# An error related with matrix exponential

1. Feb 26, 2012

### 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

Last edited: Feb 27, 2012
2. Feb 26, 2012

### 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.

Last edited: Feb 27, 2012
3. Mar 1, 2012

### 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

4. Mar 1, 2012

### 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

5. Mar 1, 2012

### Staff: Mentor

Done.
No, it's fine to use the Report button for things like this, despite the misleading directions.

6. Mar 4, 2012

### 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

Last edited: Mar 4, 2012