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Symmetric, irreducible, tridiagonal matrix: Eigenvalues

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A) Let A be a symmetric, irreducible, tridiagonal matrix. Show that A cannot have a multiple eigenvalue.
    B) Let A be an upper Hessenberg matrix with all its subdiagonal elements non-zero. Assume A has a multiple eigenvalue. Show that there can only be one eigenvector associated with it.



    2. Relevant definitions
    Symmetric: A = AT
    Irreducible: not a good definition for it, but essentially it's well-connected
    Tridiagonal: obvious
    Upper Hessenberg: http://en.wikipedia.org/wiki/Hessenberg_matrix Essentially, it's a mix between a tridiagonal and a upper triangular. It's tridiagonal with non-zero elements above its 3rd subdiagonal.



    3. The attempt at a solution
    A) It seems contradiction would be the best combined with using the Jordan canonical form of A-[itex]\lambda[/itex]I.
    Jordan canonical form: http://en.wikipedia.org/wiki/Jordan_normal_form
    So, we know that we can get the Jordan canonical form of A which has all [itex]\lambda[/itex]'s in the diagonal.
    Thus, is there anyway we can look at the canonical form of A-[itex]\lambda[/itex]I and deduce from it's rank that it has no multiple eigenvalues?
    B) The hint says to use a similar strategy as A
     
  2. jcsd
  3. Feb 12, 2012 #2
    Just a bump.
     
  4. Feb 13, 2012 #3
    Just another bump. This is a tough problem.
     
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