# About spectrum of a multiplication operator ON the Hilbert-Schmidt space

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

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 $$T(M_T^{-1}Fy)u = [M_TM_T^{-1}(Fy)]u = (Fy)u = y$$.