Generalized eigenspace invariant?

  • Thread starter mind0nmath
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  • #1
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Hey,
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.
thanks
 

Answers and Replies

  • #2
matt grime
Science Advisor
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What does invariant mean? Try to prove it.
 
  • #3
19
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got it. thanks.
 

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