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An error related with matrix exponential

  1. Feb 26, 2012 #1
    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–Campbell–Hausdorff_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
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 26, 2012 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    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.
     
    Last edited: Feb 27, 2012
  4. Mar 1, 2012 #3
    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
     
  5. Mar 1, 2012 #4
    [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
     
  6. Mar 1, 2012 #5

    Mark44

    Staff: Mentor

    Done.
    No, it's fine to use the Report button for things like this, despite the misleading directions.
     
  7. Mar 4, 2012 #6
    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
     
    Last edited: Mar 4, 2012
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