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I am reading a paper which repetitively uses the following statement, but I don't know why this is true:

Statement Let $A$ be a symmetric $n$-by-$n$ matrix, and $B$ be a $n$-by-$n$ matrix, $A\geq 0$ is equivalent to $\text{Tr}(A^TB)\geq 0$, $\forall B$.

Update:

I am not sure in the above statement on whether $B$ is an arbitrary matrix. Actually, in the paper, the matrix $B$ takes values:

(1) $B=QQ^T$, where $Q\in\mathbb{R}_{n\times 2}$ and $\|Q_{i:}\|_2^2=1$, for $1\leq i\leq n$.

(2) $B=(u\circ q)(u \circ q)^T$, where $u$ is an arbitrary vector, and $q\in\{-1,+1\}^n$

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    $\begingroup$ Sorry, your statement is incomple! $\endgroup$
    – Dieter Kadelka
    Jan 24, 2023 at 9:17
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    $\begingroup$ I assume you mean for the trace to be nonnegative and for B to also be psd. If so, this is just saying that the psd cone is self dual. math.stackexchange.com/questions/3358916/… $\endgroup$
    – Dustin G. Mixon
    Jan 24, 2023 at 9:28
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    $\begingroup$ You mean trace of $A^TB$ bigger than $0$ for every $B$? $\endgroup$
    – Beni Bogosel
    Jan 24, 2023 at 9:28
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    $\begingroup$ Note that the matrices of the form (2) are precisely the matrices of the form $xx^T$, in which case $\operatorname{Tr}(A^TB)=x^TAx$. Matrices of the form (1) contain all $xx^T$ with $x$ of norm $\sqrt{2}$ by taking the columns of $Q$ to be equal. (I'm assuming you mean $1\leq i \leq 2$.) The equivalence follows. $\endgroup$
    – Dustin G. Mixon
    Jan 24, 2023 at 10:03
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    $\begingroup$ Well, compiling the comments you have an answer. Taking $B = xx^T$ you obtain $x^TAx\geq 0$ and apply this to every eigenvector of $A$. Conversely, for any positive definite matrix $B$, write its Choleski decomposition $B=LL^T$ and use the fact that $A$ is psd for every column of $L$. $\endgroup$
    – Beni Bogosel
    Jan 24, 2023 at 12:42

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