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In related questions (here, here and here), it was shown that if $A$ and $B$ commute, then $e^A$ and $e^B$ also commute (and incidentally $e^A e^B = e^{(A+B)}$). Here, the commutative property of A and B is a sufficient condition.

Is it also a necessary condition? If no, is there another necessary condition for $e^A$ and $e^B$ to commute?

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    $\begingroup$ It is a necessary condition if we require that $e^{tA}$ commute with $e^{tB}$ for all $t \in \Bbb R$. I don't know about the weaker requirement, though. $\endgroup$ – Ben Grossmann Aug 22 '16 at 10:52
  • $\begingroup$ Related. I'll leave it to others to decide whether this question is a duplicate. $\endgroup$ – Ben Grossmann Aug 22 '16 at 11:44
  • $\begingroup$ It appears this is a duplicate indeed. I'm really sorry, I thought I searched enough! Thanks for your answers though. $\endgroup$ – user9037 Aug 22 '16 at 11:46
  • $\begingroup$ don't worry about it; duplicates like these make it easier for future potential askers to find what they need. $\endgroup$ – Ben Grossmann Aug 22 '16 at 11:48
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It's not necessary. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute?

It is known, however, that if the spectra of both $A$ and $B$ are $2\pi i$-congruence-free (i.e. if the difference of every pair of eigenvalues in $A$ is not an integer multiple of $2\pi i$, and similarly for $B$), then $e^A$ and $e^B$ commute if and only if $A$ and $B$ commute. See

Edgar M.E. Wermuth, Two remarks on matrix exponentials, Linear Algebra and Its Applications, 117(1):127-132, 1989.

Edgar M.E. Wermuth, A remark on commuting operator exponentials, Proceedings of the American Mathematical Society, 125(6): 1685-1688, 1997.

(Remark. In the first paper above, while the author has imposed the condition that the matrices have algebraic entries, this is just a sufficient condition for the matrices to have $2\pi i$ congruence-free spectra. In the second paper, the conclusion is generalised to bounded operators with $2\pi i$-congruence-free spectra on a Banach space.)

It follows that if $e^A$ and $e^B$ commute, it's almost surely true that $A$ and $B$ commute.

See also a related discussion (and loup blanc's answer in particular) on the question Is $\exp:\overline{\mathbb{M}}_n\to\mathbb{M}_n$ injective?

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  • $\begingroup$ Thank you for the second bit (and reference) about the congruence. It was very helpful in my case. $\endgroup$ – user9037 Aug 22 '16 at 11:49
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It is not a necessary condition. In particular, we can take $$ A = \pmatrix{2 \pi i & 0\\0&0}, \quad B = \pmatrix{0&1\\0&0} $$ It's clear that $e^A = I$ commutes with $e^B$, but $A$ does not commute with $B$.

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