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An error related with matrix exponential |
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| Feb26-12, 05:00 PM | #1 |
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An error related with matrix exponential
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 |
| Feb26-12, 08:33 PM | #2 |
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Recognitions:
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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. |
| Mar1-12, 07:55 PM | #3 |
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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 |
| Mar1-12, 08:51 PM | #4 |
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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 |
| Mar1-12, 11:57 PM | #5 |
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Mentor
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| Mar4-12, 09:03 PM | #6 |
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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 |
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