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

q=iI,jPαi,jeiej+iI,jPβi,jeiej+
+iI,jPδi,jeiej+iI,jPγi,jeiej.

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.

figj:VWk;
(figj)(vw)=fi(v)gj(w).

Furthermore we have

(figj)(q)=γi,j.

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

γi,j=0.

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

q=iI,jPαi,jeiejAB,

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