Show that $ e^{A+B}=e^A e^B$

If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that

$$ e^{A+B}=e^A e^B$$

Note that $A$ and $B$ do NOT have to be diagonalizable.

Solution 1:

$$\begin{align*}e^{A}e^{B} &= \left(\sum \frac{A^{n}}{n!}\right)\left(\sum\frac{B^{n}}{n!}\right)\\ &=\sum^{\infty}_{m=0}\sum^{\infty}_{n=0}\frac{A^{m}B^{n}}{m!n!}\\ &=\sum^{\infty}_{l=0}\sum^{l}_{m=0}\frac{A^{m}B^{l-m}}{m!(l-m)!}\\ &=\sum^{\infty}_{l=0}\frac{1}{l!}\sum^{l}_{m=0}\frac{l!}{m!(l-m)!}A^{m}B^{l-m}\\ &=\sum^{\infty}_{l=0}\frac{(A+B)^{l}}{l!}\\ &= e^{A+B}\end{align*}$$

Note:A and B have to commute. Also, I set l=m+n. I did this because we want to use the binomial theorem to simplify this.

Solution 2:

Here is a different-ial way, just because it is significantly different from the standard Cauchy product way.

Given a square matrix $M$, the function $X(t):=e^{tM}$ is the unique solution of the linear differential equation: $X'=MX$ and $X(0)=I$.

Now set $X(t):=e^{tA}e^{tB}$ and observe that the factors commute with each other, as well as they commute with $A$ and $B$. It follows that $$ X'(t)=Ae^{tA}e^{tB}+e^{tA}Be^{tB}=(A+B)e^{tA}e^{tB}=(A+B)X(t). $$ And since $X(0)=e^0e^0=I$, it follows from the uniqueness above that $$ X(t)=e^{tA}e^{tB}=e^{t(A+B)}\qquad\forall t\in\mathbb{R}. $$ Set $t:=1$ to get the desired formula.