Let \(V\) be a finite-dimensional vector space and let \(T\) be a linear operator on \(V\). Suppose that \(\operatorname{rank}(T^2) = \operatorname{rank}(T)\). Prove that the range and null space of \(T\) are disjoint, i.e., have only the zero vector in common.
Solution
Since \(\ker T\subseteq\ker T^2\), rank-nullity and the hypothesis give \[ \dim\ker T =\dim V-\operatorname{rank}T =\dim V-\operatorname{rank}T^2 =\dim\ker T^2. \] Therefore \[ \ker T=\ker T^2. \] Now let \(y\in\operatorname{range}T\cap\ker T\). Write \(y=T\alpha\). Since \(T(y)=0\), \[ T^2\alpha=0, \] so \(\alpha\in\ker T^2=\ker T\). Hence \(y=T\alpha=0\). Thus \[ \operatorname{range}T\cap\ker T=\{0\}. \]