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Given: $A$ is positive definite, $B$ is symmetric and $\operatorname{tr}(B)\geqslant 0$, what could be a minimal additional condition, so that $\operatorname{tr}(AB)\geqslant 0$? ("$B$ is positive definite" is too strong)

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Let $(0\le)\lambda^\uparrow_1(A)\le\ldots\le\lambda^\uparrow_n(A)$ be the eigenvalues of $A$ and $\lambda^\downarrow_1(B)\ge\ldots\ge\lambda^\downarrow_n(B)$ be the eigenvalues of $B$. The minimum possible value of $\operatorname{trace}(AB)$ is then $t=\sum_{i=1}^n\lambda^\uparrow_i(A)\lambda^\downarrow_i(B)$. So, one parsimonious sufficient condition for the trace to be nonnegative is that $t\ge0$.

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