$\newcommand{\h}{\mathcal H}$$\newcommand{\tr}{\mbox{tr}}$

In brief : Treating every bounded linear operator of a Hilbert space, as a multiplication operator which is an element of $B_2(\h)$ (Hilbert-Schmidt operators), what can you say about its spectrum?

Consider a (complex) Hilbert space $\h$ and the bounded linear operators on $\h$, namely $B(\h)$. We have the usual subspace of Hilbert Schmidt operators $B_2(\h)$, which are all $T \in B(\h)$ such that $\tr(T^*T) < \infty$. Along with the Hilbert Schmidt inner product $\langle T,S\rangle_2 = \tr(T^*S)$ and the associated norm $\|\cdot\|_2$, we get a Hilbert space structure on $B_2(\h)$. In fact, not only a subspace, it is also a two sided self adjoint ideal.

Now, it can be easily shown that any bounded linear operator $T \in B(\h)$ gives an element of $B(B_2(\mathcal H))$ via multiplication ($S \to TS$), which is bounded because one can check that the usual $\|T\|$ works.

However, if $T$ is invertible as a multiplication operator, then it is sufficient that there exist $S$ such that $ST = TS = I$ on only the subspace $B_2(\h)$, therefore one expects the spectrum of $T$ as a multiplication to be a subset of its usual spectrum.

I have tried to find the spectrum of $T$ as a multiplication operator on $B_2(\mathcal h)$ and have had no progress primarily because all I know about $B_2(\h)$ is that it is a subset of the compact operators. I would like some guidance in this direction.


Let $\cal H$ be a Hilbert space and let $HS(\cal H)$ denote the set of Hilbert-Schmidt operators. Furthermore, for $T\in B(\cal H)$ let $M_T$ be the operator of multiplication in $HS(\cal H)$ with respect to $T$, that is, $M_TX = TX$ for all $X\in HS(\cal H)$. We shall also fix some $u\in\cal H$ with $\|u\|=1$ and define $F : {\mathcal H}\to HS(\cal H)$ by $Fy := (\cdot,u)y$. Let us show the following:

(a) If $T$ is invertible, then so is $M_T$ with $M_T^{-1} = M_{T^{-1}}$.

(b) If $M_T$ is invertible, then so is $T$ with $T^{-1} = (M_T^{-1}F\cdot)u$.

Then from $M_{S+T} = M_S+M_T$ and $M_{ST} = M_SM_T$ it follows that $$ \sigma(M_T) = \sigma(T). $$ Ad (a): This is clear: $M_{T^{-1}}M_T = M_{T^{-1}T} = M_I = I$ and $M_TM_{T^{-1}} = M_{TT^{-1}} = M_I = I$.

Ad (b): Note that $FT = M_TF$. Hence, for $y\in\cal H$ we get $$ (M_T^{-1}FTy)u = (M_T^{-1}M_TFy)u = (Fy)u = y $$ and also $T(M_T^{-1}Fy)u = [M_TM_T^{-1}(Fy)]u = (Fy)u = y$.

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    $\begingroup$ Very nice! Thank you, that was helpful. $\endgroup$ Mar 26 '19 at 6:08

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