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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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    $\begingroup$ Where do you see that on Wikipedia? What do you mean by positive definite? $\endgroup$ – Jonas Meyer Feb 27 '12 at 4:56
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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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    $\begingroup$ Why is $A^{1/2}BA^{1/2}$ positive semidefinite? $\endgroup$ – bcf Nov 11 '15 at 20:45
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    $\begingroup$ @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 $B$. $\endgroup$ – angryavian Jan 23 '16 at 23:41
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    $\begingroup$ If $S$ and $T$ are $n \times m$ and $m \times n$ matrices, $ST$ and $TS$ have the same nonzero eigenvalues: $u$ is an eigenvector of $ST$ iff $Tu$ is an eigenvector of $TS$, with the same nonzero eigenvalue. Apply this to $S = A^{1/2}B$, $T =A^{1/2}$. $\endgroup$ – Robert Israel Oct 28 '16 at 18:01
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    $\begingroup$ Also, if $S$ and $T$ are both $n \times n$, the characteristic polynomials $\det(ST-\lambda I)$ and $\det(TS-\lambda I)$ are the same. $\endgroup$ – Robert Israel Oct 28 '16 at 18:05
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    $\begingroup$ @ProcrastinationSage This is precisely my point: if $A$ and $B$ are symmetric and positive semidefinite, then despite the fact that $AB$ might not be symmetric, its nonzero eigenvalues are the same as those of $A^{1/2} B A^{1/2}$ and therefore are real and nonnegative. $\endgroup$ – Robert Israel Jun 12 '17 at 23:26
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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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  • $\begingroup$ Do you know if this result can be extended to three matrices? I.e. let $A$, $B$, $C$ all be PSD. Is $ABC$ PSD if $(ABC)^TABC=ABC(ABC)^T$? $\endgroup$ – user_lambda Jun 27 '17 at 8:14
  • $\begingroup$ I guess here by "PSD" you mean all eigenvalues are non-negative, right? Or else one will get into the contradiction pointed out by @RobertIsrael $\endgroup$ – gradstudent Aug 25 '18 at 5:57
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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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    $\begingroup$ Welcome to math.SE! Can you elaborate on that? Currently your answer basically sounds like "because I said so", which is not exactly convincing... $\endgroup$ – Tobias Kienzler Jul 1 '13 at 15:26
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    $\begingroup$ "for sure, it is not positive definite." - II is PSD. $\endgroup$ – conjectures Oct 19 '13 at 8:49
  • $\begingroup$ 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 $AB$. $\endgroup$ – philmcole Apr 21 '18 at 9:06

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