An error related with matrix exponential


by umut_caglar
Tags: error, exponential, matrix
umut_caglar
umut_caglar is offline
#1
Feb26-12, 05:00 PM
P: 7
Hi guys I have a problem in finding my error in a calculation, I will be glad if you help me to find the error that I am doing

ok the problem is basically about matrix exponentials, here we go:

A, B, U, P are matrices
n is a natural number
t and T are rational numbers and T=n*t

now in general ## e^{t(A+B)}≠e^{tA}*e^{tB} ##
but can be represented by using The Zassenhaus formula

## e^{t(A+B)}=e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots ##

one can find the details of the formula from http://en.wikipedia.org/wiki/Baker%E...sdorff_formula

now I begin by writing the formula

## e^{t(A+B)}=e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots ##

and then I take the 'n'th power of both sides

## \left(e^{t(A+B)}\right)^n=\left(e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots\right)^n ##


if I define U=A+B I will get

## \left(e^{t U}\right)^n=\left(e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots\right)^n ##

by using the equality of (e^A)^n=e^[aN] i will obtain

## e^{t*n*U}=\left(e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots\right)^n ##

by using the definitions T=nt and U=A+B i will get

## e^{T*(A+B)}=\left(e^{tA}*e^{tB}*e^{(t^2/2)*(AB-BA)}*\ldots\right)^n ##

for the right side I define the P matrix as P=AB-BA

## e^{T*(A+B)}=\left(e^{tA}*e^{tB}*e^{(t^2/2)*P}*\ldots\right)^n ##

and by using the equality ## \left(e^A*e^B\right)^n=e^{(nA)}*e^{(nB)} ## recursively, I obtain

## e^{T*(A+B)}=e^{t*n*A}*e^{t*n*B}*e^{(t^2/2)*n*P}*\ldots ##

Again by using the definitions of T=tn and P=AB-BA I obtain

## e^{T*(A+B)}=e^{T*A}*e^{T*B}*e^{T*(t/2)*(AB-BA)}*\ldots ##

but, if expand the left side of the equation by using Zassenhaus formula, I end up with

## e^{TA}*e^{TB}*e^{(T^2/2)(AB-BA)}*...≠e^{TA}*e^{TB}*e^{(T*t/2)(AB-BA)}*\ldots ##


which is not clearly equal to the right side; so where is my mistake.


*****************************

As a side note; If you also show the correct version of the calculation I will be glad;

For example; I expect my error is asuuming the equality of ## \left(e^A*e^B\right)^n=e^{(nA)}*e^{(nB)} ##
if this is the mistake, could you show the correct relation?

Thanks for the help
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
Stephen Tashi
Stephen Tashi is offline
#2
Feb26-12, 08:33 PM
Sci Advisor
P: 3,173
Edit: You got the LaTex working, so I'll delete my version of it.

Another suggestion: You might get an answer quicker if you request this thread be moved to the Linear and Abstract Algebra section of the forum. I don't really know the proper way to make such a request. I suppose you could PM a moderator. The "report" feature might also be used, but the directions for using it make it sound like it's only to report offensive material.
umut_caglar
umut_caglar is offline
#3
Mar1-12, 07:55 PM
P: 7
Hi everybody I finally understand the problem, I will put a note to here in case someone else might need it

A and B are matrices n is a natural number

say ##AB-BA=\phi## then we will have ##(AB)^2=ABAB## which is equal to

##A(AB-\phi)B##

then we get

##=AABB-A\phi B##

finally get

##(AB)^2=A^2B^2-A\phi B##

so my initial guess was correct and ##(AB)^n≠A^nB^n##

thanks to everybody who tries to solve it

genericusrnme
genericusrnme is offline
#4
Mar1-12, 08:51 PM
P: 615

An error related with matrix exponential


[itex]\left(e^A*e^B\right)^n=e^{(nA)}*e^{(nB)}[/itex]

Yep, this is only true if A and B commute, that is AB-BA=0
Mark44
Mark44 is offline
#5
Mar1-12, 11:57 PM
Mentor
P: 20,988
Quote Quote by Stephen Tashi View Post
Another suggestion: You might get an answer quicker if you request this thread be moved to the Linear and Abstract Algebra section of the forum.
Done.
Quote Quote by Stephen Tashi View Post
I don't really know the proper way to make such a request. I suppose you could PM a moderator. The "report" feature might also be used, but the directions for using it make it sound like it's only to report offensive material.
No, it's fine to use the Report button for things like this, despite the misleading directions.
ajkoer
ajkoer is offline
#6
Mar4-12, 09:03 PM
P: 34
As a sidebar, it is important to know (at least when finding the root of a transition matrix) that if A= P'EP and B =R'SR where E and S are a diagonal matrix of eigenvalues. The P and R matrices are composed of the respective eigenvectors, and PP' = RR' = I. Then:

A^n = P'(E^n)P and B^n = R'(S^n)R

Note, A*A = P'(E)PP'(E)P = P'(E)I(E)P = P'(E^2)P

so one only needs to raise the diagonal elements (the eigenvalues) to the power of n (where n can also be a fraction, that is, taking a root).

As such, (A^n)*(B^n) = P'(E^n)P*R'(S^n)R


Register to reply

Related Discussions
Proving a matrix exponential determinant is a exponential trace Calculus & Beyond Homework 13
a question related to an exponential random variable Set Theory, Logic, Probability, Statistics 0
matrix exponential Linear & Abstract Algebra 3
Continuous inverse scale parameter in error function (Phase Bi-exponential & Laplace) Set Theory, Logic, Probability, Statistics 2
Matrix Exponential Calculus & Beyond Homework 4