I need your help for understanding how to compute jordan base for this matrix:
$\begin{pmatrix} 2 &1 &2 \\ -1 &0 &2 \\ 0&0 & 1 \end{pmatrix}$
This is what I tried to do:
I found the Minimal polynomial: $(x-1)^{3}$,
so I know that the Jordan normal form is: $\begin{pmatrix} 1 &0 &0 \\ 1& 1 & 0\\ 0&1 & 1 \end{pmatrix}$
Now I remember that I need to find: $\ker (A-I), \ker (A-I)^{2}, \ker(A-I)^{3}$.
I found that $\ker (A-I)$ is: $\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}$ and $\ker (A-I)^{2}$ is: $\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}$
and obivously $\ker(A-I)^{3}$ is all $\mathbb{R}^{3}$, What should I do from here? I can't remember the whole algorithem, and I can't find it anywhere.
Thank you, Have a great week-end.