How to show matrix $A$ diagonalizable iff $A^k$ is diagonalizable for $k\ge 2$?

Question

In order for a matrix to be diagonalizable, $\dim E_{\lambda i}=$ multiplicity of $\lambda i$. $Ax=\lambda x\implies $ $A^kx={\lambda}^kx$. That's all I know for the question. Could someone give some insight?

Answer 1

Consider:

$$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$

Then, $A$ is not diagonalizable, but $A^k = 0$ for all $k \ge 2$ is.

For the other direction, here is a hint:

$$P^{-1} A^k P = (P^{-1} A P)^k$$

Answer 2

As mentioned in the comments, only one direction holds. Indeed, there are a multitude of matrices for which in fact $A^k$ is already diagonal and $A$ is not diagonalizable. (HINT: consider a nontrivial nilpotent matrix)