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.

Let $v$ be a non-zero vector and consider the sequence
$$
v, Av, A^2v, ..., A^kv, 0, 0, ...
$$
where $k < n$ is the largest integer such that $A^kv \ne 0$. We will see that the non-zero elements of the sequence are linearly independent. Therefore, in a three-dimensional space, at most three, i.e. $v, Av, A^2v$, can be non-zero. Thus, $A^3v = 0$ for every $v$.
To prove linear independence, note first that the last non-zero element of the sequence, i.e. $A^kv$ lies in the kernel of $A$, but the other vectors, i.e. $v, Av, A^2v, ..., A^{k-1}v$ do not. Therefore none of the vectors
$v, Av, A^2v, ..., A^{k-1}v$ can be a multiple of $A^kv$. Similarly, the last two non-zero vectors of the sequence, i.e. $A^{k-1}v, A^kv$ lie in the kernel of $A^2$, but the preceding vectors do not. Therefore none of the preceding vectors is a linear combination of $A^{k-1}v, A^kv$.
Continuing this reasoning we see that $A^iv, ..., A^kv$ lie in the kernel of $A^{k-i+1}$ while none of $v, Av, ..., A^{i-1}v$ do. Therefore, none of $v, Av, ..., A^{i-1}v$ is a linear combination of $A^iv, ..., A^kv$. Therefore, the non-zero vectors in the sequence $v, Av, A^2v, ..., A^kv, ...$ are linearly independent.
Therefore there cannot be more than three of them, so $A^3 = 0$.
