TITLE: Commuting in Matrix Exponential
QUESTION [11 upvotes]: Let $A$, and $B$ be commuting $n\times n$ matrices, i.e. $A.B = B.A$.
Let
\begin{equation}
\exp(A) = \sum_{i=0}^\infty\frac{1}{i!} A^i
\end{equation}
show that $\exp(A+B) = \exp(A).\exp(B)$.
Edit: To those who choose to vote for closing of this question and similar ones as being a duplicate. I think the relative timing of the two questions are actually important! When there are two duplicate questions, I think it makes highly more sense to check the relative timings of the two questions and in general, tend to vote to closing of the question which is asked after (Interestingly, for this question, the duplicate is asked 5 years later than this one). Please note that if their research was done properly or the StackExchenge's duplicate detection worked perfectly at the time, they would not have been even able to post. Now by unnecessarily closing well attended questions of earlier contributors (here me) I am wondering what behavior is actually being rewarded?

REPLY [16 votes]: By Cauchy product we have
$$\exp(A)\exp(B)=\left(\sum_{i=0}^\infty\frac{1}{i!} A^i\right)\left(\sum_{i=0}^\infty\frac{1}{i!} B^i\right)=\sum_{i=0}^\infty c_i$$
where
$$c_i=\sum_{k=0}^i\frac{1}{k!} A^k\frac{1}{(i-k)!} B^{i-k}=\frac{1}{i!}\sum_{k=0}^i{i\choose k}A^k B^{i-k}=\frac{1}{i!}(A+B)^i$$
hence
$$\exp(A)\exp(B)=\sum_{i=0}^\infty\frac{1}{i!}(A+B)^i=\exp(A+B)$$