Let $A$ be a five-diagonal matrix:
$$ A = \begin{bmatrix} 1 & a & b & & \cdots & & & & \\ a & 1 & a & b& \cdots & & & & \\ b & a & 1& a & \cdots & & & & \\ & b & a & 1 & \cdots & & & & \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ & & & & \cdots & 1 & a & b & \\ & & & & \cdots & a & 1 & a & b \\ & & & & \cdots & b & a & 1 & a \\ & & & & \cdots & & b & a & 1 \\ \end{bmatrix}_{n \times n} $$
- $a, b \in \mathbb{R}$
- $0 \le b \le a < \dfrac{1}{2}$
- $A$ is symmetric
- $A$ is positive definite $\Longrightarrow$ all $\lambda > 0$
- $A$ is a Toeplitz matrix
Question: What is the maximum condition number of $A$ for all $n$?
$$K(a, b) = \max_{n \in \mathbb{N}^{*}} \kappa(a, b, n) = \max_{n \in \mathbb{N}^{*} } \dfrac{\lambda_{max}(a, b, n)}{\lambda_{min}(a, b, n)}$$
EDIT:
Using a numerical method with many pairs $(a, \ b)$, I found that $\kappa(a, \ b, \ n)$ is a monotonically non-increasing function (with respect to $n$).
$$\kappa(a, b, 1) \le \kappa(a, b, 2) \le \cdots \le \kappa(a, \ b, \ n) \le \kappa(a, \ b, \ n+1) \le \cdots$$
$$K(a, \ b) = \lim_{n \to \infty} \kappa(a, \ b, \ n)$$
Some values of conditions number are:
$$A = \begin{bmatrix}1\end{bmatrix} \Rightarrow \kappa(a, \ b, \ 1) = 1$$ $$A = \begin{bmatrix}1 & a \\ a & 1\end{bmatrix} \Rightarrow \kappa(a, \ b, \ 1) = \dfrac{1+a}{1-a}$$ $$A = \begin{bmatrix}1 & a & b \\ a & 1 & a \\ b & a & 1\end{bmatrix} \Rightarrow \kappa(a, \ b, \ 1) = \dfrac{1+\frac{1}{2}\left(b+\sqrt{b^2+8a^2}\right)}{1+\frac{1}{2}\left(b-\sqrt{b^2+8a^2}\right)}$$
Like shown here, when $b=0$: $$\lambda_k = 1+2a\cos \left(\dfrac{k\pi}{n+1}\right) \ \ \ k = 1, \ \cdots, \ n$$ $$\kappa(a, \ 0, \ n) = \dfrac{1+2a \cos \left(\frac{\pi}{n+1}\right)}{1-2a\cos \left(\frac{\pi}{n+1}\right)}$$ $$K(a, \ 0) = \dfrac{1+2a}{1-2a}$$ I also found that for $a=0$
$$\lambda_k = 1+2b \cdot \cos \phi_k$$ $$\kappa(0, \ b, \ n) = \dfrac{1+2b \ \cos \phi_{min}}{1-2b \ \cos \phi_{min}}$$ $$K(0, \ b) = \dfrac{1+2b}{1-2b}$$