Help with proving that $A^p \equiv A$ mod $p$ does not mean that $A$ is diagonalizable

Question

I'm working on a matrix extension of Fermat's Little Theorem, but I'm stuck on trying to show that if $A^p \equiv A$ mod $p$, then $A$ does not have to be diagonalizable.

Any help would be appreciated!

Edit:: I would like to either find a reason why $A$ does not have to be diagonalizable, or somehow be able to categorize the matrices that are / are not diagonalizable in a way that would suggest a pattern. An example would be that if $A^p \equiv A$ mod $p$, then $A$ is diagonalizable when xxx or not diagonalizable when xxx.

Answer 1

You haven't been clear about your base field. Let $k=\Bbb R$. Consider $p=2$. Then $\begin{pmatrix}1&2\\0&1\end{pmatrix}$ is congruent to its square modulo two. However, it is not diagonalizable over $\Bbb R$, say, since it is a Jordan block.

Answer 2

What people generally think of as the generalization of Fermat's Little Theorem is as follows:

Let $p$ be prime and $A\in GL_n(\mathbb{Z})$. Then $tr(A^p)=tr(A)$ mod $p$, where $tr(A)$ denotes the trace of the matrix $A$. In fact, this holds even more generally: $tr(A^{p^k})=tr(A^{p^{k-1}})$ mod $p^k$ where the first statement is the case for $k=1$

The first statement was proven by V.I. Arnold (who also conjectured the second). The second was proven by Alexander Zarelua in 2008.