FALL 2011, SOLUTION SET 10 LAST REVISION: NOV 27, 9:45 AM. (T c f)(x) = f(x c).

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1 FALL 2011, SOLUTION SET 10 LAST REVISION: NOV 27, 9:45 AM TRAVIS SCHEDLER (1) Let V be the vector space of all continuous functions R C. For all c R, let T c L(V ) be the shift operator, which sends a function f V to the function T c f V defined by (T c f)(x) = f(x c). Let i be the imaginary number, i 2 = 1. (a) Show that the function f(x) = e kx is an eigenvector of T c with eigenvalue e kc. In particular, for e ikx = cos(kx) + i sin(kx) is an eigenvector of T c of eigenvalue e ikc = cos(kc) i sin(kc). (b) Now suppose that f(x) is a nonzero eigenvector under T c for all c R. Our goal is to prove the converse: f(x) = ae kx for some a, k C. Here, show the following: (b.1) f(x) is nonzero for all x; (b.2) If we rescale f so that f(0) = 1, then f(x + y) = f(x)f(y) for all x, y. (c) Now suppose also that, in addition to f being an eigenvector under T c for all c, also f is differentiable at 0; continue to assume f(0) = 1 as in (b.1). Show that, by f(x + y) = f(x)f(y) from part (b.2), it follows that f is differentiable everywhere, and f (x) = f (0)f(x). Then, by existence and uniqueness of solutions to differential equations (18.03 material) it follows that f(x) = e f (0)x, which proves the converse. (Note: in fact one can show that f(x) = ae kx without assuming that f is differentiable, but this is harder.) Solution: (a) T c (f)(x) = e k(x c) = e kc e kx = e kc f(x). The second statement follows immediately. (b) We need to assume f is a nonzero eigenvector: I forgot to write that in the problem. First (b.1): If f(x 0 ) = 0 for some x 0, then for all x 1, let λ x1 x 0 be the eigenvalue of f under T x1 x 0 ; then f(x 1 ) = T x1 x 0 (f)(x 0 ) = λ x1 x 0 f(x 0 ) = 0. So f(x) = 0 for all x, i.e., f is zero. (b.2) Rescale f as indicated. Then (T x f)(0) = f(x) = f(x)f(0). Since f is an eigenvector of T x, its eigenvalue is f(x), and hence (T x f)(y) = f(x)f(y) for all y. But (T x f)(y) = f(x + y), which gives the statement. (c) Differentiating f(x + y) = f(x)f(y) with respect to y, we get f (x + y) = f(x)f (y). Plug in y = 0, and we get f (x) = f (0)f(x), as desired. (2) Now let V be the vector space of functions {0, 1 n,..., n 1 n } C. Let us use the standard inner product on these functions: f, g = 1 n 1 f(j/n)g(j/n). n j=0 1

2 Let us continue to let i denote the imaginary number, i 2 = 1. Recall the (normalized!) discrete Fourier transform F : given f V, the function F (f) V is given by F (f)(k) = 1 n 1 ( ) j e 2kπi(j/n) f. n n j=0 This is slightly modified from PS 7, to replace n with n: we need this for part (c) below (this is the proper normalization). (a) Let δ m/n be the Kronecker delta function at m/n, i.e., δ m/n (k/n) = 0 if m k and δ m/n (k/n) = 1 if m = k. Show that F (δ m/n ) is the function 1 n e 2πimx/n, i.e., F (δ m/n )(k) = 1 n e 2πimk/n. (b) Use PS 7, #9.(a) (and show that ( nδ m/n ) is an orthonormal basis), to show that F is an isometry. (c) Verify the Fourier inversion formula: F 2 (f) is the function F 2 (f)(x) = f(1 x). I.e., f(1 (l/n)) = F 2 (f)(l/n) = 1 n 1 n 1 e 2lπik/n e 2kπij/n f n k=0 j=0 Hint: It is enough to do this for the case f = δ m/n for every m. Then the above amounts to showing that e 2lπix/n, e 2mπix/n = 1 if m = n l (i.e., if e 2mπix/n = e 2lπix/n ) and 0 otherwise. (d) From part (c), conclude that F 4 = I. From part (b), conclude that F has an orthonormal eigenbasis with eigenvalues in {i, i, 1, 1}, i.e., fourth roots of unity. Solution: (a) F (δ m/n )(k/n) = 1 n j e 2kπij/n δ m/n (j/n) = 1 n e 2πimk/n. (b) Note first that nδ l/n, nδ m/n = n 1 j=0 δ l/n(j/n)δ m/n (j/n), which is 1 if l = m and 0 otherwise. So ( nδ m/n : m {0, 1,..., n 1}) is indeed an orthonormal basis. Now in PS 7 # 9.(a), we verified that (e 2πimx/n : m {0, 1,..., n 1}) was an orthonormal basis (there we had m instead of m, but the two are the same since e 2πimx/n = e 2πi(n m)x/n for x {0, 1 n,..., n 1 n }). So F takes an orthonormal basis to an orthonormal basis, and hence (by a result from class and the book) it is an isometry. Remark 0.1. Finding a nice eigenbasis for the discrete Fourier transform is actually quite an interesting question! (These should resemble Gaussians, as happens for the continuous Fourier transform). There has been recent research on this area, with the idea being analogous to the shift operator problem: rather than find an eigenbasis just for the Fourier transform, one can take an eigenbasis which is simultaneously an eigenbasis under all the natural k-th roots of the Fourier transform, which eliminates most of the choices of eigenbasis involved and defines a special choice of eigenbasis. (Note that, in general, there are many different k-th roots of a normal operator, obtained by any k-th roots of the eigenvalues under any eigenbasis, but in this case it turns out that one can define in particular, special k-th roots for certain values of k which has close to distinct eigenvalues, i.e., whose eigenspaces are mostly 1-dimensional). (3) (i) Let V be a finite-dimensional complex inner product space. Show that the following are equivalent for T L(V ): (a) T is an isometry; (b) T is normal and T satisfies a polynomial whose roots all have absolute value one. 2 ( j n ).

3 (ii) Deduce from part (i) that, if T m = I for some m 1, then T is an isometry if and only if it is normal. Solution: (a) (b): If T is an isometry on a finite-dimensional inner product space, then T 1 = T (as we saw in class), and hence T T = T T = I. So T is normal. Also, we proved that T has an orthonormal eigenbasis (e 1,..., e n ) of eigenvalues (λ 1,..., λ n ) of absolute value one. Let p(x) = n i=1 (x λ i). Then p(t )(e i ) = 0 for all i, and hence p(t ) = 0. (b) (a): If T is normal then it has an orthonormal eigenbasis, say (e 1,..., e n ) of eigenvalues (λ 1,..., λ n ). If p(t ) = 0, then p(t )(e i ) = 0, so λ i must be a root of p(t ). Hence, if p(x) has only roots of absolute value one, then all the eigenvalues of T are of absolute value one. But we saw in class (for F = C on finite-dimensional inner product spaces) that an isometry is the same thing as an operator which has an eigenbasis of eigenvalues of absolute value one. (4) Positive operators! Let V be a finite-dimensional inner product space and S, T L(V ). (a) Show that, if S and T are positive invertible operators, then S + T is invertible. (Hint: See the lecture notes from class where we discussed the null space of S + T in this case). (b) Give a counterexample to the statement if S and T are only self-adjoint, i.e., give invertible self-adjoint S and T such that S + T is not invertible. (c) Give a counterexample to the statement that, if S and T admit eigenbases with positive (nonzero) eigenvalues, then S + T is invertible. Hint: Try two by two matrices: let S = T A and T = T B ; without loss of generality A ( could ) be diagonal, and it better not 1 0 be a multiple of the identity, so let us take A =. We want to find a B such 0 2 that (A + B)v = 0 for some nonzero v. Note that if v is an eigenvector ( of A, then it 1 won t happen that (A + B)v = 0. So find a B such that (A + B) = 0. For this, 1) ( ( ) 1 1 set v = and w = Av =. So we want B such that Bv = w. Show that 1) 2 there exists such a B with ( positive eigenvalues. Namely, M(T B, (v, w)) can be any 0 matrix with first column. Now fill in the second column so that both eigenvalues 1) are positive and distinct, using the characteristic polynomial. For this choice of matrix C = M(T B, (v, w)) (i.e., for B = (vw)c(vw) 1, where (vw) is the matrix with columns v and w), it follows that (A + B)v = 0, i.e., T A + T B is noninvertible, even though T A and T B both admit eigenbases with positive eigenvalues. In particular, part (c) shows that eigenbases do not behave the same as orthonormal eigenbases: if S and T both had orthonormal eigenbases, then they would be positive operators (since they have positive eigenvalues), and then part (a) would apply to show that S + T were invertible. Solution: (a) In class, we showed that null(s + T ) = null(s) null(t ), if S and T are positive. Hence, if null(s) = {0} or if null(t ) = {0}, then null(s + T ) = {0}. Therefore, if even only one of S or T is injective, so is S + T. Since V is finite-dimensional, S + T must be invertible. (b) Let S = I and T = I (and assume V {0}). Then S + T = 0 is not invertible, but S and T are both self-adjoint (since v, w = v, w for all v, w V ). 3

4 ( ) 1 0 (c) We follow the hint. Let A = and let B be such that M(T 0 2 B, (v, w)) = ( ) ( ( ) , where v = and w =. Note that the characteristic polynomial of this 1 3 1) 2 matrix is x 2 3x + 1, so its eigenvalues are 3± 5 2 which are positive and distinct. Also (A + B)(v) = Av + w = w + w = 0. We can explicitly compute B: ( ) ( ) ( ) 1 ( B = = ( ) 5 5 Note that A + B =, which indeed is noninvertible (5) Trace! For any matrix A Mat(n, n, F), with A = (a ij ), let its trace be defined as the sum of its diagonal entries, i.e., tr(a) = n i=1 a ii. (a) Prove that tr(ab) = tr(ba). [Note: this is in Axler, Ch. 10, if you want to read it there...] (b) Conclude that tr(bab 1 ) = tr(a) for all invertible B and all A. Conclude from the change-of-basis formula that, if T L(V ) for any finite-dimensional vector space V, then tr(m(t )) does not depend on the choice of basis. So tr(t ) := tr(m(t )) makes sense! (c) Now suppose that M(T ) is upper-triangular. Then the diagonal entries are the eigenvalues of T, with some multiplicities. We are going to prove in class that the number of times each eigenvalue λ occurs is independent of the choice of basis, and equals the dimension of the generalized eigenspace V (λ). For now, conclude from (b) that at least the sum of all the diagonal entries of M(T ) (i.e., the eigenvalues with multiplicity occuring in the matrix) is tr(t ), which is independent of the choice of basis. So for most choices of eigenvalues, there will be only one choice of multiplicities (the number of times each occurs on the diagonal of M(T )) that will add up to tr(t ). Solution: (a) This is an explicit verification: if A = (a ij ) and B = (b ij ), then tr(ab) = n n i=1 j=1 a ijb ji = tr(ba). Alternatively, we could note that A, B := tr(ab) is the same as the usual dot product of the matrices A and B considered as lists of elements. Then tr(ab) = A, B = B, A = tr(ba). (b) By (a), tr(bab 1 ) = tr((ab 1 )B) = tr(a) for all invertible B. The change-of-basis formula says that, if (w 1 w n ) = (v 1 v n )B for two bases (v 1,..., v n ) and (w 1,..., w n ) of V, then M(T, (w 1,..., w n )) = BM(T, (v 1,..., v n ))B 1. So this means that tr(m(t )) is independent of the choice of basis. (c) The trace of a matrix is the sum of its diagonal entries, so by (b) this makes sense for a linear transformation regardless of the choice of basis. In particular this is true for all choices of basis for which the matrix is upper-triangular. (6) Singular value decomposition and compression! Take a look at In particular, look at slides 5 12: these show you the approximation of an image by writing the image as a matrix, taking its singular value decomposition, and setting to zero all the singular values except for the largest, then the largest 10, then the largest 20, etc., as indicated. We are going to explain why this works. See also Muddy Responses 19, #8. Suppose that T L(V ), for dim V = n, is a transformation with SVD (e i ), (s i ), and (f i ). Now, let T be the same transformation, except with SVD (e i ), (s i ), and (f i), where s i = s i 4 ).

5 for the largest k singular values, and s i = 0 for the smallest n k values. Suppose that the smallest n k singular values are all bounded by ε 0. We are going to prove that T and T are very close, precisely, that the sum of the squares of the entries of any matrix for T T in an orthonormal basis is at most ε 2 (n k). On the other hand, the information required to store T is much less than that for T, since one need only store the top k s i and the corresponding e i and f i. For any matrix A = (a ij ), let A = i,j a ij 2. This is the direct analogue of the norm of vectors, as we will see. (a) Show that, in any orthonormal basis, A = tr(a t A). Thus, by the problem on traces (5) above, this makes sense for transformations: S := M(S), independent of the choice of orthonormal basis used. (b) Show that A 2 is the sum of the squares of the norms of the columns of A, and also the sum of the squares of the norms of the rows of A. (c) Show that, if S is an isometry and T any transformation, then ST = T = T S. Hint: explain why this is the same as showing that UA = A = AU where A is any matrix and U is a unitary matrix. Note that the columns of UA all have the same norm as the columns of A, and similarly that the rows of AU have the same norm as the rows of A. Conclude the result (using (b)). (d) Conclude that, if T has singular value decomposition (e i ), (s i ), and (f i ), then T = i s2 i. (e) Deduce that, in the original situation of the problem, T T ε n k. This says, informally, that T and T are very close. Solution: (a) The i-th diagonal entry of A t A is the dot product of the i-th column of A with itself ( n j=1 a jia ji ). The sum of all of these, over i, is i,j a ij 2 = A 2. (b) We already explained the fact about the columns in (a). For the rows, the same proof works: the square of the norm of the i-th row is n j=1 a ij 2, so the sum of this over i is A 2. (c) Since Su = u when S is an isometry and u any vector, taking orthonormal bases shows that Uv = v for all unitary matrices U (a unitary matrix is the same thing as the matrix of an isometry in an orthonormal basis). Now, UA 2 is the sum of the squares of the norms of the columns of UA, and this is the same as the sum of the squares of the norms of the columns of A, i.e., A 2. The same works for AU by noting that the norms of the rows of AU are the same as the norms of the rows of A. Now, T = M(T ) in any orthonormal basis (by definition) and since matrices of isometries in orthonormal bases are unitary, and M(ST ) = M(S)M(T ) for all S and T, we deduce the result. (d) If T = SP where S is an isometry and P is positive, then by part (c), T = P. Now since P is positive, there is an orthonormal basis in which P is diagonal with nonnegative entries, and these entries are defined to be the singular values. Thus P 2 = i s2 i, where the s i are the singular values. (e) In the original situation, T T has singular values which are the lowest n k singular values of T (all of absolute value at most ε) and k zeros. So, by (d), T T 2 (n k)ε 2, and taking a square root gives the desired result. (7) Block upper triangular matrices! 5

6 Let F = R. We will show that, for the 4 4 matrix A below, T A has no basis in which it is block diagonal: A = (a) Show that the complex eigenvalues of A are ±i. (b) Show that the complex eigenspaces of each of the eigenvalues, i and i, are onedimensional. (c) Now suppose that in some (real) basis, M(T A ) were block diagonal. Conclude that, in this case, BAB 1 must be block diagonal with 2 2 blocks, each of which have complex eigenvalues ±i. Conclude that BAB 1, and hence A, has eigenspaces of i and i of dimension two, not one. This contradicts (b). Solution: (a) Since this matrix is block upper-triangular, its eigenvalues are the eigenvalues of the diagonal blocks, which in this case we already computed many times to be ±i (or one can apply the characteristic polynomial, which for this 2 2 matrix is x 2 + 1). (b) This means showing that null(a ii) and null(a+ii) are one-dimensional. Note that ther are at least one-dimensional, since (1, i, 0, 0) t is in the first null space, and (1, i, 0, 0) t is in the second null space. We claim that all null vectors are multiples of these. Since the eigenspaces of the upper-left 2 2 block are one-dimensional, all null vectors ending in two zeros are of this form. For a contradiction, suppose we had a null vector whose last two entries were not both zero. The last two entries must be in the null space of the bottom right 2 2 block, so up to scaling the last two entries are (1, i) (since they are not both zero). Now, for (a, b, 1, i) to be in the null space, we need that ( b ia + 1, a ib i) = (0, 0). This is impossible, since it would require that i( b ia + 1) + (a ib i) = 0, but this is 2i 0. So the null spaces are one-dimensional. (c) If, in some real basis, A could be put in block diagonal form, then the two blocks would have to still have eigenvalues ±i, since these are all the eigenvalues of A, and any two by two real matrix with nonreal eigenvalues has complex eigenvalues which are conjugates of each other. But then, since the new matrix A is block diagonal, its eigenspaces of eigenvalue i and i would each be two-dimensional, since if (a, b) were a nonzero eigenvector of eigenvalue i of the first block, and (c, d) a nonzero eigenvector of eigenvalue i of the second block, then (a, b, 0, 0) and (0, 0, c, d) would be linearly independent eigenvectors of the matrix A. This can t happen, since the eigenspaces of A must be the same dimension as those of A (these matrices can both be realized as matrices of the same linear transformation, and the dimension of the eigenspaces of a linear transformation make sense independent of the choice of basis). (8) Minimal polynomial! Given a transformation T L(V ) for V finite-dimensional, the minimal polynomial is defined to be the monic polynomial p(x) of minimal degree (monic means that the leading coefficient is one) such that p(t ) = 0. (a) Show that this definition makes sense: if p(x) and q(x) were two monic polynomials of the same minimal degree such that p(t ) = q(t ) = 0, show that p(x) = q(x) (Hint: consider h(x) := p(x) q(x), and note that h(t ) = 0.) (b) Now, suppose that p(x) is the minimal polynomial of T, and q(t ) = 0. Then show that p(x) q(x), i.e., q(x) = h(x)p(x) for some polynomial h(x). (Hint: using long division of polynomials, write q(x) = h(x)p(x) + r(x) where the degree of r(x) is less than the degre of p(x), and then conclude that r(x) = 0 since q(t ) = p(t ) = 0.) 6

7 (c) Next, suppose that T has an eigenbasis and its (distinct) eigenvalues are λ 1,..., λ k. Show that p(x) = (x λ 1 ) (x λ k ). (d) In the case F = C, recall from class the decomposition theorem that V = λ V (λ) where each V (λ) is the generalized eigenspace of λ, i.e., the span of vectors v such that (T λ) m v = 0 for some m 1. Now for each λ, explain why there exists a minimal m λ such that (T λ) m λv = 0 for all v V (λ) (hint: take a basis of V (λ) and take the maximum value of m among the basis). (e) Finally, explain that the minimal polynomial in the situation of (d) is p(x) = λ (x λ) m λ. Solution: (a) Indeed, following the hint, if p(x) and q(x) were monic polynomials of the same degree and p(t ) = 0 = q(t ), then h(x) := p(x) q(x) would have strictly lower degree, and also h(t ) = 0. So if p and q had minimal degree for nonzero polynomials with the properties p(t ) = 0 = q(t ), then we would conclude h(x) = 0, i.e., p(x) = q(x), as desired. (b) By long division, we can write q(x) = h(x)p(x) + r(x), for some polynomial h(x), and for r(x) a polynomial of degree less than the degree of p. Since q(t ) = 0 and p(t ) = 0, we conclude that r(t ) = 0. Since p is the minimal polynomial, this can only imply that r(x) = 0 (otherwise we could rescale r to be monic and get a contradiction to the definition of p). So q(x) = h(x)p(x), i.e., p(x) q(x). (c) Let (v 1,..., v n ) be the eigenbasis mentioned. It is clear that p(t )v i = 0 for all i, since p(t )v i = p(λ)v i = 0 where λ is the eigenvalue of v i. Conversely, suppose q(x) is a polynomial such that q(t )v i = 0 for all i. Then q(λ)v i = 0 for the eigenvalue λ of v i, so we conclude that λ must be a root of q. Hence p(x) q(x), since p(x) is just the product of the linear factors (x λ j ), where λ j ranges over all eigenvalues. This implies that the degree of p(x) is less than or equal to the degree of q(x), so p(x) is the minimal polynomial of T. (d) As explained in class, take a basis of V (λ), and let m be large enough so that each of the basis vectors are in null(t λi) m (this must exist since there are finitely many basis vectors, and each of them are in the null space of some power of T λi since they are generalized eigenvectors of eigenvalue λ. Now, we can let m λ be the minimal possible value of m with this property. (e) Now, (T λi) m λ is zero on V (λ) for all λ, so the given polynomial p(x) indeed satisfies p(t ) = 0, since p(t ) V (λ) = 0 for all λ and V = λ V (λ). For the converse, suppose that p(x) were not minimal. Since p(t ) = 0, then p(x) = h(x)q(x) by (b) for q(x) the minimal polynomial and h(x) some product of powers of the (x λ) s. So q(x) = λ (x λ)m λ where m λ m λ, but they are not all equal. Let λ be an eigenvalue for which m λ < m λ. Then (T λi) m λ is not zero on V (λ). On the other hand, as shown in class, for µ λ, (T µi) is an isomorphism on V (λ), since V (λ) cannot have any nonzero eigenvectors of eigenvalue µ (otherwise such an eigenvector v would satisfy (T λi) m λv = (µ λ) m λv 0, a contradiction). So since q(x)/(x λ) m λ is a product of factors of the form (x µ) for µ λ, we conclude that q(t ) is nonzero on V (λ). (9) Square roots! (cf. Axler p. 177): We will characterize those linear transformations T L(V ), for F = C and V finitedimensional, such that T admits a square root S such that S 2 = T. (a) By the decomposition theorem, V = λ V (λ) where the V (λ) are generalized eigenspaces. Show that T admits a square root if and only if T V (λ) does for all λ. Hint: If Sλ 2 = T V (λ) for all λ, then λ S2 λ = T, where here Sλ 2 means Sλ 2(v) = λ S2 λ (v). 7

8 Conversely, if S 2 = T, show that S(V (λ)) V (λ) for all λ, since ST = T S and hence S(T λi) m = (T λi) m S for all λ. Deduce that, if S 2 = T, then S 2 V (λ) = T V (λ). (b) Now, we reduce to the case that V = V (λ), i.e., T λi is nilpotent (in some basis, it is upper-triangular with zeros on the diagonal, so that taking it to the power of the size of the matrix will be zero). In other words, we may assume T = λi + N, where N is nilpotent, i.e., N m = 0 for some m 1. Suppose λ 0. Let µ C be a complex square root of λ, i.e., µ 2 = λ. Then T = µ 2 I + N. Let N = N/µ 2, so T = µ 2 (I + N ). Now, let f(x) be the Taylor polynomial of 1 + x up to degree m + 1, i.e., f(x) = x 1 8 x x3 ( 1) + m 1 (2(m 1))! x m 1 (cf. (1 2(m 1))((m 1)!) 2 4 m 1 org/wiki/square_root#properties). In particular, f(x) 2 = 1 + x + terms of degree m (you don t have to prove this, but it follows from the fact that f(x) is the Taylor expansion of 1 + x). Show that the following matrix S satisfies S 2 = T : S := µ f(n ) and don t forget that (N ) m = 0. Hint: Use that f(x) 2 = 1 + x + a multiple of x m. (c) Finally, let us suppose that λ = 0, i.e., T itself is nilpotent. In this case, some but not all T admit a square root: give one nonzero example where T admits a square root, and one example where T does not admit a square root. (Hint: look for upper-triangular matrices of size 3 3 or smaller.) (a) We follow the hint. Suppose that S λ L(V (λ)) satisfy Sλ 2 = T V (λ) for all λ. Let S = λ S λ L(V ) denote the operator which takes λ v λ to λ S λ(v λ ) for all choices of v λ V (λ). This makes sense by the decomposition theorem: V = λ V (λ). Then, S 2 (v λ ) = T (v λ ) for all v λ V (λ) for all λ, and hence since V = λ V (λ), we conclude that S 2 = T. Conversely, suppose that S 2 = T. Then S 2 V (λ) = (S V (λ) ) 2 = T V (λ) for all λ. So each of T V (λ) has a square root as well. (b) Clearly, it suffices to show that f(n ) 2 = 1 + N. But f(n ) 2 = g(n ) where g(x) = f(x) 2. And, as pointed out, g(x) = 1 + x+ a multiple of x m. Thus g(n ) = I+N +(N ) m ( ) where is something. Since (N ) m = 0, this proves that g(n ) = I+N, as desired. ( ) 0 1 (c) The matrix N = has no square root: if it did, call it S, then S = N 0, but S 4 = 0. However, since S 4 = 0, S could not be injective. Since S 2 0, also S 0. So rk(s) = 1. But rk(s 2 ) = rk(n) = 1 as well. So this would imply that range(s) = range(s 2 ) is one-dimensional, and hence that S range(s) is an isomorphism. But then all (positive) powers of S would have the same range, range(s), which is nonzero. So this contradicts S 4 = 0. On the other hand, we can easily come up with a nilpotent matrix that does admit a square root, by taking the square of a nilpotent matrix which does not square to zero. For example, = So the nilpotent matrix does admit a square root

9 (10) Determinants and characteristic polynomial! For the following definition, see also p. 226 of Axler. Definition 0.2. A permutation of {1,..., n} is a bijective function σ : {1,..., n} {1,..., n}. Let S n be the set of all permutations of {1,..., n}: so we say σ S n if σ is a permutation of {1,..., n}. Equivalently, a permutation is a list (σ(1),..., σ(n)) of positive integers that contains each positive integer between 1 and n inclusive exactly once. Caution: Axler uses the notation (m 1,..., m n ) = (σ(1),..., σ(n)). I am opting for the more standard notation, but the two notions are equivalent. For the following definition, see also p. 228 of Axler. Definition 0.3. The sign of a permutation σ S n, denoted by sign(σ), is defined to be ( 1) o(σ), where o(σ) is the number of pairs (i, j) with 1 i < j n such that σ(i) > σ(j). In other words, the sign of a permutation is the number of pairs i < j that appear in reverse order in the list (σ(1),..., σ(n)). We will need the following proposition, which will be discussed in class: Proposition 0.4. If σ, τ S n, then sign(σ) sign(τ) = sign(σ τ). Note that the proof of this proposition boils down to Lemma Please use this proposition, without proving it, in the below. (You are, however welcome to try to prove it if you like and/or to see Axler.) For the following definition, see also Axler, 10.25: Definition 0.5. Let us define the determinant of a matrix A = (a i,j ) by the formula (0.6) det(a) = σ S n sign(σ)a σ(1),1 a σ(2),2 a σ(n),n. Note that the sum is over n! different things, since S n has n! elements. Prove the following properties: (a) If A has two identical columns, then det(a) = 0. (b) Suppose A has columns (v 1 v n ) and B has columns (v 1 v k 1 w k v k+1 v n ), i.e., A and B have the same columns except for the k-th column. Finally, let C := (v 1 v k 1 (v k + w k )v k+1 v n ), i.e., the same as A and B except in the k-th column, where one takes the sum of the columns v k and w k. Prove: det(a) + det(b) = det C. (b ) Note also that, if instead we had w k = cv k for some c F, then det(b) = c det(a). (c) Deduce from (a) and (b) (and (b )) that, if A is not invertible, then det(a) = 0. Hint: Write one of the columns, say the k-th column, as a linear combination of the other columns, and apply (b) iteratively as well as (a). (d) Conclude that, if λ is an eigenvalue of A, then det(a λi) = 0. In other words, λ must be a root of the characteristic polynomial of A, defined as det(a xi). Equivalently, if λ is not a root of the characteristic polynomial, then A λi is invertible. Solution: 9

10 (a) First, we deduce Lemma from the proposition (or you could just cite the lemma). If τ is obtained from σ by swapping the values of i and i + 1, i.e., σ(i + 1), if j = i τ(j) = σ(i), if j = i + 1 σ(j), if j / {i, i + 1} then τ = σ κ i,i+1 where κ i,i+1 is the permutation which swaps i and i + 1 and leaves everything else fixed. Since, o(κ i,i+1 ) = 1, it follows that sign(κ i,i+1 ) = 1, and by Proposition??, sign(τ) = sign(σ). More generally, if τ is obtained from σ by swapping the values of i and j, then τ = σ κ i,j where κ i,j is the permutation which swaps i and j and leaves everything else fixed. Note that κ i,j is conjugate to κ i,i+1 : for instance, κ i,j = κ i+1,j κ i,i+1 κ i+1,j, where κ i+1,j = κ 1 i+1,j since its square is the identity. Hence, sign(κ i,j) = sign(κ i+1,j ) 2 sign(κ i,i+1 ) = 1 for all i and j. So, the argument above shows that sign(τ) = sign(σ), as desired. Now, using Lemma 10.23, the result follows easily: if column i equals column j in the matrix A, then every term sign(σ)a σ(1),1 a σ(2),2 a σ(n),n in the expression of det A cancels with sign(τ)a τ(1),1 a τ(2),2 a τ(n),n, τ = σ κ i,j. So det A = 0. (b) Let (b ij ) and (c ij ) be the entries of B and C. Then, det C = σ S n sign(σ)c σ(1),1 c σ(2),2 c σ(n),n = sign(σ)a σ(1),1 a σ(k 1),k 1 (a σ(k),k + b σ(k),k )a σ(k+1),k+1 a σ(n),n σ S n = sign(σ)a σ(1),1 a σ(n),n + sign(σ)b σ(1),1 b σ(n),n. σ S n σ S n (b ) This is similar: all of the summands of det B have one term from the k-th column in them, so each summand is c times the corresponding summand of det A. So det B = c det A. (c) If A is not invertible, then one of the columns must be in the span of the other columns. By parts (b) and (b ), det A is then a linear combination of determinants of matrices in which this column is replaced by one of the other columns, i.e., two columns are identical. By part (a), each of these determinants is zero. Hence, det A = 0. (d) If λ is an eigenvalue of A, then A λi is noninvertible. By (c), this implies that det(a λi) = 0. Hence, λ must be a root of the polynomial det(a xi). 10