some facts about injective and surjective linear maps


Let k be a field and V,W be vector spacesMathworldPlanetmath over k.

PropositionPlanetmathPlanetmath. Let f:V→W be an injectivePlanetmathPlanetmath linear map. Then there exists a (surjectivePlanetmathPlanetmath) linear map g:W→V such that g∘f=idV.

Proof. Of course Im⁢(f) is a subspacePlanetmathPlanetmathPlanetmath of W so f:V→Im⁢(f) is a linear isomorphism. Let (ei)i∈I be a basis of Im⁢(f) and (ej)j∈J be its completion to the basis of W, i.e. (ei)i∈I∪J is a basis of W. Define g:W→V on the basis as follows:

g⁢(ei)=f-1⁢(ei),if ⁢i∈I;
g⁢(ej)=0,if ⁢j∈J.

We will show that g∘f=idV.

Let v∈V. Then

f⁢(v)=∑i∈Iαi⁢ei,

where αi∈k (note that the indexing set is I). Thus we have

(g∘f)⁢(v)=g⁢(∑i∈Iαi⁢ei)=∑i∈Iαi⁢g⁢(ei)=∑i∈Iαi⁢f-1⁢(ei)=
=f-1⁢(∑i∈Iαi⁢ei)=f-1⁢(f⁢(v))=v.

It is clear that the equality g∘f=idV implies that g is surjective. □

Proposition. Let g:W→V be a surjective linear map. Then there exists a (injective) linear map f:V→W such that g∘f=idV.

Proof. Let (ei)i∈I be a basis of V. Since g is onto, then for any i∈I there exist wi∈W such that g⁢(wi)=ei. Now define f:V→W by the formulaMathworldPlanetmathPlanetmath

f⁢(ei)=wi.

It is clear that g∘f=idV, which implies that f is injective. □

If we combine these two propositions, we have the following corollary:

Corollary. There exists an injective linear map f:V→W if and only if there exists a surjective linear map g:W→V.

Title some facts about injective and surjective linear maps
Canonical name SomeFactsAboutInjectiveAndSurjectiveLinearMaps
Date of creation 2013-03-22 18:32:22
Last modified on 2013-03-22 18:32:22
Owner joking (16130)
Last modified by joking (16130)
Numerical id 6
Author joking (16130)
Entry type DerivationPlanetmathPlanetmath
Classification msc 15A04