Find Jordan form of matrix 6x6 Find the Jordan form of matrix $$A = \begin{bmatrix} 
2 & 0 & 1 & 0 & -1 & 0\\
0 & 2 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 5 & 0 & -9\\
2 & 0 & 1 & 0 & 3 & 0\\
0 & 0 & 0 & 1 & 0 & -1\\ 
\end{bmatrix}$$
I found $\lambda$ from $|A-\lambda I|= 0$ and  $(\lambda -2)^6=0$
$$B=(A-2I) = \begin{bmatrix} 
0 & 0 & 1 & 0 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 & 1 & 0\\
0 & 0 & 0 & 3 & 0 & -9\\
2 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & -3\\ 
\end{bmatrix}$$
$rankB=3 \Rightarrow$ solving $Bx=0$ I got:
$$x_1=[0,1,0,0,0,0]^T, \hspace{5pt} x_2=[0,0,0,3,0,1]^T, \hspace{5pt} x_3=[0,0,1,0,-1,0]^T$$
$$B^2 = \begin{bmatrix} 
-2 & 0 & -2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
2 & 0 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
2 & 0 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\ 
\end{bmatrix}$$
Solving $B^2x=0$ I got: $$x'_1=[1,0,-1,0,0,0]^T, \hspace{3pt} x'_2=[0,1,0,0,0,0]^T, \hspace{3pt} x'_3=[0,0,0,1,0,0]^T, \hspace{3pt} x'_4=[0,0,0,0,1,0]^T, \hspace{3pt} x'_5=[0,0,0,0,0,1]^T$$
So I got only 5 vectors but I'm in $\mathbb{R}^6$ and $B^3$ is null matrix. So I don't know what to do next

Since $\operatorname{rank} B = 3$, you have three linearly independent eigenvectors associated to the (only) eigenvalue $\lambda = 2$ so there are three Jordan blocks in the Jordan form of $A$. Since $B^3 = 0$ while $B^2 \neq 0$, the minimal polynomial of $A$ is $(x - 2)^3$ which means that the largest Jordan block in the Jordan form of $A$ must be a $3 \times 3$ block. This gives you only one option for the Jordan form of $A$:
$$ \begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 1 & 0 \\
0 & 0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 2\end{pmatrix}. $$
