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I'm asked to show that $e^{A+B} = e^{A}e^{B}$ for commuting square matrices. This question has been answered before here and the accepted answer is this one. Intuitively I agree with the proof given but I can't prove it, in particular the following identity: $$\sum^{\infty}_{m=0}\sum^{\infty}_{n=0}\frac{A^{m}B^{n}}{m!n!} =\sum^{\infty}_{l=0}\sum^{l}_{m=0}\frac{A^{m}B^{l-m}}{m!(l-m)!}$$

I tried to get an intuition behind it using partial sums but the identity doesn't seem to hold there. Although it seems to make sense that the indexes are eventually the same when $l$ goes to infinity, the statement is still kind of mysterious to me. How does one get that identity?

Rodrigo de Azevedo
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Rodrigo Meireles
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1 Answers1

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Consider the first sum with indices $n,m$ and set $l:=n+m$. Instead of summing over both $n$ and $m$, we sum over $l$ and then consider all possible $(n,m)$ with $n+m=l$: $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{A^mB^n}{m!n!}=\sum_{l=0}^{\infty}\sum_{n+m=l}\frac{A^mB^n}{m!n!}.$$ Now, $l=n+m$ is equivalent with $n=l-m$ and if $m+n=l$, we know that $0\le m\le l$, so $$\sum_{n+m=l}\frac{A^mB^n}{m!n!}=\sum_{m=0}^l\frac{A^mB^{l-n}}{m!(l-n)!}.$$ Plugging this back in gives the desired identity.

Mastrem
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    I kind of understand susbtituting $l = n + m$ but now that you put more details on the equality, my question is how do I know that summing over all $l$ and then over all possible $(n,m)$ with $n+m = l$ would not give me repeated indexes? I think my question is: how do I know there is a bijection between the first set of indexes and the second after using $l$? – Rodrigo Meireles Sep 18 '20 at 15:48
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    The bijection is $(n,m)\mapsto (n+m,m)$ and its two-sided inverse is $(l,m)\mapsto (l-m,m)$. – Mastrem Sep 18 '20 at 15:56
  • While searching for an answer before posting this question I came across this wikipedia arcticle of the Cauchy product: https://en.wikipedia.org/wiki/Cauchy_product. In there we have the Merten's theorem statement and some proof, which got me puzzled even further: why does it need a proof if there is a bijection between the the indexes (i.e. you're summing the same thing on both sides)? Wouldn't that already imply the theorem? – Rodrigo Meireles Sep 18 '20 at 16:47
  • @RodrigoMeireles When reordering an infinite sum, the value doesn't always remain the same. The proof on wikipedia is mostly concerned with questions of convergence – Mastrem Sep 19 '20 at 06:58
  • but wouldn't that kind of proof be necessary here too? Doesn't the equality sign mean that both sides converge to the same number? – Rodrigo Meireles Sep 19 '20 at 16:02
  • Well, yes, but the (first) equality isn't as obvious as I make it sound. Perhaps this isn't a good answer and I should delete it. The point is that rearranging the terms of an infinite series doesn't always keep the value the same. In this specific case, the value does say the same, which one proves with Mertens' theorem. – Mastrem Sep 19 '20 at 19:23
  • I see! Thank you very very much for your attention! Your answer is very complete now, given the follow-up comments. – Rodrigo Meireles Sep 20 '20 at 01:26