1
$\begingroup$

$\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.

$\endgroup$
0
2
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Very nice! Thank you, that was helpful. $\endgroup$ Mar 26 '19 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.