Can the matrix $A=\begin{bmatrix} 0 & 1\\ 3 & 3 \end{bmatrix}$ be diagonalized over $\mathbb{Z}_5$? Im stuck on finding eigenvalues that are in the field please help. 
Given matrix:
$$
A=  \left[\begin{matrix}
0 & 1\\
3 & 3
\end{matrix}\right]
$$
whose entries are from $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$, find, if possible, matrices $P$ and $D$ over $\mathbb{Z}_5$ such that $P^{−1} AP = D$.
I have found the characteristic polynomial: $x^2-3x-3=0$
Since its over $\mathbb{Z}_5$, $x^2-3x-3=x^2+2x+2=0$.
But from there I'm not sure how to find the eigenvalues, once I get the eigenvalues that are in the field it will be easy to find the eigenvectors and create the matrix $P$. 
 A: Hint: $x^2-3x-3\equiv x^2-3x+2$.
A: yes over $\Bbb Z_5$  because:
$\lambda^2 -3\lambda-3=o$ at Z_5 we will have $\Delta=9+12=4+2=6$ (9~4 and 12~2 at Z_5)
so $\Delta=1$
and so $\lambda_1=\frac{3+1}{2}=2$ and $\lambda_2=\frac{3-1}{2}=1$ 
about:
$\lambda_1$ we have :$ ( \left[\begin{matrix}
0 & 1\\
3 &3
\end{matrix}\right]-\left[\begin{matrix}
2 & \\
0 &2
\end{matrix}\right] )\left[\begin{matrix}
x\\
y
\end{matrix}\right]=0$
$$-2x+y=0 $$ & $$( 3x+y=0 ~ -2x+y=0 ) $$ and so $$ y=2x  $$ 
is our space of eigen value of $ \lambda_1 =\{(2,4),(0,0)(1,2)\} $ => (dim =1) base={(1,2)}
about $\lambda_2$:
$ ( \left[\begin{matrix}
0 & 1\\
3 &3
\end{matrix}\right]-\left[\begin{matrix}
1 & 0\\
0 &1
\end{matrix}\right] )\left[\begin{matrix}
x\\
y
\end{matrix}\right]=0$ and so $y=x$ is our answer 
 and eigenvector space of  $\lambda_2=\{(0,0)(1,1)(2,2)(3,3)(4,4)\} \implies $ $(\dim=1)$
base ={(1,1)}
matrix at base of$ \{(1,1),(1,2)\}$ will be diagonalizable
$\left[\begin{matrix}
2 & 0\\
0 &1
\end{matrix}\right] $
