# Linear mapping/Kernel/Injectivity criterion/Section

## Definition

Let ${\displaystyle {}K}$ denote a field, let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote ${\displaystyle {}K}$-vector spaces, and let

${\displaystyle \varphi \colon V\longrightarrow W}$

denote a ${\displaystyle {}K}$-linear mapping. Then

${\displaystyle {}\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,}$
is called the kernel of ${\displaystyle {}\varphi }$.

The kernel is a linear subspace of ${\displaystyle {}V}$.

The following criterion for injectivity is important.

## Lemma

Let ${\displaystyle {}K}$ denote a field, let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote ${\displaystyle {}K}$-vector spaces, and let

${\displaystyle \varphi \colon V\longrightarrow W}$

denote a ${\displaystyle {}K}$-linear mapping. Then ${\displaystyle {}\varphi }$ is injective if and only if ${\displaystyle {}\operatorname {kern} \varphi =0}$ holds.

### Proof

If the mapping is injective, then there can exist, apart from ${\displaystyle {}0\in V}$, no other vector ${\displaystyle {}v\in V}$ with ${\displaystyle {}\varphi (v)=0}$. Hence, ${\displaystyle {}\varphi ^{-1}(0)=\{0\}}$.
So suppose that ${\displaystyle {}\operatorname {kern} \varphi =0}$, and let ${\displaystyle {}v_{1},v_{2}\in V}$ be given with ${\displaystyle {}\varphi (v_{1})=\varphi (v_{2})}$. Then, due to linearity,

${\displaystyle {}\varphi (v_{1}-v_{2})=\varphi (v_{1})-\varphi (v_{2})=0\,.}$

Therefore, ${\displaystyle {}v_{1}-v_{2}\in \operatorname {kern} \varphi }$, and so ${\displaystyle {}v_{1}=v_{2}}$.

${\displaystyle \Box }$