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Let $Q$ be an $n \times n$ symmetric positive definite matrix, $\vec{a}, \vec{b} \in \Bbb R^n$ be two random vectors. Prove that $a^TQ(ba^T-ab^T)Qb$ is non-positive.


Since $Q$ has $n$ independent eigenvectors with positive eigenvalues, I've tried expressing $\vec{a}$ and $\vec{b}$ as linear combinations of those eigenvectors, but it didn't work out. Really appreciate any help.

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    $\begingroup$ Interestingly, $a^TQ(ba^T-ab^T)b$ (with the $Q$ near the right side of the original expression removed) is also non-positive. $\endgroup$
    – user1551
    Jul 29, 2020 at 9:23

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By Cauchy-Schwarz $$|\langle \sqrt{Q}b,\sqrt{Q}a\rangle|^2=|a^TQb|^2\le \langle \sqrt{Q}b,\sqrt{Q}b\rangle \langle \sqrt{Q}a,\sqrt{Q}a\rangle=b^TQba^TQa$$

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    $\begingroup$ Thank you. I didn't know that a symmetric positive definite matrix can be square-rooted. $\endgroup$
    – Polo Li
    Jul 30, 2020 at 0:04

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