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.