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Homework Help: Eigenvalues of an operator in an inner product space

  1. Mar 20, 2006 #1
    "Suppose [tex]V[/tex] is a (real or complex) inner product space, and that [tex]T:V\rightarrow V[/tex] is self adjoint. Suppose that there is a vector [tex]v[/tex] with [tex]||v||=1[/tex], a scalar [tex]\lambda\in F[/tex] and a real [tex]\epsilon >0[/tex] such that

    [tex]||T(v)-\lambda v||<\epsilon[/tex].

    Show that T has an eigenvalue [tex]\lambda '[/tex] such that [tex]|\lambda -\lambda '| < \epsilon[/tex]."

    Since T is self adjoint, there exists an orthonormal basis [tex](e_1,...,e_n)[/tex], with corresponding eigenvalues [tex]\lambda_1,...,\lambda_n[/tex]. Suppose [tex]v=x_1e_1+...+x_ne_n[/tex] for some [tex]x_1,...,x_n\in F[/tex]. Then,


    Since the basis is orthonormal, it follow that


    At this point I am unable to deduce the conclusion.
    Last edited: Mar 20, 2006
  2. jcsd
  3. Mar 20, 2006 #2


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    Suppose [itex]|\lambda _i - \lambda| \geq \epsilon[/itex] for all i, then:

    [tex]\epsilon^2 > |(\lambda_1-\lambda)x_1|^2+...+|(\lambda_n-\lambda)x_n|^2 = \sum _{k=1} ^n |\lambda _k - \lambda|^2|x_k|^2 \geq \sum \epsilon ^2|x_k|^2 = \epsilon ^2\sum |x_k|^2 = \epsilon ^2 ||v|| = \epsilon ^2[/tex]
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