Show that if $AB = BA$, then $\exp(A) \exp(B) = \exp(B) \exp(A)$

Question

Show that if $AB = BA$, then $$\exp(A) \exp(B) = \exp(B) \exp(A)$$

here is what i got so far: $$\begin{align}\exp(A) \exp(B) &= (I + A + {\dfrac{1}{2}} A^2 + {\dfrac{1}{6}}A^3 + \ldots)(I + B + {\frac{1}{2}}B^2 + {\frac{1}{6}}B^3 + \ldots )\\ &= I + (A + B) + {\frac{1}{2}}(A^2 + 2AB + B^2) + {\frac{1}{6}}(A^3 + 3A^2 B + 3AB^2 + B^3 ) + \ldots\\ &= I + (A + B) + {\frac{1}{2}}(A + B)^2 + {\frac{1}{6}}(A + B)^3 + \ldots \end{align}$$

am i going the right direction so far...if so, how to get to show the rest until $\exp(B)\exp(A)$.

Thank you

Answer 1

You shouldn't do it this way. Work with the formal power series and show that $\exp(A)\exp(B) = \exp(A+B)$.

Edit:

To see how this is the case, note

$$\exp(A+B) \equiv \sum_{n=0}^{\infty}\frac{1}{n!}(A+B)^n.$$

Then, since $A$ and $B$ commute, we have

$$\exp(A+B) = \sum_{n=0}^{\infty}\frac{1}{n!}\sum_{m=0}^n\binom{n}{m}A^{n-m}B^m.$$

Simplify this a bit and use Cauchy products and you'll have it.

Answer 2

Since $AB=BA$, and for real numbers $x,y$ we have that $e^xe^y=e^ye^x$, the set generated by addition and multiplication of the elements of the set $\{A,B\}$ can be identified with the set generated by addition and multiplication of the elements of the set $\{x,y\}$ through the identification $A\to x$, $B\to y$, so you know that since it is true for real numbers it must be true for commuting matrices as well.

In other words, $x$ and $y$ interact with each other in the same way $A$ and $B$ do.