$A^{n+1}=0\Rightarrow A^n=0$

Question

A real $n\times n$-matrix $A$ satisfying $A^{n+1}=0$ must necessarily satisfy $A^n=0$.

One way to see this is by looking at the Jordan Normal Form of $A$, another is by an argument involving the Cayley-Hamilton theorem and minimal polynomials. Both pieces of theory seem to be unnecessarily advanced for this particular problem, but I cannot think of another argument.

Could this be solved by a student equipped with essentially nothing but the rank-nullity theorem (and its prerequisites) or is this indeed a deeper problem?

Answer 1

Clearly, $\ker A^{k}\subseteq \ker A^{k+1}$. Also, if for some $k$ we have $\ker A^k=\ker A^{k+1}$ then $\ker A^m=\ker A^{m+1}$ for all $m\ge k$. Thus the sequence $\dim \ker A^k$, $k=0,1,2,\ldots$ is eventually stationary and is strictly increasing before that. Therefore it either never reaches $n$ at all, or it reaches $n$ no later than $\dim \ker A^n$.

Answer 2

Let $A$ be nilpotent. Let $m$ be minimal with $A^m=0$ and $A^{m-1}v \neq 0$, one easily sees that $$v,Av, \dotsc, A^{m-1}v$$ are linear independent. These are $m$ vectors in a $n$-dimensional vector space, hence we obtain $m \leq n$.