Let matrix A have compnents Aij, and B have components Bij. Then AB has components AikBkj, where singular occurances of an index are free and repeated ones are summed over (a convention which saves time). So:

Tr(AB)=AikBki=BkiAik=Tr(BA)

And that's all there is to it (remember that Tr(C)=Cii, 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-1A)=Tr(A)

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

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