$e^A\cdot e^B=e^{A+B}$ if $AB=BA$ for matrices $A$ and $B$

Question

Everyone says the proof is obvious but I haven't actually found it anywhere!

I know that

$$\exp(A) := \sum_{n=0}^\infty \frac{A^n}{n!}$$

Thanks!

Answer 1

When $AB=BA$ one has the binomial formula (this is valid in any ring and you prove it by induction using the Pascal formula for binomial coefficients)

$$(A+B)^n=\sum_{k=0}^n{n\choose k}A^k B^{n-k}$$

Now write

$$\exp(A+B)=\sum_{n=0}^\infty {(A+B)^n\over n!}=\sum_{n=0}^\infty{1\over n!}\sum_{k=0}^n{n!\over k!(n-k)!}A^kB^ {n-k}=\sum_{k=0}^\infty {A^k\over k!}\sum_{j=0}^\infty {B^j\over j!}$$

Answer 2

Let $x_1(t) = e^{At} e^{Bt}$, $x_2(t) = e^{(A+B)t}$. Note that $x_1(0) = x_2(0) = I$.

Note that $A,B$ commute with $e^{At},e^{Bt}$.

Then $x_1'(t) = Ae^{At} e^{Bt} + e^{At} B e^{Bt}= (A+B) e^{At} e^{Bt} = (A+B) x_1(t)$ and $x_2'(t) = (A+B) e^{(A+B)t} = (A+B)x_2(t)$.

Hence $x_1,x_2$ satisfy the same smooth ode. with the same initial condition. Hence $x_1(t) = x_2(t) $ for all $t$. Hence $x_1(1) = x_2(1)$.

Answer 3

For each fixed $n \in \mathbb{N}$,the property $AB=BA$ allows us to use the binomial theorem for matrices, which is given by: $$(A+B)^{n} = \sum_{k=0}^{n}\binom{n}{k}A^{k}B^{n-k}$$ Now, we have $$e^{A+B} = \sum_{n=0}^{\infty}\frac{(A+B)^{n}}{n!} = \sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}A^{k}B^{n-k} = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{1}{k!(n-k)!}A^{k}B^{n-k} = \sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{1}{k!(n-k)!}A^{k}B^{n-k} \hspace{2cm} (1)$$ Now, define $m = n-k$, so if $k \le n < \infty $ implies $0 \le n-k =m < \infty$. Thus, (1) becomes: $$\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{1}{k!m!}A^{k}B^{m} = \bigg{(}\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}\bigg{)}\bigg{(}\sum_{m=0}^{\infty}\frac{1}{m!}B^{m}\bigg{)} = e^{A}e^{B}$$