I've came with a very simple question, which asks to show that $A^3 = 0$, given that:
$$A(x,y,z) = (ay+bz, cz, 0)$$
However, I don't comprehend how to multiply this $A$ times itself. I tried to multiply as a row vector times a column vector but it didn't get the same result.
Let $B$ be the matrix representation of the linear operator $A$.
$$B\begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} ay+bz\\cz\\0 \end{pmatrix}$$ $$B=\begin{pmatrix} 0&a&b\\ 0&0&c \\ 0&0&0 \end{pmatrix}$$ $$B^3=O$$ $$A^3=0$$
alternatively,
$$A^2(x,y,z)=(acz,0,0)$$ $$A^3(x,y,z)=(0,0,0)$$