proof of the dimension theorem for subspaces

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.

Title	proof of the dimension theorem for subspaces
Canonical name	ProofOfTheDimensionTheoremForSubspaces
Date of creation	2013-03-22 16:35:17
Last modified on	2013-03-22 16:35:17
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	5
Author	yark (2760)
Entry type	Proof
Classification	msc 15A03