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Let $A$ be the $n \times n$ tridiagonal Toeplitz matrix of the form

$$A = \left[ \begin{array}{cccccc} 2 & -1 & 0 & \dots & \dots & 0 \\ -1 & 2 & -1 & 0 & \dots & 0 \\ 0 & -1 & 2 & -1 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & -1 & 2 & -1 \\ 0 & 0 & ... & 0 & -1 & 2 \\ \end{array} \right].$$

Verify that the eigenvalues of the above matrix $A$ are

$$2 - \cos\left(\frac{i\pi}{n + 1}\right)$$

for $i = 1,2,\dots,n$ and then determine the spectral radius of $A$.

In the context of this problem the spectral radius of a matrix $A$ will be denoted as $\rho(A) = $max$|\lambda_i(A)|$, where $\lambda_i$ is the $i$-th eigenvalue of an $n \times n$ matrix $A$ for $i = 1,2,...,n$. So if all the eigenvalue are the same and equal to $2 - \cos\left(\frac{i\pi}{n + 1}\right)$ then the 2nd part of the problem of finding the spectral radius would seem to follow automatically based on the definition of the spectral radius. So any advice on how to show that the eigenvalues of $A$ are $2 - \cos\left(\frac{k\pi}{n + 1}\right)$ would be appreciated, thank you.

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  • $\begingroup$ Expand $|A-\lambda|$ by the first row, and the second term by the first column to get a recurrence relation, $D_n(\lambda)=(2-\lambda)D_{n-1}(\lambda)+D_{n-2}(\lambda)$. $\endgroup$ Nov 2, 2019 at 7:33

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Your question is wrong, the eigenvalues are not given by $2-\cos(i\pi/(n+1))$ but by $2-2\cos(i\pi/(n+1))$. It can be shown recursively or by using the associated eigenvectors. The spectral radius is given by $2-2\cos(n\pi/(n+1))$.

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