Let $\displaystyle V$ be a 3-D space over a field $\displaystyle k$ with basis $\displaystyle B = (v_1,v_2,v_3)$ and consider linear operators $\displaystyle x,y,z:V\rightarrow V$ whose matrices relative to $\displaystyle B$ are:

$\displaystyle X=\left(\begin{array}{ccc}0&1&0\\0&0&0\\0&1&0\end{ array}\right), Y=\left(\begin{array}{ccc}0&2&0\\1&0&-1\\0&2&0\end{array}\right), Z=\left(\begin{array}{ccc}1&0&-1\\0&0&0\\1&0&-1\end{array}\right)$

Find vectors $\displaystyle u_1,u_2,u_3\inV$ such that

$\displaystyle x(u_1)=y(u_1)=z(u_1)=0$

$\displaystyle x(u_2),y(u_2),z(u_2)\in \ V_1=ku_1$ with $\displaystyle u_2\notin \ V_1$

$\displaystyle x(u_3),y(u_3),z(u_3)\in \ V_2=ku_1\oplus \ ku_2$ with $\displaystyle u_3\notin V_2$

I have been asked to find these vectors so that the matrices of $\displaystyle x,y,z$ with respect to $\displaystyle u_1,u_2,u_3$ are stricly upper triangular.

My inital procedure was to first find a common eigenvector for $\displaystyle x,y,z$ and take that as $\displaystyle u_1$, but I couldn't find a common eigenvector that worked, when I calculated $\displaystyle u_2, u_3$, the respective matrices for $\displaystyle x,y,z$ were not stricly upper triangular. I then tried $\displaystyle u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right )$ but that didn't work. Any help would greatly appreciated.