Proving $3\times3$ nilpotent matrix satisfies $A^3=0$ Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.

There are a few ways to do that. If you know about minimal polynomials, then you can use that the minimal polynomial of $A$ has degree at most $3$ and divides $A^n$.
A less advanced way: Note that for each $k$, the range of $A^k$ is a subspace of ${\mathbf R}^3$, and $range(A^3) \subset range(A^2)\subset range(A) \subset {\mathbf R}^3$. The dimensions must correspondingly be nonincreasing as you go down the list. If $range(A^3)$ weren't of dimension zero, then you'd have $range(A^k) = range(A^{k+1}) \neq \emptyset$ for some $k \leq 2$. Then by induction $range(A^l) = range(A^k)$ for all $l \geq k$. This would include $l = n$, a contradiction.
