Showing a linear combination of matrices is nilpotent for any constants So I have three linear operators in a $3$-dimensional vector space $V$ over field $\Bbb k$ whose matrices w.r.t basis of $V$ are
$$X=
\left(\begin{matrix}1 & 0 & 1\\
1 & 0 & 1\\
-2 &1 &-1
\end{matrix}\right)$$
$$Y=
\left(\begin{matrix}1 & -1 & 0\\
1 & -1 & 0\\
-2 &2 &0
\end{matrix}\right)$$
$$Z=
\left(\begin{matrix}1 & 0 & 1 \\
1 & 0 & 1\\
-1 &0 &-1
\end{matrix}\right)$$
I have to show that any linear combination of these matrices (with coefficients in $\Bbb k$) is nilpotent. What kind of approach should I be taking here? Thanks
 A: $\newcommand{\Span}[1]{\langle #1 \rangle}$This is a drastic simplification of my previous solution - should have thought of it before, as there is a sound theoretical reason for it. 
Let us rewrite $X, Y, Z$ in terms of the basis
$$
e_1, e_1 + e_2 - 2 e_3, e_1 + e_2 - e_3.
$$
We have
$$
X : e_{1} \mapsto e_1 + e_2 - 2 e_3 \mapsto e_1 + e_2 - e_3 \mapsto 0
$$
$$
Y : \begin{cases}
e_1 \mapsto e_1 + e_2 - 2 e_3 \mapsto 0\\
e_1 + e_2 - e_3 \mapsto 0
\end{cases}
$$
$$
Z : \begin{cases}
e_1 \mapsto e_1 + e_2 - e_3 \mapsto 0\\
e_1 + e_2 - e_3 \mapsto 0
\end{cases}
$$
So rewriting the matrices $X', Y', Z'$ with respect to this new basis we have
$$
X' = \begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{bmatrix},
\quad
Y' = \begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix},
\quad
Z' = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0\\
\end{bmatrix}.$$
So these three matrices span the space of strictly lower triangular matrices
$$
\left\{\  \begin{bmatrix}
0 & 0 & 0\\
a + b & 0 & 0\\
c & a & 0\\
\end{bmatrix} : a, b, c \in \mathbb{k}\ \right\}
=
\left\{\  \begin{bmatrix}
0 & 0 & 0\\
s & 0 & 0\\
u & t & 0\\
\end{bmatrix} : s, t, u \in \mathbb{k}\ \right\},
$$
which is well-known to consist of nilpotent matrices.
So they point is (and I should have thought of it immediately) that the three matrices can be put simultaneously in strictly lower triangular form.
A: Edit: Answer may be incomplete. I will revisit when I have time. 
Note that this isn't a very elegant answer. Perhaps there is a better way.
Taking a cue from Jean-Claude Arbaut's comment, I've used octave to verify that $Y^2 = 0$ and $Z^2 = 0$.
I've also verified that $X^2 Y = 0$, $X^2 Z = 0$ and $X Y Z = 0$.
Note also that $X^3 = Y^3 = Z^3 = 0$, since the operators have to be nilpotent.
Order of the terms don't matter, since the null spaces grow regardless of the order to take up the entire 3 dimensions.
The above shows that the expansion of $(aX+bY+cZ)^3$ has to be zero, since the expansion involves all of the above terms multiplied by a matrix and/or constant.
