(Linear Algebra) Distinct Eigenvalues of a Matrix

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The discussion centers on the distinct eigenvalues of a nearly full rank, symmetric matrix. It highlights that a symmetric matrix A of rank n-1 has one eigenvalue equal to zero, but the assertion that all eigenvalues are distinct is questioned. A counterexample is presented with a specific matrix that has repeated eigenvalues despite having rank n-1. Ultimately, the original poster expresses understanding after further clarification. The importance of symmetry in determining eigenvalue distinctness is acknowledged.
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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.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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