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.
 A: Let $$T=\begin{pmatrix}1&0&0&0&0\\
0&1&0&0&0\\
0&0&0&0&0\\
0&0&0&0&0\\
\end{pmatrix}$$
$$S=\begin{pmatrix}0&0&0&0&0\\
0&0&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
\end{pmatrix}$$
Both have a nullity of $3$ but
$$T+S=\begin{pmatrix}1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
\end{pmatrix}$$ has a nullity of only one.
A: 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.
A: The proposed solution defines linear transformations $S$ and $T$ by giving their values on the standard basis for $\mathbb{R}^5$, and says:

Let $S(e_1)=S(e_2)=S(e_3)=\mathbf{0}$, $S(e_4)=f_1$, $S(e_5)=f_2$.
Let $T(e_1)=T(e_2)=T(e_4) = \mathbf{0}$, $T(e_3)=f_3$, $T(e_5)=f_4$.

It then argues that $S+T$ has nullity two, by noting that $(S+T)(e_3) = f_3$, $(S+T)(e_4) = f_1$, and $(S+T)(e_5) = f_2+f_4$ (while $(S+T)(e_1)=(S+T)(e_2)=\mathbf{0}$).
You ask: do you really need to assume that $f_1,\ldots,f_4$ are a basis for this to work?
In order to conclude that the rank of $S+T$ is three (so that the dimension of the nullspace is exactly two), you need $f_1$, $f_3$, and $f_2+f_4$ to be linearly independent. This will definitely hold if $f_1,f_2,f_3,f_4$ are linearly independent (that is, if you start with a basis), but may fail otherwise. For instance, let's say that you pick $f_1=(1,0,0,0)$, $f_2=(1,1,0,0)$, $f_3=(0,1,1,0)$, and $f_4=(0,0,1,0)$. Then $f_1+f_3=f_2+f_4$, so $(S+T)(e_3+e_4-e_5)=\mathbf{0}$, and so the nullspace has dimension at least $3$. (Or pick $f_4=-f_2$; or all of them to be zero, etc).
It's possible for it to work out by chance even if these vectors are not linearly independent: for example, if $f_1=(1,0,0,0)$, $f_2=(0,1,0,0)$, $f_3=(0,0,1,0)$, and $f_4=(1,2,1,0)$, then they are not linearly independent, but it still the case that $f_1=(1,0,0,0)$, $f_3=(0,0,1,0)$, and $f_2+f_4 = (1,3,1,0)$ are linearly independent, and so you'd be fine.
But in order to guarantee it works without giving specific values, then you really want to assume that $f_1,f_2,f_3,f_4$ are linearly independent, and thus a basis for $\mathbb{R}^4$.
