Upper Bounds on the Largest Eigenvalue of Jacobi Matrices

Suppose I have a symmetric tridiagonal (Jacobi) matrix in the following form:

$$\begin{pmatrix} 1 & a_{1} & 0 & ... & 0 \\\ a_{1} & 1 & a_{2} & & ... \\\ 0 & a_{2} & 1 & ... & 0 \\\ ... & & ... & & a_{n-1} \\\ 0 & ... & 0 & a_{n-1} & 1 \end{pmatrix},$$

where all $$0< a_i < 1$$ for $$i = 1\ldots n-1$$ but the matrix is not necessarily diagonally dominant. I am interested in finding a tight upper-bound for the largest eigenvalue of this matrix.

Unfortunately, Eigenvalues of Symmetric Tridiagonal Matrices doesn't have the answer that I am looking for.

The entries are nonnegative, so the dominant eigenvector has all entries positive, and its eigenvalue is an increasing function of the $$a_i$$. If each $$a_i = 1$$ then that eigenvalue is $$1 + 2 \cos\frac\pi{n+1}$$ if I did this right; since you allow only $$a_i < 1$$, this bound $$1 + 2 \cos\frac\pi{n+1}$$ is not attained, but it is still the supremum of eigenvalues of such matrices.
• Thanks. Based on your argument, I will use a slightly tighter bound $1+2a_{\max}\cos\frac{\pi}{n+1}$ where $a_{\max} = \max_{i=1\ldots n-1}a_{i}$. – Taha Oct 2 '18 at 20:27