tensor product of subspaces of vector spaces
Proof. The inclusion ,,” is obvious. We will show the inclusion ,,”.
Let and be bases of and respectively. Moreover let be a completion of given basis of to the basis of , i.e. is a basis of . Analogously let be a completion of a basis of to the basis of . Then each element can be uniquely written in a form
Furthermore we have
Note that since , then and thus
Similarly we obtain that all and are equal to . Thus
which completes the proof.
|Title||tensor product of subspaces of vector spaces|
|Date of creation||2013-03-22 18:49:16|
|Last modified on||2013-03-22 18:49:16|
|Last modified by||joking (16130)|