TITLE: On the proof: $\exp(A)\exp(B)=\exp(A+B)$ , where uses the hypothesis $AB=BA$?
QUESTION [8 upvotes]: I was seeing the proof that $\exp(A)\exp(B)=\exp(A+B)$ on link
Show that $ e^{A+B}=e^A e^B$
where uses the hypothesis $AB=BA$?
Thanks!

REPLY [8 votes]: The step that claims $$\sum_{m=0}^l{\frac{l!}{m!(l-m)!}A^{m}B^{l-m}}=(A+B)^l$$ For example, we would usually say that $$(A+B)^2=A^2+AB+BA+B^2$$ Only because $$AB=BA$$ can this be simplified.

REPLY [5 votes]: The commutativity of $A$ and $B$ is used where the double summation becomes a single summation.  Consider as an example the term in that sum with $\ell =2$:  $(A+B)^2 = A^2+AB+BA+B^2$.

REPLY [4 votes]: The binomial formula
$$(A+B)^l=\sum_{k=0}^l{l\choose k}A^kB^{l-k}$$
is valid  if $AB=BA$.