(Linear Algebra) Distinct Eigenvalues of a Matrix

  • Thread starter Gg199
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  • #1
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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.
 

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  • #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.
 
  • #3
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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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