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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 ?

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    $\begingroup$ See also this post. Any such matrix $A$ also answers the question there (about the component-wise exponential). $\endgroup$ – Loïc Teyssier May 12 '16 at 12:21
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    $\begingroup$ 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. $\endgroup$ – Neil Strickland May 12 '16 at 12:28
  • $\begingroup$ I indeed found a description, but wanted to know whether this is known or not. $\endgroup$ – Paul Broussous May 12 '16 at 13:29
  • $\begingroup$ @Loïc Not exactly because I want equality for $k\geqslant 1$, not $k\geqslant 0$. $\endgroup$ – Paul Broussous May 12 '16 at 13:31
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    $\begingroup$ 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 $A$. $\endgroup$ – Mark Meckes May 12 '16 at 14:08
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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

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  • $\begingroup$ Wonderful !! Thanks a lot Carlo, this is exactly what I was looking for !! $\endgroup$ – Paul Broussous Jun 5 '16 at 13:32

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