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I want to show that for two commuting matrices, say $A$ and $B$, we have $e^{A+B}=e^Ae^B.$

So far I have \begin{align*} e^{A+B}&=\sum_{k=0}^\infty \frac{1}{k!}(A+B)^k \\ &=\sum_{k=0}^\infty \frac{1}{k!}\sum_{j=0}^k \binom{k}{j} A^jB^{k-j} \\ &= \sum_{k=0}^\infty \sum_{j=0}^k\frac{A^jB^{k-j}}{(k-j)!j!} \end{align*}

I don't really know where to go now though. What do I need to do next?

Alex M.
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MHW
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1 Answers1

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In the last expression, permute the sums in order to obtain

$$\sum _{j=0} ^\infty \sum _{k=j} ^\infty \frac {A^j B^{k-j}} {(k-j)! j!}$$

and finally make the change of summation variable $m = k-j$ to obtain

$$\sum _{j=0} ^\infty \sum _{m=0} ^\infty \frac {A^j B^m} {m! j!}$$

which is easily seen to be

$$\sum _{j=0} ^\infty \frac {A^j} {j!} \sum _{m=0} ^\infty \frac {B^m} {m!} = \Bbb e ^A \Bbb e ^B.$$

Alternatively, given that $A$ and $B$ commute, there is a basis in which both are diagonal, and working in this basis makes computations really easy. Try it!

Alex M.
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  • Well working in a diagonal basis still requires the corresponding proof for numbers. – ziggurism Oct 28 '16 at 15:21
  • What do you mean by 'permute the sums...'? – MHW Oct 28 '16 at 15:22
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    @sksks52: I meant to interchange the order of summation: take the inner sum to the front, and put the outer sum inside. – Alex M. Oct 28 '16 at 16:19
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    why is the sum suddenly up to infinity? Newton's binom says its up to the power, so up to K, permuting m=k-j only makes the counting go backward, but still only up to k..

    please elaborate

     – Yarden Cohen Feb 13 '20 at 19:20