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Thread: self-adjoint linear operator

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    self-adjoint linear operator

    If T:V->V is a self-adjoint linear transformation, V is a finite-dimensional inner product space over C, how can I prove that T has n distinct eigenvalues?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by guin View Post
    If T:V->V is a self-adjoint linear transformation, V is a finite-dimensional inner product space over C, how can I prove that T has n distinct eigenvalues?

    That is false, choose for example T=I (the identity map) and n\geq 2 .


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