Show that $\{T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) : \text{dim}(\text{null}(T)) > 2\}$ is not a subspace I am working my way through Axler's Linear Algebra Done Right. I attempted the problem and found a solution here (problem is $3$.B $4$): https://linearalgebras.com/3b.html.
The solution is as follows:
Let $U = \{T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) : \text{dim}(\text{null}(T)) > 2\}$.
Let $e_1, ..., e_5$ be a basis of $R^5$ and $f_1, ..., f_4$ be a basis of $R^4$. Define $S_1$ by $S_1e_i = 0$ for $i = 1, 2, 3$, $S_1e_4 = f_1$, and $S_1e_5 = f_2$. Define $S_2$ by $S_2e_i = 0$ for $i = 1, 2, 4$, $S_2e_3 = f_3$, and $S_2e_5 = f_4$.
Then, $T_1, T_2 \in U$. However,
$$(S_1 + S_2)(e_1) = 0, (S_1 + S_2)(e_2) = 0$$
and
$$(S_1 + S_2)(e_3) = f_3, (S_1 + S_2)(e_4) = f_1, (S_1 + S_2)(e_5) = f_2 + f_4$$
Then, $\text{dim}(\text{null}(T_1 + T_2)) = 2$ and $T_1 + T_2 \notin U$. Thus, $U$ is not closed under addition which implies $U$ is not a subspace of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$ as desired.
My solution is exactly the same except that I say let $f_1, f_2, f_3, f_4$ be arbitrary vectors in $R^4$ instead of let $f_1, f_2, f_3, f_4$ be a basis of $R^4$. Does this actually affect the validity of my solution? I do not believe so, but other solutions I found all specify the $f$'s being a basis.

Let $T(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,x_3,0,0), S(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,x_5)$ Then $T$ and $S$  are in $U$ but $T-S$ is not.
