tensor product of subspaces of vector spaces
Proposition. Let V,W be vector spaces
over a field k. Moreover let A⊆V, B⊆W be subspaces
. Then V⊗B∩A⊗W=A⊗B.
Proof. The inclusion ,,⊇” is obvious. We will show the inclusion ,,⊆”.
Let {ei}i∈I and {e′j}j∈P be bases of A and B respectively. Moreover let {ei}i∈I′ be a completion of given basis of A to the basis of V, i.e. {ei}i∈I∪I′ is a basis of V. Analogously let {e′j}j∈P′ be a completion of a basis of B to the basis of W. Then each element q∈V⊗W can be uniquely written in a form
q=∑i∈I,j∈Pαi,jei⊗e′j+∑i∈I′,j∈Pβi,jei⊗e′j+ |
+∑i∈I,j∈P′δi,jei⊗e′j+∑i∈I′,j∈P′γi,jei⊗e′j. |
Assume that q∈V⊗B∩A⊗W. Let i∈I′ and j∈P′. Consider the following linear map: fi:V→k such that fi(et)=1 if i=t and fi(et)=0 if i≠t. Analogously we define gj:W→k. Then we combine these two mappings into one, i.e.
fi⊗gj:V⊗W→k; |
(fi⊗gj)(v⊗w)=fi(v)gj(w). |
Furthermore we have
(fi⊗gj)(q)=γi,j. |
Note that since q∈V⊗B, then (fi⊗gj)(q)=0 and thus
γi,j=0. |
Similarly we obtain that all βi,j and δi,j are equal to 0. Thus
q=∑i∈I,j∈Pαi,jei⊗e′j∈A⊗B, |
which completes 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 |