For two matrices $A,B\in M_n(\mathbb{C})$, if $AB=BA,$ then $e^{A+B}=e^A\ e^B$ by power series calculation. I don't know if the converse is true. Perhaps in Lie group theory can give a clear description on this question. I appreciate any idea on this problem.
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1https://math.stackexchange.com/questions/165074/a-weak-converse-of-ab-ba-implies-eaeb-ebea-from-topics-in-matrix-analysis – Eric Towers Oct 25 '21 at 03:34
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1See (1) A weak converse of $AB=BA\implies e^Ae^B=e^Be^A$ from "Topics in Matrix Analysis" for matrices of algebraic numbers (2) If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? (3) If $\exp(t(A + B)) = \exp(tA) \exp(tB)$ for all $t \geq 0$ then $A,B$ commute (4) If $|A|, |B|$ are sufficiently small, does $e^{A+B} = e^A e^B$ imply that $AB = BA$? – user1551 Oct 25 '21 at 05:08
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I see, the conserve is not true in general, this is a problem deserving further thinking. – Tsoshamry Oct 25 '21 at 15:11
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In general, it is not true that $e^{A+B}=e^A e^B$ implies that $A$ and $B$ commute. There are different additional conditions that guarantee this is the case. One simple one states that if $e^{t(A+B)}=e^{tA} e^{tB}$ for all $t\in\mathbb{R}$, then $A$ and $B$ commute. See
- Clément de Seguins Pazzis, On commuting Matrices and Exponentials, Proceedings of the American Mathematical Society, 2012.
for a summary of some of the results.
Artemy
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