View source for A linear map is injective if and only if the image of every non-zero vector is a non-zero vector

{{Stub page|grade=A|msg=Flesh out, put on a todo-list as the proof is important! I've done it several times but keep forgetting}}
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==Statement==
Let {{M|\mathbb{K} }} be a [[field]], let {{M|(U,\mathbb{K})}} and {{M|(V,\mathbb{K})}} be [[vector spaces]] over that field and let {{M|f\in}}[[the space of all linear maps between spaces|{{M|L(U,V)}}]] (that is to say: let {{M|f:U\rightarrow V}} be any [[linear map]] between {{M|U}} and {{M|V}}). Then{{rFAGBaLN}}:
* {{M|f:U\rightarrow V}} is [[injective]] {{iff}} {{M|1=\forall u\in U[u\ne 0\implies f(u)\ne 0]}}
** Note that {{M||1=\big(\forall u\in U[u\ne 0\implies f(u)\ne 0]\big)\iff\big(\forall u\in U[f(u)\eq 0\implies u\eq 0]\big)}} so we could just as well say:
*** {{M|f:U\rightarrow V}} is [[injective]] {{iff}} {{M|1=\forall u\in U[f(u)\eq 0\implies u\eq 0]}}
** We could also say "{{M|f:U\rightarrow V}} is injective {{iff}} the {{link|kernel|linear map}} is trivial"
==Proof==
{{Requires proof|easy=true|grade=B|msg=Good proof to do, however easy and very routine}}
==Notes==
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==References==
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{{Theorem Of|Linear Algebra}}