# $\operatorname{ker}L=\{0\}$ 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 $0$. ∎

Title $\operatorname{ker}L=\{0\}$ if and only if $L$ is injective operatornamekerL0IfAndOnlyIfLIsInjective 2013-03-22 14:44:46 2013-03-22 14:44:46 Mathprof (13753) Mathprof (13753) 11 Mathprof (13753) Theorem msc 15A04