proof of $e^{(A+B)t}=e^{At}e^{Bt}$

Question

I am trying to understand what happens to the second line of this proof(p8), where the some sketchy combination of exponential terms are performed, which seems to be using the property $$e^{(A+B)t}=e^{At}e^{Bt}$$, we are trying to prove. So this proof seems to be invalid, in my opinion. What am I missing here?

Answer 1

@user14042 has it right in their comment: "there's a mistake in the document you linked, the second line should be $$(A+B-(A+B))e^{t(A+B)}e^{-tA}e^{-tB}."$$

(Honestly, I just assumed that's what it said without bothering to look too deeply at the exponentials. @user14042 has good noticing glasses!)

With that fixed maybe you can see the real work that is being done in line two, which is a few applications of your Lemma 14. For instance, in the second term $e^{t(A+B)}A$ is replaced with $Ae^{t(A+B)}$, which is permitted as $A$ and $t(A+B)$ commute. (It's a bit confusing as your Lemma 14 uses the same variables as your Theorem 15, but with different meanings, so you have to think things like "... so we'll let $B$ be $t(A+B)$...".)