## For commuting matrices, the exponential of a sum is the product of exponentials

Let $A$ and $B$ be commuting matrices. Prove that $\mathsf{exp}(A+B) = \mathsf{exp}(A)\mathsf{exp}(B)$, where $\mathsf{exp}(x) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!}x^k$ is a formal power series over a field with characteristic zero.

We will work in the ring $F[[x]][[y]]$ of formal power series over $y$ with coefficients which are formal power series over $x$. (Note that $x$ and $y$ commute.) Note that if $G(x)$ is a formal power series and $p(x)$ a polynomial, then $G(p(x))$ is the formal power series obtained by ‘collecting like terms’. Remember the binomial theorem.

 $\mathsf{exp}(x+y)$ = $\sum_{t \in \mathbb{N}} \dfrac{1}{t!}(x+y)^t)$ = $\sum_{t \in \mathbb{N}} \dfrac{1}{t!} \sum_{k=0}^t {t \choose k} x^ky^{t-k}$ = $\sum_{t \in \mathbb{N}} \sum_{k = 0}^t \dfrac{1}{k!}x^k \cdot \dfrac{1}{(t-k)!}y^{t-k}$ = $\sum_{t \in \mathbb{N}} \sum_{h+k=t} \dfrac{1}{h!}x^h \cdot \dfrac{1}{k!}y^{k}$ = $\sum_{h \in \mathbb{N}} \sum_{k \in \mathbb{N}} \dfrac{1}{h!}x^h \cdot \dfrac{1}{k!}y^k$ = $\left( \sum_{h \in \mathbb{N}} \dfrac{1}{h!}x^h \right) \left( \sum_{k \in \mathbb{N}} \dfrac{1}{k!}y^{k} \right)$ = $\mathsf{exp}(x) \mathsf{exp}(y)$.

In particular, if $A$ and $B$ commute, then $\mathsf{exp}(A+B) = \mathsf{exp}(A)\mathsf{exp}(B)$.