(Linear Algebra) Distinct Eigenvalues of a Matrix

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SUMMARY

The discussion centers on the distinct eigenvalues of a symmetric matrix A of size n x n with rank n-1. It is established that for such a matrix, all eigenvalues are distinct, except for one which is zero. The symmetry of the matrix is crucial for this conclusion, as it ensures that the eigenvalues are real and can be analyzed using properties of symmetric matrices. The example provided, a rank-deficient matrix, illustrates the point that while one eigenvalue is zero, the others must be distinct due to the matrix's properties.

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Gg199
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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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Something doesn't make sense here. What about
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}

If the rank is n-1, the only thing you can say is that exactly one eigenvalue is zero.
 
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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