每週問題 March 18, 2013

這是可交換矩陣的矩陣指數證明問題。

Let A and B be n\times n matrices. Prove the following statements.
(a) If AB=BA, then e^{A+B}=e^Ae^B.
(b) If e^{(A+B)t}=e^{At}e^{Bt} for all t, then AB=BA.

 
參考解答:

(a) 可交換矩陣,即 AB=BA,滿足二項式定理 (A+B)^m=\sum_{k=0}^m\binom{m}{k}A^{k}B^{m-k}。使用 e^A 的定義式,推導過程如下:

\begin{aligned}\displaystyle  e^{A+B}&=\sum_{m=0}^\infty\frac{(A+B)^m}{m!}=\sum_{m=0}^\infty\sum_{k=0}^m\frac{\binom{m}{k}A^kB^{m-k}}{m!}\\  &=\sum_{k=0}^\infty\sum_{m=k}^\infty\frac{A^k}{k!}\frac{B^{m-k}}{(m-k)!}=  \sum_{k=0}^\infty\sum_{l=0}^\infty\frac{A^k}{k!}\frac{B^{l}}{l!}\\  &=e^Ae^B.  \end{aligned}

(b) 考慮 e^{(A+B)t}e^{At}e^{Bt} 的展開式:

\begin{aligned}\displaystyle  e^{(A+B)t}&=I+(A+B)t+(A+B)^2\frac{t^2}{2}+\cdots\\  &=I+(A+B)t+(A^2+AB+BA+B^2)\frac{t^2}{2}+\cdots  \end{aligned}

\begin{aligned}\displaystyle  e^{At}e^{Bt}&=\left(I+At+A^2\frac{t^2}{2}+\cdots\right)\left(I+Bt+B^2\frac{t^2}{2}+\cdots\right)\\  &=I+(A+B)t+(A^2+2AB+B^2)\frac{t^2}{2}+\cdots.\end{aligned}

令上面兩式相等,可得

A^2+AB+BA+B^2=A^2+2AB+B^2

AB=BA

PowSol-March-18-13

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