# Finding Jordan basis of $5 \times 5$ nilpotent matrix

I have $$5 \times 5$$ real matrix, which 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).

• Title:"nilpotent matrix" (not nillpotent). – Dietrich Burde Dec 7 '18 at 20:00