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Characteristic and minimal polynomials

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Let V be a finite dimensional complex vector space and T be the linear operator of V. Prove that the following are equivalent

    a V has a basis consisting of eigenvectors of T.
    b T can be represented by a diagonal matrix.
    c all the eigenvalues of T have multiplicity one.
    d. the Characteristic polynomial of T equals the minimal polynomial of T.


    2. Relevant equations

    Not really applicable

    3. The attempt at a solution

    Ok so I proved A implies B, However, I feel that B does not imply C.

    I just want to see if my argument is valid. because the identity matrix, is definitely diaganol however, it does not have a multiplicity of One. Can I assume wlog that T has distinct eigenvalues along the main diaganol?
     
  2. jcsd
  3. Nov 27, 2011 #2

    Dick

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    You are quite right. But I wouldn't assume wlog. That looses generality. B doesn't imply C period. I think whoever wrote the problem up had a memory lapse or forgot to state an assumption. You could assume they forgot that assumption and proceed from there.
     
    Last edited: Nov 27, 2011
  4. Nov 28, 2011 #3

    HallsofIvy

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    ??? There are two "multiplicities", the geometric multiplicity (number of independent eigenvectors corresponding to the eigenvalue) and algebraic multiplicity (multiplicity as a root of the characteristic polynomial). But neither of those must be one in order that T can be represented by a diagonal matrix, only that, for each eigenvalue, the two multiplicities be the same.

     
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