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ST and TS have the same eigenvals

  1. Nov 2, 2006 #1

    I need to prove that for S,T linear operators on V. ST and TS have the same eigenvalues. I've gotten as far as (say g is the eigenvalue and u is a nonzero vector): STu=gu so TS(Tu)=g(Tu). So TS has eigenvalue g corresponding to eigenvector Tu. But I don't know how to guarantee that Tu is nonzero. (Or how to resolve the possibility that it is zero)

    Thanks for any help.
  2. jcsd
  3. Nov 2, 2006 #2


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    If Tu is zero, what's STu?
  4. Nov 2, 2006 #3
    If Tu=0

    STu=0=gu, but u is nonzero, so g=0?

    So does zero as an eigenvalue become a special case? Then would I do the nonzero eigenvalue case separately?
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