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).

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