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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.

- Thread starter Gg199
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- #1

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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

AlephZero

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[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.

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