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If we have square matrices $A$ and $B$ that commute (i.e. $AB=BA$), then we have $e^{A+B} = e^Ae^B$. In general this isn't true without the condition that $A$ and $B$ commute. I would like to know if this is "if and only if", or whether $A$ commuting with $B$ is just a necessary condition.

In other words, do there exist square matrices $A$ and $B$ such that $AB\ne BA$ but $e^{A+B} = e^A e^B$?

Not that it really matters because this question already got a very good answer, but I don't think this should be considered a duplicate of Does $e^{a+b}=e^{a}e^{b}$ implies that $ab=ba$ for banach algebras?, because that question is about Banach algebras whereas mine is only about ordinary finite-dimensional matrices.

Neither is it a duplicate of If $e^A$ and $e^B$ commute, do $A$ and $B$ commute for finite dimensional matrices? (of which Does $e^{a+b}=e^{a}e^{b}$ implies that $ab=ba$ for banach algebras? itself is closed as a duplicate), because that is clearly just a different question.

N. Virgo
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    I believe the linked questions and answers do actually answer your question, however I voted to reopen this because It requires knowledge to see the connection which is beyond the level required to understand this question. – Graviton Apr 20 '23 at 06:05
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    “...this is not a duplicate of Does $e^{a+b}=e^{a}e^{b}$ implies that $ab=ba$ for banach algebras?, because that question is about Banach algebras and the only answer there is infinite-dimensional.” This is not true. The answerer explicitly stated that “Let the Banach algebra be the space of operators on a finite-dimensional* Hilbert space”. In fact, his example is comprised of two $2\times2$ matrices $A=\pmatrix{2\pi i&0\ 0&0}$ and $B=\pmatrix{2\pi i&1\ 0&0}$. – user1551 Apr 20 '23 at 06:46
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    @user1551 you are correct - I had just realised my mistake and will correct it. Still, I agree with Graviton's comment: someone looking for a simple result about matrices is likely to be mystified by that answer. (If that results in this being closed again I don't really mind, but I do think the answer here is valuable.) – N. Virgo Apr 20 '23 at 06:53

1 Answers1

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It is possible to represent quaternions by $2\times2$ complex matrices or $4\times4$ real matrices (Wikipedia).

For $A=2\pi(3\mathbf{i})$ and $B=2\pi(4\mathbf{j})$ we have $e^A=e^B=e^{A+B}=1$ because $(3,4,5)$ is a Pythagorean triple, even though $A$ and $B$ do not commute (indeed, they anticommute: $BA=-AB$).

This gives the explicit counterexample $$ A = 2\pi\begin{pmatrix} 0 & 3 & 0 & 0 \\ -3 & 0 & 0 & 0 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 3 & 0 \\ \end{pmatrix},\qquad B = 2\pi\begin{pmatrix} 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \\ -4 & 0 & 0 & 0 \\ 0 & -4 & 0 & 0 \\ \end{pmatrix}. $$

N. Virgo
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