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Confusion about eigenvalues of an operator

  1. Feb 21, 2016 #1
    Suppose ##V## is a complex vector space of dimension ##n## and ##T## an operator in it. Furthermore, suppose ##v\in V##. Then I form a list of vectors in ##V##, ##(v,Tv,T^2v,\ldots,T^mv)## where ##m>n##. Due to the last inequality, the vectors in that list must be linearly dependent. This implies that the equation
    $$
    0=a_0v+a_1Tv+a_2T^2v+\ldots+a_mT^mv
    $$
    are satisfied by some nonzero coefficients. For a particular case, assume that the above equation is satisfied by some choice of coefficients where all of them are nonzero.
    Now since the equation above forms a polynomial, I can write the factorized form
    $$
    0=A(T-\mu_1)\ldots(T-\mu_m)v
    $$
    The last equation suggests that ##T## has ##m## eigenvalues. But this contradicts the fact that ##T## is an operator in a vector space of dimension ##n<m##. Where is my mistake?
     
    Last edited: Feb 21, 2016
  2. jcsd
  3. Feb 21, 2016 #2
    ##T## is an operator whereas ##\mu_i## is a scalar. Instead you should write ##T-\mu_iI_n##.

    Suppose ##m=n+1##, and take ##\mu_1, \ldots, \mu_n## to be the eigenvalues of ##T##. Then clearly the final equation holds for an arbitrary choice of ##\mu_{n+1}##. So your mistake is in concluding that ##T## must have more than ##n## eigenvalues, which is a non sequitur.
     
  4. Feb 21, 2016 #3
    Why does it have to be ##m=n+1##? I can take ##m## any value I want such that it is bigger than ##n##.
     
  5. Feb 21, 2016 #4
    Indeed, you can--I was simply providing a counterexample to get you started.
    The eigenvalues of ##T## will always be contained in some proper subset of ##\{\mu_1,\ldots,\mu_m\}##, so at least 1 of the choices for ##\mu_i## will be arbitrary.
     
  6. Mar 1, 2016 #5

    Demystifier

    User Avatar
    Science Advisor

    No it doesn't. From the equation above it does not follow that
    $$(T-\mu_i)v=0$$
     
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