Proof that if powered matrices are equal to $E$, then they are diagonalizable

Question

I need help with this:

Prove this: If $A^k=E$, then $A$ is diagonalizable.

Could anyone help me please?

Answer 1

Layout:

1. Prove that the eigenvalues of $A$ are $k$-roots of $1$ (not necessarily all of the roots, but all of them are $k$-roots of the unity).
2. Consider the jordan normal form of $A$ and take its $k$ power. On the one hand it is the identity matrix and on the other... (check link above).

Answer 2

The key observation here is that if $J$ is a Jordan block corresponding to a non-zero eigenvalue, then $J^k$ is diagonal iff $J$ is diagonal (that is, the size of the Jordan block is one).

Then if $A^k = \alpha I$, where $\alpha \neq 0$, and $A' = U^{-1} A U$ is the Jordan form of $A$, then we have $(A')^k = U^{-1} A^k U = \alpha I$, and hence all Jordan blocks of $A$ have size one, and so $A'$ is diagonal.