Problem
Prove that there does not exist a linear map $T: \mathbb{R}^5 \to \mathbb{R}^5$ such that $$\textrm{range} \thinspace T = \textrm{null} \thinspace T$$
Solution
The condition $\textrm{range} \thinspace T = \textrm{null} \thinspace T$ implies $\dim \textrm{range} \thinspace T = \dim \textrm{null} \thinspace T$, which by the fundamental theorem of linear maps implies: $$5 = \dim \mathbb{R}^5 \stackrel{T3.22}{=} \dim \textrm{range} \thinspace T + \dim \textrm{null} \thinspace T = 2\dim \textrm{null} \thinspace T$$ Implying $\dim \textrm{range} \thinspace T = \dim \textrm{null} \thinspace T = 5/2$ a contradiction.