Simplicity conjecture for the homotopy eigenvalues

Let ${{\boldsymbol{M}}}_n$ and ${\boldsymbol{S}}_n$ be the matrices described above, and define

$${\boldsymbol{T}}({\varepsilon}) = \begin{bmatrix} {\varepsilon}{\boldsymbol{e}}_1{\boldsymbol{e}}_1^T & 0 & {\boldsymbol{S}}_n\\ 0 & 0 & {\varepsilon}{\boldsymbol{e}}_1^T \\ {\boldsymbol{S}}_n & {\varepsilon}{\boldsymbol{e}}_1 & 0 \end{bmatrix}.$$

Here ${\boldsymbol{T}}({\varepsilon})$ is the symmetric homotopy matrix and $\varepsilon$ is the continuation parameter. Simplicity conjecture. The eigenvalues of ${\boldsymbol{T}}({\varepsilon})$ are simple on $0 \leq \varepsilon$. This conjecture is motivated by numerical discriminant calculations: the calculated discriminants have strictly positive coefficients as polynomials in $\varepsilon$.