Math 2.0

Linear and multilinear algebra; matrix theory Theorem

An injective endomorphism of a finite-dimensional space is surjective

Builds on Rank–nullity theorem

Statement

Let FF be a field, let VV be a finite-dimensional vector space over FF, and let T ⁣:V→VT \colon V \to V be a linear map. Then these conditions are equivalent:

  1. TT is injective.
  2. TT is surjective.
  3. TT is bijective.

Remarks

The proof uses the rank–nullity theorem: ker⁡T=0\ker T = 0 if and only if dim⁡im⁡T=dim⁡V\dim \im T = \dim V. It fails in infinite dimensions: on F[x]F[x], multiplication by xx is injective and not surjective. In characteristic 00, the derivative on F[x]F[x] is surjective and not injective. For square matrices, this is the statement that Ax=0Ax = 0 has only the trivial solution if and only if Ax=bAx = b has a solution for each bb.

Reading path

To understand this page from the ground up, read these 16 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.
  • no evidence level
  1. Vector space definition
  2. Linear combination definition
  3. Linear map definition
  4. Linear subspace definition
  5. Linear independence definition
  6. Image of a linear map definition
  7. Kernel of a linear map definition
  8. Linear span definition
  9. Spanning set definition
  10. Basis definition
  11. Finite-dimensional vector space definition
  12. Dimension definition
  13. Rank of a linear map definition
  14. Nullity definition
  15. Rank–nullity theorem statement
  16. An injective endomorphism of a finite-dimensional space is surjective statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Rank–nullity_theorem, Reformulations and generalizations
About this page
Tag0912EEEVCT
URIm2://0912EEEVCT
Kindstatement, theorem
WritingW0 stub
MSC15A04
StatusThis copy of the site has no build data. Run m2 status 0912EEEVCT.
UsesRank–nullity theorem

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