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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...

Rodrigo de Azevedo
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nullUser
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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.

Robert Israel
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    Why is $A^{1/2}BA^{1/2}$ positive semidefinite? – bcf Nov 11 '15 at 20:45
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    @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 '16 at 23:41
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    I don't understand the claim that $\sigma\left(A^{1/2}BA^{1/2}\right) = \sigma(AB)$in the case that $A$ is positive semi-definite. If $A$ is pd, it's easy because $A^{1/2}$ is p.d. so it's invertible, and thus, it itself is the similarity transformation. But if $A$ is not p.d., $A^{1/2}$ may not be invertible, and I am struggling to show that $AB$ and $\left(A^{1/2}BA^{1/2}\right)$ have the same eigenvalues. – user96966 Oct 28 '16 at 15:59
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    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 $S = A^{1/2}B$, $T =A^{1/2}$. – Robert Israel Oct 28 '16 at 18:01
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    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. – Robert Israel Oct 28 '16 at 18:05
  • Ugh. Simple. I just didn't see it before. Thanks for the reply. – user96966 Oct 28 '16 at 20:00
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    @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. – Robert Israel Jun 12 '17 at 23:26
  • Sorry for the trivial question @RobertIsrael . Why are A and B not positive semidefinite real symmetric matrices? – Euclean May 22 '18 at 08:00
  • What do you mean? By assumption, $A$ and $B$ are positive semidefinite real symmetric matrices but their product $AB$ might not be symmetric. – Robert Israel May 22 '18 at 15:38
  • For positive definite matrices, I thought the definition that $x^T A x > 0 \forall x \neq 0$ was equivalent to the definition that all eigenvalues of $A$ are positive. This isn't true? – Rylan Schaeffer Mar 07 '20 at 20:14
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    These are equivalent for a real symmetric matrix. They are not equivalent if $A$ is not a real symmetric matrix. – Robert Israel Mar 08 '20 at 00:41
  • Why are the two not equivalent if A is not a real symmetric matrix? – Rylan Schaeffer Mar 08 '20 at 18:48
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    For example, try $\pmatrix{1 & -3\cr 0 & 1\cr}$ whose eigenvalues are both $1$, but $x^T A x < 0$ for $x = \pmatrix{1\cr 1}$. – Robert Israel Mar 08 '20 at 21:53
  • How can a matrix $A$ with positive eigenvalues have a vector $x$ such that $x'Ax < 0$? I thought $\min_{x} x'Ax = \min \lambda(A)$, where the minimum over $x$ goes through unit vectors, and $\lambda(A)$ are the eigenvalues of $A$. Where am I failing? – a06e May 02 '23 at 13:07
  • @a06e Are you assuming that $A$ is symmetric? – Rodrigo de Azevedo May 02 '23 at 13:28
  • @RodrigodeAzevedo No. So maybe the variational characterisation of eigenvalues only applies for symmetric matrices? – a06e May 02 '23 at 14:41
  • @a06e The skew-symmetric part contributes zero to the quadratic form – Rodrigo de Azevedo May 02 '23 at 14:42
  • ... yes, but the eigenvalues do depend on the skew-symmetric part, which explains the mismatch. Thanks. – a06e May 02 '23 at 14:44
  • @RobertIsrael Are there "simple" conditions under which we can claim $AB + BA$ is posdef? – a06e May 02 '23 at 14:45
  • @Plainview answer below is close to what I am asking, but not quite, I think. – a06e May 02 '23 at 14:46
  • @RodrigodeAzevedo Yes I see now, I was confusing the two ideas. Thanks for clearing it up. – a06e May 02 '23 at 14:54
  • Moved my question here: https://math.stackexchange.com/q/4690923/10063. – a06e May 02 '23 at 14:56
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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.

Arseny Nerinovsky
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Plainview
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  • 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 $(ABC)^TABC=ABC(ABC)^T$? – user_lambda Jun 27 '17 at 08:14
  • 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 – gradstudent Aug 25 '18 at 05:57
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    Can we get an explanation of this proof? – Rylan Schaeffer Mar 08 '20 at 18:26
  • @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 https://math.stackexchange.com/questions/913093/a-normal-matrix-with-real-eigenvalues-is-hermitian). Hence, it is a symmetric positive semidefinite matrix. The converse is obvious. – R. W. Prado Feb 14 '23 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.

Jog
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  • The proof you give is unclear to me. Can you clean it up? – Rylan Schaeffer Mar 08 '20 at 18:25
  • 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. – Jog Mar 10 '20 at 03:01
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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.

user84615
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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 01 '13 at 15:26
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    "for sure, it is not positive definite." - II is PSD. – conjectures Oct 19 '13 at 08:49
  • 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 $AB$. – mdcq Apr 21 '18 at 09:06