I need to find the Jordan Normal form $J$ and a matrix $S$ such that $J = S^{-1} AS$.
The matrix is $$ M = \left( \begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 \\ \end{matrix} \right) $$
This matrix is over the finite field $\mathbb{Z}_3$.
I have found the characteristic polynomial. It is $x^4$, so the only eigenvalue is $x = 0$. I found two linearly independent eigenvectors.
$$ v_1 =\left( \begin{matrix} 1 \\ 0 \\ 1 \\ 0 \\ \end{matrix} \right)\qquad v_2 = \left( \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{matrix} \right) $$
So I need two more vectors to form the matrix $S$. My first thought was two find 2 generalized eigenvectors, but since there are 2 eigenvectors, I am not sure how to do this. Is this the only way to go? Or is there another way of getting the 2 vectors that I need?
Also, when I have 4 vectors, how do I know which order to put them in the matrix $S$?
Welcome! You have to find vectors $w_1, w_2$ such that $\;Mw_1=v_1$, $\;Mw_2=v_2$. In the basis $\mathcal J=(v_1, w_1,v_2,w_2)$ the matrix will have the Jordan form $$\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}$$