Rank of Linear Transformation $T^2=0$

Question

let $T:\mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation such that $T^2=0$ .Show that the rank ($r$) of the linear transformation obeys the inequality $r\le n/2$

Answer 1

Recall the rank-nullity theorem for finite dimensional spaces $$ \dim(\operatorname{Im}(T)) + \dim(\ker(T)) = n \ .$$ The condition $T^2 = 0$ implies that $\operatorname{Im}(T) \subset \ker(T)$. This set inclusion implies that $$\dim(\operatorname{Im}(T)) \leq \dim(\ker(T)) \ .$$ Combining the rank-nullity theory and this inequality, we have $$ n = \dim(\operatorname{Im}(T)) + \dim(\ker(T)) \geq 2 \dim(\operatorname{Im}(T)) \iff \dim(\operatorname{Im}(T)) \leq \frac{n}{2} \ ,$$ as desired.