TITLE: Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ then $R(T) \cap N(T) = \{0\}$
QUESTION [8 upvotes]: Let $V$ be a finite-dimensional vector space and let $T:V\to V$ be linear.
Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, then $R(T) \cap N(T) = \{0\}$.
I don't see this implication, at all. Please give hints and explain conceptually.

REPLY [9 votes]: We have that $T^2(V)\subseteq T(V)$, so $R(T)=R(T^2)$ implies that $T^2(V)=T(V)$. 
Now, for the rank nullity theorem we know that $\dim V= R(T) + N(T)=R(T^2)+N(T^2)$, so $N(T)=N(T^2)$, but similarly to before, $\ker T \subseteq \ker T^2 $ and $\ker(T)=\ker(T^2)$.
Let $x \in \ker (T) \cap T(V)$. The fact that  $x\in T(V)$ implies that there exists $x' \in V$ such that $T(x')=x$, then $T(T(x'))=T(x)=0$ (because $x \in \ker (T)$). 
Then $x' \in \ker (T^2)=\ker (T)$, so $T(x')=0=x$.
We conclude that $\ker (T) \cap T(V)=\{ 0\}$.