of p. . + am T m−1 )T = I.30 with the Taylor polynomial of √ 3 1 + x and copy the proof of the lemma and theorem 8. Thus N has an upper-triangular matrix in the basis (e1 . Continuing in the proof of lemma 8. . m i 20. . . λm are the eigenvalues of T . en ). Thus. . However. Let p(x) = i=0 ai x be the minimal polynomial of T .λ1 . the whole polynomial is the only factor that annihilates A. . so there 17. 22. e1 e1 − . . . By lemma 8. so that N ei+1 = N vi+1 − vi+1 . 0 Hence setting q(x) = −a−1 (a1 + . ei ei ∈ span(e1 . Hence a0 = 0. Choose the matrix 1 0 A= 0 0 0 1 0 0 0 1 1 0 0 0 . . vn ) such that N has an upper-triangular matrix. − vi+1 . . Just replace the Taylor-polynomial in lemma 8. 26 . . . ei+1 ).32. . so z(z − 1)2 is the minimal polynomial. e1 e1 − . 0 21. . . . . . . If a0 = 0. So that √ 1 1 1 5 N4 I + N = I + N − N2 + N3 − 2 8 16 128 19. . . ei ei vi+1 − vi+1 . ei ). − vi+1 . 18. 0 0 It’s easy to see that A(A − I)2 = 0. Apply Gram-Schmidt orthogonalization to the basis. . so solving for I in p(T ) we get −a−1 (a1 + . . . . then p(x) = m m i−1 = 0 which contradicts i−1 . m i=1 dim null(T − λi I) < dim V . λm be the distinct eigenvalues corresponding to the 1 × 1 blocks. vn ) be a Jordan-basis of T . so that 0 = q(T )v = s(T )p(T )v + r(T )v = r(T )v. so that the minimal polynomial divides p and hence has no double root.23. we can multiply the both sides with the inverse of the highest coefficient of r yielding a monic polynomial r2 of degree less than p such that r2 (T )v = 0 contradicting the minimality of p. dm . Now clearly p = m (z − λi ) annihilates T . . Now let λ1 . . Now clearly p(A) = 0. Clearly this polynomial has a negative discriminant. Let p(z) = m (z − λi ). Then we get i=1 p(T )vi = j=i (T − λj I) (T − λi I)vi = 0. Now assume that V is a real vector space. . If V is a complex vector space. Assuming r = 0. . . Since A was the largest Jordan-block it follows that all basis vectors are eigenvalues (corresponding to a 1 × 1 matrix). so that A is a 1 × 1 matrix and the basis vector corresponding to A is an eigenvalue. 27 . so the minimal polynomial of A is (z − λ)m+1 . . 24.25 we can choose the 2 × 2 blocks to be of the form a −b b a where b > 0 and it’s easy to see that a −b b a − 2a a −b b a + (a2 + b2 ) 1 0 0 1 = 0. However. so that the i=1 minimal polynomial being a factor of p doesn’t have a double root. Let the minimal polynomial be p and let (v1 . . Then p(z) = m (z − λi ) annihilates all but the 2 × 2 blocks of the matrix. . Hence r = 0 and p divides q. . the minimal polynomial of (A − λI) is z m+1 where m is the length of the longest consecutive string of 1’s that appear just above the diagonal. Assume that V has a basis of eigenvectors (v1 . . . Let λ1 . by exercise 29. . . . . Let q be the minimal polynomial. λm be the eigenvalues of T with multiplicity d1 . then V has a basis of eigenvectors. By theorem 7. Write q = sp + r. vn ). Hence m = 0. Now it’s i=1 sufficient to show that each 2 × 2 block is annihilated by a polynomial which doesn’t have real roots. Now assume that the minimal polynomial has no double roots. . Let A be the largest Jordan-block of T . where deg r < deg p. . . λm be the distinct eigenvalues. . . . Let λ1 . so the claim follows. By theorem 7. . . 25. .25 we can find a basis of T such that T has a block diagonal matrix where each block is a 1 × 1 or 2 × 2 matrix. Hence the minimal polynomial has degree n. . . 28. . so p is the minimal polynomial. Now clearly N m+1 = 0.. 0 1 0 . . vm ) be the vectors corresponding to all but the last Jordan-block and (vm+1 . . . . Now p(T )(e1 ) = 0. Thus for any non-zero polynomial p of degree less than n we have p(ei ) = 0. Assume that V can’t be decomposed into two proper subspaces.. − an−1 T n−1 . 29. vi+m be the basis elements corresponding to the biggest Jordan-block. . . 0 0 0 0 but A(A − I)(A − 3I) = 0. . so that the minimal polynomial equals the characteristic polynomial. . . . Now from the matrix of T we see that T n (ei ) = −a0 − a1 T (e1 ) − . . . . 0 0 0 0 so the it’s minimal polynomial is z(z − 1)2 (z − 3) which by definition is also the characteristic polynomial. Then N m ei+m = ei = 0. . Then T has only one eigenvalue. Hence (e1 . . . If there is more than one Jordan-block. It’s easy to see that no proper factor of z(z − 1)(z − 3) annihilates the matrix 3 1 1 1 0 1 1 0 A= 0 0 1 0 . . en ) is linearly independent. vn ) the 28 . We see that T (ei ) = ei+1 for i < n. then Let (v1 . . 30. The biggest Jordan-block of N is of the form 0 1 0 1 . 27. . T e1 . . so the minimal polynomial is z m+1 . so the claim follows. However. It’s easy to see that no proper factor of z(z − 1)2 (z − 3) annihilates the matrix 3 1 1 1 0 1 1 0 A= 0 0 1 0 .26. + an−1 z n−1 + z n . so the minimal polynomial divides z m+1 . . so by exercise 25 p divides the minimal polynomial. . Set p(z) = a0 + a1 z + . T n−1 e1 ) = (e1 . the minimal polynomial is monic and has degree n. . . Let vi . Thus. . Now assume that T has minimal polynomial (z −λ)dim V . Hence deg p1 p2 = deg p1 + deg p2 ≥ dim V . . . . If V = U ⊕W . The claim follows. so that Ai − λI is invertible for each i. so that (z − λ)dim V divides p1 p2 . . First let λ be an eigenvalue of A. Hence the characteristic polynomial has a root if and only if (a − d)2 + 4bc ≥ 0. Reversing the Jordan-basis simply reverses the order of the Jordan-blocks and each block needs to be replaced by its transpose.vectors corresponding to the last Jordan-block. −1 0 (Am − λI) 29 . This means that p1 (z) = (z − λ)dim U and p2 (z) = (z − λ)dim W . vn ). . The characteristic polynomial of the matrix is p(x) = (x − a)(x − d) − bc. This means that (T − λI)n = 0 contradicting the fact that (z − λ)dim V is the minimal polynomial. dim W } we have (T|U − λI)n = (T|W − λI)n = 0. though in this special case the proof is trivial. but deg p1 + deg p2 ≤ dim U + dim W = dim V . 4. Then clearly V = span(v1 . let p1 be the minimal polynomial of T|U and p2 the minimal polynomial of T|W . Now if n = max{dim U. 31. See the proof of the next problem. Hence T has minimal polynomial (z − λ)dim V . Then the matrix must have the form a 1−a 1−a a and it’s easy to see that 1 1 is an eigenvector corresponding to the eigenvalue 1. 3. Assume that λ is not an eigenvalue of A1 . 9 Operators on Real Vector Spaces 1. If p(x) = (x − λ1 )(x − λ2 ). then p(T ) = 0 implies that either A − λ1 I or A − λ2 I is not invertible hence A has an eigenvalue. Now (p1 p2 )(T ) = 0. so that v = 0. . .. Hence A has no eigenvalues. Thus deg p1 p2 = dim V . . Now let B be the matrix (A1 − λI)−1 ∗ . vm ) ⊕ span(vm+1 . If p doesn’t have a root then assuming that Av = λv we get 0 = p(A)v = p(λ)v. we have only one Jordan-block and T − λI is nilpotent and has minimal polynomial z dim V by the previous exercise. . . Let a be the value in the upper-left corner. . Now the discriminant equals (a + d)2 − 4(ad − bc) = (a − d)2 + 4bc. Am . . 2. . so that (T 2 + αT + βI)k = 0.10. . 7. .9 as 7/2 is not a whole number. 6. . . Hence T is not injective. Assuming that such an operator existed we would have basis (v1 . 1) dim null(T 2 + T + I)dim V = 7/2 2 times on the diagonal. 9. . vn ) such that T has a matrix of the form in theorem 9. ej−1 . . Then the operator corresponding to the matrix λI clearly is an example. en ) be the standard basis. + ak ej+k . . λ is an eigenvalue of some Ai . Set v = a1 ej + . . this is impossible. . . . Choose a basis (v1 . . . Let (a1 . so that λ is an eigenvalue of A. then span(v1 . . . Let T be the operator in L(R2k ) corresponding to the block diagonal matrix where each block has characteristic polynomial x2 + αx + β. Then by theorem 9. The equation x2 + x + 1 = 0 has a solution in C. Thus. vj ) is an invariant subscape. . Let Ai correspond to the columns j. j + k and let (e1 . . . . 8. Let T be the operator corresponding to A − λI in L(Fn ). 30 . 5. . Let (v1 . . . . However. Now let λ be an eigenvalue of Ai . . 10. It follows that dim null(T 2 + αT + βI)2k = dim null(T 2 + αT + βI)k and the claim follows. . This would contradict theorem 9. so that span(v1 . ak ) be the eigenvector of Ai corresponding to λ. ej−1 ). . . 2 so that dim null(T 2 + αT + βI)2k is even.9 we have k= dim null(T 2 + αT + βI)2k . v) to the space span(e1 . Now if vj corresponds to the second vector in a pair corresponding to a 2 × 2 matrix or corresponds to a 1 × 1 matrix. . . Then it’s easy to see that T maps the space span(e1 . .It’s now easy to verify that B(A−λI) is a diagonal matrix with all 1s on the diagonal. the minimal polynomial of T divides (T 2 + αT + βI)2k and has degree less than 2k. . . . The second posibility means that vj corresponds to the first vector in a pair corresponding to a 2 × 2 matrix. since A − λI is not. . Applying Gram-Schmidt to the basis the matrix trivially has the same form. . Let λ be a solution. vn ) be the basis with the respect to which T has the given matrix. . . It follows that such an operator doesn’t exist. vj+1 ) is an invariant subspace. The proof is identical to the argument used in the solution of exercise 2. . v7 ) such that the matrix of T would have a matrix with eigenpair (1. . . However. . hence it’s invertible. . β) is an eigenpair and from the nilpotency of T 2 + αT + βI and theorem 9. so that T − λ1 I = 0. 12. 13. If neither T − λ1 I or T − λ2 I is injective. Hence we get c = b = 0 and a = d = λ1 .9 we have dim null(T 2 + αT + βI)dim V = (dim V )/2 2 it follows that dim V is even. For the second part we know from the Cayley-Hamilton theorem that the minimal polynomial p(x) of T has degree less than dim V .11. Assume then that p(z) = (z − λ1 )(z − λ2 ) which implies that p(T ) = (T − λ1 I)(T − λ2 I) = 0. Hence x2 + αx + β must equal the characteristic polynomial of some Ai . so that they are isometries.5 dim null T n−1 ≥ n − 1. Assume then that F = R. 31 . Thus we have p(x) = (x2 + αx + β)k and from deg p ≤ dim V we get k ≤ (dim V )/2. 15. Hence p(z) = (z − λ1 )2 . We see that T 2 + αT + βI is not injective if and only if A2 + αAi + βI is not injective i for some 2 × 2 matrix Ai . We see that A satisfies the polynomial p(z) = (z − a)(z − d) − bc no matter if F is R or C. so that p is the characteristic polynomial by definition. From S ∗ S = I we see that each block Ai satisfy A∗ Ai = I. Assume then that T − λ2 I is invertible. This is impossible as the diagonal has only length 3. S is normal. We see that (α. Then dim null(T − λ1 I) = 2. Like in the previous exercise we know that the matrix [0] is at least n − 1 times on the diagonal and it’s impossible to fit a 2 × 2 matrix in the only place left.9 we can choose a basis such that the matrix of T has the form of 9. If F = C.25). has degree 2 and annihilates A it follows that p is the characteristic polynomial. Now if A has no eigenvalues. then p is the characteristic polynomial by definition. By proposition 8. then because p is monic. Assuming that T has an eigenpair we would have at least one 2 × 2 matrix on the diagonal. By theorem 9. Now we have the matrices [5] and [7] at least once on the diagonal. Hence a 2 × 2 block is of the form i cos θ − sin θ sin θ cos θ . so by the previous exercise it’s of the form (x − cos θ)(z − cos θ) + sin2 θ = x2 − 2x cos θ + cos2 θ + sin2 θ so that β = cos2 θ + sin2 θ = 1. so there’s an orthonormal basis of V such that S has a block diagonal matrix with respect to the basis and each block is a 1 × 1 or 2 × 2 matrix and the 2 × 2 blocks have no eigenvalue (theorem 7. 14.10. Hence (T + αT + βI)(dim V )/2 = 0. then by definition p is the minimal polynomial. vn )) = M(S. . (v1 . . which implies (a21 v2 + . . (v1 . . 3. . (v1 . . (v1 . . un ). . . . . . . . so that T v1 = a11 v. . (v1 . .10 Trace and Determinant 1. . . . . . . + an1 vn = 2(a21 v2 + . . . . . The map T → M(T. . . Let (v1 . . 2v2 . Both matrices represent an operator in L(Fn ). . . . We thus get a21 v2 + . The claim now follows trivially. . Now (v1 . = an1 = 0. 7. Then we can find an operator T ∈ L(Cn ) such that M(T. . + an1 vn = 0. (u1 . + an1 vn ). (v1 . Let (e1 . . The claim now follows from exercise 3. vn )) = I. . . . By theorem 10. Let T be the operator corresponding to the matrix 0 −1 1 0 . . 4. 0 −1 1 0 2 = −1 0 0 −1 . . vn )). . + λ2 ≥ 0 by theorem 10. . Now T has an upper-triangular matrix corresponding to some basis (v1 . en )) = B. . . Hence ST = I ⇔ M(ST. . .11. . . vn )).11 we have trace(T 2 ) = −2 < 0. . . (v1 . . . en ) be the standard basis for Cn . 0 λn Clearly trace(T 2 ) = λ2 + . . By linear independence we get a21 = . Follows directly from the definition of M(T. . . 5. 2. vn )) is bijective and satisfies ST → M(ST. . Then A−1 BA = M(T. . 2vn ) is also a basis and we have by our assumption T v = a11 v1 + 2(a21 v2 + . + an1 vn . . . (e1 . . . . .23. . . . Choose a basis (v1 . λn be the corresponding eigenvalues. vn ) be a basis of eigenvectors of T and λ1 . vn ))M(T. . . . . en )). . Let A = M((v1 . . vn ))M(T. . (v1 . . . . The claim clearly follows. . vn ) and let A = (aij ) be the matrix corresponding to the basis. 6. . vn ) of V . vn )) which is upper-triangular. . n 1 32 . . . Clearly A is an invertible square matrix. vn ).. . Then T v1 = a11 v1 + . . . . Then T has the matrix λ1 0 . (e1 . . vn )) = M(S. . . . (v1 . . . + an1 vn ). . Hence P v = P 2 w = P w = v. a1 . . . Hence trace P = dim range P ≥ 0. ei . 12. . . an . Now let ai. . .j . en )). Choose a basis of (v1 . v w. . . Then it’s easy to see that in BA the only non-zero diagonal element is ai. . . . (v1 . .j . . 9. . (v1 . ei = 0. so that A = 0 and hence T = 0.j = trace(ST ) = 0. Extend v/ v to an orthonormal basis (v/ v . . . an denote the diagonal elements of A. . Then clearly the matrix for P in this basis consists of a diagonal matrix with 1s on part of the diagonal corresponding to the vectors (v1 . en ) of V and let A = M(T. Let B be the matrix with 1 in row j column i and 0 elsewhere. . Then A = M(T. The trace of T is the sum of the eigenvalues. 14. . (v1 . Thus trace(BA) = ai.j be the element in row i. Hence λ − 48 + 24 = 51 − 40 + 1. v/ v = v/ v . . . Then we have ai = T ei . From exercise 5. vn ) for V and let A = M(T. By the spectral theorem we can find a basis (v1 . . + an = trace(T ). vn ) be some basis of V and let A = M(T. .21 we have V = null P ⊕range P . Now choose a basis (v1 . v w. an be the diagonal elements of A. so that A = 0. . . . . . . Let S be the operator corresponding to B. 33 . + can = c(a1 + . Then we have trace(cT ) = ca1 + . . + an = 0 implies a1 = . Now let v ∈ range P . vm ) and 0s on the rest of the diagonal. column j of A. . an are positive. . (v/ v . . e1 . Let a1 . . . 15. . = an = 0. By theorem 10. . Then the matrix of cT is cA. . so that the diagonal elements of A∗ are a1 . . . Let a. . . un ) of null P to get a basis for V . . . . + an ) = ctrace(T ). . ei = 0. vm ) for range P and extend it with a basis (u1 . Then v = P u for some u ∈ V . . . vn ) of eigenvectors of T . . an be the diagonal elements of A. . 11. . A positive operator is self-adjoint. . . v . . . . . . . . Let a1 . . . Hence a1 + . By proposition 6. . . It’s sufficient to prove that A = 0. . vn )) is a diagonal matrix and by positivity of T all the diagonal elements a1 . . . . + an = a1 + . Let (v1 . . . vn )). . . ei = so that trace(T ) = a = T v/ v .11 we have trace(T ∗ ) = a1 + . so that λ = 36. e1 . vn ) of V and let A = M(T. The example in exercise 6 shows that this is false. vn )). . v/ v = w. . . so that P is the identity on range P . . For another example take S = T = I. . Choose a basis (v1 . 13. (v1 . .8. vn )). . . . . It follows that ai. . 10.47 T ∗ has the matrix A∗ . . . 22. Positivity and definiteness follows from exercise 16 and additivity in the first slot from corollary 10. where |bii |2 = |λi |2 . . . . Now the ith element on the diagonal of B ∗ B is n bji bji = n |bji |2 . so T is normal.1 · · · caσ(n). . |λi |2 ≤ i=1 i=1 j=1 |bji |2 = trace(B ∗ B) = trace(A∗ A) = k=1 j=1 |ajk |2 . 19. . Choose a basis (v1 . Let T ∗ T ei = a1 e1 + . . so that trace(T ∗ T ) = T e1 The second assertion follows immediately. . Let T ∈ L(Cn ) be the operator corresponding to the matrix A. . Clearly ai is the ith diagonal element in the matrix M(T ∗ T. . . vi+k ). ei = T ei 2 by the ortogonality of (e1 . . 34 . . . vi+k ). . T ∗ v = (T ∗ T − T T ∗ )v. v . because the formulas only depends on the elements of the matrix. . Hence T ∗ T − T T ∗ is a positive operator (it’s clearly self-adjoint). . Let S = I and T = −I. T v − T ∗ v. en ). ei+k ). .16. so after doing this for all blocks we can assume that A is upper-triangular. . . . We can use the result of the next exercise which means that we only need to prove this for the complex case. . . (e1 . . spanning the same subspace such that Aj in this new basis is upper-triangular. These can be replaced with another set of vectors. + T en 2 . 23. Hence T ∗ T − T T ∗ = 0 by the spectral theorem. . λn on the diagonal. . . It follows that the formula defines an inner-product. . 20. .n = cn det A = cn det T 21. Then we have that ai = T ∗ T ei . The claim follows immediately. . 17. This follows trivially from the formula of trace and determinant for a matrix. Now each block Aj on the diagonal corresponds to some basis vectors (ei . vn ) such that T has an upper triangular matrix B = (bij ) with eigenvalues λ1 . . en )). Homogeneity in the first slot is exercise 13 and by exercise 10 S. S . We have T (span(vi . + an en . Hence j=1 j=1 n n n n n 2 + . so all the eigenvalues are 0. vi+k )) ⊂ span(vi . but trace(T ∗ T − T T ∗ ) = 0. T = trace(ST ∗ ) = trace((ST ∗ )∗ ) = trace(T S ∗ ) = T. . . Then we have det cT = det cA = σ caσ(1). (vi . .12. We have 0 ≤ T v. . 18. Let dim V = n and write A = M(T ). Then we have 0 = det(S + T ) = det S + det T = 1 + 1 = 2. . cz). Let Ω = {(x.n = det B = det T ∗ .1 · · · bσ(n). Then we have det T = det A = σ aσ(1). The second claim follows immediately from theorem 10. y. The first equality on the second line follows easily that every term in the upper sum is represented by a term in the lower sum and vice versa. y. It’s easy to see that T (Ω) is the ellipsoid. Let dim V = n and let A = M(T ). 35 .31.σ(1) · · · an. Hence 4 4 | det T |volume(Ω) = abc × π = πabc.σ(n) σ = σ a1. z) ∈ R3 | x2 + y 2 + z 2 < 1} and let T be the operator T (x. 3 3 which is the volume of an ellipsoid.σ(n) = = σ bσ(1). by. z) = (ax. 25.24.e.σ(1) · · · an.n a1. Now Ω is a circle with radius 1. bij = aji . so that B = M(T ) equals the conjugate transpose of A i.1 · · · aσ(n).
Report "Solutions to Axler Linear Algebra Done Right"