TITLE: Commutativity in Binomial Theorem
QUESTION [0 upvotes]: I have a question in a math exam that is asking us to prove that for matrices $A$ and  $B$ that
$e^{A+B}=e^Ae^B$.  Our professor has asked us to prove this and has been emphatic about "having to show exactly where commutativity is required".  I want to double check that I have given the right answer.
I found a way to prove $e^{A+B}=e^Ae^B$ using the power series of $e^{A+B}$ and getting the binomial theorem out of it i.e. you end up with $$\sum_{n=0}^\infty \frac{1}{n!}\sum_{k=0}^n \binom{n}{k}A^nB^{n-k} $$ and that second sum is just the binomial theorem.  In there by the principle of multiplication we have to be able to choose A or B anyway we want.  And there is where commutativity is required.  I don't need to explain the rest of the proof I know it is correct but I'm pretty sure this step in particular is what my proff is referring to as "where commutativity is required".

REPLY [2 votes]: Hint: By definition of matrix' multiplication and its properties one gets $(A+B)^2=A^2+AB+BA+B^2$ which unless $AB=BA$ that is  not equals to
$A^2+2AB+B^2$.
And with respect to exponentiation $e^{A+B}\neq e^Ae^B$ in general. If $AB=BA$ then  $e^{A+B}=e^Ae^B$.
You can verify this easily with $2\times 2$ matrices chosen at random.
{"set_name": "stack_exchange", "score": 0, "question_id": 3842883}