# Hilbert Schmidt Operators

by GSpeight
Tags: hilbert, operators, schmidt
 P: 31 Hi there, Can anyone give me an hint/idea of how to prove Hilbert-Schmidt operators are compact? More specifically, if X is a seperable Hilbert space and T:X->X is a linear operator such that there exists an orthonormal basis $(e_{n})$ such that $\sum_{n} ||T(e_{n})||^{2}<\infty$ then show that T is compact. It looks like an easy exercise given that both definitions are given in terms of sequences but I'm being quite stupid so I'm having trouble. Thanks for any help.
 Sci Advisor HW Helper P: 2,020 There is more than one way you can do this; perhaps the easiest is to show that T is a limit of finite ranks. Can you think of suitable ones?
 P: 31 Thanks for the help - sounds good. Obviously I'd have to show they converge in operator norm. Trying your hint now!
P: 31

## Hilbert Schmidt Operators

Done it - thanks so much for the hint :)

I have my functional analysis exam next Saturday (after complex analysis and manifolds on Monday and Friday) so I might post a few more questions some time if I have more trouble.

 Sci Advisor HW Helper P: 2,020 Good luck!
 P: 31 Thanks! Could anyone give me a hint as to how to prove the Hilbert-Schmidt norm, $||T||_{HS}=(\sum_{n\geq 1}||Te_{n}||^{2})^{1/2}$ is independent of the choice of orthonormal basis. I've tried taking another orthonormal basis $w_{n}$, writing $e_{n}=\sum_{k=1}^{\infty}(w_{k},e_{n})w_{k}$ so that $Te_{n}=\sum_{k=1}^{\infty}(w_{k},e_{n})Tw_{k}$ and considering $||Te_{n}||^2$ but only got that this is less than or equal to $\sum_{k=1}^{\infty}||Tw_{k}||^2$ which is obviously not helpful. Clearly I've applied too many inequalities (triangle inequality and cauchy-schwarz). Anyone have a better approach? Thanks for any help.
 Sci Advisor HW Helper Thanks P: 24,975 The Hilbert-Schmidt norm can be written as sqrt(trace(A.A*)) (where * is hermitian conjugate). Does that help you to prove it's basis independent?
 Sci Advisor HW Helper P: 2,020 I don't think this will solve the problem fully, because the usual definition of "trace" for an arbitrary operator on a (separable) Hilbert space will call out an o.n. bases and thus we're going to have to prove that this new value is independent of this choice of basis, see here. [GSpeight: consider this an extra exercise!] Alternatively, we can modify post #6 slightly: instead of writing $$e_{n}=\sum_{k=1}^{\infty}(e_{n},w_{k})w_{k},$$ write $$Te_{n}=\sum_{k=1}^{\infty}(Te_{n},w_{k})w_{k}.$$ Try to see if you can take it from here. No inequalities will be needed.
 Quote by morphism I don't think this will solve the problem fully, because the usual definition of "trace" for an arbitrary operator on a (separable) Hilbert space will call out an o.n. bases and thus we're going to have to prove that this new value is independent of this choice of basis, see here. [GSpeight: consider this an extra exercise!] Alternatively, we can modify post #6 slightly: instead of writing $$e_{n}=\sum_{k=1}^{\infty}(e_{n},w_{k})w_{k},$$ write $$Te_{n}=\sum_{k=1}^{\infty}(Te_{n},w_{k})w_{k}.$$ Try to see if you can take it from here. No inequalities will be needed.
 P: 31 Sorry for the slow reply. I've been busy revising a different topic and haven't really encountered the generalisation of trace to operators on Hilbert spaces. Firstly sorry if my order of writing terms in the inner product was confusing - for some reason my lecturer prefers the inner product to be linear in the second argument (even though the first time I met inner product spaces and the usual seems to be linear in the first argument). If we have $Te_{n}=\sum_{k=1}^{\infty}(Te_{n},e_{k})w_{k}$ then clearly it follows that $||Te_{n}||^{2}=\sum_{k=1}^{\infty}|(Te_{n},w_{k})|^{2}$ but even with adding up all the terms and interchanging the order of summation I don't see where factors $||Tw_{n}||^2$ come from, unless we assume the operator is self adjoint or something. Thanks for the help so far to both of you.