Proof-U=e^A ny unitary matrix U

[bor0000]

Proof: Any unitary matrix U in C(nxn) can be expressed as e^A, where A is skew-symmetric in C(nxn).
Hint: U=Qdiag(m1...,mn)Q* and the absolute value of the eigenvalues of U is 1.

thanks!

 

[Hurkyl]

Do you know what e^A looks like for a typical matrix A?

 

[bor0000]

not really. I mean i could break it into A^n/n! but i don't see the point. i guess in case A is skew symmetric, it could be something like combinations of 2x2 blocks of e^a*(rotation matrix), where a is the same constant for 2 consecutive diagonal elements?

in the soln that i have available, the next line in the proof, after the '|mi|=1' was
"Hence mi=e^(iQi) ", where mi is the index of the eigenvalue of the U matrix. i don't know how to get that. i.e. i think it refers to mi*v=e^A*v for some v. but how is e^A=e^(iqi)

 

[Hurkyl]

I mean i could break it into A^n/n! but i don't see the point

(I presume you mean the sum over all nonnegative n of that)

Well, it's what e^A means, so it might help us to understand it. Does the infinite sum simplify if A is diagonalizable?

 

[LeBrad]

A^2 = AA = UQU*UQU* = ?
A^3 = AAA = UQU*UQU*UQU* = ?
A^n = ?

If U is orthogonal and Q is diagonal, what does this turn out to be?

 

[bor0000]

Thanks!
Hurkyl, i don't know, how if a matrix is diagonalizable, it will be easier to break it into that summation. unless the matrix is something like 2x2 instead of nxn.

LeBrad, A^n=QD^nQ*
elements of D are eigenvalues ik with k being various constants
e^A=Qe^DQ*. U=Q1DQ1* with elements of D having abs. value 1. and i need to prove somehow e^ik multiplied by some factor will give |1|, but i wouldn't know that since i don't know the value of Q.

 

[Don Aman]

So U = QDQ^{-1} with D diagonal and with every diagonal element of D having modulus 1 and Q unitary. Any complex number whose modulus is equal to one can be written e^{ia}, and the exponential of a diagonal matrix is just the new diagonal containing so D can be written e^A, with A a diagonal matrix with pure imaginary entries. Also, e^{QAQ^{-1}}=Qe^AQ^{-1}, and if A is diagonal with pure imaginary, then QAQ^-1 is skew adjoint.

 

[bor0000]

thanks! so e^(ia) is always equal to +-1??

 

[Don Aman]

thanks! so e^(ia) is always equal to +-1??

only if a is a multiple of pi

 

[bor0000]

Thanks!...

 

[steveurkell]

I use Taylor expansion of e^x where x now is a matrix,
e^A= I + A +A^2/2!+A^3/3!+...
taking the determinant of both sides
det(e^A)=det(I+...) where the right hand side will just be sum of determinants of all terms
since for skew-symmetric matrix A transpose = -A, then on can show det(A)=0
hence all determinants on RHS vanish, except det(I);
proven that e^A = I
regards,

 

[Hurkyl]

(1) The determinant is not a linear operation. det(A + B) is usually not equal to det(A) + det(B).

(2) Most skew-symmetric matrices of even dimension (e.g. 6x6) don't have determinant zero.

(3) det(A) = det(B) does not imply A = B.