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Eigenvalue kind of nilpotent problem

  1. Jun 29, 2008 #1
    If a vector [tex]v\in V[/tex] and a linear mapping [tex]T:V\to V[/tex] are fixed, and there exists numbers [tex]\lambda_1\in\mathbb{C}[/tex], [tex]n_1\in\mathbb{N}[/tex] so that

    [tex]
    (T - \lambda_1)^{n_1}v = 0,
    [/tex]

    is it possible that there exists some [tex]\lambda_2\neq\lambda_1[/tex], and [tex]n_2\in\mathbb{N}[/tex] so that

    [tex]
    (T - \lambda_2)^{n_2}v = 0?
    [/tex]

    (Here complex numbers are interpreted as multiplication operators [tex]V\to V[/tex], [tex]v\mapsto \lambda v[/tex], as usual.)
     
  2. jcsd
  3. Jun 29, 2008 #2

    morphism

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    That happens iff v=0. This is the statement that the intersection of different generalized eigenspaces is {0}.
     
  4. Jun 30, 2008 #3

    gel

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    Furthermore, if f,g are any two polynomials with no common factor and f(T)v=g(T)v=0 then you can conclude that v=0.
     
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