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

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Flesh out, put on a todo-list as the proof is important! I've done it several times but keep forgetting

## 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:

• [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"