Linear and multilinear algebra; matrix theory Theorem
An injective endomorphism of a finite-dimensional space is surjective
Builds on Rank–nullity theorem
Statement
Let be a field, let be a finite-dimensional vector space over , and let be a linear map. Then these conditions are equivalent:
- is injective.
- is surjective.
- is bijective.
Remarks
The proof uses the rank–nullity theorem: if and only if . It fails in infinite dimensions: on , multiplication by is injective and not surjective. In characteristic , the derivative on is surjective and not injective. For square matrices, this is the statement that has only the trivial solution if and only if has a solution for each .
Reading path
To understand this page from the ground up, read these 16 pages in order. The foundations come first.
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- no evidence level
- Vector space definition
- Linear combination definition
- Linear map definition
- Linear subspace definition
- Linear independence definition
- Image of a linear map definition
- Kernel of a linear map definition
- Linear span definition
- Spanning set definition
- Basis definition
- Finite-dimensional vector space definition
- Dimension definition
- Rank of a linear map definition
- Nullity definition
- Rank–nullity theorem statement
- An injective endomorphism of a finite-dimensional space is surjective statement
Further reading
- Web https://en.wikipedia.org/wiki/Rank–nullity_theorem, Reformulations and generalizations
About this page
| Tag | 0912EEEVCT |
|---|---|
| URI | m2://0912EEEVCT |
| Kind | statement, theorem |
| Writing | W0 stub |
| MSC | 15A04 |
| Status | This copy of the site has no build data. Run m2 status 0912EEEVCT. |
| Uses | Rank–nullity theorem |
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