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Let $A\in M_n(K)$ be a matrix over algebraic closed field $K$, where $n>1.$

When $A^i=A^j$ for $i,j\geq 0$ such that $i\neq j$?

I tried solve it by Jordan form of matrix $A;$ is is sufficient to answer when $J^i=J^j,$ where $J$ is Jordan-form matrix and $i\neq j.$ I know formula for a power of Jordan blocks, but I had problem to make computations for non-diagonal entries of blocks.

I tried also solve it by minimal polynomial $A$ (here $A$ satisfy a polynomial $f(x)=x^i-x^j.$)

Could anyone help me work out this problem? (I would be grateful for a hint; I don't want a full solution.)

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  • $\begingroup$ Well, clearly, if $A$ is either nilpotent, or a root of the identity, then it satisfies $x^i-x^j=x^j(x^{i-j}-1)$. The interesting part is finding the rest. $\endgroup$ – Arthur Apr 6 at 10:04
  • $\begingroup$ Yes, I know it, it's obvious. I'm interested in another answers. $\endgroup$ – jpatrick Apr 6 at 10:05
  • $\begingroup$ But I would like to know also how looks a root of the identity in matrix monoid. $\endgroup$ – jpatrick Apr 6 at 10:08
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    $\begingroup$ My gut says that for a matrix to be a root of the identity, all eigenvalues must be a root of unity in $K$ and all Jordan blocks must be diagonal. But there might be other solutions as well. $\endgroup$ – Arthur Apr 6 at 10:11
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Let $i>j$. Up to a change of basis, we may assume that

$A=diag(N,U)$ where $N$ is nilpotent, $U$ is invertible, $N^i=N^j$ and $U^i=U^j$.

Thus $U^k=I$ where $k=i-j$ is a positive integer. Then $U$ is diagonalizable and $U=PDP^{-1}$ where $D$ is diagonal and $d_{l,l}^k=1$.

Using the Jordan form of $N$, it is easy to see that the NS condition is $N^j=0$. .

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