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(Linear Algebra) Distinct Eigenvalues of a Matrix

  1. Nov 29, 2011 #1
    I am reading through a proof and one line of it is not immediately obvious to me, despite it's simplicity. It relates to eigenvalues of a (nearly) full rank, symmetric matrix.

    Say we have a symmetric matrix A(nxn) that has rank=n-1. Why is this enough to say that all eigenvalues of A are distinct? Note that the symmetry is important for the result to hold, but I don't understand why.

    Thank you in advance.
     
  2. jcsd
  3. Nov 29, 2011 #2

    AlephZero

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    Something doesn't make sense here. What about
    [tex]\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}[/tex]

    If the rank is n-1, the only thing you can say is that exactly one eigenvalue is zero.
     
  4. Nov 30, 2011 #3
    Sorry I didn't give enough details at all. I think I understand it now though, thank you for the help anyway.
     
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