Problem
Show that $$\lbrace T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) | \dim \textrm{null} \thinspace T > 2 \rbrace$$ is not a subspace of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$
Solution
Let $U = \lbrace T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) | \dim \textrm{null} \thinspace T > 2 \rbrace$, then $T_1 \in U$ defined by $T_1(x_1, x_2, x_3, x_4, x_5) = (x_1,x_2,0,0)$ and $T_2 \in U$ defined by $T_2(x_1, x_2, x_3, x_4, x_5) = (0,0,x_3,x_4)$, where $x_1, \dots, x_5 \in \mathbb{R}$. But their sum: $$(T_1 + T_2)(x_1, x_2, x_3, x_4, x_5) \stackrel{D3.6}{=} T_1(x_1, x_2, x_3, x_4, x_5) + T_2(x_1, x_2, x_3, x_4, x_5)$$ $$= (x_1, x_2,0,0) + (0,0,x_3,x_4) \stackrel{D1.12}{=} (x_1,x_2,x_3,x_4)$$ Therefore $\dim \textrm{range} \thinspace T = \dim \mathbb{R}^4 = 4$ which by the fundamental theorem of linear maps implies $\dim \textrm{null} \thinspace (T_1 + T_2) = 5 - 4 = 1$, therefore $U$ is not closed under addition.