Proof of commutative property in exponential matrices using power series

  • Thread starter Paalfaal
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  • #1
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I'm trying to prove eA eB = eA + B using the power series expansion eXt = [itex]\sum_{n=0}^{\infty}[/itex]Xntn/n!

and so
eA eB = [itex]\sum_{n=0}^{\infty}[/itex]An/n! [itex]\sum_{n=0}^{\infty}[/itex]Bn/n!


I think the binomial theorem is the way to go: (x + y)n = [itex]\displaystyle \binom{n}{k}[/itex] xn - k yk = [itex]\displaystyle \binom{n}{k}[/itex] yn - k xk, ie. it's only true for AB = BA.



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

  • #2
kai_sikorski
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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

[tex] \sum_{n = 0}^\infty \frac{(A + B)^n}{n!} = \sum_{n = 0}^\infty \frac{1}{n!} \sum_{k=0}^n \left(
\begin{array}{c}
n \\
k \\
\end{array}
\right) A^k B^{n-k} [/tex]
[tex] = \sum_{n = 0}^\infty \sum_{k=0}^n \frac{1}{(n-k)!(k)!}A^k B^{n-k}[/tex]

Now what I'll do is pull out all the terms where [itex]k=0[/itex] from the sum

[tex]= 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}[/tex]

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

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

Notice that after [itex]k \ge 2[/itex] we need to start pulling out factors of [itex]\frac{1}{k!}[/itex]. 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 [itex]\ldots[/itex] I used? Once you figure that out I think you'll be able to prove your result.
 
  • #3
kai_sikorski
Gold Member
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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.
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
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412
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.
$$\bigg(\sum_{n=0}^\infty a_n\bigg)\bigg(\sum_{k=0}^\infty b_n\bigg)=\sum_{n=0}^\infty\sum_{k=0}^n a_k b_{n-k}.$$
 
  • #5
13
0
[tex] + \frac{A}{2} \left( I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \ldots \right) [/tex]
I think it should be an A2 instead of A this line..

However, it made sense now. Thank you so much! Very helpful :smile:
 
  • #6
kai_sikorski
Gold Member
162
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I think it should be an A2 instead of A this line..

However, it made sense now. Thank you so much! Very helpful :smile:
You are correct.
 

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