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?
Best Answer
Your calculations are fine. However, by definition of kernels, an element of the kernel of $(A+4E)^2$ vanishes when applying $A+4E$ to it twice, so you should not be surprised. You just made the wrong conclusion. The eigenvector $(3,0,1)$ together with the generalized eigenvector $(1,0,0)$ form part of a Jordan basis giving you a Jordan block of size $2$. All you need to do is add another eigenvector which is linear independent to $(3,0,1)$, for example $(1,1,2)$.
Then $$ \begin{pmatrix} 3 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 3 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$