Proof exp(A)exp(B)=exp(A+B) if and Only if A and B Commute

Theorem
For matrices  
\[A, \: B\]
,  
\[e^A e^B =e^{A+B}\]
  if and only if  
\[AB=BA\]

\[A+B=B+A \rightarrow e^{A+B}=e^{B+A} \]

Then  
\[e^A e^B=e^{A+B}=e^{B+A}=e^Be^A\]
  so  
\[A, \: B\]
  commute.
Hence if  
\[e^A e^B=e^{A+B}\]
  then  
\[A, \: B\]
  commute. Now suppose that  
\[A, \: B\]
  commute, so that in particular  
\[(A+B)^n= {}^n C_k A^{n-k}B^k\]

\[e^A=I+ \frac{A}{1!}+\frac{A^2}{2!}+...+ \frac{A^n}{n!}+...= \sum^{\infty}_{n=0} \frac{A^n}{n!}\]

\[e^B=I+ \frac{B}{1!}+\frac{B^2}{2!}+...+ \frac{B^n}{n!}+...= \sum^{\infty}_{n=0} \frac{B^n}{n!}\]

\[\begin{equation} \begin{aligned} e^Ae^B &= \sum^{\infty}_{n=0} \frac{A^n}{n!} \sum^{\infty}_{m=0} \frac{B^m}{m!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{A^{n-m}}{(n-m)!} \frac{B^m}{m!} \end{aligned} \end{equation}\]

\[\begin{equation} \begin{aligned} e^{A+B} &= I+ \frac{A+B}{1!}+\frac{(A+B)^2}{2!}+...+ \frac{(A+B)^n}{n!}+...\\ &= \sum^{\infty}_{n=0} \frac{(A+B)^n}{n!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{{}^n C_m A^{n-m}B^m}{n!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{A^{n-m}}{(n-m)!} \frac{B^m}{m!} \end{aligned} \end{equation}\]

Hence  
\[a^Ae^B=e^{A+B}\]
.

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