$\ker L=\{0\}$ 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 $0$. ∎

Title	$\ker L=\{0\}$ if and only if $L$ is injective
Canonical name	operatornamekerL0IfAndOnlyIfLIsInjective
Date of creation	2013-03-22 14:44:46
Last modified on	2013-03-22 14:44:46
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	11
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 15A04