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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.

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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.

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    $\begingroup$ Thanks. Based on your argument, I will use a slightly tighter bound $1+2a_{\max}\cos\frac{\pi}{n+1}$ where $a_{\max} = \max_{i=1\ldots n-1}a_{i}$. $\endgroup$ – Taha Oct 2 '18 at 20:27

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