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Math Help - Spectrum

  1. #1
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    Question Spectrum

    If  \cal{X} is a Banach space and  T: \cal{X} \rightarrow \cal{X} is an isometry , then either  \sigma(T) \subseteq \partial \cal{D} or  \sigma(T) = cl \cal{D}

    where
     \partial\cal{D} = {z in C: |z| = 1}

    Case 1: isometry is bijective
    When  \sigma(T) \subseteq \partial \cal{D} all points of the specrum are on the boundary

    Case 2: the isometry is not surjective.
     \sigma(T) = cl \cal{D}

    the spectrum is the whole disk.
    Last edited by Nusc; March 25th 2009 at 04:55 PM.
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  2. #2
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    Quote Originally Posted by Nusc View Post
    If  \cal{X} is a Banach space and  T: \cal{X} \rightarrow \cal{X} is an isometry , then either  \sigma(T) \subseteq \partial \cal{D} or  \sigma(T) = cl \cal{D}

    where
     \partial\cal{D} = {z in C: |z| = 1}

    Case 1: isometry is bijective
    When  \sigma(T) \subseteq \partial \cal{D} all points of the specrum are on the boundary

    Case 2: the isometry is not surjective.
     \sigma(T) = cl \cal{D}
    the spectrum is the whole disk.
    LaTeX note: use \mathcal, not \cal, in formulas (\cal works globally, and converts the rest of the formula to gibberish).

    Use the theorem which says that the boundary of the spectrum is contained in the approximate point spectrum. Suppose that \lambda is an approximate eigenvalue for T, with approximate eigenvector x. Then x is a unit vector, and Tx\approx\lambda x. But \|Tx\| =1, and it follows that |\lambda|=1.

    Thus \partial\sigma(T)\subseteq\partial\mathcal{D}. It then follows on purely geometrical grounds that either \sigma(T) \subseteq\partial\mathcal{D} or \sigma(T) = \overline{\mathcal{D}}. The first alternative occurs when 0\notin\sigma(T) (in other words, when T is unitary), and the second alternative in the case when 0\in\sigma(T) (in other words, when T is a proper isometry).
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