1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$ J = V^{-1} AV = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.