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For the $N \times N$ tridiagonal Toeplitz matrix

$$A_N = \left[{\begin{array}{*{20}{c}}2&{ - 1}&0& \cdots &0&0\\{ - 1}&2&{ - 1}& \cdots &0&0\\0&{ - 1}&2& \cdots &0&0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\0&0&0& \cdots &2&{ - 1}\\0&0&0& \cdots &{ - 1}&2\end{array}} \right]$$

We want to prove that there exists a constant $c>0$ (independent of $N$) such that

$$\frac{c}{N^2}x^T x \leq x^T A \,x$$

for all $x \in \mathbb R^N$. Also, can we find the largest value of $c$?

Many thanks.

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Since $\mathrm A \in \mathbb R^{n \times n}$ is tridiagonal and Toeplitz, its $n$ real eigenvalues are given by [0]

$$\lambda_k (\mathrm A) = 2 + 2 \cos \left(\frac{k \pi}{n+1}\right)$$

for $k \in \{1,2,\dots,n\}$. Hence,

$$\lambda_{\min} (\mathrm A) = 2 - 2 \cos \left(\frac{\pi}{n+1}\right)$$

and, thus,

$$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \left(2 - 2 \cos \left(\frac{\pi}{n+1}\right)\right) \|\mathrm x\|_2^2$$

It is now easy to find the largest value of $c$.


[0] Silvia Noschese, Lionello Pasquini, and Lothar Reichel, Tridiagonal Toeplitz Matrices: Properties and Novel Applications, 2006.

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  • $\begingroup$ Many thanks Rodrigo. Also, the reference paper was helpful. $\endgroup$ – Mago Mohsen Sep 21 '16 at 2:22

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