Math 2.0

statement theorem

Every spanning set contains a basis

Statement

Let FF be a field, let VV be a vector space over FF, and let S⊆VS \subseteq V be a spanning set of VV. Then there is a subset B⊆SB \subseteq S that is a basis of VV.

Remarks

For finite SS, remove from SS, one at a time, a vector that is a linear combination of the other vectors. When no such vector remains, the set is linearly independent and still spans VV. In general, take a maximal linearly independent subset of SS, by Zorn's lemma. For example, {(1,0),(0,1),(1,1)}\{(1, 0), (0, 1), (1, 1)\} spans R2\R^2 and contains the basis {(1,0),(1,1)}\{(1, 0), (1, 1)\}.

Explanations

L1 human

If a set of vectors reaches the whole space, then some part of it is a basis. We throw away vectors that the others can already build. When nothing redundant remains, the remaining vectors form a basis.

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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 spanning set contains a basis statement

Sources

  1. Web https://en.wikipedia.org/wiki/Basis_(linear_algebra), Properties