Kernel of successive powers of a matrix For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}) = \ker(A^{n+3}) = \dots$ ?
If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case?
I know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.
But is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $m,k > 0$?
 A: This is true.  To my knowledge, there is no name for this theorem.
You can think of this as a consequence of Jordan canonical form.  In particular, we can always write
$$
A = S[N \oplus P]S^{-1}
$$
Where $N$ is nilpotent and $P$ has full rank.  It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank.
A: An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$.
To show this, it would be sufficient to show that $\ker(A^{k+2}) = \ker(A^{k+1})$, and the rest would follow from a simple inductive argument.
Note that, we have $\ker(A^{k+1}) \subseteq \ker(A^{k+2})$, and thus it is enough to show that $\ker(A^{k+1}) \supseteq \ker(A^{k+2})$.
For this, consider a vector $v$ such that $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(A^{k+1})$ because $A^{k+1}(Av) = 0$.
Since $\ker(A^{k+1}) = \ker(A^k)$, we have $Av \in \ker(A^k)$.
Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \in \ker(A^{k+1})$.
Clearly, $\ker(A^{k+2}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{k+1})$
