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Homework Help: Unitary Matrices Problem

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    U is a unitary matrix. Show that
    ||UX|| = ||X|| for all X in the complex set.

    Also show that |λ| = 1 for every eigenvalue λ of U.


    2. Relevant equations



    3. The attempt at a solution
    I'm not sure where to start. So I looked up the definition of a unitary matrix. It satisfies one of these conditions:
    U-1 = UH
    The rows of U are an orthonormal set in the complex set
    The columns of U are an orthonormal set in the complex set

    Say X = [x1 x2 ... xn]

    Now I know that ||X||2 = <X, X> = |x1|2 ... |xn|2

    I'm not sure where to go from here. Can anyone help?
     
  2. jcsd
  3. Mar 29, 2010 #2

    Dick

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    Look at <UX,UX>. You want to show that's equal to <X,X>. Use U^H=U^(-1).
     
  4. Mar 29, 2010 #3
    I don't get how U^H = U^(-1) comes in here.

    Maybe (UX)^H = (X^H)U^(-1), does that help?
     
  5. Mar 29, 2010 #4

    Dick

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    The hermitian adjoint satisfies <X,AY>=<A^H X,Y>. Move the U on the right over.
     
  6. Mar 29, 2010 #5
    Sorry just for clarification, A^H and X are multiplied together in the above, correct? I can't find that in my textbook...
     
  7. Mar 29, 2010 #6

    Dick

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    Yes, A^H and X are multiplied together.
     
  8. Mar 29, 2010 #7
    Move the U on the right over to get [(UX)^H]U = X^H?
     
  9. Mar 29, 2010 #8
    Wait, I think I got it!

    <UX, UX > = <U^H UX, X> = <U^-1 UX, X> = <X, X>!
     
  10. Mar 29, 2010 #9

    Dick

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    No. No conjugate on X. <UX,UX>=<U^H UX,X>.
     
  11. Mar 29, 2010 #10
    Okay, that makes sense. So now we have that's equal to <X, X>, so obviously their lengths are the same.

    The only thing it says in my book though is that <VZ, W> = <Z, VW>... I wonder if I'm allowed to use the fact you gave me. Would it work using that instead?
     
  12. Mar 29, 2010 #11

    Dick

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    <VZ,W>=<Z,VW> is only true if V is hermitian, i.e. V^H=V. That's not generally true for unitary matrices. Are you sure you are reading everything in the book, or are you just picking out formulas?
     
  13. Mar 29, 2010 #12
    I missed the part where it said V has to be hermitian. I'm reading it but I guess I need to read more carefully...

    I'm trying the next part now, showing that |λ| = 1.

    Starting with UX = λX, I think I should be able to manipulate this to give λ = 1
     
  14. Mar 29, 2010 #13
    I think I got the last part.
    UX = λX
    ||UX|| = ||λX||
    ||UX|| = |λ| ||X||
    ||UX|| / ||X|| = |λ|

    ||X|| = ||UX|| so |λ| = 1

    Going back to the first part of the question, the reason hermitian adjoint doesn't look familiar is because I haven't learned hilbert spaces yet. It's definitely not in there. Just out of interest, is there another way to do it?

    If it's true that <UX,UX> = (UX)^H (UX), then I know how to do it.
     
    Last edited: Mar 29, 2010
  15. Mar 29, 2010 #14

    Dick

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    Sure, that shows |λ| = 1. The definition of <X,Y> is X^H Y. Since X^H is the conjugate transpose. So sure. <UX,UX> = (UX)^H (UX).
     
  16. Mar 29, 2010 #15
    Okay, great. Then I think the first part can also be proved this way:
    <UX, UX> = (UX)^H UX = X^H U^H U X = X^H U^(-1) U X = X^H X = <X,X>.

    And necessarily if <UX, UX> = <X,X> their lengths are the same.

    It's possible though for 2 vectors to have the same length but not be equal, isn't it?
     
  17. Mar 29, 2010 #16

    Dick

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    You KNOW two vectors can have the same length and not be equal, right?
     
  18. Mar 29, 2010 #17
    Whoops, I got confused. I forgot that <UX, UX> was directly related to the length of UX, and thought for a second that I was just proving that UX = X.

    But yeah, [0 1] and [1 0] aren't equal but they have the same length.
     
  19. Mar 29, 2010 #18

    Dick

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    Right. Not trying to prove UX=X.
     
  20. Mar 29, 2010 #19
    I just read somewhere that X*X = ||X||^2. I was thinking all along that X*X = ||X||, but I guess it doesn't affect my work.

    * just means the same thing as H, right?
     
  21. Mar 29, 2010 #20

    Dick

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    '*' means a lot of things. I'd stick with '^H'. And when you use it say 'I mean H is hermitian conjugate' (i.e. transpose and conjugate). The usual symbol is a superscript dagger. Always say what you think your symbols mean, especially if you aren't using TeX.
     
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