A linear map is injective if and only if the image of every nonzero vector is a nonzero vector
From Maths
Stub grade: A
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Flesh out, put on a todolist as the proof is important! I've done it several times but keep forgetting
Contents
Statement
Let [ilmath]\mathbb{K} [/ilmath] be a field, let [ilmath](U,\mathbb{K})[/ilmath] and [ilmath](V,\mathbb{K})[/ilmath] be vector spaces over that field and let [ilmath]f\in[/ilmath][ilmath]L(U,V)[/ilmath] (that is to say: let [ilmath]f:U\rightarrow V[/ilmath] be any linear map between [ilmath]U[/ilmath] and [ilmath]V[/ilmath]). Then^{[1]}:
 [ilmath]f:U\rightarrow V[/ilmath] is injective if and only if [ilmath]\forall u\in U[u\ne 0\implies f(u)\ne 0][/ilmath]
 Note that [ilmath]\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)[/ilmath] so we could just as well say:
 [ilmath]f:U\rightarrow V[/ilmath] is injective if and only if [ilmath]\forall u\in U[f(u)\eq 0\implies u\eq 0][/ilmath]
 We could also say "[ilmath]f:U\rightarrow V[/ilmath] is injective if and only if the kernel is trivial"
 Note that [ilmath]\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)[/ilmath] so we could just as well say:
Proof
Grade: B
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This proof has been marked as an page requiring an easy proof
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Good proof to do, however easy and very routine
This proof has been marked as an page requiring an easy proof
Notes
References
