Linear and multilinear algebra; matrix theory Vector spaces Theorem
Every vector space has a basis
Builds on Basis
Statement
Let be a field and let be a vector space over . Then has a basis.
Remarks
For a space with a finite spanning set, the proof is elementary: remove vectors from the spanning set until it is linearly independent. In general, the proof uses Zorn's lemma: a maximal linearly independent subset of is a basis. The statement for all vector spaces is equivalent to the axiom of choice, over set theory without choice. For example, the axiom of choice gives a basis of as a vector space over , a Hamel basis.
Reading path
To understand this page from the ground up, read these 8 pages in order. The foundations come first.
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- Vector space definition
- Linear combination definition
- Linear subspace definition
- Linear independence definition
- Linear span definition
- Spanning set definition
- Basis definition
- Every vector space has a basis statement
Where this leads
2 pages build on this one.
Further reading
- Web https://en.wikipedia.org/wiki/Basis_(linear_algebra), Proof that every vector space has a basis
About this page
| Tag | 00MS9VMQ9Z |
|---|---|
| URI | m2://00MS9VMQ9Z |
| Kind | statement, theorem |
| Writing | W0 stub |
| MSC | 15A03, 03E25 |
| Status | This copy of the site has no build data. Run m2 status 00MS9VMQ9Z. |
| Uses | Basis |
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