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Math Help - Compact operators

  1. #1
    Senior Member Dinkydoe's Avatar
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    Compact operators

    I'm struggeling with the following questions:

    Prove or disprove the following statements:
    (a) If \mathcal{H} is a complex Hilbert-space and T\in B(\mathcal{H}) be a compact operator such that  \sigma(T) = \left\{0\right\} (the spectrum) then T= 0
    (b) If \mathcal{H} is a complex Hilbert-space and T\in B(\mathcal{H}) such that T^2 is compact, then the set \left\{\lambda\in\sigma(T):|\lambda|>\delta \right\} is finite for each \delta>0.

    I suspect (a) is false: Taking a Hilbert-space of finite dimension, and a nilpotent matrix should be a counterexample.
    I suspect (b) is false. If we find a compact operator T such that \sigma(T) is (not countable) infinite, this should be a counter example for (b). However I'm not even sure such T exists.

    Any help would be appreciated!
    Last edited by Dinkydoe; December 22nd 2009 at 02:22 PM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    I'm struggeling with the following questions:

    Prove or disprove the following statements:
    (a) If \mathcal{H} is a complex Hilbert-space and T\in B(\mathcal{H}) be a compact operator such that  \sigma(T) = \left\{0\right\} (the spectrum) then T= 0
    (b) If \mathcal{H} is a complex Hilbert-space and T\in B(\mathcal{H}) such that T^2 is compact, then the set \left\{\lambda\in\sigma(T):|\lambda|>\delta \right\} is finite for each \delta>0.

    I suspect (a) is false: Taking a Hilbert space of finite dimension, and a nilpotent matrix should be a counterexample. (Correct.)
    I suspect (b) is false. If we find a compact operator T such that \sigma(T) is (not countable) infinite, this should be a counter example for (b). However I'm not even sure such T exists.
    You can get a counterexample to (a) on an infinite-dimensional space by taking T to act nilpotently on a finite-dimensional subspace, and to be zero on the orthogonal complement.

    But (b) is true. Hint: T^2-\lambda^2 I = (T+\lambda I)(T-\lambda I). If T^2-\lambda^2 I is invertible then so is T-\lambda I.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    So if I'm right, and please correct me where I'm wrong:

    There's a Theorem in my book that more or less states: For compact operators we have \sigma_p(T)=\sigma(T)(Pointspectrum = Spectrum). And another Theorem states that (*)for any t > 0 the set of all distinct eigenvalues \lambda of T with \lambda \geq t is finite.

    Thus given your hint: if \lambda = a^2 is such that T^2-\lambda = (T-a)(T+a) is invertible iff T\pm a are invertible

    Thus \lambda\in \rho(T^2) \Leftrightarrow \pm a \in \rho(T). And since T^2 is compact we have (*), thus since \sigma(T)= \mathbb{C}\setminus \rho(T) should follow that for any \delta >0, we have \left\{a\in\sigma(T):|a|\geq \delta \right\} is finite.
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