Show $V = range(T^i) \oplus null(T^i)$ Show $V = range(T^i) \oplus null(T^i)$ where $dim V = n$, $T \in L(V)$ and $i \geq n$
My idea (1):
Show that if $dim(null(T^i)) = n-k$ for $k \geq 0$ then $dim(range(T^i)) = k$. This way if the dimensions add to $n$ exactly then we have a direct sum.
My idea (2):
Another idea is to show set containment. It is clear that for any $v \in range(T^i) \oplus null(T^i), v \in V$. However showing the other way is more difficult ie why does $v \in V$ imply $v \in range(T^i) \oplus null(T^i)$
I am not sure if my approach is correct. Any guidance would be greatly appreciated.
 A: Note that, $T$ annhilates its characteristics polynomial $x^n+a_{n-1}x^{n-1}+....+a_1x+a_0$, i.e. $T^nv+a_{n-1}T^{n-1}v+....+a_1Tv+a_0v=0,\forall v\in V$. Now let $k$ be the smallest integer for which $a_k\not=0$. Then, $T^nv+a_{n-1}T^{n-1}v+....+a_kT^kv=0,\forall v\in V$. 
Let $i\geq n$. Now note that, $\text{range}(T^i)+\text{null}(T^i)$ is a subspace of $V$ with $\text{dim}\big(\text{range}(T^i)\big)+\text{dim}\big(\text{null}(T^i)\big)=\text{dim}(V)$. 
So to prove the direct sum enough to show, $\text{range}(T^i)\cap\text{null}(T^i)=\{0\}$. So let $w\in \text{range}(T^i)\cap\text{null}(T^i)$, then, $w=T^iv$ for some $v\in V$ and $T^iw=0$. If possible let $m$ be the smallest positive integer for which $T^mw=0$, then $T^{m+i}v=0$ but, $T^{m+i-1}v\not=0$, but $m+i-1\geq n$ and $$T^iv+a_{i-1}T^{i-1}v+....+a_kT^kv=0$$$$\implies T^{m+i-1-k}\big(T^iv+a_{i-1}T^{i-1}v+....+a_kT^kv\big)=0$$$$\implies a_k T^{m+i-1}v=0\implies T^{m+i-1}v=0\text{ as }a_k\not=0,$$ contradiction. So there is no smallest positive integer for which $T^mw=0$. That is $0=T^0w=w$. So we are done.
A: Since $V \supset \operatorname{range}(T) \supset \operatorname{range}(T^2) \supset \operatorname{range}(T^3) \supset \ldots \supset \operatorname{range}(T^n)$, $n = \operatorname{dim} V \ge \operatorname{dim} \operatorname{range}(T) \ge \operatorname{dim} \operatorname{range}(T^2) \ge \operatorname{dim} \operatorname{range}(T^3) \ge \ldots \ge \operatorname{dim} \operatorname{range}(T^n)$. So there exists $0 \le i \le n$ such that $\operatorname{dim} \operatorname{range}(T^i) = \operatorname{dim} \operatorname{range}(T^{i + 1}) = \ldots = \operatorname{dim} \operatorname{range}(T^n)$, hence $\operatorname{range}(T^i) = \operatorname{range}(T^{i + 1}) = \ldots = \operatorname{range}(T^n)$. Also by rank-nullity theorem $\operatorname{null}(T^i) = \operatorname{null}(T^{i + 1}) = \ldots = \operatorname{null}(T^n)$. Let $i$ be the least integer satisfying the condition. If $i = n$ $\operatorname{range}(T^n) = (0)$ then it is trivial. Let's assume that $i < n$. Then $T$ is bijective on $\operatorname{range}(T^i)$. $T^i$ is also bijective, so  $\operatorname{range}(T^i) \cap \operatorname{null}(T^i) = (0)$. With rank-nullity theorem $V = \operatorname{range}(T^i) \oplus \operatorname{null}(T^i) = \operatorname{range}(T^n) \oplus \operatorname{null}(T^n)$. For $T^m, m > n$, note that $\operatorname{range}(T^n) = \operatorname{range}(T^{n + 1}) = \ldots$ 
