Proof of commutative property in exponential matrices using power series

[Paalfaal]

I'm trying to prove e^A e^B = e^{A + B} using the power series expansion e^Xt = $\sum_{n=0}^{\infty}$Xⁿtⁿ/n!

and so
e^A e^B = $\sum_{n=0}^{\infty}$Aⁿ/n! $\sum_{n=0}^{\infty}$Bⁿ/n!


I think the binomial theorem is the way to go: (x + y)ⁿ = $\displaystyle \binom{n}{k}$ x^{n - k} y^k = $\displaystyle \binom{n}{k}$ y^{n - k} x^k, ie. it's only true for AB = BA.



I'm really bad at manipulating series and matrices. Could I please get some hints?

 

[kai_sikorski]

You are correct that the binomial theorem will be useful. The theorem you're trying to prove IS only valid if AB = BA, so the fact that the binomial theorem breaks down otherwise is not a concern. Personally I think I would find it easier to start from the opposite direction than you

$$\sum_{n = 0}^\infty \frac{(A + B)^n}{n!} = \sum_{n = 0}^\infty \frac{1}{n!} \sum_{k=0}^n \left(<br /> \begin{array}{c}<br /> n \\<br /> k \\<br /> \end{array}<br /> \right) A^k B^{n-k}$$
$$= \sum_{n = 0}^\infty \sum_{k=0}^n \frac{1}{(n-k)!(k)!}A^k B^{n-k}$$

Now what I'll do is pull out all the terms where $k=0$ from the sum

$$= I \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right) + \sum_{n = 1}^\infty \sum_{k=1}^n \frac{1}{(n-k)!(k)!}A^k B^{n-k}$$

Can you follow what I did there? Notice the lower bound on the sums changed. Let me do it for two more terms so you definitely get the idea

$$= I \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right)$$
$$+ A \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right)$$
$$+ \frac{A}{2} \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right)$$
$$+ \sum_{n = 3}^\infty \sum_{k=3}^n \frac{1}{(n-k)!(k)!}A^k B^{n-k}$$

Notice that after $k \ge 2$ we need to start pulling out factors of $\frac{1}{k!}$. Okay now can you see how this process would continue? How would you write down the above idea using the sum notation instead of the $\ldots$ I used? Once you figure that out I think you'll be able to prove your result.

 

[kai_sikorski]

By the way if you're trying to write a rigorous proof, I would concentrate on proving the first statement I made where I pulled out some terms. Then you can easily pull out an A factor, and shift indices; and then work by induction.

 

[Fredrik]

It might be a good idea to first prove the formula for products of series and then use that result along with the binomial theorem to prove the result you want.
$$\bigg(\sum_{n=0}^\infty a_n\bigg)\bigg(\sum_{k=0}^\infty b_n\bigg)=\sum_{n=0}^\infty\sum_{k=0}^n a_k b_{n-k}.$$

 

[Paalfaal]

$$+ \frac{A}{2} \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right)$$

I think it should be an A² instead of A this line..

However, it made sense now. Thank you so much! Very helpful

 

[kai_sikorski]

I think it should be an A² instead of A this line..

However, it made sense now. Thank you so much! Very helpful

You are correct.