An error related with matrix exponential

[umut_caglar]

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

 

[Stephen Tashi]

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]

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]

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

Yep, this is only true if A and B commute, that is AB-BA=0

 

[Mark44]

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.

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]

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