Solution
= Solution
A map between <vector spaces> over the same <field> is linear when $L(av+bw)=aL(v)+bL(w)$ for every $v,w$ and scalars $a,b$. Its <kernel> is $\ker L=\{v\in V:L(v)=0\}$. It contains zero because $L(0)=0$. If $v,w\in\ker L$, then
$$
L(av+bw)=aL(v)+bL(w)=0.
$$
Thus it is closed under linear combinations and is a <vector subspace> of $V$. No injectivity or equality of the dimensions of $V,W$ is needed.