## 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{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}

\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}

Hence
$a^Ae^B=e^{A+B}$
.