Does $e^{AB}=e^{BA}$ imply $AB=BA$?

Question

Let $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times n}$. Is it true that $e^{AB}=e^{BA}$ implies $AB=BA$? If not, can you provide a counter example?

Answer 1

Not necessarily. You can find $A$, $B$, $2\times 2$ matrices with real entries such that $$AB - BA = \left(\begin{matrix} 0 & - 2 \pi \\ 2\pi & 0 \end{matrix} \right)$$

Then $$e^{AB-BA} = \left(\begin{matrix} \cos 2\pi & - \sin 2 \pi \\ \sin 2\pi & \cos 2 \pi \end{matrix} \right)= I_2$$ and so $e^{AB} = e^{BA}$ ( cheating a bit here, since we need $AB$, $BA$ commuting, but see details below).

$\bf{Added:}$ Here are details:

Take $$A= \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)\\ B= \left(\begin{matrix} 0 & -\pi \\ -\pi & 0 \end{matrix} \right)\\ $$

Then $$AB = \left(\begin{matrix} 0 & -\pi \\ \pi & 0 \end{matrix} \right)\\ BA= \left(\begin{matrix} 0 & \pi \\ -\pi & 0 \end{matrix} \right)\\ $$

We have $e^{AB} = e^{BA}= -I_2$.