Editing Tensor product (section)

== Linear maps as tensors ==
Given two finite dimensional vector spaces {{math|''U''}}, {{math|''V''}} over the same field {{math|''K''}}, denote the [[dual space]] of {{math|''U''}} as {{math|''U*''}}, and the {{math|''K''}}-vector space of all linear maps from {{math|''U''}} to {{math|''V''}} as {{math|Hom(''U'',''V'')}}. There is an isomorphism:
<math display="block">U^* \otimes V \cong \mathrm{Hom}(U, V),</math>
defined by an action of the pure tensor <math>f \otimes v \in U^*\otimes V</math> on an element of {{tmath|1= U }},
<math display="block">(f \otimes v)(u) = f(u) v.</math>

Its "inverse" can be defined using a basis <math>\{u_i\}</math> and its dual basis <math>\{u^*_i\}</math> as in the section "[[#Evaluation map and tensor contraction|Evaluation map and tensor contraction]]" above:
<math display="block">\begin{cases} \mathrm{Hom} (U,V) \to U^* \otimes V \\ F \mapsto \sum_i u^*_i \otimes F(u_i). \end{cases}</math>

This result implies:
<math display="block">\dim(U \otimes V) = \dim(U)\dim(V),</math>
which automatically gives the important fact that <math>\{u_i\otimes v_j\}</math> forms a basis of <math>U \otimes V</math> where <math>\{u_i\}, \{v_j\}</math> are bases of {{math|''U''}} and {{math|''V''}}.

Furthermore, given three vector spaces {{math|''U''}}, {{math|''V''}}, {{math|''W''}} the tensor product is linked to the vector space of ''all'' linear maps, as follows:
<math display="block">\mathrm{Hom} (U \otimes V, W) \cong \mathrm{Hom} (U, \mathrm{Hom}(V, W)).</math>
This is an example of [[adjoint functor]]s: the tensor product is "left adjoint" to Hom.