Problem
Suppose $V$ and $W$ are finite-dimensional with $\dim V \geq \dim W \geq 2$. Show that $\lbrace T \in \mathcal{L}(V,W) | T$ is not surjective $ \rbrace$ is not a subspace of $\mathcal{L}(V,W)$
Solution
We can copy the setup in the solution to P3B7:
Let $v_1, v_2, \dots, v_n$ be a basis for $V$ and let $w_1, w_2, \dots, w_m$ be a basis for $W$, then define $T_1, T_2 \in \mathcal{L}(V,W)$ by $T_1v_{i} = 0$, $T_1v_{j} = w_{\textrm{min}(j,m)}$ and $T_2v_{i} = w_{i}$, $T_2v_{j} = 0$ and $T_2v_{i} = w_{\textrm{min}(i,m)}$. Where $i = 1,3,5,\dots$ and $j = 2,4,6,\dots$
Then $\textrm{range} \thinspace T_1 = \textrm{span}(v_2,v_4,\dots,v_{2k})$ and $\textrm{range} \thinspace T_2 = \textrm{span}(v_1,v_3,\dots,v_{2k \pm 1})$, but the sum of $T_1$ and $T_2$ is: $$(T_1 + T_2)(v) = T_1v + T_2v = w_{even} + w_{odd} = w$$ I.e. $T_1 + T_2$ maps to $\textrm{span}(w_1, \dots, w_m) = W$ and since we can find a list of vectors $u_1, \dots, u_m \in V$ such that $T_1 + T_2$ maps these to any $w \in W$ of our choosing, by construction, it is surjective. Implying that $\lbrace T \in \mathcal{L}(V,W) | T$ is not surjective $ \rbrace$ is not closed under addition.