What happens to dimension of kernel when squaring non-invertible matrix? I need to show that given $A \in \mathbb{M}_{(3 \times 3)}$ $A^2 \neq [0]$ and $A^3 = [0]$, there exists $\vec{v}$ s.t. $\{\vec{v}, A\vec{v}, A^2\vec{v}\}$ is a basis in $\mathbb{R}^3$.
I was hoping to do that if I could show that $\dim{\ker{A}} < \dim{\ker{A^2}}$, i.e. since $A$ is not invertible, that would mean $\dim{\ker{A}} = 2$ and $\dim{\ker{A^2}} = 1$.  My "intuition" for this is that if $\dim{\ker{A^2}} = \dim{\ker{A}} = 1$, then this would be projection of a plane on a plane, which cannot be a point.  But I don't know what to do with the possibility that $\dim{\ker{A^2}} = \dim{\ker{A}} = 2$. I.e. a projection of a line on a line (which can be a point).
 A: More generally, for any $n\times n$ matrix $A$, one has
$$\ker A\varsubsetneq \ker A^2\varsubsetneq\dots\varsubsetneq \ker A^k= \ker A^{k+1}=\dotsm$$
for some $k,\enspace0\le k\le n$.
Added:
First observe the chain of $\ker u^k$ cannot be a strictly ascending chain as the dimensions of the kernels is at most $n$. So the must be $k$s for which $\ker u^k=\ker u^{k+1}$.
So, to prove the assertion, it is enough to prove that, as soon as $\ker u^k=\ker u^{k+1}$  for some $k$, one also has  $\ker u^{k+1}=\ker u^{k+2}$.
So let $x\in \ker u^{k+2}$. Rewrite $u^{k+2}(x)=u^{k+1}(u(x))=0$, so $u(x)\in\ker u^{k+1}=\ker u^k$ and  finally $u^k(u(x))=u^{k+1}(x)=0$, which proves $\ker u^{k+2}\subseteq\ker u^{k+1}$.The reverse inclusion is obvious.
A: Expanding on Bernard's answer, in this specific case we must have dim ker $A^k$ = dim ker $A^{k+1}$ for all $k \ge 3$ since the dimension of a subspace of R^3 clearly cannot be greater than 3.
Since A is non-invertible, we have dim ker $A \ge 1$. Hence ker $A \not=$ ker $A^3$.
Therefore there are two possibilities: ker $A^2$ = ker $A^3$, which would imply that dim ker $A^2$ = dim ker $A^3$ = 3, which contradicts $A^2 \not= [0]$.
Thus ker $A^2$ is also a proper subset of ker $A^3$, and thus k=3 is the smallest possible integer for which that condition holds (and the second possibility, namely dim ker $A^2$=2, holds).
Anyway for v, take any vector such that $A^2v \not=0$; such a vector must exist since dim ker $A^2 \not = 3$. Then argue by contradiction that the iterates of v (i.e. $Av, A^2v$) must be linearly independent of it, or else v would lie in the kernel of either of A or $A^2$, contradicting the assumption that $A^2v \not= 0$.
