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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

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  • $\begingroup$ I guess $P^{-1}$ is the ordinary matrix inverse? Matrix algebras which are also closed under the Hadamard product are called association schemes. There is a monograph by Bannai-Ito on them (a special case is given by algebras generated by strongly regular graphs, there is also a monograph on them by Brouwer, .... which contains chapters on association schemes). Perhaps you should also have a look at Terwilliger pairs which might be relevant for your problem. $\endgroup$
    – Roland Bacher
    May 27, 2010 at 16:18
  • $\begingroup$ Yes, P^{-1} is the ordinary matrix inverse, sorry for the confusion. Thanks for the references, I'll take a look at them. $\endgroup$
    – Malik Younsi
    May 27, 2010 at 17:10
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    $\begingroup$ Related question: mathoverflow.net/questions/63027/…. I'd start from the book suggested in the answer, Horn and Johnson's Topics in Matrix Analysis (not to be confused with Matrix Analysis by the same authors). $\endgroup$
    – Federico Poloni
    May 22, 2013 at 8:44
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    $\begingroup$ Just terminology. The matrix $P^{-T}\circ P$ is the gain array matrix associated with $P$. It was studied by C. R. Johnson & H. Shapiro. $\endgroup$
    – Denis Serre
    May 22, 2013 at 11:44

3 Answers 3

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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.

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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.

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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

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