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Symmetric Matrice Proof

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose H is an n by n real symmetric matrix. v is a real column n-vector and H^(k+1)v = 0. Prove that Hv = 0

    3. The attempt at a solution

    Since H is a real symmetric matrice we can find an orthogonal matrix Q to diagnolize it:

    M = Q transpose.

    MA^(k+1)Qv = 0


    A^(k+1)Qv = 0

    This is where I'm stuck i'm not sure how to proceed.

    I'm pretty sure its not possible to somehow get Q remove from the equation because that implies v or A would have to be 0 but this does not follow since I can easily cook up an example were there is a symmetric matrix to a power were H^(K+1)v=0 and v != 0. Thus any hints would be appreciated.
  2. jcsd
  3. Sep 10, 2007 #2


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    If you work in the basis where H is diagonal then replace the phrase "n by n real symmetric matrix" in your premise with "n by n DIAGONAL matrix". Can you prove that one?
  4. Sep 10, 2007 #3


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    Assume v is not zero. Then, since Q is invertible, Qv is not zero. So Qv must be in the nullspace of A^(k+1). What is the null space of a diagonal matrix, and how is this related to the nullspace of its powers?
  5. Sep 11, 2007 #4
    That was the proof shown in class where you chose to represent the matrix using a basis consisting of eigen vectors, though I didnt start my proof that way and I believe that the diagonalization method should work if I could finish the last step.

    Statusx: It would imply that Qv is in the nullspace of A itself since A^(k+1) simply exponentiates each entry by k+1 (diagonal matrice), though I'm unsure how that shows Av = 0.

    You would get AQv = 0

    Transpose of Qv (easier as a row vector when typing) would be something along the lines:

    [0,0,0,....,a_i,.....,0] where a_i can by any real number corresponding to a 0 in the diagonal entry a_ii in the diagonal matrice A.
  6. Sep 12, 2007 #5


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    Right, and so MAQv = Hv = 0.
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