Let $K$ be a commutative field and ${\rm M}_n (K)$ be the ring of $n\times n$ square matrices with coefficients in $K$ ($n\geqslant 1$ is an integer). For $k\geqslant 1$ and $A =(a_{ij})_{1\leqslant i,j\leqslant n}\in {\rm M}_n (K)$, define: $A^{[k]} =(a_{ij}^k )_{1\leqslant i,j\leqslant n}$.

Is the description of all matrices $A\in {\rm M}_n (K)$ satisfying $A^k =A^{[k]}$, for all $k\geqslant 1$, known? If yes do you have a reference ?

  • 1
    $\begingroup$ See also this post. Any such matrix $A$ also answers the question there (about the component-wise exponential). $\endgroup$ May 12, 2016 at 12:21
  • 1
    $\begingroup$ It is not clear from your question whether (a) you are looking for a description, or (b) you have found a description but are unsure whether it is already known. Please clarify. $\endgroup$ May 12, 2016 at 12:28
  • $\begingroup$ I indeed found a description, but wanted to know whether this is known or not. $\endgroup$ May 12, 2016 at 13:29
  • $\begingroup$ @Loïc Not exactly because I want equality for $k\geqslant 1$, not $k\geqslant 0$. $\endgroup$ May 12, 2016 at 13:31
  • 1
    $\begingroup$ I don't have an answer, but in searching the literature it may be useful to know that $A^{[k]}$ is called a "Hadamard power" of $A$. $\endgroup$ May 12, 2016 at 14:08

1 Answer 1


This is problem 50.4 by Moubinool Omarjee from volume 50 of The Bulletin of the International Linear Algebra Society, with solutions in volume 51 by Eugene Herman and Bojan Kuzma, and further work by Roman Drnovsek in When powers of a matrix coincide with its Hadamard powers.

For real matrices the solution is

  • $\begingroup$ Wonderful !! Thanks a lot Carlo, this is exactly what I was looking for !! $\endgroup$ Jun 5, 2016 at 13:32

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.