Subspace of linear transformation proof legit? Let V and W be vectorspaces over a body $\mathbb{K}$ and U a subspace of W.
$f:V \rightarrow W$ is a linear transformation.
Proof that $S=\{v \in V |f(v) \in U\}$ is a subspace of V.
Because we can assume that the transformation is linear, the criteria for linearity always holds, right?
Let $v, v‘ \in V | f(v),f(v‘) \in U$
$f:V \rightarrow W$ be a linear transformation.
Thus any vector v,v‘ satisfies $f(v + v‘)=f(v)+f(v‘)$ and $f(av)=af(v)$.
Is the question proofed just by relying on the fact that the transformation is linear to begin with?
 A: Yes, it relies on the fact that $f$ is a linear transformation. For $S$ to be a subspace of $V$, it has to satisfy three conditions.
$1.$ The additive identity of $V$ denoted $0$ is an element of $S$.
$2.$. $S$ is closed under addition.
$3. $ S is closed under scalar multiplication.
You have proved both of them apart from the first one. The first one is easy as a linear transformation always takes $0$ to $0$. Thus, $0\in V$ as $f(0)=0\in W$. Here the zero on the right side is the additive identity of $W$.
You have showed that it closed under addition and closed under scalar multiplication. Thus, $S$ is a subspace of $V$.
But I would word that part of the proof a bit differently. Let $u, v\in V$. We have to show that $u+v\in V$. Given that $u, v\in V$, we know that $f(u)$ and $f(v)$ are in $W$. Because $W$ is a vector space, it is closed under addition. Therefore $f(u)+f(v)\in W$. Now use the fact that $f$ is a linear transformation to deduce that $f(u+v)\in W$ and therefore $u+v\in V$.
The same goes for proving that $S$ is closed under scalar multiplication. Otherwise, the proof is good. Well done!
