tensor product of subspaces of vector spaces

PropositionPlanetmathPlanetmath. Let V,W be vector spacesMathworldPlanetmath over a field k. Moreover let AV, BW be subspacesPlanetmathPlanetmath. Then VBAW=AB.

Proof. The inclusion ,,” is obvious. We will show the inclusion ,,”.

Let {ei}iI and {ej}jP be bases of A and B respectively. Moreover let {ei}iI be a completion of given basis of A to the basis of V, i.e. {ei}iII is a basis of V. Analogously let {ej}jP be a completion of a basis of B to the basis of W. Then each element qVW can be uniquely written in a form


Assume that qVBAW. Let iI and jP. Consider the following linear map: fi:Vk such that fi(et)=1 if i=t and fi(et)=0 if it. Analogously we define gj:Wk. Then we combine these two mappings into one, i.e.


Furthermore we have


Note that since qVB, then (figj)(q)=0 and thus


Similarly we obtain that all βi,j and δi,j are equal to 0. Thus


which completesPlanetmathPlanetmathPlanetmath the proof.

Title tensor product of subspaces of vector spaces
Canonical name TensorProductOfSubspacesOfVectorSpaces
Date of creation 2013-03-22 18:49:16
Last modified on 2013-03-22 18:49:16
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 15A69