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Thread: spectrum of adjoint operator

  1. #1
    Member Mauritzvdworm's Avatar
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    spectrum of adjoint operator

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

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

    here's an idea

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

    is this on the right track?
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  2. #2
    Junior Member
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    Lisbon
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    I would suggest you think of it this way: if $\displaystyle \lambda \not \in \sigma (T)$, i.e., if $\displaystyle \lambda \in \rho (T)$, then there exists $\displaystyle S\in B(H)$ such that

    $\displaystyle T_\lambda S=ST_\lambda=I,$

    where $\displaystyle T_\lambda \equiv T-\lambda I$ (or ($\displaystyle \lambda I-T)$, doesn't matter). That's just the definition of resolvent. Can you see how this translates to in terms of $\displaystyle (T_\lambda S)^*$ and $\displaystyle (ST_\lambda)^*$?
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