On the proof: $\exp(A)\exp(B)=\exp(A+B)$ , where uses the hypothesis $AB=BA$?

Question

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!

Answer 1

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.

Answer 2

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 $\ell =2$: $(A+B)^2 = A^2+AB+BA+B^2$.

Answer 3

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