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Unitary operators

  1. May 9, 2007 #1
    1. The problem statement, all variables and given/known data

    why is the spectrum of the unitary operator the unit circle?

    2. Relevant equations

    i know that U^(-1)=U* and i know this makes U normal
    i also know that normal means UU*=U*U


    3. The attempt at a solution

    i know that from spectral theory there is some lambda in the spectrum
    such that abs(lambda)=1, but i dont understand why ALL of them are on
    the unit circle. (i understand the operator, but spectrums are confusing to me.

    thanks
     
  2. jcsd
  3. May 9, 2007 #2
    If x is an eigenvector of U, and [itex]\lambda[/itex] is its eigenvalue, then what is the length of Ux?

    [itex]Ux = \lambda x[/itex]
     
  4. May 9, 2007 #3
    hmmm, i know that, so i have Ux = Lx and L is 1....so then Ux = x...? i'm just getting lost
     
  5. May 9, 2007 #4

    matt grime

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    Just so we're happy: you're only talkiong about operators on finite dimensional spaces, right? Because, in general, spectrum and 'set of eigenvalues' are not the same thing.
     
  6. May 9, 2007 #5
    heh, sorry about that. a unitary operator in a Hilbert space is what i'm working with
     
  7. May 10, 2007 #6
    L may not be 1, and you don't know what L is. But what is the length of Ux?
     
  8. May 10, 2007 #7

    matt grime

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    Let x be an e-vector of U with e-value L as above.


    What do we know?

    <x,x>=<Ux,Ux>

    because U is unitary. If you don't see that then consider the intermediate steps:

    <x,x>=<Ix,x>=<U*Ux,x>=<Ux,Ux>

    Note we've just used the unitariness of U. So now we've got to use the fact that x is an e-vector

    <x,x>=<Ux,Ux>=<Lx,Lx>=.....?
     
  9. May 10, 2007 #8
    and <Lx, Lx> is the inner product of an e-vector with its e-value....so do i get one? or am i using L=1?
     
  10. May 10, 2007 #9
    L is a number. There is a formula for pulling a number multiplier out of an inner product.
     
  11. May 10, 2007 #10
    a formula? isn't <Lx, Lx> = ||Lx||^2?
    and i know that L<x, y> = <Lx, y> ....
     
  12. May 10, 2007 #11
    What about <x,Ly>?
    What about <Lx,Lx>?
     
  13. May 10, 2007 #12
    Raven, could you satisfy my curiosity? Are you taking a course, or reading a book on your own? What is the name and level of the course or the name of the book?
     
  14. May 10, 2007 #13
    ahhh i see what you mean! thanks!
     
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