# Proving an equality associated with symmetric positive definite matrices

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.

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

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