# Find Jordan basis for the following nilpotent matrix

I need to find the Jordan basis and Jordan form for this nilpotent matrix:

$$\begin{pmatrix} -2i & 1 & 10+5i \\ 4 & 2i & -10+19i \\ 0 & 0 & 0 \\ \end{pmatrix}$$

The matrix is nilpotent so all eigenvalues are 0, therefore $Av_1=0$ and from that I got that $v_1=\begin{pmatrix} 1 \\ 2i \\ 0 \\ \end{pmatrix}$ Now $Av_2=v_1$ so $v_2=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix}$. This is where I'm stuck there seems to be no $v_3$ such that $Av_3=v_2$. How do I continue or what am I doing wrong?

## 1 Answer

You have it a bit backwards. We have $A^3 = 0$ but $A^2 \neq 0.$ It appears you want the columns of the change of basis matrix, call that $V,$ as you want column vectors $v_1, v_2,v_3.$

Well, $$A^2 = \left( \begin{array}{rrr} 0 & 0 & -i \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right)$$ In order to arrange $A^2 v_3 \neq 0,$ take $$v_3 = \left( \begin{array}{rr} 0 \\ 0 \\ 1 \end{array} \right)$$ Next $$v_2 =A v_3 = \left( \begin{array}{rr} 10 + 5i \\ -10 + 19i \\ 0 \end{array} \right)$$ $$v_1 = Av_2 =A^2 v_3 = \left( \begin{array}{rr} -i \\ 2 \\ 0 \end{array} \right)$$ we have set up $A v_1 = A^3 v_3 = 0 v_3 = 0.$

Put the columns in order, we have $$V = \left( \begin{array}{rrr} -i & 10+5i & 0 \\ 2 & -10 + 19i & 0 \\ 0 & 0 & 1 \end{array} \right)$$ Note $\det V = -1,$ so finding $V^{-1}$ is not that bad. Then $$J = V^{-1} AV = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)$$