Smallest eigenvalue of a certain Toeplitz Hermitian matrix

Question

Let $G\in\mathbb{C}^{n\times n}$ such that $G_{k\ell}=\frac{1}{1+i(k-\ell)\varepsilon}$ (here $i=\sqrt{-1}$ while $k,\ell$ are indices). For example, if $n=3$ we obtain $$ G=\begin{bmatrix}1 & \frac{1}{1-i\varepsilon} & \frac{1}{1-i2\varepsilon}\\ \frac{1}{1+i\varepsilon} & 1 & \frac{1}{1-i\varepsilon}\\ \frac{1}{1+i2\varepsilon} & \frac{1}{1+i\varepsilon} & 1\end{bmatrix}$$ I would like to determine a closed form expression for the smallest eigenvalue of $G$, or to have a lower bound for it. Any thoughts related would be helpful.

Answer 1

Fourier transformation of the Toeplitz matrix elements in the infinite-matrix limit gives $$w(\theta)=\sum_{m=-\infty}^\infty \frac{e^{im\theta}}{1+im\epsilon}=\frac{i}{\epsilon} \left[e^{-i \theta} \Phi \left(e^{-i \theta},1,\frac{\epsilon+i}{\epsilon}\right)-\Phi \left(e^{i \theta},1,-\frac{i}{\epsilon}\right)\right],$$ with $\Phi$ the Hurwitz-Lerch transcendent. _{(Beware of a $2\pi/\epsilon$ discontinuity at $\theta=0$ mod $2\pi$.)}
The Fourier transformed matrix is diagonal, so the estimate for the eigenvalues of $G$ for $n\gg 1$ is $$\lambda_k=w(2\pi k/n),\;\; k=1,2,\ldots n.$$

This is a monotonically decreasing function of $k$. For the smallest eigenvalue I find $$\lim_{n\rightarrow\infty}\lambda_{\rm min}=\lim_{\theta\uparrow 2\pi}w(\theta)=\frac{\pi}{\epsilon} \left(\coth \left(\frac{\pi }{\epsilon}\right)-1\right).$$

A numerical check, for $n=100$, shows this is quite accurate:

Red curve is $w(2\pi)$ as a function of $\epsilon$, the blue dots are the smallest eigenvalue of the $100\times 100$ Toeplitz matrix.

Numerically, I find that for finite $n$ the smallest eigenvalue remains above the large-$n$ limit, indicating that $(\pi/\epsilon)[\coth(\pi/\epsilon)-1]$ is a lower bound (but I have no proof for that).

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Above I compared the smallest eigenvalue, the full set of eigenvalues also agrees nicely with $w(2\pi k/n)$. Here is a comparison for $\epsilon=1$ and $n=100$.

Red curve: $w(\theta)$ for $\epsilon=1$; blue data points: $\lambda_k$ as a function of $2\pi k/n$ for $n=100$.

For the $\epsilon$-dependence of the largest eigenvalue $\lambda_{\rm max}$ I find $$\lim_{n\rightarrow\infty} \lambda_{\rm max}=\lim_{\theta\downarrow 0}w(\theta)=\frac{\pi}{\epsilon} \left(\coth \left(\frac{\pi }{\epsilon}\right)+1\right).$$ Here is a comparison of $w(0)$ with $\lambda_{\rm max}$ for $n=500$:

Answer 2

To add onto Carlo Beenakker's answer, the infinite-$n$ result is indeed a lower bound, because the lowest eigenvalue is a decreasing function of $n$.

For any nonzero vector $x$ and fixed $n$, the Rayleigh quotient gives an upper bound on $\lambda_{\text{min}}$: $$ \frac{x^{\ast} G x}{x^{\ast} x} \geq \lambda_{\text{min}} = \frac{v_{\text{min}}^{\ast} G v_{\text{min}}}{v_{\text{min}}^{\ast} v_{\text{min}}} $$ Here, $v_{\text{min}}$ is the corresponding eigenvector. For any larger $m > n$, we can simply pad $v_{\text{min}}$ with zeros without changing the Rayleigh quotient, showing that $\lambda_{\text{min}}$ is non-increasing.