Finding Jordan basis of 5x5 nilpotent matrix

[Juicer]

Hi, guys!
I have 5x5 dim matrix, it is nilpotent:

A = \begin{bmatrix}
-2 & 2 & 1 & 3 & -1 \\
3 & -8 & -2 & -9 & 3 \\
-2 &-8&0 & -6 & 2 \\
-4 & 8 & 2 & 9 & -3 \\
-4 & -4& 0 &-3 & 1
\end{bmatrix}

I have successfully found the Jordan normal form, it is:

J = \begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}

Now I am trying to find a Jordan basis.
First of all, I square A, obtaining

A^2 = \begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
-2 & 2 & 1 & 3 & -1 \\
-4 & 4 & 2 & 6 & -2 \\
4 & -4 & -2 & -6 & 2 \\
4 & -4 & -2 &-6 & 2
\end{bmatrix}

The cube equals zero: A^3 = 0. Now we see one column of A^2 basis. It is a third column : \vec{e}_1 = \vec{z^{(2)}_1} = (0, 1, 2, -2, -2)^T (by \vec{z^{(j)}_i} I am noting vectors, that I am looking for, where i - number of vector in R^j, R^j is a space induced by columns of matrix (A - \lambda E) ^ j).


Next I need to find basis vectors of columns from A and also to find \vec{z^{(1)}_1}. Then, because of A has dimension 3, we need to find \vec{z^{(1)}_2} also to complete the basis for R^1.


Already at this step I don't know, how to find vectors $\vec{z^{(j)}_i}
Please, help me. Also I would be grateful, if you could tell me all of the way to finding Jordan basis (untill the end, i.e. fifth vector).