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Generalized eigenspace invariant?

  1. Jun 7, 2008 #1
    Is the generalized eigenspace invariant under the operator T? Let T be finite dimensional Linear operator on C(complex numbers).
    My understanding of the Generalized Eigenspace for the eigenvalue y is:
    "All v in V such that there exists a j>=1, (T-yIdenitity)^j (v) = 0." plus 0.
  2. jcsd
  3. Jun 7, 2008 #2

    matt grime

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    What does invariant mean? Try to prove it.
  4. Jun 8, 2008 #3
    got it. thanks.
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