We have just covered Hilbert-Schmidt operators in class (which I missed) and I am having a hard time understanding them. I know the definition:

If $\;H,F\;$ are Hilbert spaces and $\;T \in \mathcal B(H,F)\;$ then $\;T\;$ is Hilbert-Schmidt if there is some orthonormal basis of $\;H\;$ such that: $\; \sum_{n=1}^{\infty} {\vert \vert T(e_n)\vert \vert}^2 <∞\;$

An easy calculation shows that this definition does not depend on the choice of basis.

I understand why $\sum_{n=1}^{\infty} {\vert \vert T(e_n)\vert \vert}^2 \;=\;\sum_{k=1}^{\infty} {\vert \vert T^*(f_k)\vert \vert}^2 \;$ where $\; \{ e_n \} \;,\; \{ f_k \}\;$ are orthonormal basis on $\;H,F\;$ respectively. But I have trouble understanding why this shows that the operator is well defined.

I believe I should have $\sum_{n=1}^{\infty} {\vert \vert T(e_n)\vert \vert}^2 \;=\;\sum_{n=1}^{\infty} {\vert \vert T(e_n)'\vert \vert}^2 \;$ where $\; \{ e_n \} \;,\; \{ e_n \}'\;$ are orthonormal basis on $\;H\;$ in order to claim the independence of the choice of basis.

Is this equal to the above statement somehow? I would appreciate any help!

Thanks in advance

  • $\begingroup$ We are assuming the operator is already defined, and then it is called a Hilbert-Schmidt operator under these conditions. $\endgroup$ – Jason May 7 '17 at 16:44
  • $\begingroup$ @Jason Ok but I still don't undestand why the definition does not depend on the choice of basis... I 'll edit the title of my post $\endgroup$ – kaithkolesidou May 7 '17 at 16:47

Let me give a somewhat analogous example:

Definition A sequence of real numbers $(x_n)$ is called convergent if there exists a real number $x$ such that for every $\varepsilon>0$, there exists $n_0$ such that $|x_n-x|<\varepsilon$ for all $n>n_0$. The number $x$ is called a limit of the sequence $(x_n)$.

A simple result is that limits are unique, so we can call $x$ the limit of $(x_n)$, but this is not necessary to the definition of convergence. Indeed, there are spaces where limits need not be unique. If all we are interested in is discussing the convergence or non-convergence of a sequence, the particular limit is unimportant - all we care about is that some limit exists.

Now, how does that relate here?

Definition An operator $T\in\mathcal B(H,F)$ is called a Hilbert-Schmidt operator if there exists an orthonormal basis $(e_n)$ of $H$ such that $\sum_n\|Te_n\|^2<\infty$.

We could continue that definition with something along the lines of "The basis $(e_n)$ is called a Schmidt basis of $T$." (Note: I have made up the term Schmidt basis.) However, rather than uniqueness of Schmidt bases, we have the sort-of opposite result: if $T$ is a Hilbert-Schmidt operator, then every basis $(f_n)$ of $H$ is a Schmidt basis of $T$. (Hence the term Schmidt basis becomes redundant.)

There is no issue with well-definedness; an operator $T$ is a Hilbert-Schmidt operator if it satisfies the given property for some orthonormal basis, but it is a straightforward result that if $T$ satisfies the property for some basis, then it satisfies it for every basis.

EDIT: Let us show that $\sum_{i\in I}\|Te_i\|^2=\sum_{i\in I}\|Tf_i\|^2$ for any two bases $(e_i)$, $(f_i)$. First note that we may write the left hand side as $\sum_i\langle T^*Te_i,e_i\rangle=:\operatorname{Tr}(T^*T)$. Note that if $A,B\in\mathcal B(H)$ are operators such that both $AB$ and $BA$ are self-adjoint and positive definite, then $$\operatorname{Tr}(AB)=\sum_i\left\langle A\left(\sum_j\langle Be_i,e_j\rangle e_j\right),e_i\right\rangle=\sum_{i,j}\langle Ae_j,e_i\rangle\langle Be_i,e_j\rangle=\operatorname{Tr}(BA)$$ by symmetry. If $U\in\mathcal B(H)$ is unitary, then both $T^*T$ and $U^*T^*TU$ are self-adjoint and positive definite, and thus $$\operatorname{Tr}(U^*T^*TU)=\operatorname{Tr}(T^*TUU^*)=\operatorname{Tr}(T^*T).$$ In particular, if $U$ is the unitary operator such that $f_i=Ue_i$ for each $i$, this implies $$\sum_i\|Tf_i\|^2=\sum_i\langle TUe_i,TUe_i\rangle=\operatorname{Tr}(U^*T^*TU)=\operatorname{Tr}(T^*T)$$ as claimed.

  • $\begingroup$ First of all, thanks a lot for your time and your answer... I can say you convinced me about the "well-definedness" of the operator. However I feel a bit unsure about the straightforward result you mentioned. I guess I'm really silly but if this holds $\;\sum_{n=1}^{\infty} {\vert \vert T(e_n)\vert \vert}^2 \;=\;\sum_{k=1}^{\infty} {\vert \vert T^*(f_k)\vert \vert}^2 \;$ then how can I claim that $\;T:H \rightarrow F\;$ satisfy the property for every basis on $\;H\;$? All I have in this case is a basis on $\;F\;$ and the adjoint of $\;T\;$, not $\;T\;$.. I hope I didn't confuse you. $\endgroup$ – kaithkolesidou May 7 '17 at 18:38
  • $\begingroup$ Ah, I see. I assumed when you wrote "An easy calculation shows..." that you had already done this. I will add it to my answer. $\endgroup$ – Jason May 8 '17 at 12:48
  • $\begingroup$ Your edit was extremely helpful. However I've got one more question.. How do I know $\;T^*T\;$ and $\;U^*T^*TU\;$ are self adjoint? Shouldn't be $\;T^*TU\;$ and $\;U^*\;$ instead of the ones above? It's a bit unclear to me.. Thank you $\endgroup$ – kaithkolesidou May 10 '17 at 17:36
  • $\begingroup$ Calculate the adjoint! Recall that $(AB)^*=B^*A^*$. In particular, $(A^*A)^*=A^*(A^*)^*=A^*A$. Apply this with $A=T$ and $A=TU$. (The only reason I used the positive definite, self-adjoint assumption is that it guarantees the series are defined; it is not really necessary.) $\endgroup$ – Jason May 10 '17 at 18:14
  • $\begingroup$ I omitted the obvious... Calculations! You've been very helpful!! Thanks $\endgroup$ – kaithkolesidou May 10 '17 at 19:56

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