some facts about injective and surjective linear maps
Let be a field and be vector spaces![]()
over .
Proposition. Let be an injective
linear map. Then there exists a (surjective
) linear map such that .
Proof. Of course is a subspace of so is a linear isomorphism. Let be a basis of and be its completion to the basis of , i.e. is a basis of . Define on the basis as follows:
We will show that .
Let . Then
where (note that the indexing set is ). Thus we have
It is clear that the equality implies that is surjective.
Proposition. Let be a surjective linear map. Then there exists a (injective) linear map such that .
Proof. Let be a basis of . Since is onto, then for any there exist such that .
Now define by the formula![]()
It is clear that , which implies that is injective.
If we combine these two propositions, we have the following corollary:
Corollary. There exists an injective linear map if and only if there exists a surjective linear map .
| 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 | Derivation |
| Classification | msc 15A04 |