If $\operatorname{Ker}(T^k)=\operatorname{Ker}(T^{k+1})$ then in fact $\operatorname{Ker}(T^{k+c})=\operatorname{Ker}(T^k)$

Question

$\newcommand{\Ker}{\operatorname{Ker}}$Let $V$ be a finite dimensional vector space and let $T : V → V$ be a linear map. Prove that if $\Ker(T^k)=\Ker(T^{k+1})$ then in fact $\Ker(T^{k+c})=\Ker(T^k)$ for all positive integers $c$. I see that induction is likely to be the way forward - how can I prove that if if holds for some positive intger $c$, then it holds for $c+1$ also?

Answer 1

It is sufficient to prove by induction on $n \ge 0$ that $\ker T^{k+n}=\ker T^{k+n+1}$.

It's clear for $n = 0$ by hypothesis. Suppose that $\ker T^{k+n}=\ker T^{k+n+1}$. It's also clear that $\ker T^{k+n+1} \subseteq \ker T^{k+n+2}$. Now suppose that $T^{k+n+2}(x) = T^{k+n+1}(T(x))=0$. It means that $T(x) \in \ker T^{k+n+1}$ and by induction hypothesis that $T(x) \in \ker T^{k+n}$. Therefore $x \in \ker T^{k+n+1}$... and we're done.