Showing a linear combination of matrices is nilpotent for any constants

Question

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

Answer 1

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.

Answer 2

$\newcommand{\Span}[1]{\langle #1 \rangle}$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.