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Symmetric endomorphism

  1. Jun 19, 2010 #1
    Hi,

    We know that if u is a real symetric endomorphism, then u has a real eigenvalue and that u is diagonalizable.
    But can we say that u is diagonalizable with only real eigenvalues?
     
  2. jcsd
  3. Jun 19, 2010 #2

    HallsofIvy

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    Yes. Since you are talking about eigenvalues, I take it that u is an endomorphism on some vector space (a linear transformation from vector space V to itself). Specifically, if u is symmetric and [itex]\lambda[/itex] is an eigenvalue, then there exist non-zero x such that ux= \lambda x. Further, we can take x to have magnitude 1: <x, x>= 1. Then [itex]\lambda= \lambda<x, x>= <\lambda x, x>= <ux, x>= <x, ux>[/itex] (because u is symmetric). [itex]<x , ux>= \overline{<ux, x>}= \overline{\lambda x, x}= \overline{\lambda}\overline{<x, x>}= \overline{\lambda}[/itex].
    (The overline indicates complex conjugation.)

    Since [itex]\lambda= \overline{\lambda}[/itex], [itex]\lambda[/itex] is real. That is all eigenvalues are real for a symmetric endomorphism.
     
  4. Jun 19, 2010 #3
    I think you forgot an overline in the end, but I can't understand why we have:
    <x,ux>=conjugate(<ux,x>) ?
     
  5. Jun 20, 2010 #4

    HallsofIvy

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    One of the requirements in the definition of "inner product" is that [itex]<u, v>= \overline{<v, u>}[/itex]. Of course, if the vector space is over the real numbers, that is the same as "<u, v>= <v, u>" but then your question is trivial.
     
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