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Let $A,B$ be two symmetric and positive definite (posdef) real matrices.

Under what conditions is

$$AB+BA$$

also positive (semi-)definite?

Note: By posdef, I mean that $x'Ax > 0$ for any non-zero vector $x$.

a06e
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    Related: https://math.stackexchange.com/q/113842/10063 – a06e May 02 '23 at 14:56
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    Is there any particular reason you're asking (e.g. an application) or was this just idle curiosity? – Ben Grossmann May 02 '23 at 15:13
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    @BenGrossmann In an application I encountered a function whose Jacobian is a product of two posdef matrices. I'm trying to understand under what conditions this Jacobian is posdef. But describing this problem would lead too far. – a06e May 02 '23 at 15:15
  • I see, thanks. I agree, I see no need to get into the details of that here – Ben Grossmann May 02 '23 at 15:18

1 Answers1

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A nice case is when $A,B$ commute. Then $A,B$ can be diagonalised simultaneously:

$$A=VSV',\quad B=VDV'$$

where $S,D$ are diagonal and positive, and $V$ is an orthogonal matrix with the common set of eigenvectors. See Do commuting matrices share the same eigenvectors?.

It follows that

$$AB=BA=VSDV'$$

Then

$$\frac{1}{2} x'(AB+BA)x = x' VSDV' x = y' SD y > 0$$

for all non-zero $x$ because $SD$ is positive, where $y=V'x$.

Question: Are there other cases where $AB+BA$ is posdef but $A,B$ do not commute?

a06e
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  • One natural extension of this case is that $AB + BA$ will be positive definite if $A$ and $B$ are "close to" commuting. More precisely, $AB + BA$ will be positive definite if $|A^{1/2}B - BA^{1/2}|$ (spectral norm) is smaller than the smallest eigenvalue of $AB$, which can in turn be bounded in terms of the eigenvalues of $A$ and $B$. If either matrix is only semidefinite, then an approach like the one I'm suggesting would be fruitless – Ben Grossmann May 02 '23 at 15:11
  • @BenGrossmann Can you explain this a bit more? I don't understand how the smallest eigenvalue of $AB$ enters – a06e May 02 '23 at 15:14
  • If I find time later I might flesh this out into a full answer, but the idea is as follows. First, note that $AB$ and $A^{1/2}BA^{1/2}$ have the same eigenvalues, with the latter matrix necessarily positive definite. From there, note that $$ AB + BA = A^{1/2}(A^{1/2}B - BA^{1/2}) + (BA^{1/2} - A^{1/2}B)A^{1/2} + 2 A^{1/2}BA^{1/2} $$ this is a perturbation of the symmetric matrix $A^{1/2}BA^{1/2}$, so Weyl's matrix inequalities apply. – Ben Grossmann May 02 '23 at 15:25
  • I just noticed that the above can be nicely written with a repeated application of the commutator $[P,Q] = PQ - QP$. In particular, we have $$ AB + BA = [A^{1/2},[A^{1/2},B]] + 2A^{1/2}BA^{1/2}. $$ – Ben Grossmann May 02 '23 at 15:29