Suppose there are given two real symmetric matrices $\pmb A$ and $\pmb B$. I want to find a permutation matrix $\pmb P$ such that
$\mathrm{Tr}(\pmb P^T\pmb A\pmb P\pmb B) \to \max$
My current status is the following:
$\pmb A = \pmb U\pmb D_a\pmb U^T$ and $\pmb B = \pmb V\pmb D_b\pmb V^T$, where $\pmb U$ and $\pmb V$ are orthogonal matrices. $\pmb D_a$ and $\pmb D_b$ are diagonal and contain the eigenvalues of $\pmb A$ and $\pmb B$ in ascending order.
Therefore I can reformulate the problem as
$\mathrm{Tr}(\pmb X^T \pmb D_a \pmb X \pmb D_b)\quad$ with $\quad \pmb X = \pmb U^T \pmb P \pmb V$.
Since $\pmb U^T$, $\pmb P$ and $\pmb V$ are orthogonal, I know that also $\pmb X$ is.
From this post I know that in case of arbitrary orthogonal $\pmb X$ the optimal solution is $\pmb X = \pmb I$.
However, here $\pmb X$ can only be influenced through $\pmb P$.
My intuition is that $\pmb P$ should be selected to make $\pmb X$ "close" to the identity matrix, i.e.,
$\pmb X = \pmb I + \pmb N \quad$ with $\quad ||\pmb N|| \to \min\quad$ in terms of some matrix norm.
Is there any way to formally prove this?
