# $\{T\in\mathcal{L}(\mathbf{R^5},\mathbf{R^4})|\dim\operatorname{null}T>2\}$ is not a subspace of $\mathcal{L}(\mathbf{R^5},\mathbf{R^4})$

Is the following Proof Correct ?

Theorem. The set $\{T\in\mathcal{L}(\mathbf{R^5},\mathbf{R^4})|\dim\operatorname{null}T>2\}$ is not a subspace of $\mathcal{L}(\mathbf{R^5},\mathbf{R^4})$.

Proof. Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5$ and $\beta_1,\beta_2,\beta_3,\beta_4$ denote the standard basis for $\mathbf{R^5}$ and $\mathbf{R^4}$ respectively and consider the Linear Transformations $T_1$ and $T_2$ defined by

$$T_1v_j= \begin{cases} \beta_j\ \ 1\leq j\leq 2\\ 0\ \ \ 3\leq j\leq 5\\ \end{cases} \tag{1}$$ $$T_2v_j= \begin{cases} 0\ \ \ \ 1\leq j\leq 3\\ \beta_j\ \ \ 4\leq j\leq 5\\ \end{cases} \tag{2}$$

It is then evident that $\operatorname{null}T_1 = \operatorname{span}(\beta_3,\beta_4,\beta_5)$, $\operatorname{null}T_2 = \operatorname{span}(\beta_1,\beta_2,\beta_3)$ and since these lists are linearly independent it follows that $\dim\operatorname{null}T_1 = \dim\operatorname{null}T_2 = 3>2$. We then have the following definition for $T_1+T_2$.

$$(T_1+T_2)v_j= \begin{cases} \beta_j\ \ \ j\neq 3\\ 0\ \ \ \ j=3\\ \end{cases} \tag{3}$$

which implies that $\operatorname{null}(T_1+T_2) = \operatorname{span}(\alpha_3)$ and so $\dim\operatorname{null}(T_1+T_2) = 1$.

I'm not sure what you mean by the $v_j$'s in your proof, but I interpret you idea as something similar to this:
For $v = (x_1, x_2,x_3,x_4, x_5)\in \mathbb{R}^5$ put
$$T_1v = x_1\beta_1 + x_2\beta_2$$
$$T_2v = x_3\beta_3 + x_4\beta_4$$
Then $T_1$ and $T_2$ both have a null space with dimension bigger than $2$, but their sum doesn't.