Math 2.0

statement theorem

Every vector space has a 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.

Explanations

L1 human

Every vector space has a set of building blocks that is independent and reaches every vector. For finite-dimensional spaces, we can find one by hand. For very large spaces, we must use the axiom of choice, and we may not be able to write the basis down.

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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

Sources

  1. Web https://en.wikipedia.org/wiki/Basis_(linear_algebra), Proof that every vector space has a basis