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I see on Wikipedia that the product of two commuting symmetric positive definite matrices is also positive definite. Does the same result hold for the product of two positive semidefinite matrices?

My proof of the positive definite case falls apart for the semidefinite case because of the possibility of division by zero...

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    Where do you see that on Wikipedia? What do you mean by positive definite?2012-02-27

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You have to be careful about what you mean by "positive (semi-)definite" in the case of non-Hermitian matrices. In this case I think what you mean is that all eigenvalues are positive (or nonnegative). Your statement isn't true if "$A$ is positive definite" means $x^T A x > 0$ for all nonzero real vectors $x$ (or equivalently $A + A^T$ is positive definite). For example, consider $ A = \pmatrix{ 1 & 2\cr 2 & 5\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr},\ AB = \pmatrix{-1 & 3\cr -3 & 8\cr},\ (1\ 0) A B \pmatrix{1\cr 0\cr} = -1$

Let $A$ and $B$ be positive semidefinite real symmetric matrices. Then $A$ has a positive semidefinite square root, which I'll write as $A^{1/2}$. Now $A^{1/2} B A^{1/2}$ is symmetric and positive semidefinite, and $AB = A^{1/2} (A^{1/2} B)$ and $A^{1/2} B A^{1/2}$ have the same nonzero eigenvalues.

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    What do you mean? By assumption, $A$ and $B$ are positive semidefinite real symmetric matrices but their product $AB$ might not be symmetric.2018-05-22
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The product of two symmetric PSD matrices is PSD, iff the product is also symmetric. More generally, if $A$ and $B$ are PSD, $AB$ is PSD iff $AB$ is normal, ie, $(AB)^T AB = AB(AB)^T$.

Reference: On a product of positive semidefinite matrices, A.R. Meenakshi, C. Rajian, Linear Algebra and its Applications, Volume 295, Issues 1–3, 1 July 1999, Pages 3–6.

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    I guess here by "PSD" you mean all eigenvalues are non-negative, right? Or else one will get into the contradiction pointed out by @RobertIsrael2018-08-25
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The product of two positive definite matrices is not necessarily positive definite. The product in most cases is not even symmetric and for sure, it is not positive definite.

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    I guess his definition of definiteness is restricted to symmetric matrices only. And since $AB$ needn't be symmetric for two symmetric $A$ and $B$, we can't talk about the definiteness of $AB$.2018-04-21