$A$ is a square complex matrix. $A^k-A=0$ for some $k\geq 2$. Prove that $A$ is diagonalizable over $\mathbb C$

Question

$p(x)=x^k-x=x(x^{k-1}-1)$

What I want to do is to say that $(x^{k-1}-1)=(x-z_1)(x-z_2)...(x-z_{k-1})$ and therefore A is diagonalizable (because of the distinct roots in the polynomial), but i'm not sure that I don't get repeated roots here.

I know in general that there are $n$ different complex solutions for $z^n=r$ when $z$ is different than zero, but is it also true for when $r=1$?

Answer 1

You don't have repeated roots since they are the $(k-1)^{th}$ roots of unit. So you have $k$ distinct eigenvalues, then $A$ is diagonalizabile over $\Bbb C$. Moreover if $k=2,3$ $A$ is diagionalizable over $\Bbb R$.