TITLE: Prove $\ker(T^n)=\ker(T^{n+1})$ and $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$
QUESTION [5 upvotes]: Let $V$ be an $n$-dimensional vector space over a field $F$ and $T$ an operator on $V.$ Prove $\ker(T^n)=\ker(T^{n+1})$ and $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$

Suppose $v \in \ker(T^n).$ Then $T^n(v) = 0,$ implying that $T^{n+1}(v)=T(T^n(v))=T(0)=0.$ Thus $v \in \ker(T^{n+1})$ and so $\ker(T^n) \subset \ker(T^{n+1}).$ 
Question: How do I prove  $\ker(T^n) \supset \ker(T^{n+1})?$ I would also like some hints on proving $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$

REPLY [4 votes]: By contradiction, suppose that there exists $v\in\ker(T^{n+1})$ such that  $T^nv\neq0$. Then the vectors $v,Tv,...,T^nv$ are all nonzero, and you can see that they are in fact linearly independent by evaluating powers of $T$ on a linear combination $\sum_{k=0}^n\alpha_k\,T^kv=0$. But this is absurd as $\dim V=n$.
Hint for the second statement: write $v=T^nw$ and decompose $w$ as a sum of a vector in $\ker (T^n)$ and a vector in the range of $T$.
Second hint: proceeding as in the first point you can show that $\ker (T^n)\cap\operatorname{range}(T^n)=\{0\}$ (see also the answers to this this question).