# When $A^i=A^j$ for $i,j\geq 0$ such that $i\neq j$ for matrix $A$ over algebraic closed field?

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.)

• 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. – Arthur Apr 6 at 10:04
• Yes, I know it, it's obvious. I'm interested in another answers. – jpatrick Apr 6 at 10:05
• But I would like to know also how looks a root of the identity in matrix monoid. – jpatrick Apr 6 at 10:08
• 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. – Arthur Apr 6 at 10:11

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 $$N^j=0$$. .