Math 2.0

Linear and multilinear algebra; matrix theory Vector spaces Theorem

Every vector space has a basis

Builds on Basis

Statement

Let FF be a field and let VV be a vector space over FF. Then VV 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 VV 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 ℝ\R as a vector space over ℚ\Q, 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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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Vector space definition
  2. Linear combination definition
  3. Linear subspace definition
  4. Linear independence definition
  5. Linear span definition
  6. Spanning set definition
  7. Basis definition
  8. Every vector space has a basis statement

Where this leads

2 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Basis_(linear_algebra), Proof that every vector space has a basis
About this page
Tag00MS9VMQ9Z
URIm2://00MS9VMQ9Z
Kindstatement, theorem
WritingW0 stub
MSC15A03, 03E25
StatusThis copy of the site has no build data. Run m2 status 00MS9VMQ9Z.
UsesBasis

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