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Orthogonal Projections

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    P is mxm complex matrix, nonzero, and a projector (P^2=P). Show 2-norm ||P|| >= 1
    with equality if and only if P is an orthogonal projector (P=P*)

    2. Relevant equations
    Let ||.|| be the 2-norm


    3. The attempt at a solution

    a. show ||P|| >= 1
    let v be in the range of P. Pv = v. so ||Pv|| = ||v||. By definition of matrix norm ||Pv|| <= ||P||*||v||. thus ||P|| >= 1. otherwise, ||Pv|| > ||P||*||v|| which is a contradiction.

    b. show equality if and only if P = P*
    first assume P =P*
    ||Px|| ^2 = <Px,Px> = <x,P*Px>=<x,Px> <= ||Px|| * ||x|| by cauchy-schwarz inequality
    so we have ||Px||^2 <= ||Px|| * ||x||
    we know ||P|| = max ||Px|| where ||x|| = 1
    so let ||x||=1
    we have ||Px||^2 <= ||Px|| so dividing by ||Px|| givex
    ||Px|| <= 1 thus Thus 1<=||P||<=1 so ||P||=1

    This is the part of the problem that I'm having trouble with
    assume ||P|| = 1 show P = P*
    <Px,Px> = 1
    <Px,Px> = <Px,PPx> = <P*Px,Px> = 1
    <Px,Px> - <P*Px,Px> = <Px-P*Px,Px> = 0
    P-P*P = 0 so P = P*P
    does this imply that P = P*?
    is so why? I'm stuck at this point. Thanks for any help pointing me in the right direction
     
  2. jcsd
  3. Sep 11, 2009 #2

    gabbagabbahey

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    P also equals P2=PP, so P-P*P = 0 implies PP-P*P = (P-P*)P=0 and P is non-zero so ______
     
  4. Sep 11, 2009 #3
    So I understand now that (P-P*)P=0 implies P=P* since P is nonzero. I'm questioning my method for getting there now. The reason is I don't use the fact that ||P|| = 1
    I only use the fact that <Px,Px>=<P*Px,Px>
    For example
    What if ||P|| = 2 then <Px,Px> = <Px,PPx> = <P*Px,Px> = 2 and
    <Px,Px> - <P*Px,Px> = <Px-P*Px,Px> = 0 still getting me to the same place.
    then I would have P = P* and from part 1 ||P|| = 1 a contradiction.
    Does this make sense?
    I'm not sure how I can use the ||P|| = 1
     
    Last edited: Sep 11, 2009
  5. Sep 11, 2009 #4

    gabbagabbahey

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    Hmmm...yes, I just noticed that you go directly from [itex]\langle Px-P^*Px\vert Px\rangle=0[/itex] to [itex]P-P^*P=0[/itex]....how do you justify that?

    Also, your notation looks pretty suspect to me....is it supposed to be Dirac notation, or something else?
     
    Last edited: Sep 11, 2009
  6. Sep 11, 2009 #5
    Yes, I see the problem now. I was thinking I could use ||.|| = 0 only is . = 0
    but <Px-P*Px,Px> = 0 doesn't mean that the first part is zero.
    I've been playing with this and can't find how to use the fact that ||P|| = 1
    any pointers?
    Thanks
     
  7. Sep 11, 2009 #6

    gabbagabbahey

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    Hmmm... how about using [tex]\langle(P-P^*P)x,Px\rangle=\langle(P-P^*P)x,(P-P^*P+P^*P)x\rangle=\langle(P-P^*P)x,(P-P^*P)x\rangle+\langle(P-P^*P)x,P^*Px\rangle[/tex]
     
  8. Sep 11, 2009 #7
    so I have <(P-P*P)x,Px> = <(P-P*P)x,(P-P*P)x>+<(P-P*P)x,P*PX>
    I know <Px,Px> = 1 = x*P*Px
    now I write <(P-P*P)x,Px>=x*P*Px-x*PP*Px=1-x*PP*Px

    and <(P-P*P)x,(P-P*P)x> = ||(P-P*P)x||

    and <(P-P*P)x,P*PX> = x*P*P*Px-x*PP*P*Px = 1-x*PP*Px
    Then <(P-P*P)x,Px> = 1-x*PP*Px = <(P-P*P)x,(P-P*P)x>+<(P-P*P)x,P*PX>
    = ||(P-P*P)x|| + 1-x*PP*Px

    subtract 1-x*PP*Px from each sde

    then we have
    ||(P-P*P)x|| = 0 and thus P = P*P so (P-P*)P =0 P isn't 0 so thus P=P*

    This seems right but I have one concern
    If ||P|| = 2 for example then <(P-P*P)x,P*PX> = 2-x*PP*Px = ||(P-P*P)x|| + 2-x*PP*Px which gives me the same thing of P = P* but from the first part of the proof P = P* implies ||P|| = 1 a contradiction
     
  9. Sep 12, 2009 #8
    I'm still confused because we're not using the fact that ||P|| = 1 just the fact that
    <(P-P*P)x,Px> = <(P-P*P)x,P*Px>
    and that <(P-P*P)x,Px> =||(P-P*P)x|| - <(P-P*P)x,P*Px>
    so ||(P-P*P)x|| = 0 and thus P-P*P = 0 thus P = P*
    but we never use ||P|| = 1 so wouldn't this work for and finite value of ||P||?
     
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