Editing Trace (linear algebra) (section)

==Traces in the language of tensor products==
Given a vector space {{mvar|V}}, there is a natural bilinear map {{math|''V'' × ''V''<sup>∗</sup> → ''F''}} given by sending {{math|(''v'', &phi;)}} to the scalar {{math|&phi;(''v'')}}. The [[tensor product#Universal property|universal property]] of the [[tensor product]] {{math|''V'' ⊗ ''V''<sup>∗</sup>}} automatically implies that this bilinear map is induced by a linear functional on {{math|''V'' ⊗ ''V''<sup>∗</sup>}}.<ref name="kassel">{{cite book|last1=Kassel|first1=Christian|title=Quantum groups|series=[[Graduate Texts in Mathematics]]|volume=155|publisher=[[Springer-Verlag]]|location=New York|year=1995|isbn=0-387-94370-6|mr=1321145|doi=10.1007/978-1-4612-0783-2|zbl=0808.17003}}</ref>

Similarly, there is a natural bilinear map {{math|''V'' × ''V''<sup>∗</sup> → Hom(''V'', ''V'')}} given by sending {{math|(''v'', &phi;)}} to the linear map {{math|''w'' ↦ &phi;(''w'')''v''}}. The universal property of the tensor product, just as used previously, says that this bilinear map is induced by a linear map {{math|''V'' ⊗ ''V''<sup>∗</sup> → Hom(''V'', ''V'')}}. If {{mvar|V}} is finite-dimensional, then this linear map is a [[linear isomorphism]].<ref name="kassel" /> This fundamental fact is a straightforward consequence of the existence of a (finite) basis of {{mvar|V}}, and can also be phrased as saying that any linear map {{math|''V'' → ''V''}} can be written as the sum of (finitely many) rank-one linear maps. Composing the inverse of the isomorphism with the linear functional obtained above results in a linear functional on {{math|Hom(''V'', ''V'')}}. This linear functional is exactly the same as the trace.

Using the definition of trace as the sum of diagonal elements, the matrix formula {{math|tr('''AB''') {{=}} tr('''BA''')}} is straightforward to prove, and was given above. In the present perspective, one is considering linear maps {{mvar|S}} and {{mvar|T}}, and viewing them as sums of rank-one maps, so that there are linear functionals {{math|''&phi;''<sub>''i''</sub>}} and {{math|''&psi;''<sub>''j''</sub>}} and nonzero vectors {{math|''v''<sub>''i''</sub>}} and {{math|''w''<sub>''j''</sub>}} such that {{math|''S''({{mvar|u}}) {{=}} Σ''&phi;''<sub>''i''</sub>(''u'')''v''<sub>''i''</sub>}} and {{math|''T''({{mvar|u}}) {{=}} Σ''&psi;''<sub>''j''</sub>(''u'')''w''<sub>''j''</sub>}} for any {{mvar|u}} in {{mvar|V}}. Then
:<math>(S\circ T)(u)=\sum_i\varphi_i\left(\sum_j\psi_j(u)w_j\right)v_i=\sum_i\sum_j\psi_j(u)\varphi_i(w_j)v_i </math>
for any {{mvar|u}} in {{mvar|V}}. The rank-one linear map {{math|''u'' ↦ ''&psi;''<sub>''j''</sub>(''u'')''&phi;''<sub>''i''</sub>(''w''<sub>''j''</sub>)''v''<sub>''i''</sub>}} has trace {{math|''&psi;''<sub>''j''</sub>(''v''<sub>''i''</sub>)''&phi;''<sub>''i''</sub>(''w''<sub>''j''</sub>)}} and so
:<math>\operatorname{tr}(S\circ T)=\sum_i\sum_j\psi_j(v_i)\varphi_i(w_j)=\sum_j\sum_i\varphi_i(w_j)\psi_j(v_i).</math>
Following the same procedure with {{mvar|S}} and {{mvar|T}} reversed, one finds exactly the same formula, proving that {{math|tr(''S'' ∘ ''T'')}} equals {{math|tr(''T'' ∘ ''S'')}}.

The above proof can be regarded as being based upon tensor products, given that the fundamental identity of {{math|End(''V'')}} with {{math|''V'' ⊗ ''V''<sup>∗</sup>}} is equivalent to the expressibility of any linear map as the sum of rank-one linear maps. As such, the proof may be written in the notation of tensor products. Then one may consider the multilinear map {{math|''V'' × ''V''<sup>∗</sup> × ''V'' × ''V''<sup>∗</sup> → ''V'' ⊗ ''V''<sup>∗</sup>}} given by sending {{math|(''v'', ''&phi;'', ''w'', ''&psi;'')}} to  {{math|''&phi;''(''w'')''v'' ⊗ ''&psi;''}}. Further composition with the trace map then results in {{math|''&phi;''(''w'')''&psi;''(''v'')}}, and this is unchanged if one were to have started with {{math|(''w'', ''&psi;'', ''v'', ''&phi;'')}} instead. One may also consider the bilinear map {{math|End(''V'') × End(''V'') → End(''V'')}} given by sending {{math|(''f'', ''g'')}} to the composition {{math|''f'' ∘ ''g''}}, which is then induced by a linear map {{math|End(''V'') ⊗ End(''V'') → End(''V'')}}. It can be seen that this coincides with the linear map {{math|''V'' ⊗ ''V''<sup>∗</sup> ⊗ ''V'' ⊗ ''V''<sup>∗</sup> → ''V'' ⊗ ''V''<sup>∗</sup>}}. The established symmetry upon composition with the trace map then establishes the equality of the two traces.<ref name="kassel" />

For any finite dimensional vector space {{mvar|V}}, there is a natural linear map {{math|''F'' → ''V'' ⊗ ''V''{{'}}}}; in the language of linear maps, it assigns to a scalar {{mvar|c}} the linear map {{math|''c''⋅id<sub>''V''</sub>}}. Sometimes this is called ''coevaluation map'', and the trace {{math|''V'' ⊗ ''V''{{'}} → ''F''}} is called ''evaluation map''.<ref name="kassel" /> These structures can be axiomatized to define [[categorical trace]]s in the abstract setting of [[category theory]].