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!
Best Answer
Your claim that $x^3-1$ and $x-1$ are the same polynomial is the mistake. Yes, as functions over $\mathbb{Z_2}$ they are the same. But a polynomial is not a function. A polynomial is a formal sum of the form $\sum_{i=0}^n a_ix^i$ where the coefficients are elements in the field. So $x^3-1$ is actually not a product of distinct linear factors.
Another way to see this is a mistake is to note that if $x-1$ was the minimal polynomial of $A$ then it would mean that $A-I=0$ and hence $A=I$. But $A$ is not the identity matrix.