On the spectral properties of real anti-tridiagonal Hankel matrices

TL;DR

This paper investigates the spectral properties of real anti-tridiagonal Hankel matrices by expressing their eigenvalues as zeros of rational functions and deriving their eigenvectors based on these eigenvalues.

Contribution

It introduces a novel method to determine eigenvalues and eigenvectors of these structured matrices using rational functions, expanding understanding of their spectral characteristics.

Findings

Eigenvalues are characterized as zeros of specific rational functions.

Eigenvectors can be derived once eigenvalues are known.

Provides a new analytical approach for spectral analysis of structured matrices.

Abstract

In this paper we express the eigenvalues of real anti-tridiagonal Hankel matrices as the zeros of given rational functions. We still derive eigenvectors for these structured matrices at the expense of prescribed eigenvalues.

Equations

$$\mathbf{H}_{n+2}=\left[\begin{array}[]{ccccccc}0&\ldots&\ldots&\ldots&0&a&c\\ \vdots&&&\iddots&a&c&b\\ \vdots&&\iddots&\iddots&c&b&0\\ \vdots&\iddots&\iddots&\iddots&\iddots&\iddots&\vdots\\ 0&a&c&\iddots&\iddots&&\vdots\\ a&c&b&\iddots&&&\vdots\\ c&b&0&\ldots&\ldots&\ldots&0\end{array}\right],$$

$$\omega :={\mathrm{e}}^{\frac{2\pi \mathrm{i}}{n+2}},$$

$${\lambda }_k:=b+a{\omega }^{-nk}+c{\omega }^{-(n+1)k},k=0,1,\dots ,n+1$$

$${\theta }_k:=\arg ({\lambda }_k),k=0,1,\dots ,n+1$$

$$F_n(t;\alpha ,\beta ):=\frac{1}{t-{\lambda }_0}+2\sum _{k=1}^n\left[\frac{\cos \left(\frac{{\theta }_k}{2}-\alpha k\right)\cos \left(\frac{{\theta }_k}{2}-\beta k\right)}{t-|{\lambda }_k|}+\frac{\sin \left(\frac{{\theta }_k}{2}-\alpha k\right)\sin \left(\frac{{\theta }_k}{2}-\beta k\right)}{t+|{\lambda }_k|}\right]$$

$$G_n(t;\alpha ,\beta ):=\frac{1}{t-{\lambda }_0}+\frac{1}{t-{\lambda }_{n+1}}+2\sum _{k=1}^n\left[\frac{\cos \left(\frac{{\theta }_k}{2}-\alpha k\right)\cos \left(\frac{{\theta }_k}{2}-\beta k\right)}{t-|{\lambda }_k|}+\frac{\sin \left(\frac{{\theta }_k}{2}-\alpha k\right)\sin \left(\frac{{\theta }_k}{2}-\beta k\right)}{t+|{\lambda }_k|}\right]$$

$$n$$

$$\begin{array}{rl}& f(t)=1+\frac{a}{n+2}{F}_{\frac{n+1}{2}}\left(t;\frac{2\pi }{n+2},\frac{2\pi }{n+2}\right)+\frac{b}{n+2}{F}_{\frac{n+1}{2}}(t;0,0)+\\ & \frac{ab}{(n+2{)}^2}{F}_{\frac{n+1}{2}}\left(t;\frac{2\pi }{n+2},\frac{2\pi }{n+2}\right)F_{\frac{n+1}{2}}(t;0,0)-\frac{ab}{(n+2{)}^2}{F}_{\frac{n+1}{2}}^2\left(t;0,\frac{2\pi }{n+2}\right).\end{array}$$

$$d_k+\min \left\{0,-a,-b\right\}⩽{\mu }_k⩽d_k+\max \left\{0,-a,-b\right\},k=1,\dots ,n+2.$$

$$n$$

$$\begin{array}{rl}& g(t)=1+\frac{a}{n+2}{G}_{\frac{n}{2}}\left(t;\frac{2\pi }{n+2},\frac{2\pi }{n+2}\right)+\frac{b}{n+2}{G}_{\frac{n}{2}}(t;0,0)+\\ & \frac{ab}{(n+2{)}^2}{G}_{\frac{n}{2}}\left(t;\frac{2\pi }{n+2},\frac{2\pi }{n+2}\right)G_{\frac{n}{2}}(t;0,0)-\frac{ab}{(n+2{)}^2}{\left[G_{\frac{n}{2}}\left(t;0,\frac{2\pi }{n+2}\right)-\frac{2}{t-{\lambda }_{\frac{n}{2}+1}}\right]}^2.\end{array}$$

$$d_k+\min \left\{0,-a,-b\right\}⩽{\nu }_k⩽d_k+\max \left\{0,-a,-b\right\},k=1,\dots ,n+2.$$

$$\mathbf{u}(\mu_{m})=\left[\begin{array}[]{c}F_{\frac{n+1}{2}}\left(\mu_{m};0,\frac{2\pi}{n+2}\right)-\frac{b\,F_{\frac{n+1}{2}}\left(\mu_{m},0,\frac{2\pi}{n+2}\right)F_{\frac{n+1}{2}}\left(\mu_{m};0,0\right)}{b\,F_{\frac{n+1}{2}}(\mu_{m},0,0)+n+2}\\[10.0pt] F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{b\,F_{\frac{n+1}{2}}\left(\mu_{m},0,\frac{2\pi}{n+2}\right)F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2\pi}{n+2},0\right)}{b\,F_{\frac{n+1}{2}}(\mu_{m},0,0)+n+2}\\[5.0pt] \vdots\\[10.0pt] F_{\frac{n+1}{2}}\left(\mu_{m};\frac{2(1-k)\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{b\,F_{\frac{n+1}{2}}\left(\mu_{m},0,\frac{2\pi}{n+2}\right)F_{\frac{n+1}{2}}\left(\mu_{m};\frac{2(1-k)\pi}{n+2},0\right)}{b\,F_{\frac{n+1}{2}}(\mu_{m},0,0)+n+2}\\[10.0pt] \vdots\\[10.0pt] F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2(n+1)\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{b\,F_{\frac{n+1}{2}}\left(\mu_{m},0,\frac{2\pi}{n+2}\right)F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2(n+1)\pi}{n+2},0\right)}{b\,F_{\frac{n+1}{2}}(\mu_{m},0,0)+n+2}\end{array}\right]$$

$$\begin{split}&\mathbf{v}(\nu_{m})=\left[\begin{array}[]{@{\hspace{1pt}}c@{\hspace{1pt}}}\hskip 1.0pt\lx@intercol\hfil G_{\frac{n}{2}}\left(\nu_{m};0,\frac{2\pi}{n+2}\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}-\frac{b\,G_{\frac{n}{2}}\left(\nu_{m};0,0\right)\left[G_{\frac{n}{2}}\left(\nu_{m};0,\frac{2\pi}{n+2}\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]}{b\,G_{\frac{n}{2}}(\nu_{m};0,0)+n+2}\hfil\hskip 1.0\\[10.0pt] \hskip 1.0pt\lx@intercol\hfil G_{\frac{n}{2}}\left(\nu_{m};-\frac{2\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{b\left[G_{\frac{n}{2}}\left(\nu_{m};-\frac{2\pi}{n+2},0\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]\left[G_{\frac{n}{2}}\left(\nu_{m};0,\frac{2\pi}{n+2}\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]}{b\,G_{\frac{n}{2}}(\nu_{m};0,0)+n+2}\hfil\hskip 1.0\\[5.0pt] \hskip 1.0pt\lx@intercol\hfil\vdots\hfil\hskip 1.0\\[10.0pt] \hskip 1.0pt\lx@intercol\hfil G_{\frac{n}{2}}\left(\nu_{m};\frac{2(1-k)\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{1-(-1)^{k}}{\nu_{m}-\lambda_{\frac{n}{2}+1}}-\frac{b\left[G_{\frac{n}{2}}\left(\nu_{m};\frac{2(1-k)\pi}{n+2},0\right)-\frac{1+(-1)^{k}}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]\left[G_{\frac{n}{2}}\left(\nu_{m};0,\frac{2\pi}{n+2}\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]}{b\,G_{\frac{n}{2}}(\nu_{m};0,0)+n+2}\hfil\hskip 1.0\\[10.0pt] \hskip 1.0pt\lx@intercol\hfil\vdots\hfil\hskip 1.0\\[10.0pt] \hskip 1.0pt\lx@intercol\hfil G_{\frac{n}{2}}\left(\nu_{m};-\frac{2(n+1)\pi}{n+2},\frac{2\pi}{n+2}\right)-\frac{b\left[G_{\frac{n}{2}}\left(\nu_{m};-\frac{2(n+1)\pi}{n+2},0\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]\left[G_{\frac{n}{2}}\left(\nu_{m};0,\frac{2\pi}{n+2}\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\right]}{b\,G_{\frac{n}{2}}(\nu_{m};0,0)+n+2}\hfil\hskip 1.0\end{array}\right]\end{split}$$

$$\mathbf{u}(\mu_{m})=\left[\begin{array}[]{c}F_{\frac{n+1}{2}}\left(\mu_{m};0,0\right)\\ F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2\pi}{n+2},0\right)\\ \vdots\\ F_{\frac{n+1}{2}}\left(\mu_{m};-\frac{2(n+1)\pi}{n+2},0\right)\end{array}\right]$$

$$\mathbf{v}(\nu_{m})=\left[\begin{array}[]{c}G_{\frac{n}{2}}\left(\nu_{m};0,0\right)\\ G_{\frac{n}{2}}\left(\nu_{m};-\frac{2\pi}{n+2},0\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\\ \vdots\\ G_{\frac{n}{2}}\left(\nu_{m};\frac{2(1-k)\pi}{n+2},0\right)-\frac{1+(-1)^{k}}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\\ \vdots\\ G_{\frac{n}{2}}\left(\nu_{m};-\frac{2(n+1)\pi}{n+2},0\right)-\frac{2}{\nu_{m}-\lambda_{\frac{n}{2}+1}}\end{array}\right]$$

$$\mathbf{A}_{n+2}:=\left[\begin{array}[]{ccccccc}b&0&\ldots&\ldots&0&a&c\\ 0&&&\iddots&a&c&b\\ \vdots&&\iddots&\iddots&c&b&0\\ \vdots&\iddots&\iddots&\iddots&\iddots&\iddots&\vdots\\ 0&a&c&\iddots&\iddots&&\vdots\\ a&c&b&\iddots&&&0\\ c&b&0&\ldots&\ldots&0&a\end{array}\right]$$

$${\left[{\mathbf{\Omega }}_{n+2}\right]}_{k,\ell }:=\frac{{\omega }^{(k-1)(\ell -1)}}{\sqrt{n+2}}$$

$$n$$

$${\mathbf{A}}_{n+2}={\mathbf{P}}_{n+2}\mathrm{diag}\left({\lambda }_0,|{\lambda }_1|,\dots ,|{\lambda }_{\frac{n+1}{2}}|,-|{\lambda }_{\frac{n+1}{2}}|,\dots ,-|{\lambda }_1|\right){\mathbf{P}}_{n+2}^{\top }$$

$$\left[\mathbf{P}_{n+2}\right]_{k,\ell}=\left\{\begin{array}[]{l}\frac{1}{\sqrt{n+2}},\ \ell=1\\[8.0pt] \sqrt{\frac{2}{n+2}}\cos\left[\frac{\theta_{\ell-1}}{2}+\frac{2(k-1)(\ell-1)\pi}{n+2}\right],\ 1<\ell\leqslant\frac{n+3}{2}\\[8.0pt] \sqrt{\frac{2}{n+2}}\sin\left[\frac{\theta_{n+3-\ell}}{2}+\frac{2(k-1)(n+3-\ell)\pi}{n+2}\right],\ \ell>\frac{n+3}{2}\end{array}\right..$$

$$n$$

$${\mathbf{A}}_{n+2}={\mathbf{Q}}_{n+2}\mathrm{diag}\left({\lambda }_0,|{\lambda }_1|,\dots ,|{\lambda }_{\frac{n}{2}}|,{\lambda }_{\frac{n}{2}+1},-|{\lambda }_{\frac{n}{2}}|,\dots ,-|{\lambda }_1|\right){\mathbf{Q}}_{n+2}^{\top }$$

$$\left[\mathbf{Q}_{n+2}\right]_{k,\ell}=\left\{\begin{array}[]{l}\frac{1}{\sqrt{n+2}},\ \ell=1\\[8.0pt] \sqrt{\frac{2}{n+2}}\cos\left[\frac{\theta_{\ell-1}}{2}+\frac{2(k-1)(\ell-1)\pi}{n+2}\right],\ 1<\ell\leqslant\frac{n}{2}+1\\[8.0pt] \frac{(-1)^{k-1}}{\sqrt{n+2}},\ \ell=\frac{n}{2}+2\\[8.0pt] \sqrt{\frac{2}{n+2}}\sin\left[\frac{\theta_{n+3-\ell}}{2}+\frac{2(k-1)(n+3-\ell)\pi}{n+2}\right],\ \ell>\frac{n}{2}+2\end{array}\right..$$

$${\mathbf{A}}_{n+2}={\mathbf{P}}_{n+2}\mathrm{diag}\left({\lambda }_0,|{\lambda }_1|,\dots ,|{\lambda }_{\frac{n+1}{2}}|,-|{\lambda }_{\frac{n+1}{2}}|,\dots ,-|{\lambda }_1|\right){\mathbf{P}}_{n+2}^{\ast }$$

$$\mathbf{P}_{n+2}=\boldsymbol{\Omega}_{n+2}^{\ast}\left[\begin{array}[]{cc}1&\mathbf{0}\\ \mathbf{0}&\mathbf{R}_{n+1}\end{array}\right]$$

$$\mathbf{R}_{n+1}=\frac{\sqrt{2}}{2}\left[\begin{array}[]{cccccccc}\mathrm{e}^{-\frac{\mathrm{i}\theta_{1}}{2}}&0&\ldots&0&0&\ldots&0&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{1}}{2}}\\[5.0pt] 0&\mathrm{e}^{-\frac{\mathrm{i}\theta_{2}}{2}}&\ddots&\vdots&\vdots&\iddots&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{2}}{2}}&0\\[5.0pt] \vdots&\ddots&\ddots&0&0&\iddots&\iddots&\vdots\\ 0&\ldots&0&\mathrm{e}^{-\frac{\mathrm{i}\theta_{(n+1)/2}}{2}}&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{(n+1)/2}}{2}}&0&\ldots&0\\[5.0pt] 0&\ldots&0&\mathrm{e}^{\frac{\mathrm{i}\theta_{(n+1)/2}}{2}}&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{(n+1)/2}}{2}}&0&\ldots&0\\[5.0pt] \vdots&\iddots&\iddots&0&0&\ddots&\ddots&\vdots\\[5.0pt] 0&\mathrm{e}^{\frac{\mathrm{i}\theta_{2}}{2}}&\iddots&\vdots&\vdots&\ddots&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{2}}{2}}&0\\[5.0pt] \mathrm{e}^{\frac{\mathrm{i}\theta_{1}}{2}}&0&\ldots&0&0&\ldots&0&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{1}}{2}}\end{array}\right].$$

$$\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}+\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \overline{\omega}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}+\overline{\omega}^{(n+1)}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \overline{\omega}^{2}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}+\overline{\omega}^{2(n+1)}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \vdots\\ \overline{\omega}^{n+1}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}+\overline{\omega}^{(n+1)^{2}}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\end{array}\right]$$

$$\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\mathrm{i}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}-\mathrm{i}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \mathrm{i}\,\overline{\omega}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}-\mathrm{i}\,\overline{\omega}^{(n+1)}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \mathrm{i}\,\overline{\omega}^{2}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}-\mathrm{i}\,\overline{\omega}^{2(n+1)}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\\ \vdots\\ \mathrm{i}\,\overline{\omega}^{n+1}\,\mathrm{e}^{-\frac{\mathrm{i}\theta_{\ell}}{2}}-\mathrm{i}\,\overline{\omega}^{(n+1)^{2}}\,\mathrm{e}^{\frac{\mathrm{i}\theta_{\ell}}{2}}\end{array}\right]$$

$${\overline{\omega }}^k{\mathrm{e}}^{-\frac{\mathrm{i}{\theta }_{\ell }}{2}}+{\overline{\omega }}^{(n+1)k}{\mathrm{e}}^{\frac{\mathrm{i}{\theta }_{\ell }}{2}}=\cos \left(\frac{{\theta }_{\ell }}{2}+\frac{2k\ell \pi }{n+2}\right),k=0,1,\dots ,n+1$$

$$\mathrm{i}{\overline{\omega }}^k{\mathrm{e}}^{-\frac{\mathrm{i}{\theta }_{\ell }}{2}}-\mathrm{i}{\overline{\omega }}^{(n+1)k}{\mathrm{e}}^{\frac{\mathrm{i}{\theta }_{\ell }}{2}}=\sin \left(\frac{{\theta }_{\ell }}{2}+\frac{2k\ell \pi }{n+2}\right),k=0,1,\dots ,n+1$$

$${\mathbf{A}}_{n+2}={\mathbf{Q}}_{n+2}\mathrm{diag}\left({\lambda }_0,|{\lambda }_1|,\dots ,|{\lambda }_{\frac{n}{2}}|,{\lambda }_{\frac{n}{2}+1},-|{\lambda }_{\frac{n}{2}}|,\dots ,-|{\lambda }_1|\right){\mathbf{Q}}_{n+2}^{\top }$$

$$\mathbf{Q}_{n+2}=\boldsymbol{\Omega}_{n+2}^{\ast}\left[\begin{array}[]{cc}1&\mathbf{0}\\ \mathbf{0}&\mathbf{S}_{n+1}\end{array}\right]$$

$$\mathbf{S}_{n+1}=\frac{\sqrt{2}}{2}\left[\begin{array}[]{ccccccccc}\mathrm{e}^{-\frac{\mathrm{i}\theta_{1}}{2}}&0&\ldots&0&0&0&\ldots&0&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{1}}{2}}\\[5.0pt] 0&\mathrm{e}^{-\frac{\mathrm{i}\theta_{2}}{2}}&\ddots&\vdots&\vdots&\vdots&\iddots&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{2}}{2}}&0\\[5.0pt] \vdots&\ddots&\ddots&0&0&0&\iddots&\iddots&\vdots\\ 0&\ldots&0&\mathrm{e}^{-\frac{\mathrm{i}\theta_{n/2}}{2}}&0&\mathrm{i}\mathrm{e}^{-\frac{\mathrm{i}\theta_{n/2}}{2}}&0&\ldots&0\\[5.0pt] 0&\ldots&0&0&\frac{2}{\sqrt{2}}&0&0&\ldots&0\\[5.0pt] 0&\ldots&0&\mathrm{e}^{\frac{\mathrm{i}\theta_{n/2}}{2}}&0&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{n/2}}{2}}&0&\ldots&0\\[5.0pt] \vdots&\iddots&\iddots&0&0&0&\ddots&\ddots&\vdots\\[5.0pt] 0&\mathrm{e}^{\frac{\mathrm{i}\theta_{2}}{2}}&\iddots&\vdots&\vdots&\vdots&\ddots&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{2}}{2}}&0\\[5.0pt] \mathrm{e}^{\frac{\mathrm{i}\theta_{1}}{2}}&0&\ldots&0&0&0&\ldots&0&-\mathrm{i}\mathrm{e}^{\frac{\mathrm{i}\theta_{1}}{2}}\end{array}\right].$$

$$\sqrt{\frac{1}{n+2}}\left[\begin{array}[]{c}1\\ \overline{\omega}^{\frac{n}{2}+1}\\ \overline{\omega}^{2\left(\frac{n}{2}+1\right)}\\ \vdots\\ \overline{\omega}^{(n+1)\left(\frac{n}{2}+1\right)}\end{array}\right]=\sqrt{\frac{1}{n+2}}\left[\begin{array}[]{c}1\\ -1\\ (-1)^{2}\\ \vdots\\ (-1)^{n+1}\end{array}\right].$$

$$\mathbf{x}=\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\frac{1}{\sqrt{2}}\\[5.0pt] \cos\left(\frac{\theta_{1}}{2}\right)\\[5.0pt] \vdots\\[5.0pt] \cos\left[\frac{\theta_{(n+1)/2}}{2}\right]\\[5.0pt] \sin\left[\frac{\theta_{(n+1)/2}}{2}\right]\\[5.0pt] \vdots\\[5.0pt] \sin\left(\frac{\theta_{1}}{2}\right)\end{array}\right],\quad\mathbf{y}=\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\frac{1}{\sqrt{2}}\\[5.0pt] \cos\left(\frac{\theta_{1}}{2}-\frac{2\pi}{n+2}\right)\\[5.0pt] \vdots\\[5.0pt] \cos\left[\frac{\theta_{(n+1)/2}}{2}-\frac{(n+1)\pi}{n+2}\right]\\[5.0pt] \sin\left[\frac{\theta_{(n+1)/2}}{2}-\frac{(n+1)\pi}{n+2}\right]\\[5.0pt] \vdots\\[5.0pt] \sin\left(\frac{\theta_{1}}{2}-\frac{2\pi}{n+2}\right)\end{array}\right]$$

$${\mathbf{H}}_{n+2}={\mathbf{P}}_{n+2}\left[\mathrm{diag}\left({\lambda }_0,|{\lambda }_1|,\dots ,|{\lambda }_{\frac{n+1}{2}}|,-|{\lambda }_{\frac{n+1}{2}}|,\dots ,-|{\lambda }_1|\right)-b\mathbf{x}{\mathbf{x}}^{\top }-a\mathbf{y}{\mathbf{y}}^{\top }\right]{\mathbf{P}}_{n+2}^{\top }$$

$$\mathbf{x}=\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\frac{1}{\sqrt{2}}\\[5.0pt] \cos\left(\frac{\theta_{1}}{2}\right)\\[5.0pt] \vdots\\[5.0pt] \cos\left(\frac{\theta_{n/2}}{2}\right)\\[5.0pt] \frac{1}{\sqrt{2}}\\[5.0pt] \sin\left(\frac{\theta_{n/2}}{2}\right)\\[5.0pt] \vdots\\[5.0pt] \sin\left(\frac{\theta_{1}}{2}\right)\end{array}\right],\quad\mathbf{y}=\sqrt{\frac{2}{n+2}}\left[\begin{array}[]{c}\frac{1}{\sqrt{2}}\\[5.0pt] \cos\left(\frac{\theta_{1}}{2}-\frac{2\pi}{n+2}\right)\\[5.0pt] \vdots\\[5.0pt] \cos\left(\frac{\theta_{n/2}}{2}-\frac{n\pi}{n+2}\right)\\[5.0pt] -\frac{1}{\sqrt{2}}\\[5.0pt] \sin\left(\frac{\theta_{n/2}}{2}-\frac{n\pi}{n+2}\right)\\[5.0pt] \vdots\\[5.0pt] \sin\left(\frac{\theta_{1}}{2}-\frac{2\pi}{n+2}\right)\end{array}\right]$$

Full text

Key words: Anti-tridiagonal matrix, Hankel matrix, eigenvalue, eigenvector

2010 Mathematics Subject Classification: 15A18, 15B05

1 Introduction

Recently, some authors have computed the eigenvalues and eigenvectors for a sort of Hankel matrices thus obtaining its eigendecomposition (see [3], [4], [7], [11], [12], [13], [14] among others). In contrast to the tridiagonal Toeplitz matrices case where the eigenvalues and eigenvectors are well-known (see, for instance, [9]), a possible closed-form expression for the eigenvalues and eigenvectors of general anti-tridiagonal Hankel matrices is still to be found.

The aim of this short note is to give a contribution in that search. Specifically, we shall present an eigenvalue localization tool for real anti-tridiagonal Hankel matrices, providing also associated eigenvectors. To achieve our purpose, we shall use eigendecompositions of anti-circulant matrices available in [8] to ensure a decomposition for the matrices under study in a first step, and results concerning to sum of (rank one) matrices in a final stage to obtain all formulae.

The rational functions exhibited in this paper to locate eigenvalues of real anti-tridiagonal Hankel matrices, as well as the expressions for its eigenvectors, are given in explicit form which, on one hand, can be easily evaluated in computer programs, and, on the other, are useful for further theoretical investigations in this subject.

2 Main results

Let $n$ be a positive integer and consider the following $(n+2)\times(n+2)$ anti-tridiagonal Hankel matrix

[TABLE]

where $a,b,c$ are real numbers. Throughout, we shall set

[TABLE]

where $\mathrm{i}$ denotes the imaginary unit.

2.1 Eigenvalue localization for $\mathbf{H}_{n+2}$

Our first statement is an eigenvalue localization theorem for matrices of the form (2.1).

Theorem 1

Let $n$ be a positive integer, $a,b,c$ real numbers, $\omega$ given by (2.2),

[TABLE]

[TABLE]

[TABLE]

2.2 Eigenvectors of $\mathbf{H}_{n+2}$

Owning the eigenvalues of $\mathbf{H}_{n+2}$ in (2.1), we are able to determine the corresponding eigenvectors.

Theorem 2

Let $n$ be an integer, $a,b,c$ real numbers such that $a\neq 0$, and $\lambda_{k}$, $\theta_{k}$ $(k=0,1,\ldots,n+1)$, $F_{n}(t;\alpha,\beta)$, $G_{n}(t;\alpha,\beta)$ be given by (2.3a), (2.3b), (2.3c), (2.3d), respectively.

(a)* If $n$ is odd, the zeros $\mu_{1},\mu_{2},\ldots,\mu_{n+2}$ of (2.4a) are not of the form $\lambda_{0}$, $\lvert\lambda_{k}\rvert$, $-\lvert\lambda_{k}\rvert$, $k=1,\ldots,\frac{n+1}{2}$ and $b\,F_{\frac{n+1}{2}}(\mu_{m},0,0)+n+2\neq 0$, then*

[TABLE]

is an eigenvector of $\mathbf{H}_{n+2}$ associated to $\mu_{m}$, $m=1,2,\ldots,n+2$.

(b)* If $n$ is even, the zeros $\nu_{1},\nu_{2},\ldots,\nu_{n+2}$ of (2.5a) are not of the form $\lambda_{0}$, $\lvert\lambda_{k}\rvert$, $\lambda_{\frac{n}{2}+1}$, $-\lvert\lambda_{k}\rvert$, $k=1,\ldots,\frac{n}{2}$ and $b\,G_{\frac{n}{2}}(\nu_{m},0,0)+n+2\neq 0$, then*

[TABLE]

is an eigenvector of $\mathbf{H}_{n+2}$ associated to $\nu_{m}$, $m=1,2,\ldots,n+2$.

• Remark

We point out that if $a=0$ then expressions (2.6) and (2.7) should be replaced by

[TABLE]

and

[TABLE]

respectively, provided that $b\neq 0$. Of course, if $a=b=0$ then an eigenvalue decomposition of the exchange matrix is already known (see [3]).

3 Lemmas and proofs

Let $n$ be a positive integer. Consider the following real anti-circulant $(n+2)\times(n+2)$ matrix

[TABLE]

and the $(n+2)\times(n+2)$ unitary discrete Fourier transform matrix $\boldsymbol{\Omega}_{n+2}$, that is, the matrix defined by

[TABLE]

where $\omega$ is given by (2.2). Our first auxiliary result is an orthogonal decomposition for $\eqref{eq:3.1}$. We shall denote by $\ast$ the conjugate transpose of any complex matrix.

Lemma 1

Let $n$ be a positive integer, $a,b,c$ real numbers, and $\omega$, $\lambda_{k}$, $\theta_{k}$, $k=0,1,\ldots,n+1$ given by (2.2), (2.3a), (2.3b), respectively.

[TABLE]

[TABLE]

• Proof.

Let $n$ be a positive odd integer. According to Theorem 3.6 of [8] we have

[TABLE]

where

[TABLE]

with $\boldsymbol{\Omega}_{n+2}$ given by (3.2) and $\mathbf{R}_{n+1}$ the following $(n+1)\times(n+1)$ matrix

[TABLE]

Note that the first column of $\mathbf{P}_{n+2}$ has all components equal to $1/\sqrt{n+2}$; their next $(n+1)/2$ columns are

[TABLE]

for each $\ell=1,\ldots,\frac{n+1}{2}$ and the last ones are

[TABLE]

for $\ell=\frac{n+1}{2},\ldots,1$. Since

[TABLE]

we get that the entries of $\mathbf{P}_{n+2}$ are given by (3.3b) which leads to (3.3a). Supposing a positive even integer $n$, Theorem 3.7 in [8] ensures

[TABLE]

where

[TABLE]

with $\boldsymbol{\Omega}_{n+2}$ given by (3.2) and $\mathbf{S}_{n+1}$ the $(n+1)\times(n+1)$ matrix

[TABLE]

The first column of $\mathbf{Q}_{n+2}$ has all its components equal to $1/\sqrt{n+2}$. The next $n/2$ columns are given by (3.5) for $\ell=1,\ldots,\frac{n}{2}$ and the $n/2$ last ones are defined by (3.6) for each $\ell=\frac{n}{2},\ldots,1$; the $\left(\frac{n}{2}+1\right)$th column of $\mathbf{Q}_{n+2}$ is

[TABLE]

From identities (3.7), (3.8) we obtain (3.4a). The proof is completed. $\square$

The statement below is a decomposition for the matrices $\mathbf{H}_{n+2}$ and plays a central role in the main results.

Lemma 2

Let $n$ be a positive integer, $a,b,c$ real numbers, and $\omega$, $\lambda_{k}$, $\theta_{k}$, $k=0,1,\ldots,n+1$ given by (2.2), (2.3a), (2.3b), respectively.

(a)* If $n$ is odd,*

[TABLE]

then

[TABLE]

where $\mathbf{P}_{n+2}$ is the $(n+2)\times(n+2)$ matrix defined by (3.3b).

(b)* If $n$ is even,*

[TABLE]

then

[TABLE]

where $\mathbf{Q}_{n+2}$ is the $(n+2)\times(n+2)$ whose the entries are given by (3.4b).

• Proof.

We only prove (a) since (b) can be proven in the same way. Consider a positive odd integer $n$ and the following matrices

[TABLE]

From Lemma 1,

[TABLE]

where $\mathbf{P}_{n+2}$ is the matrix defined by (3.3b), $\mathbf{A}_{n+2}$ is the matrix (3.1), $\mathbf{x}$ is the first row of $\mathbf{P}_{n+2}$, i.e.

[TABLE]

and $\mathbf{y}$ is the last row of $\mathbf{P}_{n+2}$,

[TABLE]

$\square$

• Proof of Theorem 1.

Consider a positive odd integer $n$, $\mathbf{x},\mathbf{y}$ given by (3.9), and $d_{1}=\lambda_{0},d_{2}=\lvert\lambda_{1}\rvert,\ldots,$ $d_{\frac{n+3}{2}}=\lvert\lambda_{\frac{n+1}{2}}\rvert,d_{\frac{n+5}{2}}=-\lvert\lambda_{\frac{n+1}{2}}\rvert,\ldots,d_{n+2}=-\lvert\lambda_{1}\rvert$. According to Lemma 2, it should be noted that the matrix $\mathbf{H}_{n+2}$ and

[TABLE]

share the same eigenvalues. Let us adopt the notations of [1] by denoting $\EuScript{S}(k,m)$ the collection of all $k$-element subsets of $\{1,2,\ldots,m\}$ written in increasing order; additionally, for any rectangular matrix $\mathbf{M}$, we shall indicate by $\det\left(\mathbf{M}[I,J]\right)$ the minor determined by the subsets $I=\left\{i_{1}<i_{2}<\ldots<i_{k}\right\}$ and $J=\left\{j_{1}<j_{2}<\ldots<j_{k}\right\}$. Setting

[TABLE]

and

[TABLE]

we have from Theorem 1 of [1] that $\zeta$ is an eigenvalue of (3.11) if and only if

[TABLE]

provided that $\zeta$ is not an eigenvalue of $\mathrm{diag}\left(d_{1},\ldots,d_{n+2}\right)$. Since

[TABLE]

we obtain (2.4a). Let $\mu_{1}\leqslant\mu_{2}\leqslant\ldots\leqslant\mu_{n+2}$ be the eigenvalues of $\mathbf{H}_{n+2}$ and $d_{\tau(1)}\leqslant d_{\tau(2)}\leqslant\ldots\leqslant d_{\tau(n+2)}$ be arranged in non-decreasing order by some bijection $\tau$ defined in $\{1,2,\ldots,n+2\}$. Thus,

[TABLE]

for each $k=1,2,\ldots,n+2$ (see [6], page $242$). Using Miller’s formula for the determinant of the sum of matrices (see [10], page $70$), we can compute the characteristic polynomial of $-b\mathbf{x}\mathbf{x}^{\top}-a\mathbf{y}\mathbf{y}^{\top}$,

[TABLE]

because

[TABLE]

and

[TABLE]

Hence, $\mathrm{Spec}\left(-b\mathbf{x}\mathbf{x}^{\top}-a\mathbf{y}\mathbf{y}^{\top}\right)=\left\{0,-a,-b\right\}$ and (3.12) yields (2.4b). The proof of the remaining assertion is performed in the same way and so will be omitted. $\square$

• Proof of Theorem 2.

Since both assertions can be proven in the same way, we only prove (a). Let $n$ be a positive odd integer, $a\neq 0$, and $\mu_{1},\mu_{2},\ldots,\mu_{n+2}$ the eigenvalues of $\mathbf{H}_{n+2}$. We can rewrite the matricial equation $(\mu_{m}\mathbf{I}_{n+2}-\mathbf{H}_{n+2})\mathbf{z}=\mathbf{0}$ as

[TABLE]

where $\mathbf{x},\mathbf{y}$ are defined in (3.9) and $\mathbf{P}_{n+2}$ the matrix whose the entries are given by (3.3b). Thus,

[TABLE]

that is,

[TABLE]

for $\xi\neq 0$ (see Theorem 5 of [2], page $41$) and

[TABLE]

is a nontrivial solution of (3.13). Thus, choosing $\xi=1$ we conclude that the vector having components (2.6) is an eigenvector of $\mathbf{H}_{n+2}$ associated to the eigenvalue $\mu_{m}$. $\square$

Acknowledgements

This work is a contribution to the Project UID/GEO/04035/2013, funded by FCT - Fundação para a Ciência e a Tecnologia, Portugal.