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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).

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

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