<s>[INST] Is $\left(\begin{smallmatrix}0&0&1\\1&0&0\\ 0&1&0\end{smallmatrix}\right)$ diagonalizable over $\mathbb{Z}_2$? Is $A= \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$ diagonalizable over $\mathbb{Z}_2$?
I tried two approaches and got two different answers so I was hoping someone could point me to a flaw in my reasoning:
First approach:
The minimal polynomial for $A$ is easily found to be $m(x) =x^3-1$ which is the same as $x-1$ over $\mathbb{Z}_2$. Since the minimal polynomial decomposes into distinct linear factors it must be that $A$ is diagonalizable over $\mathbb{Z}_2$.
Second approach:
It follows from the minimal polynomial that $1$ is the only eigenvalue of $A$ . The eigenvector equation is
$\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix} = \begin{pmatrix} z \\ x\\ y \end{pmatrix} = 1 \times \begin{pmatrix} x \\ y\\ z \end{pmatrix}  $ and the only solution is $\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}$.
But $\mathbb{Z}_2^3$ has dimension $3$, so there is no basis for $\mathbb{Z}_2^3$ consisting of eigenvectors for $A$. $A$ can't be diagonalized over $\mathbb{Z}_2$.

What went wrong? Many thanks!
 [/INST] No, $x^3-1$ is not equal to $x-1$ over $\Bbb Z_2$, although the corresponding polynomial functions are equal indeed. On the other hand, $x^3-1=(x-1)(x^2+x+1)$ and $x^2+x+1$ is irreducible of $\Bbb Z_2$. Therefore, yes, your matrix is not diagonalizable over $\Bbb Z_2$.
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