Math 2.0

Linear and multilinear algebra; matrix theory Theorem

Dimension theorem for vector spaces

Builds on Basis

Statement

Let FF be a field and let VV be a vector space over FF. Any two bases of VV have the same cardinality.

Remarks

For finite bases, this follows from the Steinitz exchange lemma. For infinite bases, the proof uses cardinal arithmetic: each vector of one basis is a combination of finitely many vectors of the other basis. The common cardinality is the dimension of VV. The theorem holds more generally for free modules over a commutative ring that is not the zero ring.

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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, no evidence levelVector spacedefinitionLinear combination, definition, no evidence levelLinear combinationdefinitionLinear subspace, definition, no evidence levelLinear subspacedefinitionLinear independence, definition, no evidence levelLinear independencedefinitionLinear span, definition, no evidence levelLinear spandefinitionSpanning set, definition, no evidence levelSpanning setdefinitionBasis, definition, no evidence levelBasisdefinitionDimension theorem for vector spaces, statement, no evidence levelDimension theorem for…statement
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.
  • no evidence level
  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. Dimension theorem for vector spaces statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces, lead section
About this page
Tag0WK0Y8RH1M
URIm2://0WK0Y8RH1M
Aliasesinvariance-of-dimension
Kindstatement, theorem
WritingW1 stated
MSC15A03
StatusThis copy of the site has no build data. Run m2 status 0WK0Y8RH1M.
UsesBasis
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