tensor product and dual spaces
Let be a field and be a vector space over . Recall that
denotes the dual space of (which is also a vector space over ).
Proposition. Let and be vector spaces. Consider the map such that
where is a basis of , is a basis of and . Then for any and we have:
Since is arbitrary then we can write this equality in the form:
and since are linearly independent we obtain that for all . Again since was arbitrary we obtain that for all . Now since are linearly independent we obtain that for all . Thus .
Now assume that . Let be a basis of and let be an induced basis of . Moreover let be a basis of . We wish to show that is onto, so let be an element of . Define by the formula:
where is such that . Then for any from and for any from we have:
and thus .
Remark. The map from the previous proposition is very important in studying algebras and coalgebras (more precisly it is an essence in defining dual (co)algebras). Unfortunetly does not have to be an isomorphism in general. Nevertheless, the spaces and are always isomorphic (see this entry (http://planetmath.org/TensorProductOfDualSpacesIsADualSpaceOfTensorProduct) for more details).
|Title||tensor product and dual spaces|
|Date of creation||2013-03-22 18:31:51|
|Last modified on||2013-03-22 18:31:51|
|Last modified by||joking (16130)|