= Solution
The two square products can have different sizes, but their <matrix traces> are both the same finite double sum:
$$
\operatorname{tr}(XY)=\sum_{i=1}^{l}(XY)_{ii}
=\sum_{i=1}^{l}\sum_{j=1}^{k}X_{ij}Y_{ji}
=\sum_{j=1}^{k}\sum_{i=1}^{l}Y_{ji}X_{ij}
=\operatorname{tr}(YX).
$$
Interchanging two finite sums and commuting their scalar factors is valid over the real numbers. No commutation of the <matrix> products themselves is asserted. This proves the <cyclic property of the trace> in the rectangular-factor case.
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