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I see on wikipedia that the product of two symmetric positive definite matrices is also positive definite (Edit: the matrices must commute or else this is false; I leave the false statement here so that the existing answers still make sense) 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? – Jonas Meyer Feb 27 '12 at 4:56
This fact is definitely true only for commuting positive definite matrices. – darij grinberg Mar 27 '12 at 22:29
@Sally, what I had intended to ask is lost to me now. I was confused about the definition of positive definite, as I had only seen it in the symmetric case where it is characterized by eigenvalues. – nullUser Feb 22 at 6:24
up vote 22 down vote accepted

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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Thanks, I'll edit. – Robert Israel Oct 10 '12 at 22:31
Why is $A^{1/2}BA^{1/2}$ positive semidefinite? – bcf Nov 11 '15 at 20:45
@bcf $v^\top A^{1/2} B A^{1/2} v = (A^{1/2} v)^\top B (A^{1/2} v) \ge 0$, where the equality is due to symmetry of $A^{1/2}$ and the inequality is due to positive semidefiniteness of $B$. – angryavian Jan 23 at 23:41

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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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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Welcome to math.SE! Can you elaborate on that? Currently your answer basically sounds like "because I said so", which is not exactly convincing... – Tobias Kienzler Jul 1 '13 at 15:26
"for sure, it is not positive definite." - II is PSD. – conjectures Oct 19 '13 at 8:49

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