Eigenvalues of tridiagonal matrix

Question

on page 13 of the paper here there is a proof in theorem 4 that all eigenvalues of this tridiagonal matrix, which has strictly positive entries down the subdiagonals, are simple. Unfortunately, I don't get the argument. Apparently, it is almost immediate to the editor that $ker(J-\lambda I)$ must be one-dimensional for eigenvalues $\lambda$, where $J$ is the special tridiagonal matrix.

Answer 1

If you delete the first row and last column from an irreducible $n\times n$ tridiagonal matrix $T$, the resulting submatrix is triangular with non-zero diagonal entries. Hence it is invertible, and it follows that $\mathbb{rank}(T-\lambda I)$ is always at least $n-1$.