need help Jordan base Need help, how to find Jordan base for matrix:
$A=\begin{pmatrix}
-1&-1 &-2  &4 \\ 
1&-3 &1  &-2 \\ 
0&0&2&-8\\ 0&0  & 2&-6
\end{pmatrix}$
I found the Minimal polynomial: $(x+2)^3$
and normal Jordan is $Aj=\begin{pmatrix}
-2&0 &0  &0 \\ 
0&-2 &1  &0 \\ 
0&0&-2&1\\ 0&0  & 0&-2
\end{pmatrix}$
Then I have problem with find $v_1, v_2, v_3, v_4 $that:
$(A+2I)v_1=0 $  $ (A+2I)v_2=0 $ $ (A+2I)v_3=v_2$ $ (A+2I)v_4=v_3$  
when i'm trying to find it take $v_1=(1,1,0,0) $ $ v_2=(0,0,2,1)$ and later i get contradiction 
 A: Ok i have as jordan basis 
$$\frac{1}{3} \cdot \begin{pmatrix}
2 \\ 2\\ -4 \\ -2 \end{pmatrix}; \qquad \begin{pmatrix} 6 \\ 6\\0 \\ 0\\ \end{pmatrix}; \qquad
\begin{pmatrix} 4 \\ -2 \\ -8 \\ -4 \end{pmatrix}; \qquad \begin{pmatrix} 0 \\ 0\\ 0\\ 1 \end{pmatrix} $$
Got them with that ugly algorithm.
At first we compute $(A-\lambda I)^k$ for $k=1,2,3$ which is $(A+2 I)^k$
For $k=1$ this is not so difficult we have 
$$\begin{pmatrix}
1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8  \\ 0 & 0 & 2 & -4 \end{pmatrix}$$
As next we calculate $(A+2 I)^2$
$$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0  & 0  & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix}$$
We see that $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$$ is not in the kernel of $(A+2I)^2$ (this is great!). So one jordan chain is 
$$\begin{pmatrix}
1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8  \\ 0 & 0 & 2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\-8 \\ -4 \end{pmatrix}$$
and 
$$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0  & 0  & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1  \end{pmatrix}= \begin{pmatrix} 6 \\ 6 \\ 0 \\ 0\\ \end{pmatrix}$$ 
As we only got one other eigenvector $$\begin{pmatrix} 0 \\ 0 \\2 \\ 1\\ \end{pmatrix}$$ and the minimalpolynom is $(A+ \lambda I)^3$ this is the only chain we have to calculate (what a luck). 
