We know that if two matrix $A$ and $B$ commutes then $\exp(A+B)=\exp(A)\exp(B)$. I am trying to find two matrix that does not commute but $\exp(A+B)=\exp(A)\exp(B)$ is true for them.
Can anybody give exact example. Thanks
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2See here: http://math.stackexchange.com/questions/638627/expab-and-baker-campbell-hausdorff – Cameron L. Williams Jul 28 '14 at 02:02
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Let $A = \begin{bmatrix}0&6\pi&0\\-6\pi&0&0\\0&0&0\end{bmatrix}$ and $B = \begin{bmatrix}0&0&0\\0&0&8\pi\\0&-8\pi&0\end{bmatrix}$.
Using the formula I derived here, $e^A = e^B = e^{A+B} = I$. Hence, $e^{A+B} = e^Ae^B$.
However, $AB-BA = \begin{bmatrix}0&0&48\pi^2\\0&0&0\\-48\pi^2&0&0\end{bmatrix} \neq 0$, and so, $AB \neq BA$.
JimmyK4542
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