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 $\alpha\leq -1$…
Best Answer
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 $${\lambda_\max\over \lambda_\min}=\begin{cases} \displaystyle {2\over 3}h^{-\alpha} +O(1) &\alpha <2\\ \displaystyle {4\over 6+\pi^2}h^{-2}+O(1) & \alpha=2\\ \displaystyle {4\over \pi^2}h^{-2} +O(1) & \alpha>2 \end{cases}$$