Is $\mbox{tr}(XY) \geq 0$ for all $X, Y \in \Bbb S_+$?
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1See http://math.stackexchange.com/q/113842/27978. If $X,Y$ are symmetric, then yes. – copper.hat Aug 06 '14 at 02:54
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This also follow from http://math.stackexchange.com/questions/168470/if-a-ge-0-and-b-ge-0-then-sigmaab-subset-mathbbr. – Jonas Meyer Aug 06 '14 at 03:09
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Yes, because $\mathrm{tr}(XY)=\mathrm{tr}(X\sqrt Y\sqrt Y)=\mathrm{tr}(\sqrt Y X\sqrt Y)\geq 0$.
Positive semidefinite matrices have positive semidefinite square roots. The trace satisfies $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. If $A$ and $X$ are positive semidefinite, then so is $AXA$. The trace of a positive semidefinite matrix is nonnegative.
Jonas Meyer
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1I did not understand how $AXA$ is positive semidefinite, given $A$ and $X$ are positive semidefinite? – Anindita Sarkar Apr 27 '24 at 10:47
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@AninditaSarkar $A=C^T\Sigma_1 C$, $X=D^T\Sigma_2 D$,then $$AXA=(C^T\Sigma_1 C)(D^T\Sigma_2 D)(C^T\Sigma_1 C)=(DC^T\Sigma_1 C)^T\Sigma_2 (DC^T\Sigma_1 C)$$ – HeroZhang001 Jul 21 '24 at 03:40
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Let $n \times n$ matrices $\rm A$ and $\rm B$ be symmetric and positive semidefinite. Hence, there exists an $n \times r$ matrix $\rm Q$ such that $\rm B = Q Q^\top$, and
$$\mbox{tr} \left( {\rm A} {\rm B} \right) = \mbox{tr} \left( {\rm A} {\rm Q} {\rm Q}^\top \right) = \mbox{tr} \left( {\rm Q}^\top {\rm A} {\rm Q} \right) = \sum_{i=1}^r \underbrace{{\rm q}_i^\top {\rm A} \, {\rm q}_i}_{\geq 0} \color{blue}{\geq 0}$$
where ${\rm q}_i$ is the $i$-th column of $\rm Q$ and ${\rm q}_i^\top {\rm A} \, {\rm q}_i \geq 0$ follows from the positive semidefiniteness of $\rm A$.
Rodrigo de Azevedo
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The other direction also holds: if $X$ satisfies $\operatorname{tr}(XY) \geq 0$ for all $Y \in \mathbb{S}^{n}_{+}$, then $X \in \mathbb{S}^{n}_{+}$.
Suppose $X \notin \mathbb{S}^{n}_{+}$, in particular $\exists v \in \mathbb{R}^{n}$ such that $v^{\top}Xv < 0$. Let $Y = vv^{\top}$, then $\operatorname{tr}(XY) = \operatorname{tr}(Xvv^{\top}) = \operatorname{tr}(v^{\top}Xv) = v^{\top}Xv < 0$.
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This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – 311411 Mar 28 '22 at 02:43
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5@311411 I've seen just so many much upvoted answers that don't answer the question exactly, where the question asks about one direction of implication and the answer is solely about the other direction (of course at the same time there are other good enough answers that are actually about the original question). – Arnold Nightingale Mar 28 '22 at 14:05
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3Your answer solves my confusion of the other direction. P.s. I think it is good to provide just a bit extension of the original question, and this should not be considered to be downvoted. – chloe Jan 07 '23 at 12:55