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Let $X,Y,Z$ be positive definite matrices such that $XYZ$ is hermitian . How can we show that $XYZ$ is also a positive definite matrix?

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Somewhat relevant :… – Inquest Dec 5 '12 at 15:39
up vote 2 down vote accepted

Since by hypothesis $XYZ$ is Hermitian, then it is enough to show that its eigenvalues are positive. But the product of positive-definite matrices, even though not hermitian in general, always has positive eigenvalues. See a proof in my answer to this question: Product of positve definite matrix and seminegative definite matrix

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Denote $H=XYZ$. Let $X^{1/2}$ be a positive definite square root of $X$ (it always exists -- think unitary diagonalization) and \begin{align} y &= X^{1/2}YX^{1/2},\\ z &= X^{-1/2}YX^{-1/2},\\ h &= X^{-1/2}HX^{-1/2} = X^{1/2}YZX^{-1/2} = yz. \end{align} Then $y$ and $z$ are positive definite and $h$ is Hermitian. Also, $H$ is positive definite iff $h$ is positive definite. Now, let $y^{1/2}$ be a positive definite square root of $y$. Then $h$ is similar to $y^{-1/2}hy^{1/2}=y^{1/2}zy^{1/2}$, which is positive definite. Therefore $h$ and in turn $H$ are positive definite.

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