How to calculate Jordan basis when the $(A-\lambda I)^2=0$

Question

\begin{align}A&=\begin{pmatrix}3&1&1\\-3&-1&-3\\2&2&4\end{pmatrix}\\\\ J&=\begin{matrix}2&1&0\\0&2&0\\0&0&2\end{matrix}\end{align}

I want to know about "Jordan basis" of this problem I already calculate Jordan Form. and then, Use the fact that $AQ = QJ$

\begin{align}(A-2I)V_1 &= 0\\ (A-2I)V_2 &= V_1\\ (A-2I)V_3 &= 0\end{align}

so I finally get $V_1$ and $V_3$. $$V_1 = (-1, 1, 0)\\ V_3 = (-1, 0, 1)$$

but $V_2$ is not clear because $$(A-2I)^2 \times V_2 = (A-2I)V_1 = 0$$

In this situation how can I get the $V_2$?

Answer 1

I think the problem is that you made an unfortunate choice of $V_1$. Had you instead used $V_1=(1,-3,2)$ (which is $(-3)(-1,1,0)+(2)(-1,0,1)$) you would have had no difficulty finding $V_2$ such that $(A-2I)V_2=V_1$. Then you can still use $V_3=(-1,0,1)$.