证明: 当 $[A,B]=\lambda I_n$ 时, $e^A e^B=e^B e^A e^{[A,B]}$.
设 $A,B\in\mathcal{M}_{n\times n}(\mathbb{F})$, $\lambda\in\mathbb{F}$. 这里 $\mathbb{F}=\mathbb{R}$, 或 $\mathbb{F}=\mathbb{R}$. 证明:
(1) 当 $AB=BA$ 时, $e^A e^B=e^{A+B}$.
(2) 当 $[A,B]=\lambda I_n$ 时, $e^A e^B=e^B e^A e^{[A,B]}$.
Example:
(1) 中条件如果不满足时, 则有反例. 若令
\[A=\left(\begin{array}{cc} 0 & 1 \\ 0 &0 \\ \end{array}\right),\quad\quad B=\left(\begin{array}{cc} 0 & 0 \\ 1 &0 \\ \end{array}\right)\]
则我们得到
\[[A,B] = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right)\]
\[e^A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right ),\quad\quad e^B = \left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array}\right),\quad\quad e^{[A,B]} = \left(\begin{array}{cc} e & 0 \\ 0 & e^{-1} \\ \end{array}\right)\]
\[e^A e^B = \left(\begin{array}{cc} 2 & 1 \\ 1 & 1 \\ \end{array}\right)\]
\[e^B e^A = \left(\begin{array}{cc} 1 & 1 \\ 1 & 2 \\ \end{array}\right)\]
从而
\[e^B e^A e^{[A,B]}=\left(\begin{array}{cc} 1 & 1 \\ 1 & 2 \\ \end{array}\right)\left(\begin{array}{cc} e & 0 \\ 0 & e^{-1} \\ \end{array}\right) = \left(\begin{array}{cc} e & e^{-1} \\ e & 2e^{-1} \\ \end{array}\right)\neq e^A e^B\]
Reference