Linear mapping/Kernel/Injectivity criterion/Section

Definition

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

$\varphi \colon V\longrightarrow W$

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

${}\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,$

is called the kernel of ${}\varphi$.

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

The following criterion for injectivity is important.

Lemma

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

$\varphi \colon V\longrightarrow W$

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

holds.

Proof

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

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

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

$\Box$