Commuting matrix exponentials: necessary condition. 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?

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 $B$.
