TITLE: What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ?
QUESTION [8 upvotes]: Let $P,Q$ be $n$ by $n$ invertible matrices. Suppose further that $P$ and $Q$ satisfies the following equation :
$$(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$$
where $\circ$ denotes the Hadamard matrix product, which is simply the entrywise product. 
Then what can be said about $P$ and $Q$? More precisely, I want to know if there are additional relations between $P$ and $Q$. For example, one can show that the condition $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ implies
$$tr(P^{-1}DPE) = tr(Q^{-1}DQE)$$ for all diagonal matrices $D$ and $E$.
References in the litterature about matrices of the form $(P^{-1})^T \circ P$ would help too.
Thank you,
Malik

REPLY [3 votes]: Actually, if $D, E$ are diagonal and $Q=EPD$, then $(P,Q)$ is such a pair. A natural question is whether every pair such that $P^{-T}\circ P=Q^{-T}\circ Q$ is of the form above. But even this is false, because if $P$ is triangular, then $P^{-T}\circ P=I_n$.
I take the occasion to mention an open question: define $\Phi(P):=P^{-T}\circ P$ (the gain array). What are the matrices $P$ such that $\Phi^{(k)}(P)\rightarrow I_n$ as $k\rightarrow+\infty$ ? According to Johnson & Shapiro, this is true at least for

Strictly diagonally dominant matrices
Symmetric positive definite matrices

On the contrary, $\Phi$ has fixed points, for instance the mean $\frac12(P+Q)$ of two permutation matrices. See Exercises 335, 336, 342, 343 of my blog about Exercises on matrix analysis.