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

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

    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 $\displaystyle T$ such that $\displaystyle \sigma(T)$ 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: $\displaystyle T^2-\lambda^2 I = (T+\lambda I)(T-\lambda I)$. If $\displaystyle T^2-\lambda^2 I$ is invertible then so is $\displaystyle T-\lambda I$.
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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 $\displaystyle \sigma_p(T)=\sigma(T)$(Pointspectrum = Spectrum). And another Theorem states that (*)for any t > 0 the set of all distinct eigenvalues $\displaystyle \lambda$ of $\displaystyle T$ with $\displaystyle \lambda \geq t$ is finite.

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

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