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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, 2012 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, 2015 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, 2016 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, 2016 at 18:01
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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, 2017 at 23:26
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    $\begingroup$ For example, try $\pmatrix{1 & -3\cr 0 & 1\cr}$ whose eigenvalues are both $1$, but $x^T A x < 0$ for $x = \pmatrix{1\cr 1}$. $\endgroup$
    – Robert Israel
    Mar 8, 2020 at 21:53
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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, 2017 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, 2018 at 5:57
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    $\begingroup$ Can we get an explanation of this proof? $\endgroup$
    – Rylan Schaeffer
    Mar 8, 2020 at 18:26
  • $\begingroup$ @Rylan Schaeffer If $AB$ is normal, then it is diagonalizable by an orthogonal matrix. Since $AB=A^{1/2} A^{1/2} B$ and $A^{1/2} B A^{1/2}$ have the same nonzero eigenvalues, all eigenvalues of $AB$ are positive and real. Now, since $AB$ is diagonalizable by an orthogonal matrix, and it has only real non-negative eigenvalues, it can be diagonalizable by real orthogonal eigenvectors (see math.stackexchange.com/questions/913093/…). Hence, it is a symmetric positive semidefinite matrix. The converse is obvious. $\endgroup$
    – R. W. Prado
    Feb 14 at 21:05
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Actually, one has to be vary careful in the way one interprets the results of Meenakshi and Rajian (referenced in one of the posts above). Symmetry is inherent in their definition of positive definiteness. Thus, their result can be stated very simply as follows: If $A$ and $B$ are symmetric and PSD, then $AB$ is PSD iff $AB$ is symmetric. A direct proof for this result can be given as follows. If $AB$ is PSD, it is symmetric (by Meenakshi and Rajian's definition of PSD). If it is symmetric, it is PSD since the eigenvalues of $AB$ are non-negative. To summarize, all the stuff about normality in their paper is not required (since normality of $AB$ is equivalent to the far simpler condition of symmetry of $AB$ when $A$ and $B$ are symmetric PSD). The most important point here is that if one adopts a more general definition for PSD ($x^TAx\ge 0$) and if one now considers cases where the product $AB$ is unsymmetric, then their results do not go through.

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  • $\begingroup$ The proof you give is unclear to me. Can you clean it up? $\endgroup$
    – Rylan Schaeffer
    Mar 8, 2020 at 18:25
  • $\begingroup$ Meenakshi and Rajian assume symmetry in their definition of positive semi-definiteness (PSD). Thus, if AB is PSD it is automatically symmetric (by their definition). Conversely, if AB is symmetric, then since the eigenvalues of AB are nonnegative, AB is PSD. $\endgroup$
    – Jog
    Mar 10, 2020 at 3:01
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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, 2013 at 15:26
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    $\begingroup$ "for sure, it is not positive definite." - II is PSD. $\endgroup$
    – conjectures
    Oct 19, 2013 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 $AB$. $\endgroup$
    – mdcq
    Apr 21, 2018 at 9:06

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