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