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Eigenvalues of a unitary matrix

  1. Dec 11, 2007 #1
    Q: Prove htat if a matrix U is unitary, then all eigenvalues of U have absolute value 1.

    My try:
    Suppose U*=U^-1 (or U*U=I)

    Let UX=(lambda)X, X nonzero
    => U*UX=(lambda) U*X
    => X=(lambda) U*X
    => ||X||=|lambda| ||U*X||
    => |lambda| = ||X|| / ||(U^-1)X||

    And now I am really stuck and hopeless, what can I do?

    Thanks for helping!
     
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2

    Dick

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    If Ux=(lambda)x then (x*)(U*)=(lambda*)(x*). Multiply those together.
     
  4. Dec 12, 2007 #3

    Multiply which of them together??
     
  5. Dec 12, 2007 #4

    quasar987

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    The two equations.
     
  6. Dec 12, 2007 #5
    OK, so
    Ux (x*)(U*) = (lambda)x (lambda*)(x*)
    But how does this prove the statement?
     
  7. Dec 12, 2007 #6

    morphism

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    If U is unitary, then ||Ux||=||x||.
     
  8. Dec 12, 2007 #7
    Why?

    Also, I know nothing about isomertry. Is there any way to prove the statement without this concept? Can somone please show me how?
     
  9. Dec 12, 2007 #8

    HallsofIvy

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    Your initial definition of "unitary" was U*U= I, right? If [itex]\lambda[/itex] is an eigenvalue of U, with corresponding eigenvector v, then [itex]Uv= \lambda v[/itex]. But then [itex]U*U(v)= \lambda U*(v)[/itex] and since U*U= I, U*U(v)= v. That is, for any eigenvalue [itex]\lambda[/itex] of U and corresponding eigevector v, [itex]\lambda U*(v)= v[/itex]. That means that v is also an eigenvector of U* with eigenvalue [itex]1/\lambda[/itex].
     
  10. Dec 12, 2007 #9

    morphism

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    On second thought, maybe you should stick to what the other guys are suggesting. (For me, unitary = special kind of operator on an inner product space. But if you're working strictly with matrices, this might not be helpful. [For the sake of completeness: ||Ux||^2 = <Ux,Ux> = <x,U*Ux> = <x,x> = ||x||^2.])

    Just thought I'd also mention lambda is never going to be zero (because U is invertible), so we can safely say things like 1/lambda!
     
    Last edited: Dec 12, 2007
  11. Dec 12, 2007 #10

    Dick

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    Multiply them in the OTHER ORDER, so you get a (U*)(U).
     
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