<s>[INST] Find Jordan's normal form It is first time I try normally to figure out how it works.
So there's the matrix:
$$A=\begin{pmatrix} 0 & 6 & 4 & 0 \\ 0 & 4 & 0 & 2 \\ 0 & -1 & 2 & -1 \\1 & -3 & -2 & 2 \end{pmatrix}$$
1.) First of all we have to find eignevalue:
$$\mathrm{det}|A-\lambda{I}|= \begin{vmatrix} 0 - \lambda & 6 & 4 & 0 \\ 0 & 4-\lambda & 0 & 2 \\ 0 & -1 & 2-\lambda & -1 \\ 1 & -3 & -2  & 2-\lambda \end{vmatrix} = (\lambda-2)^4$$
So eigenvalue $\lambda = 2$
2.) Substitute $\lambda$ value into the matrix, got:
$$A_{\varphi}=\begin{pmatrix} -2 & 6 & 4 & 0 \\ 0 & 2 & 0 & 2 \\ 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 \end{pmatrix}$$
3.) As far as I understand I have to find rank of $A_{\varphi}$. Therefore $\mathrm{rank}A_{\varphi}=2$
Because $$A_{\varphi}=\begin{pmatrix} 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 
\end{pmatrix}$$
I do not know how to procced from here...
 [/INST] You can easily, though annoyingly carefully, check that $\;(A-2I)^3=0\;$ and that means the minimal polynomial of your matrix is $\;(x-2)^3\;$ , which means its Jordan form must have at least (and, in this case, obviously also at most) a Jordan block of size three, so
$$J_A=\begin{pmatrix}
2&1&0&0\\
0&2&1&0\\
0&0&2&0\\
0&0&0&2\end{pmatrix}$$ 
</s>