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[parent] proof of the dimension theorem for subspaces (Proof)

Let $S$ and $T$ be subspaces of a vector space. By the rank-nullity theorem and the second isomorphism theorem (for modules) we have

$\displaystyle \operatorname{dim}(S+T)$ $\displaystyle = \operatorname{dim}S + \operatorname{dim}((S+T)/S)$    
  $\displaystyle = \operatorname{dim}S + \operatorname{dim}(T/(S\cap T)).$    

Therefore
$\displaystyle \operatorname{dim}(S+T) + \operatorname{dim}(S\cap T)$ $\displaystyle = \operatorname{dim}S + \operatorname{dim}(T/(S\cap T)) + \operatorname{dim}(S\cap T)$    
  $\displaystyle = \operatorname{dim}S + \operatorname{dim}T,$    

by the rank-nullity theorem again.



"proof of the dimension theorem for subspaces" is owned by yark.
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Keywords:  vector space, dimension, vector subspace

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Cross-references: modules, second isomorphism theorem, rank-nullity theorem, vector space, subspaces

This is version 2 of proof of the dimension theorem for subspaces, born on 2007-01-17, modified 2007-01-17.
Object id is 8782, canonical name is ProofOfTheDimensionTheoremForVectorSubspaces.
Accessed 2778 times total.

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AMS MSC: 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank)
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