I'm trying to prove that for two commutative matrices $N$ and $M$, that $e^{N+M}=e^Ne^M$.

I wrote using the binomial expansion and commutativity:

$$e^{M+N}=\sum_{k=0}^{\infty}\frac{1}{k!}(M+N)^k=\sum_{k=0}^{\infty}\frac{1}{k!}\left(\sum_{i=0}^{k}\frac{k!}{i!(k-i)!}M^{i}N^{k-i}\right)$$ Now I'm stuck, as I don't know if I can use commutativity of double sums considering the second sum depends on the first, although the result looks somewhat promising, since I can kind of make out what I'm trying to prove.

  • $\begingroup$ The key step is in Marc van Leeuwen’s answer below: the double sum you have is equivalent to summing over all non-negative $i,j$. $\endgroup$ – amd Dec 24 '15 at 20:02

I think the easiest approach is to show that $\exp(X+Y)=\exp(X)\exp(Y)$ is an identity in the ring $\Bbb Q[[X,Y]]$ of formal power series in commuting variables $X,Y$. The necessary transformations of sums are now trivially satisfied, as long as the sums make sense for formal power series, notably: $$\exp(X+Y)= \sum_{k=0}^{\infty}\left(\sum_{i=0}^{k}\frac{1}{i!(k-i)!}X^{i}Y^{k-i}\right) =\sum_{i,j=0}^\infty\frac{X^i}{i!}\frac{Y^j}{j!}=\exp(X)\exp(Y). $$ Then since in the series $\exp$ the coefficients tend to zero faster than exponentially with the degree, the series have infinite radius of convergence, so the identity remains valid after substitution of any pair of commutaing values for $X,Y$.


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