Results 1 to 2 of 2

Math Help - spectrum of adjoint operator

  1. #1
    Member Mauritzvdworm's Avatar
    Joined
    Aug 2009
    From
    Pretoria
    Posts
    122

    spectrum of adjoint operator

    Show that \sigma(T^{*})=\{ \overline{\lambda}: \lambda\in \sigma(T)\}

    where \sigma(T) is the spectrum of T\in B(H) and B(H) is the space consisting of all linear operators mapping elements fromthe Hilbert space H to itself

    here's an idea

    \lambda \in \rho(T^{*})<=>(\lambda\mathbb{I} - T^{*}) exists, and \rho(T^{*}) is the resolvant of T^{*}
    <=>(\lambda - T^{*})x\neq 0 \ ,\forall x\in H
    <=>\lambda x \neq T^{*}x, \forall x\in H
    <=>\overline{\lambda}x\neq x^{*}T, \forall x^{*}\in H

    is this on the right track?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2010
    From
    Lisbon
    Posts
    51
    I would suggest you think of it this way: if \lambda \not \in \sigma (T), i.e., if \lambda \in \rho (T), then there exists S\in B(H) such that

    T_\lambda S=ST_\lambda=I,

    where T_\lambda \equiv T-\lambda I (or ( \lambda I-T), doesn't matter). That's just the definition of resolvent. Can you see how this translates to in terms of (T_\lambda S)^* and (ST_\lambda)^*?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A property of the spectrum of a compact operator
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 16th 2010, 12:09 PM
  2. Adjoint operator
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: February 24th 2010, 01:51 AM
  3. adjoint operator
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 4th 2010, 08:01 PM
  4. self-adjoint operator
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: October 18th 2009, 04:02 AM
  5. Spectrum of Linear Operator
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 24th 2009, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum