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Conjugate homogeneity

  1. Jul 22, 2009 #1
    (aT)∗ = [tex]\bar{a}[/tex]T∗ for all a ∈ C and T ∈ L(V,W);
    This doesn't make much sense to me. Isn't a supposed to be=x+iy and
    [tex]\bar{a}[/tex]=x-iy? Not a fan of complex numbers. And
    this proof also confuses me.


    7.1 Proposition: Every eigenvalue of a self-adjoint operator is real.

    Proof: Suppose T is a self-adjoint operator on V. Let λ be an
    eigenvalue of T, and let v be a nonzero vector in V such that Tv = λv.
    Then
    λllvll2 = <λv,v>
    = <Tv,v>
    = <v,Tv>
    = <v,λv>
    = [tex]\bar{λ}[/tex]llvll2 that 955 is lambda
    so im guessing that for <Tv, v> to=<v, Tv>, the eigenvalue
    should not have a conjugate of the form x-iy but just a real number.
    I mean that makes sense I guess. In other words, the matrix for T* should not be a conjugate
    transpose since the matrix of T must=the matrix of T* for self adjointness.
     
    Last edited: Jul 22, 2009
  2. jcsd
  3. Jul 22, 2009 #2

    tiny-tim

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    Hi evilpostingmong! :smile:

    (it's \bar{\lambda} :wink:)
    Shouldn't that be

    = <v,Tv>*
    = <v,λv>*
    = λ*<v,v>* ? :redface:
     
  4. Jul 22, 2009 #3
    yes it should
     
  5. Jul 22, 2009 #4

    HallsofIvy

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    But, of course, <v, v> is a positive real number (part of the definition of "inner product") so that is [itex]\lambda<v,v>= \lambda^*<v,v>^*= \lambda^*<v,v>[/itex].
     
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