given $T:V\longrightarrow V$ a linear map ,$U=\{S\in Hom_{\mathbb{F}}(V,V):\ S\circ T=0\}$, prove: $U$ is a subspace of $Hom_{\mathbb{F}}(V,V)$

Question

let $T:V\longrightarrow V$ be a linear map (transformation). let $$U=\{S\in Hom_{\mathbb{F}}(V,V):\ S\circ T=0\}$$

1. prove that $U$ is a subspace of $Hom_{\mathbb{F}}(V,V)$.

2. if $dimV=2$, what is $dimU$?

for section 1 ,someone send the following proof:

let $S_{1},S_{2}\in U\subset Hom_{\mathbb{F}}(V,V)$. let $\alpha_{1},\alpha_{2}\in\mathbb{F}$. by definition of $U$, $S_{1}\circ T=S_{2}\circ T=0$. Hence, $(\alpha_{1}S_{1}+\alpha_{2}S_{2})\circ T=(\alpha_{1}S_{1})\circ T+(\alpha_{2}s_{2})\circ T=0+0=0$. Hence, $\alpha_{1}S_{1},\alpha_{2}S_{2}\in U$.

I can't understand why does it prove what needed to prove, as the proof started with: let ${S_{1},S_{2}\in U\subset Hom_{\mathbb{F}}(V,V)}$.

I'd be very happy if someone could please explain me why is this a sufficient prove or if they could suggest an alternative proof.

for section 2, Intuitively it seems that $dim U=0$ but my thoughts are that this is probably not the answer for all cases of $S \circ T\in U$.

since $dimV=2$ then, $dimHom_{\mathbb{F}}(V,V)=4$, and that's the point where I got stock.

Answer 1

as been commented,

a criteria that can be used to show that a subset of a vector space is a subspace. One way would be to show that $U≠∅$ and that for every $v,w∈U$ and $a,b∈F$, we have $av+bw∈U$. It seems that someone used this criterion (but left out the $U≠∅$ part) hence, that proves what needed to be proved.