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If two matrices $A$ and $B$ commute, then $e^{A+B} = e^A e^B$ by rearrangement of the $A$'s and $B$'s in the sum. But would the converse be true?

So far I've tried to find a counterexample by considering the cases where $A$ and $B$ are 2x2 real upper triangular matrices, but those don't work.

Any hint is appreciated, thanks.

Bio
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    You'll find a counterexample, and some further discussion, here – lulu Feb 24 '20 at 15:29
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    Counterexamples can also be constructed using similar matrices $A$ and $B$ such that their exponentials are scalar. I used the idea here, but that is somewhat unlikely to be the first appearance. – Jyrki Lahtonen Feb 24 '20 at 16:02
  • Thanks for the feedback. The example you linked to is perfect @lulu. As an extension, what if we only want real matrices A and B? – Bio Feb 24 '20 at 17:22
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    here is a duplicate in which a real counterexample is given. – lulu Feb 24 '20 at 19:05
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    If you stick with Hermitian (or if you prefer real symmetric) $A$ and $B$, then the statement is true. The result is implied e.g. by the equality conditions of the Golden-Thompson Inequality. – user8675309 Feb 24 '20 at 21:22

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