The matrix diagonalizable coefficients in a field of characteristic different from $2$

Question

I read a property that I did not understand:

Let $M\in GL(2,\mathbb{F}_3)$ the matrix such that $M= I_2$ or $M^2=I_2$. As it has coefficients in a field of characteristic different from $2$, it is diagonalizable.

Is it a theorem? Can you help me prove this result

Answer 1

If $M=I_2$ then $M$ is already diagonal. Same if $M = -I_2$. If $M^2 = I_2$ with $M \ne \pm I_2$ (so that no linear polynomial annihilates $M$) then $p(M) = M^2 - I_2 = (M-I)(M+I) = 0$ so $p(x) = (x-1)(x+1)$ is the minimal polynomial of $M$, and in a field of characteristic different from 2, the two factors $x-1$ and $x+1$ are distinct. This implies that $M$ is diagonalizable.

Answer 2

If not already diagonal, the matrix has trace zero.

When the characteristic is not two, we have $2=1+1 \neq 0$

$$ \frac{1}{2} \; \left( \begin{array}{cc} 2&b\\ 0&1\\ \end{array} \right) \left( \begin{array}{cc} 1&b\\ 0&-1\\ \end{array} \right) \left( \begin{array}{cc} 1&-b\\ 0&2\\ \end{array} \right)= \left( \begin{array}{cc} 1&0\\ 0&-1\\ \end{array} \right) $$

For the next one, we need $a^2 + bc = 1 \; .$

$$ - \frac{1}{2c} \; \left( \begin{array}{cc} c&-1-a\\ -c&a-1\\ \end{array} \right) \left( \begin{array}{cc} a&b\\ c&-a\\ \end{array} \right) \left( \begin{array}{cc} a-1&a+1\\ -c&c\\ \end{array} \right)= \left( \begin{array}{cc} -1&0\\ 0&1\\ \end{array} \right) $$