Kernel is a Subspace

For any linear map $T:V\to W$, the kernel $ker(T)$ is a subspace of $V$

Proof

1. Apply Subspace Test
2. $ker(T)$ is non-empty as $\overrightarrow{0}\in ker(T)$
3. $T(c\overrightarrow{x}+\overrightarrow{y})=cT(\overrightarrow{x})+T(\overrightarrow{y})$
4. $=c(0w)+0w$
5. $=0w+0w$
6. $=0w$
7. Therefore, $cx+y\in ker(T)$, thus its closed under linear combinations
8. Thus, $ker(T)$ is a subspace