A have compnents Aij
, and B have components Bij
. Then AB has components Aik
, where singular occurances of an index are free and repeated ones are summed over (a convention which saves time
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
so that the trace of a linear map is independent of its matrix representation.