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
dim(S+T) | =dimS+dim((S+T)/S) | ||
=dimS+dim(T/(S∩T)). |
Therefore
dim(S+T)+dim(S∩T) | =dimS+dim(T/(S∩T))+dim(S∩T) | ||
=dimS+dimT, |
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 |