# Relation between Hilbert-Schmidt inner product and tensor products

Suppose $$A$$ is a Hilbert-Schmidt operator on a Hilbert space $$\mathcal{H}$$, then when is it true that

$$\langle Ax, x \rangle_{\mathcal{H}} = \langle A, x \otimes x \rangle_{\text{HS}}, \quad \forall \, x \in \mathcal{H}$$

I know from the definition of Hilbert-Schmidt inner product and the definition of tensor product that

$$\langle A, x \otimes x \rangle_{\text{HS}} = \sum_{j \in J} \langle A e_j, \langle x, e_j \rangle_{\mathcal{H}} x \rangle_{\mathcal{H}}$$ where $$\{e_j\}_{j \in J}$$ is an arbitrary ONB of $$\mathcal{H}$$. However, I am unable to simplify further. Any help would be appreciated.

If the above result is incorrect, in general, I am interested in going from an expression of the form $$\langle Ax, x \rangle_{\mathcal{H}}$$ to an expression of the form $$\langle A, x \otimes x \rangle_{\text{HS}}$$. How could I do that?

Also, I would really appreciate some references where I can look into these kind of results in more detail.

• Shouldn't this follow simply from a resolution of the identity? You have outer products of $e_j$ in there which sum to $I$. Commented Apr 4, 2021 at 4:08
• I am sorry, but I am very new to functional analysis. Could you be more detailed about what you mean by the outer product of $e_j$ and how it follows from resolution of the identity? Let me read up on resolution of the identity till you reply. Commented Apr 4, 2021 at 4:15
• @CameronWilliams forgot to tag you. Commented Apr 4, 2021 at 4:28

What you need to do is use an orthonormal basis whose first element is $$x/\|x\|$$. Then $$\langle A,x\otimes x\rangle_{\rm HS}=\operatorname{Tr}(A(x\otimes x))=\langle A(x\otimes x)\tfrac{x}{\|x\|},\tfrac{x}{\|x\|}\rangle=\langle Ax,x\rangle,$$ since $$(x\otimes x)x=\|x\|^2\,x$$.
• This is brilliant! Just to confirm one thing though: $\langle A , x \otimes x \rangle_{\text{HS}}$ should equal $\text{Tr}(A^* (x \otimes x))$, right? and therefore we have $\langle A , x \otimes x \rangle_{\text{HS}} = \langle A^* x , x \rangle$. Or is it true that $A$ being Hilbert-Schmidt must also be self-adjoint? Commented Apr 4, 2021 at 4:45
• No, $\langle A,x\otimes x\rangle$ is $\operatorname{Tr}((x\otimes x)^*A)$. And $x\otimes x$ is selfadjoint. Commented Apr 4, 2021 at 4:55
• Okay, must be different conventions then? The Wikipedia page says $\langle A, B \rangle_{\text{HS}} = \text{Tr}(A^* B)$. Commented Apr 4, 2021 at 4:59