Let $$M=\begin{bmatrix}3 & 0 & 2 & 4 \\ 1 & 0 & 4 & 3 \\ 3 & 1 & 0 & 0 \\ 0 & 2 & 1& 2 \\ \end{bmatrix}\in(\mathbb{Z}/\mathbb{5Z})$$
I want to prove that this matrix has a Jordan canonical form and find it. When I try to calculate it, I have that the characteristic polynomial is $x^4$ and the minimal polynomial $x$, so the Jordan canonical form must be
$$J=\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0& 0 \\ \end{bmatrix}\in(\mathbb{Z}/\mathbb{5Z})$$ Is this correct?
The Jordan form of a non-zero matrix cannot possibly be zero. Note that for any invertible $S$, $S0S^{-1} = 0$. So, the only matrix similar to the zero matrix is the zero-matrix itself (a similar phenomenon occurs with the multiples of the identity matrix).
Note that the minimal polynomial of this matrix is actually $x^2$. In this case, the minimal polynomial is not sufficient to determine the Jordan form. It suffices, however, to note that $M$ has minimal polynomial $x^2$ and rank at least $2$. We can thereby deduce that the Jordan form is $$ J = \pmatrix{0&1\\&0\\&&0&1\\&&&0} $$