I have a nilpotent linear transformation $A: \mathbb{R^n} \to \mathbb{R^n}$ (i.e. $\exists \ m \ge 1$ such that $A^m = 0$). And I need to prove that $\dim\ker A^2 > \dim\ker A$, assuming $A \neq 0$.
I know that always $\ker A \subseteq \ker A^2$ hence $\dim\ker A \leq \dim\ker A^2$. But why in a nilpotent case the equality isn't possible?
Thank you a lot for any hints!
Suppose that $\ker A=\ker A^2$. Take $v\in\ker A^3$. Then $A^2.(A.v)=0$, that is, $A.v\in\ker A^2$. But then $A.v\in\ker A$, that is $A^2.v=0$. So $v\in\ker A^2$. For the same reason,$$\ker A=\ker A^2\implies(\forall n\in\mathbb{N}):\ker A=\ker A^n,\tag1$$which is impossible, since $A$ is nilpotent and therefore $\ker A^n$ is the whole space if $n$ is big enough. Unless, of course, $A=0$, in which case of course you have $\ker A=\ker A^2$.
Note that the nilpotency of $A$ is only used in the last part of the proof. That is, it is not needed in the proof of $(1)$. More generaly, you always have$$\ker A\subset\ker A^2\subset\ker A^3\subset\cdots$$and if $\ker A^k=\ker A^{k+1}$ for some $k\in\mathbb N$, then$$(\forall n\in\mathbb N):n\geqslant k\implies\ker A^n=\ker A^k,$$by the same argument that I used above.