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Trivial Eigenspace

  1. Apr 13, 2006 #1
    What's a "trivial eigenspace"?
     
  2. jcsd
  3. Apr 13, 2006 #2

    matt grime

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    Put it in context. Probably something to do with 0, and probably 0 as a vector.
     
  4. Apr 14, 2006 #3
    "Let V be a vector space of dimension n over a field F, and let T:V->V be linear. Let R={linear transformations V->V that commute with T}.

    Suppose T is diagonalisable, show that T is a commutative ring <=> all non-trivial eigenspaces of T are one-dimensional."

    I just need to know what "non-trivial" means in this case. We defined the 0 vector to NOT be an eigenvector.
     
  5. Apr 14, 2006 #4

    matt grime

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    0 is not en eigenVECTOR, agreed, but as a subvector space 0 is an eigenSPACE, the trivial eigenspace.
     
  6. Apr 15, 2006 #5
    How do I show that R is commutative <=> the minimal and characteristic polynomials of T are the same? (T might not be diagonalizable)
     
  7. Apr 16, 2006 #6

    matt grime

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    does that mean you've done the first question?

    There is no harm in passing to the algebraic closure of your field if it helps you to think about it. Put everythin in jordan form, what can you say about the commuting matrices? just to give you some idea of how to use the difference between the minimal and characeristic poly.
     
  8. Apr 16, 2006 #7
    Yes, I did the first question. I've got an idea about the second one that doesn't involve Jordan forms, but I'm still working on it.
     
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