Proof of commutative property in exponential matrices using power seriesby Paalfaal Tags: exponential matrice, infinite sums, power series method 

#1
Feb1812, 08:42 AM

P: 10

I'm trying to prove e^{A} e^{B} = e^{A + B} using the power series expansion e^{Xt} = [itex]\sum_{n=0}^{\infty}[/itex]X^{n}t^{n}/n!
and so e^{A} e^{B} = [itex]\sum_{n=0}^{\infty}[/itex]A^{n}/n! [itex]\sum_{n=0}^{\infty}[/itex]B^{n}/n! I think the binomial theorem is the way to go: (x + y)^{n} = [itex]\displaystyle \binom{n}{k}[/itex] x^{n  k} y^{k} = [itex]\displaystyle \binom{n}{k}[/itex] 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? 



#2
Feb1812, 10:47 AM

PF Gold
P: 162

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^{nk} [/tex] [tex] = \sum_{n = 0}^\infty \sum_{k=0}^n \frac{1}{(nk)!(k)!}A^k B^{nk}[/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}{(nk)!(k)!}A^k B^{nk}[/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}{(nk)!(k)!}A^k B^{nk}[/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
Feb1812, 11:01 AM

PF Gold
P: 162

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
Feb1812, 01:46 PM

Emeritus
Sci Advisor
PF Gold
P: 8,989

Proof of commutative property in exponential matrices using power series
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.
$$\bigg(\sum_{n=0}^\infty a_n\bigg)\bigg(\sum_{k=0}^\infty b_n\bigg)=\sum_{n=0}^\infty\sum_{k=0}^n a_k b_{nk}.$$ 



#5
Feb1812, 04:31 PM

P: 10

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


Register to reply 
Related Discussions  
commutative property of multiplication...  General Math  8  
Exponential Power Series Expansion  Calculus  1  
Complex exponential proof using power series  General Math  1  
power series / exponential  Calculus & Beyond Homework  7  
power series expansion of an exponential  Calculus & Beyond Homework  5 