I begin with an $n \times n$ real symmetric tridiagonal matrix. However, I replace the non-zero elements in the first and last rows with zeros, so it is no longer symmetric

$$M = \begin{bmatrix} 0 & 0 & & & \\ b_1 & a_2 & b_2 & & & \\ & & \ddots & & \\ & & b_{n-2} & a_{n-1} & b_{n-1} \\ & & & 0 & 0 \end{bmatrix}$$.

1. Is it possible to diagonalize this matrix to find $M = Q D Q^{-1}$ in $O(n^2)$ operations?
2. If the $b_i$ are all positive and the $a_i$ all negative, can we be sure that there are $n$ real negative eigenvalues?

The background is this is a numerical discretization of the Laplacian with specific boundary conditions. For this reason, in my example, the $b_i$ and $a_i$ are related approximately (not exactly) by $a_i \approx -2 b_i$.

I begin with an $n \times n$ real symmetric tridiagonal matrix. However, I replace the non-zero elements in the first and last rows with zeros, so it is no longer symmetric

$$M = \begin{bmatrix} 0 & 0 & & & \\ b_1 & a_2 & b_2 & & & \\ & & \ddots & & \\ & & b_{n-2} & a_{n-1} & b_{n-1} \\ & & & 0 & 0 \end{bmatrix}$$.

1. Is it possible to diagonalize this matrix to find $M = Q D Q^{-1}$ in $O(n^2)$ operations?
2. If the $b_i$ are all positive and the $a_i$ all negative, can we be sure that there are $n$ real (non-positive) eigenvalues?

The background is this is a numerical discretization of the Laplacian with specific boundary conditions. For this reason, in my example, the $b_i$ and $a_i$ are related approximately (not exactly) by $a_i \approx -2 b_i$.