Let

matrix A have compnents A

_{ij}, and B have components B

_{ij}. Then AB has components A

_{ik}B

_{kj}, where singular occurances of an index are free and repeated ones are summed over (

a convention which saves time). So:

Tr(AB)=A_{ik}B_{ki}=B_{ki}A_{ik}=Tr(BA)

And that's all there is to it (remember that Tr(C)=C

_{ii}, the sum of the diagonal entries).

A very useful result for one so easy to prove, and important consequence being that

conjugate matricies have the same trace

Tr(QAQ^{-1})=Tr(QQ^{-1}A)=Tr(A)

so that the trace of a linear map is independent of its matrix representation.