= Positive semidefinite trace nonnegativity
For real <positive semidefinite matrices> $A,B$, their product need not be symmetric, but its <matrix trace> is nonnegative:
$$
\operatorname{tr}(AB)=\operatorname{tr}(\sqrt A\,B\sqrt A)\geq0.
$$
The equality uses the cyclic property of the <matrix trace>, and the last <matrix> is a <positive semidefinite matrix>. Consequently the <Loewner order> inequality $A\preceq C$ implies $\operatorname{tr}(AB)\leq\operatorname{tr}(CB)$ whenever $B\succeq0$.
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