What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ? 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
 A: You might already know this, but I thought it was interesting:
The all-ones vector is always an eigenvector of $(P^{-1})^\mathrm{T}\circ P$ with eigenvalue $1$.  To see this, note that the $i$th entry of $((P^{-1})^\mathrm{T}\circ P)\mathbf{1}=(P\circ (P^{-1})^\mathrm{T})\mathbf{1}$ is precisely the $(i,i)$th entry of $PP^{-1}=I$ by the definition of matrix multiplication.
UPDATE: In hindsight, this is a special case of Theorem DMHP from Dietrich Burde's link.
A: 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.
A: Also the following can be said about $P$ and $Q$. Your relation implies
that the two matrices $PDP^{-1}$ and $QDQ^{-1}$ have the same diagonal elements for any diagonal matrix $D$. For $n=2$ this is a necessary and suffcient condition for the equality 
$(P^{-1})^T\circ P=(Q^{-1})^T\circ P$. 
Here is a Reference: http://linear.ups.edu/jsmath/0200/fcla-jsmath-2.00li101.html
