$\ker L=\{0\}$ if and only if $L$ is injective

Theorem.

A linear map between vector spaces is injective if and only if its kernel is $\mathrm{\{}\mathrm{0}\mathrm{\}}$.

Proof.

Let $L:V\to W$ be a linear map. Suppose $L$ is injective, and $L(v)=0$ for some vector$v\in V$. Also, $L(0)=0$ because $L$ is linear.Then $L(v)=L(0)$, so $v=0$. On the other hand, suppose$\ker L=\{0\}$, and $L(v)=L(v^{\prime })$ for vectors $v,v^{\prime }\in V$.Hence $L(v-v^{\prime })=L(v)-L(v^{\prime })=0$ because $L$ is linear.Therefore, $v-v^{\prime }$ is in $\ker L=\{0\}$, which means that$v-v^{\prime }$ must be $0$.∎