# 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 ProofOfTheDimensionTheoremForSubspaces 2013-03-22 16:35:17 2013-03-22 16:35:17 yark (2760) yark (2760) 5 yark (2760) Proof msc 15A03