Proof of commutative property in exponential matrices using power series

In summary: I made a mistake and it should be A2 instead of A. I apologize for the confusion. I'm glad it helped you understand the proof. In summary, the conversation discusses using the binomial theorem to prove the identity eA eB = eA + B. The conversation also suggests using the formula for products of series to simplify the proof.
  • #1
Paalfaal
13
0
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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  • #2
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
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
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
kai_sikorski said:
[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
Paalfaal said:
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.
 

1. What is the commutative property in exponential matrices?

The commutative property in exponential matrices states that the order in which two matrices are multiplied does not affect the final result. In other words, if matrix A and matrix B are multiplied, the result will be the same as if matrix B and matrix A were multiplied.

2. Why is the commutative property important in exponential matrices?

The commutative property is important because it allows us to simplify calculations and make them more efficient. It also allows us to generalize results and apply them to different situations.

3. How can proof of commutative property be shown in exponential matrices?

The proof of commutative property in exponential matrices can be shown using power series. This involves expanding the exponential function in a series of powers and using properties of power series to show that the order of multiplication does not affect the final result.

4. What is a power series?

A power series is a series of terms that involve powers of a variable. In the case of exponential matrices, the power series involves the exponential function raised to different powers.

5. Are there any exceptions to the commutative property in exponential matrices?

Yes, there are exceptions to the commutative property in exponential matrices. These exceptions occur when the matrices involved do not commute, meaning that the order of multiplication does affect the final result. In such cases, the commutative property does not hold and the proof using power series cannot be applied.

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