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Adjoint of Linear Operator

  1. Feb 25, 2012 #1
    I don't know how to start this problem. Since it's a bi-implication, I need to show each statement implies the other. I started playing around with the definitions of inner product and norm directly, but it's not going anywhere.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142910.jpg [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142929.jpg [Broken]

    If I expand the inner product on the right,

    <T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

    Of course,

    [itex]\|\vec{T(x)}\| = \sqrt{<T(x),T(x)>} = \sqrt{<x,x>}[/itex].
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 25, 2012 #2
    Look up isometries it should help you with this problem.
     
    Last edited by a moderator: May 5, 2017
  4. Feb 25, 2012 #3
    Two sections ahead in the book they talk about isometry and list the two implications in a theorem.
     
  5. Feb 25, 2012 #4
    <x,y> = <T*T(x),y> = <T(x),T**y> = <T(x),T(y)>

    <x,x> = <T*T(x),x> = <T(x),T**x> = <T(x),T(x)> = [itex]\|\vec{T(x)}\|^{2}[/itex].
     
    Last edited: Feb 25, 2012
  6. Feb 25, 2012 #5
    What do you mean by T*?
     
  7. Feb 25, 2012 #6
    Here is a hint:

    [itex]\langle T(x),T(y)\rangle = \frac{1}{4}\left(||T(x)+T(y)||^2-||T(x)-T(y)||^2\right)=\cdots[/itex]

    That is by the polarization identity.
     
  8. Feb 26, 2012 #7
    I already wrote that out.

    <T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

    The <T(x),T(x)> and <T(y),T(y)> terms cancel.
     
  9. Feb 26, 2012 #8
    T adjoint.
     
  10. Feb 26, 2012 #9
    If you use the polar identity, your next step should be

    [tex]\frac{1}{4}\left(||T(x+y)||^2-||T(x-y)||^2\right)[/tex]

    [tex]=\frac{1}{4}\left(||x+y||^2-||x-y||^2\right)=\langle x,y\rangle[/tex]
     
  11. Feb 26, 2012 #10
    How did you get to the second line from the first one, specifically T(x+y) = (x+y)? It looks like you used the left statement in the problem to do that.

    In that last line, if you set x = y, then it looks like you get the relation you want.
     
    Last edited: Feb 26, 2012
  12. Feb 26, 2012 #11
    Your question says, "Let T be a linear operator..." What does that mean?
     
  13. Feb 26, 2012 #12
    T(cx + y) = cT(x) + T(y).

    The transformation disappeared. I'm wondering where it went there. You went from the norm squared of T(x+y) to the norm squared of just (x+y). If you use the first property listed, then it makes sense to me.
     
  14. Feb 26, 2012 #13
    Because once you prove ||T(x)|| = ||x||, what does ||T(x+y)|| = ??
     
  15. Feb 26, 2012 #14
    That's what I just said twice. In order to make the step you did, you have to use that property. I suppose that satisfies the iff in the proposition.
     
  16. Feb 26, 2012 #15
    Then why did you ask if you knew that part?

    What you have as a proof doesn't look like you used the polar identity. I am not really sure what you have done to claim <T(x),T(y)>=<x,y>
     
  17. Feb 26, 2012 #16

    Deveno

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    suppose <T(x),T(y)> = <x,y> for all x and y. what happens when x = y?

    on the other hand, suppose ||T(x)|| = ||x|| for all x.

    what happens when you use x+y and x-y instead of x?
     
  18. Feb 26, 2012 #17
    I just wanted to be sure. I wasn't getting anywhere with my work.
     
  19. Feb 26, 2012 #18
    For <T(x),T(y)> = <x,y>, if x = y, you can get the norm squared of x.

    I thought we already already plugged in x+y and x-y.

    <T(x+y),T(x+y)> = <T(x),T(x)> + <T(y),T(y)> + <T(x),T(y)> + <T(y),T(x)>

    <T(x-y),T(x-y)> = <T(x),T(x)> + <T(y),T(y)> - <T(x),T(y)> - <T(y),T(x)>
     
    Last edited: Feb 26, 2012
  20. Feb 26, 2012 #19
    Okay, guys, I think I have figured it out. I'll finish it up on my paper and be finished with this assignment. Thanks for the help.
     
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