Math Solutions to homework 5

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1 Math 75 - Solutions to homework 5 Cédric De Groote November 9, 207 Problem (7. in the book): Let {e n } be a complete orthonormal sequence in a Hilbert space H and let λ n C for n N. Show that there is a bounded linear operator D on H such that De n λ n e n for all n N, if and only if {λ n } is a bounded sequence. When is D, when defined? Solution: Assume first that D is bounded. Then, λ n λ n e n De n D, hence {λ n } is a bounded sequence. Conversely, assume that {λ n } is bounded. Then, for x ne n x H, we have Dx x n λ n e n which proves that D is bounded if sup n λ n <. x n 2 λ n 2 sup n λ n x n 2 sup λ n x, n The above shows that D sup n λ n, if defined. We will prove that D sup n λ n by showing the reverse inequality. Simply observe that De n λ n e n λ n, which shows that D λ n for every n. This proves that D sup n λ n. Problem 2 (7.39 in the book): Let A, B L(E), for E a Banach space. Show that σ(ab) and σ(ba) coincide, except possibly for the point 0: that is, Give an example in which σ(ab) σ(ba). σ(ab) \ {0} σ(ba) \ {0}. Solution: Let λ 0, and suppose that λ / σ(ab), ie λi AB is invertible. Then, a simple computation shows that λ I + B(λI AB) λ A is an inverse of λi BA, hence λi BA is invertible, which means that λ / σ(ba). So, we proved that σ(ba) \ {0} σ(ab) \ {0}.

2 Inverting the roles of A and B above, we see that this inclusion is actually an equality. For an example in which σ(ab) σ(ba), take E l 2 (N) (square-summable sequences), take A to be the left shift and B to be the right shift: A(x, x 2, ) (x 2, x 3, ) B(x, x 2, ) (0, x, x 2, ). Then AB I while BA(x, x 2, ) (0, x 2, x 3, ). Hence AB is invertible but not BA, proving that 0 σ(ba) but 0 / σ(ab). Problem 3 (7.34 in the book): Find the spectrum of the operator D of Problem 7., when {λ n } is a bounded sequence. Solution: We claim that σ(d) {λ, λ 2, }. First, we clearly have {λ, λ 2, } σ(d). Indeed, for any k N, we have (λ k I D)(e k ) λ k e k λ k e k 0, proving that e k Ker (λ k I D). So, λ k I D is not invertible, hence λ k σ(d). Furthermore, since the spectrum of an operator is always closed, we must have {λ, λ 2, } σ(d). Conversely, let λ / {λ, λ 2, }. Then, λi D is the operator associated to the sequence {λ λ, λ λ 2, }. Clearly, since none of these numbers are 0, we can (set-theoretically) invert that operator: the inverse is given by D (e i ) λ λ i e i. Now, we need this inverse to be bounded. This happens if and { } only if (by problem ) the sequence λ λ n is bounded, which is the case: since λ / {λ, λ 2, }, there is some small number ε > 0 so that each λ λ n is larger than ε, hence each λ λ n < ε. This proves that λ / σ(d). 2

3 Problem 4 (8.2 in the book): Let E be an infinite-dimensional Hilbert space and let A, B L(E). Which of the following are true? (a) If AB is compact then either A or B is compact. (b) If A 2 0 then A is compact. (c) If A n I for some n N, then A is not compact. Solution: (a) False. Take E H H for some infinite-dimensional Hilbert space H. Take A to be the projection onto the first factor, and B to be the projection onto the second factor. Then AB 0 (which is clearly compact), but neither A nor B is compact: take any bounded nonconvergent sequence {h n } n in H that does not have a convergent subsequence (for example, make h n run through an orthonormal basis of H); then {(h n, 0)} is a sequence in E H H whose image under A is itself (and it has no convergent subsequence). Same for B. (b) False. Take the same example as above, and A(h, h ) (h, 0). For the same reason as above, this is not compact, but we have A 2 0. (c) True. Take an orthonormal basis {e n } of E. If {Ae n } admits a convergent subsequence, then {A n e n } {e n } should also admit one, since A is continuous (hence sends convergent sequences to convergent sequences). But {e n } does not admit a convergent subsequence. Problem 5 (8.5 in the book): Let {e n }, {f m} be complete orthonormal sequences in Hilbert spaces H, K respectively and let A L(H, K). Show that A f m 2. Deduce that the quantity Ae n 2 has the same value for every choice of the complete orthonormal sequence {e n } in H. ( The quantity Ae n 2) /2, if finite, is called the Hilbert-Schmidt norm of A. Show that it equals /2 a ij 2 i,j if A has matrix [a ij ] with respect to any pair of complete orthonormal sequences in H and K. Solution: By completeness of {e n }, we have A f m A f m, e n e n A f m 2 A f m, e n 2. 3

4 Similary, by completeness of {f m }, we have Ae n Ae n, f m f m Ae n, f m 2. Therefore: Ae n, f m 2 e n, A f m 2 A f m, e n 2 A f m 2 We used the definition of A at the second line. At the third line, we exchanged the summations and we used the fact that v, w w, v w, v. Therefore, for another complete orthonormal sequence {e n} in H, we have A f m 2 Ae 2 n, since we can repeat the argument above with {e n}. This proves that the quantity Ae n 2 does not depend on the choice of complete orthonormal sequence. As we saw above, Ae n, f m 2 ; the numbers Ae n, f m are precisely the a mn, i.e. the coefficients of the matrix of A. Problem 6: Let K be a non-empty, compact subset of C. Show that there exists a Hilbert space H and a bounded linear operator T L(H, H) such that σ(t ) K. [Hint: you may use that any subset of a separable metric space is separable.] Solution: Since K is separable, there exists points {λ, λ 2, } K such that this set is dense in K, i.e. the closure of {λ, λ 2, } in K is K. Since K is closed in C, the same is true in C. Take T to be the operator D of problem ; it is bounded because {λ n } is bounded, since K compact (and hence bounded). We proved that σ(t ) {λ, λ 2, } in problem, which proves that the spectrum of T is K. 4

5 Problem 7: Consider the bounded operator D defined in problem 7.. Show that D is compact if and only if λ n 0. Solution: The proof that D is compact if λ n 0 is done in Example 8.2.(iii) in the book. In short: approximate D by D n, where D n is the same diagonal operator as D, except that all diagonal entries λ k are 0 for k > n. Then, by problem, D D n sup k>n λ k, which goes to 0 as n if λ n 0. Furthermore, all the D n s have finite rank. This proves that one can approximate D in the norm topology by finite rank operators, hence D is compact. Conversely, assume that D is compact. Then, since the sequence {e n } is bounded in H, and since D is compact, the sequence {De n } {λ n e n } must have a convergent subsequence. But λ m e m λ n e n λ 2 m + λ 2 n can go to 0 only if λ n 0. Problem 8: Let H be a Hilbert space and {e n } a complete orthonormal sequence. Define T : H H by T (e i ) i e i+. Show that T is compact but has no eigenvalues. (So the spectral theorem is false if T is not self adjoint.) Solution: We can prove it is compact in (at least) two ways: Notice that it is the composition of the operator D of problem (with λ n n ) and the right shift. Since problem 7 shows that D is compact, and since compact operators form a two-sided ideal in the ring of all bounded operators, T must be compact as well. (this means that the composition of a compact operator with any other operator is compact; it is easy to prove). It is an Hilbert-Schmidt operator: T e n 2 <. Hence by Theorem 8.7 in the n 2 book, T is compact. The proof of this theorem is basically the same as the first paragraph of the solution to problem 7 above. Now, assume that T has an eigenvector x x ne n, with eigenvalue λ C. Then T x λx, so λ x n e n λx T x x n n e n+ n2 x n n e n, where we simply shifted the summation index in the last equality. Identifying the coefficients of each e n, we get { λx 0 λx n x n n for n >. If λ 0, then the first equation force x to be 0, and then the second family of equations forces all the x n s to be 0, by induction. If λ 0, then all the x n s are 0 by the second family of equations. In any cases, this proves that x 0, hence λ is not an eigenvalue. So, T does not have any eigenvalue. 5