# Spectrum invariant under (generalised) transpose as operator on trace class operators

For matrices $$A$$ it is well known that the spectrum is invariant under transpose $$\sigma(A^T) = \sigma(A)$$. Furthermore, the spectrum of the adjoint matrix $$\sigma(A^*) = \overline{ \sigma(A)}$$ the complex conjugation of the spectrum of the matrix $$\sigma(A)$$. The results also has immediate generalisations to operators on Hilbert or Banach spaces using the Hilbert or Banach space adjoints respectively.

Suppose now, that we have an operator on the space of trace class operators $$TC( \mathcal{H})$$ for some Hilbert space $$\mathcal{H}$$ with a given basis $$\vert i \rangle_{i \in \mathbb{N}}$$. Then $$TC( \mathcal{H})$$ is spanned by the elementary matrices $$E_{ij} = \vert i \rangle \langle j \vert_{i,j \in \mathbb{N}}$$ closed in the trace norm. Thus, I can describe any operator $$A \in \mathcal{B}( TC( \mathcal{H}))$$ in terms of its matrix elements \begin{align*} A_{(i,j), (k,l)} = \langle k \mid, A \left( \vert i \rangle \langle j \vert \right) l \rangle. \end{align*} This allows me to define a "transpose" $$\tilde A$$ operator with respect to our choice of basis by defining it on matrix elements \begin{align*} \langle k \mid, \tilde A \left( \vert i \rangle \langle j \vert \right) l \rangle = A_{ (k,l),(i,j)} \end{align*} and extending it by linearity. I would expect that $$\tilde A \in \mathcal{B}( TC( \mathcal{H}))$$ and that $$\sigma_{ \mathcal{B}( TC( \mathcal{H}))}(\tilde A) = \sigma_{ \mathcal{B}( TC( \mathcal{H}))}(A)$$, but how would I go about proving such a statement?

In other words: If I have a trace-class operator with matrix elements $$A_{(i,j), (k,l)}$$ and "transpose" the operator so that it has matrix elements $$A_{(k,l),(i,j)}$$ is the operator then still bounded operator on trace class operators?

• This is related to the results of this paper: arxiv.org/abs/2206.09879 Jun 22, 2022 at 18:34

It is not true that $$\tilde A$$ maps trace class operators to trace class operators in general. For a counterexample, consider the maps $$A:X\mapsto \mathrm{Tr}(X) \vert 1\rangle \langle 1 \vert$$. Then $$\tilde A$$ should send $$\vert 1\rangle \langle 1 \vert$$ to $$\mathrm{Id}_{\mathcal H}$$, which is not trace class.
What is true in full generality is that, if $$X$$ is a Banach space and $$A \in B(X)$$, then its adjoint $$A^*$$ is a well-defined operator on the dual of $$X$$, and that the spectrum of $$A$$ and $$A^*$$ coincide.
In your setting, the dual of the space of trace class operators is $$B(\mathcal H)$$, so your operator $$\tilde A$$ extends to a bounded map on $$B(\mathcal H)$$, which has the same spectrum as $$A$$.