Let $X$ be a separable Hilbert space (real or complex). Let $A\in\mathcal{L}\left(X\right)$, a bounded linear operator on $X$, and suppose $B\in\mathcal{L}\left(X\right)$, which is of trace-class. Then we have the property $\operatorname{Tr}AB=\operatorname{Tr}BA$. So far, so good. What I am interested in is this: do we then have the property that $\operatorname{Tr}A^\ast B=\operatorname{Tr}B^\ast A$ also? What if $X$ was assumed real? (Note: we do not assume symmetry of any of the operators.)

  • $\begingroup$ I think we generally have $\operatorname{Tr}(A^*) = \overline{\operatorname{Tr}(A)}$ $\endgroup$ – Omnomnomnom Jul 7 '15 at 11:58
  • $\begingroup$ That is correct. Under the assumptions above, we have that $\operatorname{Tr}A^\ast B=\left(\operatorname{Tr}B^\ast A\right)^\ast$. Yet, do we then have that $\operatorname{Tr}A^\ast B=\operatorname{Tr}B^\ast A$ (as in the case $\operatorname{dim}X<\infty$)? $\endgroup$ – Trevor3 Jul 7 '15 at 12:10

The trace is a positive linear functional. In particular this means that if $A \ge 0$, then $\mathrm{Tr}(A) \ge 0$. Similarly, we have $\mathrm{Tr}(B^*) = \overline{\mathrm{Tr}(B)}$ for $B$ trace-class (as Omnomnomnom pointed out in the comments).

Thus, in general we have $$\mathrm{Tr}(A^*B) = \mathrm{Tr}((B^*A)^*) = \overline{\mathrm{Tr}(B^*A)},$$ which means that $\mathrm{Tr}(A^*B) = \mathrm{Tr}(B^*A)$ if and only if $\mathrm{Tr}(A^*B) \in \mathbb{R}$. All of this holds regardless of whether or not the space is finite dimensional.

In particular, if $X$ is a real Hilbert space, then the equality you mentioned is valid. In general for complex Hilbert spaces it is not (finite dimensional or otherwise).

As an aside, notice that the map $(A,B) \mapsto \mathrm{Tr}(B^*A)$ is an inner product on the ideal of Hilbert-Schmidt operators, so it satisfies all the standard properties of an inner product, of which the displayed equation above is a straightforward consequence.

  • $\begingroup$ Good answer, many thanks. $\endgroup$ – Trevor3 Jul 7 '15 at 12:33

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