= Solution
In a finite-dimensional <basis>, interchange the two finite sums:
$$
\operatorname{Tr}(AB)=\sum_{j,k}A_{jk}B_{kj}
=\sum_{k,j}B_{kj}A_{jk}
=\boxed{\operatorname{Tr}(BA)}.
$$
This <cyclic property of the trace> does not assert $AB=BA$ or permit arbitrary reordering of three factors. For infinite-dimensional <linear operators>, the <traces> and rearrangements need hypotheses, for example one <trace-class operator> and one <bounded operator>; an arbitrary pair of unbounded <linear operators> is not covered.
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