I'm studying for a qualifying exam in algebra and I've come across the following problem:
Let $V$ be a finite dimensional vector space and let $T:V\rightarrow V$ be a linear operator on $V$. Let $\ker(T)$ denote the kernel of $T$ and $\operatorname{Im}(T)$ denote the image of $T$.
(a) Prove that $\operatorname{rank}(T)\geq \operatorname{rank}(T^2)$, and that equality holds if and only if $V=\ker(T)\oplus \operatorname{Im}(T)$.
(b) Prove that there is a positive integer $m$ such that $$V=\ker(T^m)\oplus \operatorname{Im}(T^m).$$
Here's what I have so far:
(a) Observe that $\operatorname{Im}(T)\supseteq \operatorname{Im(T^2)}$. Thus, because rank of an operator is the dimension of the image of the operator, $\operatorname{rank}(T)\geq \operatorname{rank}(T^2)$.
$(\Rightarrow)$ Suppose equality holds. It follows then that $\operatorname{Im}(T) = \operatorname{Im}(T^2)$. By the rank nullity theorem, \begin{align} \dim(V)=\operatorname{rank}(T)+\operatorname{nullity}(T)&=\operatorname{rank}(T^2)+\operatorname{nullity}(T^2)\\ \Rightarrow \operatorname{nullity}(T)&=\operatorname{nullity}(T^2)\\ \Rightarrow\ker(T)&=\ker(T^2). \end{align} Now let $$x\in \ker(T)\cap \operatorname{Im}(T).$$ Then there exists an $x'\in V$ such that $T(x')=x$. Hence, $$T^2(x')=T(T(x'))=T(x)=0,$$ as $x\in \ker(T)$. It follows that $$x'\in \ker(T^2)=\ker(T).$$ Thus, $T(x')=0$, so $x=0$. Therefore $$\ker(T)\cap \operatorname{Im}(T)=\{0\}.$$ Since any element $v\in V$ can be expressed as $v=u+w$ where $u\in \ker(T)$ and $w\in \operatorname{Im}(T)$, the desired result follows.
So for part (a), I still need the $(\Leftarrow)$ direction. This seems like it should be the same logic in reverse, but I'm struggling to actually make the argument.
I don't have any progress on part (b) yet.
I assume this is a duplicate problem, but I was only able to find questions relating to what I've already proved. Thanks in advance for any help.
may help to observe:
for $(\Leftarrow)$ direction consider what is $T(Im(T))$
similar reasoning for part (b). (with $T^0 = $Id) set $I_k = Im(T^k)$. Then $I_{k+1} \subseteq I_k$. since $V$ is finite-dimensional there must be a smallest $r$ for which $Im(T^{r+1}) = Im(T^r)$