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[parent] $\operatorname{ker} L=\{0\}$ if and only if $L$ is injective (Theorem)
Theorem 1   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')$ for vectors $ v,v'\in V$. Hence $ L(v-v') = L(v)-L(v') = 0$ because $ L$ is linear. Therefore, $ v-v'$ is in $ \operatorname{ker}L=\{0\}$, which means that $ v-v'$ must be 0. $ \qedsymbol$



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Cross-references: vector, kernel, injective, vector spaces, linear map
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This is version 8 of $\operatorname{ker} L=\{0\}$ if and only if $L$ is injective, born on 2004-10-17, modified 2006-10-27.
Object id is 6384, canonical name is OperatornamekerL0IfAndOnlyIfLIsInjective.
Accessed 1548 times total.

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AMS MSC: 15A04 (Linear and multilinear algebra; matrix theory :: Linear transformations, semilinear transformations)
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