Let $A$ and $B$ be positive definite matrices. What is known about $f(A,B)=\exp(\log A + \log B)$? Does this function have a name? This is interesting because $f(A,B) = AB$ for commuting matrices and $f(A,B)=f(B,A)$ even for non-commuting matrices.
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1You might be interested in this more general question? – J.-E. Pin Mar 28 '20 at 06:25
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1Does this distribute over $+$? – Oscar Cunningham Mar 28 '20 at 10:51
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1No, $f(a+b, c) \ne f(a,c) + f(b,c)$ (checked numerically). – Dan Stahlke Mar 28 '20 at 21:25
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The title says positive semidefinite, but the question text says positive definite. Can $f$ be defined also for positive semidefinite matrices (this is a question I just asked here)? – ViktorStein Sep 20 '22 at 19:58
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Since $\log(A)$ and $\log(B)$ are symmetric matrices, by the Golden-Thompson theorem we have $$ \text{tr}\big(f(A,B)\big) \le \text{tr}\big(\exp\big(\log(A)\big)\exp\big(\log(B)\big)\big) =\text{tr}(AB). $$
ViktorStein
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1Thanks. And equality iff $A$ and $B$ commute: Petz (1994) A survey of certain trace inequalities. That paper has some other interesting stuff. I'll leave the question open for a bit to see if there are other answers. – Dan Stahlke Mar 29 '20 at 01:35
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Petz (1994) A survey of certain trace inequalities, Lemma 4 leads to
$f(A, B) = \lim_{s \to \infty} (A^{1/s} B^{1/s})^s$
This gives a nice Lie algebra geometric interpretation: suppose Alice is trying to rotate in the $\log A$ direction and Bob is trying to rotate in the $\log B$ direction. They compromise and rotate in the $\log A + \log B$ direction, or take turns rotating an infinitesimal amount.
Dan Stahlke
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