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

###### Theorem.

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

###### 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 $\operatorname{ker}L=\{0\}$, 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 $\operatorname{ker}L=\{0\}$, which means that $v-v^{{\prime}}$ must be $0$. ∎

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## Corrections

typo by yark ✓

suppress link by Mathprof ✓

remove theorem numbering by alozano ✓

two addenda by alozano ✓

proof supplied by Wkbj79 ✓

suppress link by Mathprof ✓

remove theorem numbering by alozano ✓

two addenda by alozano ✓

proof supplied by Wkbj79 ✓