$\max \Re \, \text{tr} (U V)$, with $U$ and $V$ unitary I'm trying to find the $V$ which maximizes $$ \Re \, \text{tr}(UV) $$ where $U$ and $V$ are unitary matrices, and $U$ is given. I'm starting to expect that the solution is quite trivial, but I might be mistaken. Is it $V = U^{-1}$? Here's my reasoning: * let $W = UV$. Then $W$ itself is also unitary. * The eigenvalues $w_i$ of $W$ are all on the unit circle. * $\max \Re \, \text{tr}(W) = \max \sum_i \Re[w_i]$ * The solution to this is obtained if all $w_i = 1$, hence $W = I$ and $V = U^{-1}$ Is this argument correct? The reason I'm sceptical of this finding is because in my new research field people are coming up with very complicated algorithms to do this numerically, so can it be so simple as just to take the hermitian conjugate?

As you stated, $W$ is unitary and so, assuming your vector space have dimension $n$, $\Re \text{tr} (W) = \sum_i \Re[w_i] \leq \sum_i \lvert w_i \rvert = n$. Now using $V=U^\dagger$ you get $W=I$ and $\Re \text{tr} (W) = n$. You have a upper bound on $\Re \text{tr} (W)$ and showed a way of achieving it so you have your maximizer