问题及解答

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

http://www.physicsforums.com/showthread.php?t=255395

http://www.jstor.org/discover/10.2307/2162208?uid=3737800&uid=2129&uid=2134&uid=374415477&uid=2&uid=70&uid=3&uid=374415467&uid=60&sid=21101125311297

(1) 当 $A,B$ 可交换时, 容易看到

\[Ae^B=e^B A,\]

类似的,

\[A^i e^B=e^B A^i.\]

从而

\[e^A e^B=e^B e^A.\]

而证明 $e^A e^B=e^{A+B}$ 的关键在于将它们展开, 并注意到

\[
\frac{A^i}{i!}\frac{B^j}{j!}=\frac{\binom{n}{i}A^i B^{n-i}}{n!},
\]

这里 $n=i+j$.

\[e^B=I+\frac{B}{1!}+\frac{B^2}{2!}+\frac{B^3}{3!}+\cdots+\frac{B^m}{m!}+\cdots\]

所以

\[Ae^B=A+\frac{AB}{1!}+\frac{AB^2}{2!}+\frac{AB^3}{3!}+\cdots+\frac{AB^m}{m!}+\cdots\]

\[e^B A=A+\frac{BA}{1!}+\frac{B^2 A}{2!}+\frac{B^3 A}{3!}+\cdots+\frac{B^m A}{m!}+\cdots\]

两式相减, 得

\[
Ae^B-e^BA=\frac{1}{1!}(AB-BA)+\frac{1}{2!}(AB^2-B^2 A)+\cdots+\frac{1}{m!}(AB^m-B^m A)+\cdots
\]

Claim: $AB^m-B^m A=m\lambda B^{m-1}$, ($m\geqslant 1$).

Pf. 由 $AB-BA=\lambda I_n$ 得

\[
\begin{array}{l}
AB^2-BAB=\lambda B\\
BAB-B^2A=\lambda B
\end{array}\biggr\}
\Rightarrow AB^2-B^2 A=2\lambda B.
\]

由上面, 又可得

\[
\begin{array}{l}
AB^3-BAB^2=\lambda B^2\\
BAB^2-B^2 AB=\lambda B^2\\
B^2 AB-B^3 A=\lambda B^2
\end{array}\biggr\}
\Rightarrow AB^3-B^3 A=3\lambda B^2.
\]

再仿照上面, 对于上面三个式子, 最后一个式子两边左乘 $B$, 其余都右乘 $B$. 可得

\[AB^4-B^4 A=4\lambda B^3.\]

如此进行下去, 利用归纳法即可证明.

Q.E.D of Claim.

从而

\[
\begin{split}
Ae^B-e^BA&=\frac{1}{1!}(AB-BA)+\frac{1}{2!}(AB^2-B^2 A)+\cdots+\frac{1}{m!}(AB^m-B^m A)+\cdots\\
&=\lambda I_n+\frac{1}{2!}2\lambda B+\frac{1}{3!}3\lambda B^2+\cdots+\frac{1}{m!}m\lambda B^{m-1}+\cdots\\
&=\lambda\biggl[I_n+\frac{B}{1!}+\frac{B^2}{2!}+\cdots+\frac{B^{m-1}}{(m-1)!}+\cdots\biggr]\\
&=\lambda I_n e^B\\
&=[A,B]e^B
\end{split}
\]