Differences of matrix exponentials

Question

Let $T:V\rightarrow W$ be a linear map of inner product spaces with $T^\ast$ the dual map. I am to calculate $f(\lambda)=\operatorname{tr}e^{-\lambda T^\ast T}-\operatorname{tr}e^{-\lambda TT^\ast }$. Is my solution correct?

Solution: $TT^\ast , T^\ast T$ are both symmetric hence unitarily diagonalizable by some $P$. Hence $f(\lambda)=\operatorname{tr}(Pe^{-\lambda D^2}P^\ast-\operatorname{tr}(P^\ast e^{-\lambda D^2}P$), which is in turn zero by the cyclic property of the trace. ($D$ is the diagonal matrix corresponding to $T$.)

Answer 1

You have the right basic idea. Note, however, that $TT^*$ and $T^*T$ are not necessarily simultaneously diagonalizable. That is, we can't use the same $P$ for both matrices. They are, however, symmetric with the same eigenvalues, which is enough.

Answer 2

I am assuming that $\lambda$ is real here.

There is no need for simultaneous diagonalisation.

Note that $AB$ and $BA$ have the same eigenvalues (and multiplicities).

Hence the eigenvalues of $e^{AB}$ and $e^{BA}$ are the same.

The eigenvalues of $e^{-\lambda T T^*}$ are the same as the eigenvalues of $e^{-\lambda T^* T}$.