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Proof-U=e^A ny unitary matrix U

  1. Apr 11, 2005 #1
    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!
     
  2. jcsd
  3. Apr 11, 2005 #2

    Hurkyl

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    Do you know what e^A looks like for a typical matrix A?
     
  4. Apr 11, 2005 #3
    not really. I mean i could break it into A^n/n! but i dont 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 dont 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)
     
  5. Apr 11, 2005 #4

    Hurkyl

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    (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?
     
  6. Apr 11, 2005 #5
    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?
     
  7. Apr 11, 2005 #6
    Thanks!
    Hurkyl, i dont 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 wouldnt know that since i dont know the value of Q.
     
  8. Apr 12, 2005 #7
    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.
     
  9. Apr 12, 2005 #8
    thanks!! so e^(ia) is always equal to +-1??
     
  10. Apr 12, 2005 #9
    only if a is a multiple of pi
     
  11. Apr 13, 2005 #10
    Thanks!........
     
  12. Apr 17, 2005 #11
    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,
     
  13. Apr 17, 2005 #12

    Hurkyl

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    (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.
     
    Last edited: Apr 17, 2005
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