# Maximal and minimal eigenvalues of a symmetric tridiagonal Toeplitz matrix

Given $$m \times m$$ symmetric tridiagonal Toeplitz matrices $$M=\begin{pmatrix} 4 & 1 & & \\ 1 & 4 & \ddots & \\ & \ddots & \ddots & 1\\ & & 1 & 4\end{pmatrix}, \qquad A=\begin{pmatrix} 2 & -1 & & \\ -1 & 2 & \ddots & \\ & \ddots & \ddots & -1\\ & & -1 & 2\end{pmatrix}$$ determine the maximal and minimal eigenvalues of $$F = M + h^{-\alpha} A$$ with $$\alpha \in \mathbb{Z}, h=\frac{1}{m+1}$$ when $$\mu_j = 4 + 2\cos(jh\pi)$$ are the eigenvalues of $$M$$ and $$\eta_j = 2 - 2\cos(jh\pi)$$ are the eigenvalues of $$A$$ with $$j=1,\dots, m$$.

I guess it is $$\lambda_j=4+2\cos(jh\pi) + h^{-\alpha}(2-2\cos(jh\pi))$$ but I cannot show it. To determine the eigenvalues of

$$M=\begin{pmatrix} 4 & 1 & & \\ 1 & 4 & \ddots & \\ & \ddots & \ddots & 1\\ & & 1 & 4 \\ \end{pmatrix}$$

I used the ansatz $$\det(M-\mu_jI)=\det(J-(\mu_j-4)I)$$ with

$$J=\begin{pmatrix} 0 & 1 & & \\ 1 & 0 & \ddots & \\ & \ddots & \ddots & 1\\ & & 1 & 0 \\ \end{pmatrix}$$

for which the eigenvalues are $$\sigma_j=2\cos(jh\pi)$$ and I hope $$\mu_j=4+2\cos(jh\pi)$$ is correct! The eigenvalues of $$A$$ were already given. The maximal eigenvalue of $$M$$ is $$\mu_1$$ but the maximal eigenvalue of $$A$$ is $$\eta_m$$.

I need to show that the spectral condition $$\kappa(F)=\frac{\lambda_{max}}{\lambda_{min}}$$ in first approximation of $$h$$ (means with the big O notation) is $$\kappa(F)\sim \begin{cases} 3-6h^{-\alpha}, \alpha \leq -1\\ 1, \alpha=0 \text{ (done})\\ \frac{2}{3}h^{-1}, \alpha=1 \\ \frac{4}{6+\pi^2}h^{-2}, \alpha=2 \\ \frac{4}{\pi^2}h^{-2}, \alpha\geq 3 \end{cases}$$

Consider $$\alpha \leq -1$$, then it is $$\kappa(F)=\frac{\lambda_{\max}}{\lambda_{\min}}=\frac{4+2h^{-\alpha} +2\cos(\pi h)(1-h^{-\alpha})}{4+2h^{-\alpha} -2\cos(\pi h)\,(1-h^{-\alpha})} \sim \frac{4+2h^{-\alpha} +2(1-h^{-\alpha})}{4+2h^{-\alpha} -2\,(1-h^{-\alpha})}$$ since $$cos(h\pi)=1-O(h^2)$$ with $$\alpha \leq -1$$. I cannot show that $$\kappa(F)=3-6h^{-\alpha}$$ for $$\alpha\leq -1$$...

According to the notation $$M=4I+J,\quad A=2I-J$$ Therefore $$M+h^{-\alpha}A =(4+2h^{-\alpha})\,I +(1-h^{-\alpha})\,J$$ Every eigenvalue of $$M+h^{-\alpha}A$$ is thus of the form $$[4+2h^{-\alpha}]+(1-h^{-\alpha})\, \lambda\qquad (*)$$ where $$\lambda$$ is an eigenvalue of $$J.$$ The eigenvalues of $$J$$ are known, and are listed in the question. That's actually a nontrivial fact which follows from the trigonometric identity $$2\cos \theta\,\sin k\theta= \sin (k+1)\theta+ \sin (k-1)\theta$$ for $$\theta$$ satisfying $$\sin(m+1)\theta =0.$$

In the question it is assumed that $$\alpha\in \mathbb{N}.$$ I will consider $$\alpha>0.$$ Hence $$h^{-\alpha}>1$$ and by $$(*)$$ the maximal eigenvalue $$\lambda_{\max}$$ corresponds to the minimal eigenvalue $$\sigma_{\min}$$ of $$J,$$ and the minimal one $$\lambda_{\min}$$ to the maximal eigenvalue $$\sigma_{\max}$$ of $$J.$$ We have $$\sigma_{\max}=2\cos{\pi\over m}=2\cos \pi h,\qquad \sigma_{\min}=2\cos{\pi(m-1)\over m}=-\sigma_\max$$ The ratio $$\lambda_{\max}/\lambda_{\min}$$ is thus equal $$\displaylines{{4+2h^{-\alpha} -2\cos {\pi h}\,(1-h^{-\alpha}) \over 4+2h^{-\alpha} +2\cos {\pi h}\,(1-h^{-\alpha}) }={2+4h^\alpha +2\cos {\pi h}\,(1-h^{\alpha}) \over 2+ 4h^\alpha -2\cos {\pi h}\,(1-h^{\alpha})}\\ = {4+2h^\alpha-2(1-\cos\pi h)(1-h^\alpha )\over 6h^\alpha+2(1-\cos\pi h)(1-h^\alpha )}}$$ Applying $$2(1-\cos x)={x^2}+O(x^4),\qquad x\to 0,$$ gives $$\displaylines{4+2h^\alpha-2(1-\cos\pi h)(1-h^\alpha )=4+2h^\alpha -\pi^2h^2+O(h^{2+\alpha})+O(h^4)\\ =\begin{cases}4+2h^\alpha +O(h^2) & \alpha<2\\ 4+(2-\pi^2)h^2+O(h^4) &\alpha=2\\ 4 -\pi^2h^2+O(h^\alpha) &\alpha>2 \end{cases}}$$ Next $$\displaylines{6h^\alpha+2(1-\cos\pi h)(1-h^\alpha )=6h^\alpha +\pi^2h^2+O(h^{2+\alpha})+O(h^4)\\ =\begin{cases}6h^\alpha +O(h^2) & \alpha<2\\ (6+\pi^2)h^2+O(h^4) &\alpha=2\\ \pi^2h^2+O(h^\alpha) &\alpha>2 \end{cases}}$$ Therefore $${\lambda_\max\over \lambda_\min}=\begin{cases} \displaystyle {2\over 3}h^{-\alpha} +O(1) &\alpha <2\\ \displaystyle {4\over 6+\pi^2}h^{-2}+O(1) & \alpha=2\\ \displaystyle {4\over \pi^2}h^{-2} +O(1) & \alpha>2 \end{cases}$$

• Thank you! I used $det(F_{\alpha}-\nu_jI)=det((1-h^{-\alpha})J-(\nu_j-4-2h^{-\alpha})I)$, so it holds $(1-h^{-\alpha})\lambda_j=\nu_j-4-2h^{-\alpha} \Rightarrow \nu_j=(4+2h^{-\alpha})+(1-h^{-\alpha})\lambda_j$. It is $\lambda_{max}$ for $j=1$ and $\lambda_{min}$ for $j=m$, right?
– Uhmm
Commented May 6, 2022 at 12:40
• That's right $\lambda_\max$ corresponds to the largest eigenvalue of $J,$ i.e. to $j=1.$ Similarly $\lambda_\min$ corresponds to $j=m.$ Commented May 6, 2022 at 12:49
• I made an edit at the end of my question referring to the maximal and minimal eigenvalue. Do you know how to show this for $\kappa(F)$? I only get a long fraction with the only difference $cos(\pi h)$ in the numerator and $cos(m\pi h)$ in the denominator which I cannot reduce...
– Uhmm
Commented May 6, 2022 at 13:02
• @Uhmm Sorry, I made a mistake. As the coefficient of $J$ in the expression for $F$ is negative then $\lambda_{\max}$ corresponds to the smallest eigenvalue of $J.$ I have re-edited my answer. Commented May 6, 2022 at 13:31
• @Uhmm I got the limit $3$ for any $\alpha>0.$ It is included at the end of the answer. Hopefully there are no errors. Commented May 6, 2022 at 13:40