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

  • $\begingroup$ 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 '10 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 '10 at 17:10
  • 1
    $\begingroup$ 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 '13 at 8:44
  • 3
    $\begingroup$ 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 '13 at 11:44

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.


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.