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I see on wikipedia that the product of two 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? –  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

3 Answers 3

up vote 12 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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Very minor correction $[AB]_{21} = -3$. –  copper.hat Oct 10 '12 at 19:38
Thanks, I'll edit. –  Robert Israel Oct 10 '12 at 22:31

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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