Is the following argument correct?
Proposition. Given a finite-dimensional vector space $V$ and a linear map $T:V\to V$ such that $\operatorname{rank}(T^2) = \operatorname{rank}(T)$ prove that $\operatorname{range}T\cap\operatorname{null}T = \{0\}$.
Proof. Assume that there exists at least one non-zero vector $v\in\operatorname{range}T\cap\operatorname{null}T$ and let $U$ be the linear map formed by restricting the domain of $T$ to $\operatorname{range}T$ then the rank-nullity theorem implies $$\dim\operatorname{range}T = \dim\operatorname{null}U+\dim\operatorname{range}U = \dim \operatorname{null}U+\dim\operatorname{range}T^2$$ Now since $\operatorname{range}T\cap\operatorname{null}T\subseteq\operatorname{null}U$ it follows that $v\in\operatorname{null}U$ and thus $\dim\operatorname{null}U\ge1$ consequently $\dim\operatorname{range}T^2<\dim\operatorname{range}T$.
$\blacksquare$