Question. Let $A$ be a $3\times 3$ matrix. If $A^3=0$ and $A^2 \neq 0$, prove that $A^2v=0$ for some $v\in\mathbb{R}^3\setminus0$.
To generalize the solution I'm defining $A$ as: \begin{pmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{pmatrix}
and have calculated $A^2$ to be:
\begin{pmatrix} a_1^2+a_2a_4+a_3a_7 & a_1 a_2+a_2 a_5+a_3 a_8 & a_1 a_3+a_2 a_6+a_3 a_9\\ a_4 a_1+a_5 a_4+a_6 a_7 & a_4 a_2+a_5^2+a_6 a_8 & a_4 a_3+a_5 a_6+a_6 a_9\\ a_7 a_1+a_8 a_4+a_9 a_7 & a_7 a_2+a_8 a_5+a_9 a_8 & a_7 a_3+a_8 a_6+a_9^2 \end{pmatrix}
Just the thought of having to calculate $A^3$ gives me a headache... I've read something about diagnolization, but I couldn't apply it here.
I assume there must be some characteristic of a matrix $A$ for which $A^3=0$ I couldn't think about.
Prove that ∃v(A²v=0), v∈R³
$\endgroup$ – TastySpaceApple Apr 5 '14 at 10:51