<s>[INST] Find Jordan Decomposition of $\left(\begin{smallmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{smallmatrix}\right)$ over $\mathbb{F}_5$ 
Find the Jordan decomposition of
$$
A := \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}
\in M_3(\mathbb{F}_5),
$$
where $\mathbb{F}_5$ is the field modulo 5.

What I've done so far
The characteristic polynomial is
\begin{equation}
P_A(t) = (4 - t)(1-t)(3-t) - (1-t)
= -t^3 + 8t^2-18t+1
\equiv 4t^3 + 3t^2 + 2t + 1 \mod5.
\end{equation}
Therefore, $\lambda = 1$ is a zero of $P_A$, since $4+3+2+1 = 10 \equiv 0 \mod 5$.
By polynomial division one obtains
$$
P_A(t)
= (t + 4)(4t^2 + 2t + 4)
= (t + 4)(t + 4) (4t + 1)
\equiv 4 (t + 4)^3
$$
Therefore $\lambda = 1$ is the only eigenvalue of $A$.
To find the eigenspace we calculate the kernel of $A + 4 E_3$ and obtain
$$
\text{span}\left( \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right)
$$
Since $(A + 4 E_3)^2 = 0$, the kernel of $(A + 4 E_3)^2$ is the whole space.
Now, I choose $v := (1, 0, 0) \in \text{ker}(A + 4 E_3)^2$ such that $v \not\in \text{ker}(A + 4 E_3)$.
We calculate $(A + 4E)v = (3, 0, 1)$ and then
$$
(A + 4E)
\begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},
$$
but the zero vector can't be a basis vector of our Jordan decomposition.
Have I made a mistake in my calculations?
 [/INST] Here is a 4th answer that takes into account that we are in a particular case yielding a reduction to a $2 \times 2$ matrix.
Indeed, up to a simultaneous permutation $P$ on lines and columns, $A=PBP^{-1}$ is similar to 
$$B:= \left(\begin{array}{cc|c} 4 & 1 & 0 \\ 1 & 3 & 0 \\ \hline 0 & 0 & 1 \end{array}\right)$$
On this form, we have reduced the issue to find a Jordan form for $2 \times 2$ upper block $U$.
A quick computation shows that $1$ is a double eigenvalue of $U$. 
The Jordan form of $U$ is :
$$\left(\begin{array}{cc} 1 & 1 \\ 0 & 1  \end{array}\right)$$
because (due to a quick reasoning, that has been developed in the other answers, but with simpler computations) :
$$\underbrace{\left(\begin{array}{cc} 1 & 2 \\ 2 & 0  \end{array}\right)}_{Q}\left(\begin{array}{cc} 1 & 1 \\ 0 & 1  \end{array}\right)\underbrace{\left(\begin{array}{cc} 0 & 3 \\ 3 & 1  \end{array}\right)}_{Q^{-1}}=\left(\begin{array}{cc} 4 & 1 \\ 1 & 3  \end{array}\right).$$
The Jordan form of $B$, which is the same as the Jordan form of $A$, is thus :
$$J:= \left(\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 1 & 0 \\ \hline 0 & 0 & 1 \end{array}\right).$$
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