1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spectrum of operator's exponent

  1. May 7, 2008 #1
    Let A be a bounded operator in complex Hilbert space. Prove that [tex]\sigma(Exp(A))=Exp(\sigma(A))[/tex].

    It is known, that [tex]\sigma(P(A))=P(\sigma(A))[/tex], where P is a polynomial.
    In addition, if an operator A has a bounded inverse, then for any operator B such that [tex]||B||<1/||A^{-1}||[/tex] their sum A+B has a bounded inverse.

    I managed to prove that [tex]\sigma(Exp(A))\supseteq Exp(\sigma(A))[/tex].
    Then [tex]||Q_n(A)||\to 0[/tex].

    Let [tex]\lambda\notin\sigma(Exp(A))[/tex]. Then [tex]Exp(A)-\lambda I[/tex] is invertible, and [tex]P_n(A)-\lambda I=Exp(A)-\lambda I-Q_n(A)[/tex] is invertible for large n.
    Moreover, [tex]P_n(A)-(\lambda+\epsilon)I[/tex] is invertible for sufficiently small epsilon. So [tex]\lambda\notin Exp(\sigma(A))[/tex].
    Last edited: May 7, 2008
  2. jcsd
  3. May 12, 2008 #2
    No ideas...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Spectrum of operator's exponent