\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 172, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/??\hfil Sturm-Liouville problems]
{Similarities of discrete and continuous Sturm-Liouville problems}

\author[K. Ghanbari\hfil EJDE-2007/172\hfilneg]
{Kazem Ghanbari}

\address{Kazem Ghanbari \newline
 Department of Mathematics,
Sahand University of Technology, Tabriz, Iran}
\email{kghanbari@sut.ac.ir}

\thanks{Submitted November 7, 2006. Published December 6, 2007.}
\subjclass[2000]{15A18, 15A24, 34B25, 34B27}
\keywords{Green's function; Jacobi matrix;
Sturm-Liouville equation; \hfill\break\indent eigenvalue; eigenvector}

\begin{abstract}
 In this paper we present a study on the analogous properties
 of discrete and continuous Sturm-Liouville problems arising in
 matrix analysis and differential equations, respectively. Green's
 functions in both cases have analogous expressions in terms of the
 spectral data. Most of the results associated to inverse problems
 in both cases are identical. In particular, in both cases Weyl's
 $m$-function determines the Sturm-Liouville operators
 uniquely. Moreover, the well known  Rayleigh-Ritz Theorem in
 linear algebra can be proved by using the concept of Green's
 function in discrete case.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

First we present a brief description of the discrete
Green's functions, which are discussed in more detail in \cite{GW}.
 It is well known that a Sturm-Liouville problem
is an initial value problem of the form
\begin{equation}\label{e1}
\begin{gathered}
Ly(x)=\lambda \rho(x)y(x),\quad 0\leq x\leq \ell \\
p(0)y'(0)-hy(0)=0 \\
p(\ell)y'(\ell)+Hy(\ell)=0\\
\end{gathered}
\end{equation}
where, $p(x),q(x),\rho(x)$ are given, $\rho(x)>0$, and $L$ is a
second order differential operator of the form
\[
Ly(x)=-(p(x)y'(x))'+q(x)y.
\]
It is well known that for every Sturm-Liouville problem of the
form \eqref{e1} there is a corresponding \emph{Green's function}
$G(x,s,\lambda)$. If $\lambda$ is not an eigenvalue of \eqref{e1}
then it is well known (see \cite{COD}, chapter 7) that
$G(x,y,\lambda)$ can be constructed as follows. Let $\varphi(x)$
and $\psi(x)$ be solutions of the Sturm-Liouville equation
satisfying the first and second condition of \eqref{e1},
respectively. Then
\begin{equation}\label{e2}
[\varphi\psi](x)=p(x)\{\varphi(x)\psi'(x)-\varphi'(x)\psi(x)\}=
\text{constant.}
\end{equation}
This constant is zero if $\lambda$ is an eigenvalue of \eqref{e1}
and nonzero otherwise. In the latter case we can choose the
constant to be $-1$, and then
\begin{equation}\label{e3}
G(x,s,\lambda)=\begin{cases} \varphi(x)\psi(s),& 0\leq x\leq s \\
\varphi(s)\psi(x), & s\leq x \leq \ell.
\end{cases}
\end{equation}
The functions $\varphi, \psi$ are functions of $x,\lambda$. If
$\lambda=0$ is not an eigenvalue of \eqref{e1}; i.e.
$h,H$ are not both zero, then
\begin{equation}\label{e4}
G(x,s,\lambda)=\begin{cases}
\varphi_0(x)\psi_0(s),& 0\leq x\leq s \\
 \varphi_0(s)\psi_0(x),&  s\leq x \leq \ell.
\end{cases}
\end{equation}
where $\varphi_0,\psi_0$ denote $\varphi, \psi$, respectively, for
$\lambda=0$. Now let $a\leq b$ be two fixed points in the interval
$(0,\ell)$. We define the normalized Green's function as follows:
\begin{equation}\label{e5}
\Phi(a,b,\lambda)=\frac{G(a,b,\lambda)}{G(a,b,0)}=
\frac{\varphi(a)\psi(b)}{\varphi_0(a)\psi_0(b)}.
\end{equation}
Now if we consider a discrete form of the
Sturm-Liouville problem, we may define a similar terminology.
Using the concept of discrete derivative; i.e.,
$y'=y_{n+1}-y_n$, the discrete form of the Sturm-Liouville
equation is a system of three-term recurrence relations that can
be written in a compact form
\begin{equation}\label{e6}
Ax=\lambda Bx
\end{equation}
where $A$ is the tridiagonal matrix
$$
A=\begin{pmatrix}a_1&c_1&0&.&.&.\\ c_1&a_2&c_2&0&.&.\\ 0&c_2&a_3
&c_{3}&0&.\\.&.&.&.&.&.\\.&.&.&.&.&c_{n-1}\\.&.&.&.&c_{n-1}&a_n
\end{pmatrix}
$$
and $B=\mathop{\rm diag}(b_1,b_2,\ldots,b_n)$. For more details
see Atkinson [1]. The corresponding {\it Green's function} for
discrete case is defined by
\begin{equation}\label{e7}
G(i,j,\lambda)=(\lambda)=e_i^T(A-\lambda B)^{-1}e_j.
\end{equation}
The corresponding normalized Green's function is defined by
\begin{equation}\label{e8}
\Phi_{ij}(\lambda)=\frac{G(i,j,\lambda)}{G(i,j,0)}.
\end{equation}
Now denote the leading principal submatrix of order
$k$ of the matrix $A$ by $A_k$ and denote the trailing
principal submatrix of order $k$ of the matrix $A$ by
$A_k^R$.
By $\sigma(A,B)$ we denote the set of generalized eigenvalues
corresponding to the generalized eigenvalue problem \eqref{e6}. A
special case of Green's functions in both discrete and continuous
cases are m-functions. In the discrete case m-functions
corresponding to \eqref{e6} are defined as
$m_i(\lambda)=G(i,i,\lambda)$. In \cite{KG} we proved that for a
given basic m-function $G(1,1,\lambda)$ and a diagonal matrix $B$
we can construct a unique tridiagonal positive definite matrix $A$
such that $m_i(\lambda)$, $1\leq i\leq n$, are the corresponding
m-functions for the pair $(A,B)$, (inverse problem). There are
some interesting spectral similarities in discrete and continuous
Sturm Liouville problems in the following theorems
that have been proved in \cite{GW}.

\begin{theorem} \label{thm1.1}
Suppose $B$ is a positive definite diagonal
matrix and $A$ is a tridiagonal matrix. Let
$\sigma(A,B)=\{\lambda_k\}_1^n$ and let $\{x^{(k)}\}$ be the
corresponding eigenvectors. Then
\begin{equation}\label{e9}
G(i,j,\lambda)=\sum_{k=1}^n\frac{x_i^{(k)}x_j^{(k)}}{\lambda_k-\lambda}
\end{equation}
where $x_i^{(k)}$ is the $i^{th}$ entry of $x^{(k)}$.
\end{theorem}

\begin{remark} \label{rmk1.2} \rm
This result is the similar result to the
expansion of the Green's function in terms of eigenfunctions,
i.e.,
$$
G(t,\tau,\lambda)=\sum_{k=1}^\infty\frac{\chi_k(t)
\chi_k(\tau)}{\lambda_k-\lambda}.
$$
\end{remark}

\begin{theorem} \label{thm1.3}
Under the assumptions of Theorem \ref{thm1.1}, let
$$
\sigma(A_{i-1},B_{i-1})=\{\lambda_m^L\}_1^{i-1},~~
\sigma(A_{n-j}^R,B_{n-j}^R)=\{\lambda_m^R\}_1^{i-1}.
$$
Then
\begin{equation}\label{e10}
\Phi_{ij}(\lambda)=\frac{\prod_{m=1}^{i-1}(1-\lambda/\lambda_m^L)\prod_{m=1}^{n-j}
(1-\lambda/\lambda_m^R)} {\prod_{m=1}^{n}(1-\lambda/\lambda_m)}.
\end{equation}
\end{theorem}

\begin{theorem} \label{thm1.4}
\begin{equation}\label{e11}
\frac{x_i^{(k)}x_j^{(k)}}{G(i,j,0)}=
\frac{\lambda_k\prod_{m=1}^{i-1}(1-\lambda_k/\lambda_m^L)
\prod_{m=1}^{n-j} (1-\lambda_k/\lambda_m^R)} {\prod_{m\neq k
}^{n}(1-\lambda_k/\lambda_m)}.
\end{equation}
\end{theorem}

Therefore, the Green's function has been constructed using a pair
of spectra. There is a similar construction procedure to recover
the Green's function in continuous Sturm-Liouville as follows;
consider a pair of Sturm-Liouville problems with boundary
conditions at left $\{\lambda_n^L\}_1^\infty$,
\begin{equation}\label{e12}
\begin{gathered}
Ly(x)=\lambda \rho(x)y(x),\quad  0\leq x\leq \ell \\
p(0)y'(0)-hy(0)=0\\
y(a)=0,\quad a\in (0,\ell)\\
\end{gathered}
\end{equation}
and right $\{\lambda_n^R\}_1^\infty$,
\begin{equation}\label{e13}
\begin{gathered}
Ly(x)=\lambda \rho(x)y(x),\quad 0\leq x\leq \ell \\
p(\ell)y'(\ell)+Hy(\ell)=0 \\
y(b)=0,\quad b\in (0,\ell)\,.
\end{gathered}
\end{equation}
Let $\{\lambda_n\}_1^\infty$ be the eigenvalues of the original
Sturm-Liouville problem
\begin{equation}\label{e14}
\begin{gathered}
Ly(x)=\lambda \rho(x)y(x),\quad 0\leq x\leq \ell \\
p(0)y'(0)-hy(0)=0 \\
p(\ell)y'(\ell)+Hy(\ell)=0\,.
\end{gathered}
\end{equation}
Then we have the following analogous theorems \cite{GW}.

\begin{theorem} \label{thm1.5}
\begin{equation}\label{e15}
\Phi_{ij}(\lambda)=\frac{\prod_{n=1}^{\infty}(1-\lambda/\lambda_n^L)
\prod_{n=1}^{\infty} (1-\lambda/\lambda_n^R)}
{\prod_{n=1}^{\infty}(1-\lambda/\lambda_n)}.
\end{equation}
\end{theorem}

\begin{theorem} \label{thm1.6}
Let $\{u_r(x)\}_1^\infty$ be the corresponding
eigenfunctions to the original Sturm-Liouville problem, then
\begin{equation}\label{e16}
\frac{u_r(a)u_r(b)}{G(a,b,0)}=
\frac{\lambda_k\prod_{m=1}^{\infty}(1-\lambda_k/\lambda_m^L)
\prod_{m=1}^{\infty} (1-\lambda_k/\lambda_m^R)} {\prod_{m\neq k
}^{\infty}(1-\lambda_k/\lambda_m)}.
\end{equation}
\end{theorem}

\section{Weyl m-functions}

Another similar property of discrete and continuous
Sturm-Liouville problems is the uniqueness of the solution of
inverse problem in both cases by given weyl m-function.  Weyl
m-function for discrete case is defined by $G(1,1,\lambda)$. If we
have two given spectra $\sigma (A,B)=\{\lambda _i\}_1^N$, ~and
$\sigma (A_{[2,N]},B_{[2,N]})=\{\mu _i\}_1^{N-1}$, where
$A_{[2,N]},B_{[2,N]}$ are submatrices obtained from $A$ and $B$,
respectively, by deleting the first row and first column of $A$
and $B$ with interlacing property
$$
0<\lambda_1<\mu_1<\lambda_2<\mu_2<\ldots<\mu_{N-1}<\lambda_N,
$$
then it is well known that we can construct
the first entries of eigenvectors corresponding to $\{\lambda _i\}_1^N$
 by Lanczos algorithm. In this case if $B$
is a nonsingular diagonal matrix we can construct the m-function of the
pair $(A,B)$ as follows.

\begin{theorem} \label{thm2.1}
The Weyl function $m(\lambda)$ has the following representations in
the discrete case
\begin{gather}\label{e17}
m(\lambda)=\sum_{k=1}^n\frac{[x_1^{(k)}]^2}{\lambda_k-\lambda}, \\
\label{e18}
m(\lambda)=\frac{1}{b_1}\prod_{j=1}^{N-1}{(\mu _j-\lambda)}\prod_{j=1}^{N}
{(\lambda _j-\lambda)^{-1}},
\end{gather}
where
\begin{equation}\label{e19}
[x_1^{(k)}]^2=\frac{1}{b_1}\prod_{j=1}^{N-1}{(\mu_j-\lambda_k)}
\prod_{j\neq k}^{N} {(\lambda_j-\lambda_k)^{-1}}.
\end{equation}
\end{theorem}

\begin{proof}
 The first part is an immediate consequence of
Theorem \ref{thm1.1}. For the second part consider the equation
\begin{equation}\label{e20}
(A-\lambda B)y=e_1
\end{equation}
If $\lambda\neq \lambda_i,$ then
\begin{equation}\label{e21}
y_1=e_1^T(A-\lambda B)^{-1}e_1=m(\lambda).
\end{equation}
By applying Cramer's rule to (\ref{e21}) we find
$$
y_1=\frac{\det (A_1-\lambda B_1)}{\det (A-\lambda B)}
=\frac{P_{N-1}(\lambda)}{P_N(\lambda)}.
$$
Since $\{\mu_i\}_1^{N-1}$ and $\{\lambda_i\}_1^N$ are zeros
of $P_{N-1}(\lambda)$ and $P_N(\lambda)$,
respectively, we obtain
\begin{equation}\label{e22}
\begin{aligned}
m(\lambda)&=y_1=\frac{\det (B_1)}{\det (B)}
 \prod_{j=1}^{N-1}{(\mu_j-\lambda)}\prod_{j=1}^{N}
{(\lambda_j-\lambda)^{-1}} \\
&=\frac{1}{b_1}\prod_{j=1}^{N-1}{(\mu_j-\lambda)}\prod_{j=1}^{N}
{(\lambda_j-\lambda)^{-1}}.
\end{aligned}
\end{equation}
Multiplication of the two representations of $m(\lambda)$ by
$\lambda_k-\lambda$ and taking the limit as
$\lambda \to \lambda_k$ completes the proof.
\end{proof}

\begin{theorem} \label{thm2.2}
For a given m-function there is a unique positive definite Jacobi matrix
with the prescribed m-function, \cite{KG}.
\end{theorem}

In the continuous case, let $\varphi(x,\lambda)$ and
$\psi(x,\lambda)$ be two solutions of Sturm-Liouville equation (1)
with the boundary conditions
$\varphi(0,\lambda)=1,~p(0)\varphi'(0,\lambda)=h$ and
$\psi(\ell,\lambda)=1,~p(\ell)\psi'(\ell,\lambda)=-H,$
respectively. Suppose $\Delta(\lambda)$ be the Wronskian of
$\varphi,~\psi$, that is
$$
\Delta(\lambda)=\varphi(x,\lambda)\psi'(x,\lambda)
-\varphi'(x,\lambda)\psi(x,\lambda).
$$
It is well known \cite{FY} that $\Delta(\lambda)$ is an analytic
function in $\lambda$ whose zeros are the eigenvalues of
Sturm-Liouville problem (1). In this case the Weyl m-function is
defined by
$$
m(\lambda)=-\frac{\varphi(0,\lambda)}{\Delta(\lambda)}.
$$
In the continuous case the following results are well known, for
details see \cite{FY}.


\begin{theorem} \label{thm2.3}
The $m$-function has the  representation
\begin{equation}\label{e23}
m(\lambda)=\sum_{k=1}^\infty\frac{1}{\alpha_k(\lambda_k-\lambda)},
\end{equation}
where
$\frac{1}{\alpha_k}=Res_{\lambda=\lambda_k}m(\lambda)$.
\end{theorem}

\begin{theorem} \label{thm2.4}
The function $m(\lambda)$ uniquely determines the
Sturm-Liouville operator.
\end{theorem}

\section{Eigenvalues of Hermitian matrices}

For a general matrix $A\in M_n$ there is no way to characterize
the eigenvalues of $A$ except that they are the roots of the
characteristic polynomial of $A$. For Hermitian matrices, however,
the eigenvalues can be characterized as the solutions of a
sequence of optimization problems. In this regard we have the well
known Theorem of Rayleigh-Ritz, see \cite{HR} for more details.
Now we want to prove a similar theorem by using the concept of the
Green's function. That is to characterize the eigenvalues and
corresponding orthonormal basis of eigenvectors for a Hermitian
matrix. It is well known that every Hermitian matrix can be
reduced to a tridiagonal matrix with the same eigenvalues. Some of
the well-known methods in this regards are Householder, Givens and
Lanczos transformations, \cite{HR}. For this reason we adopt the
following method for a general Hermitian matrix instead of a
Jacobi matix. Let $A$ and $B$ be two given Hermitian matrices. As
we defined the Green's function by (1.7); i.e.,
$$
 G(i,j,\lambda)=(A-\lambda
B)^{-1}(i,j)=e_i^T(A-\lambda B)^{-1}e_j.
$$
Now consider the nonhomogeneous system
\begin{equation}\label{e24}
Ax=\lambda Bx +f,\quad f\in \mathbb{C}^n.
\end{equation}
If $\lambda\notin \sigma(A,B)$, then the unique solution of this
system is
$$
x=(A-\lambda B)^{-1}f\,;
$$
that is,
\begin{equation}\label{e25}
x_i=\sum_{j=1}^nG(i,j,\lambda)f_j, \quad i=1,\ldots,n.
\end{equation}
For simplicity we assume $\lambda=0$ is not an eigenvalue and
let $G(i,j)=G(i,j,0)$. Note that the same algorithm
can be applied for any $\lambda\notin \sigma(A,B)$.
Thus we can define a linear operator $\mathbb{G}$ on
$\mathbb{C}^n$ as follows:
\begin{equation}\label{e26}
\mathbb{G}f_i=\sum_{j=1}^nG(i,j)f_j, \quad i=1,\ldots,n.
\end{equation}
Clearly in this case indeed $\mathbb{G}=A^{-1}$. Therefore
$\mathbb{G}$ is selfadjoint; i.e.,
\begin{equation}\label{e27}
\langle f,\mathbb{G}g\rangle=\langle\mathbb{G}f,g\rangle,
\quad \forall f,g\in \mathbb{C}^n.
\end{equation}

First we prove the following results then we use them
to characterize the eigenvalues.

\begin{lemma} \label{lem3.1}
The norm of $\mathbb{G}$ satisfies
\begin{equation}\label{e28}
\|\mathbb{G}\|=\text{max}|\langle\mathbb{G}u,u\rangle|,\quad
u \in\mathbb{C}^n,\;\|u\|=1,
\end{equation}
where $\langle u,u\rangle=u^*u$ is the standard inner product on
$\mathbb{C}^n$.
\end{lemma}

\begin{proof} By (\ref{e27}) $\langle\mathbb{G}u,u\rangle$ is
real. If $\|u\|=1$, then it follows from the properties of inner
product that
$$
|\langle\mathbb{G}u,u\rangle|\leq\|\mathbb{G}u\|\|u\|
\leq\|\mathbb{G}\|,
$$
and hence
$\eta=\text{max}|\langle\mathbb{G}u,u\rangle|\leq\|\mathbb{G}\|$.
In order to prove the inverse inequality, we have
$$
\langle\mathbb{G}(u+v),u+v\rangle=\langle\mathbb{G}u,u\rangle
+\langle\mathbb{G}v,v\rangle+2\Re\langle
\mathbb{G}u,v\rangle\leq \eta\|u+v\|^2
$$
and similarly,
$$
\langle\mathbb{G}(u-v),u-v\rangle=\langle\mathbb{G}u,u\rangle
+\langle\mathbb{G}v,v\rangle-
2\Re\langle\mathbb{G}u,v\rangle\geq-\ \eta\|u-v\|^2
$$
Subtracting these equations we obtain
\begin{equation}\label{e29}
4\Re\langle\mathbb{G}u,v\rangle\leq\ 2\eta(\|u\|^2+\|v\|^2).
\end{equation}
Note that $\mathbb{G}u\neq 0$ for $u\neq 0$ since $\mathbb{G}$ is invertible.
Put $v=\mathbb{G}u/ \|\mathbb{G}u\|$ in (\ref{e29}) to obtain
$\|\mathbb{G}u\|\leq \eta$ which completes the proof.
\end{proof}


 \begin{theorem} \label{thm3.2} Either $\|\mathbb{G}\|$ or
$-\|\mathbb{G}\|$ is an eigenvalue of $\mathbb{G}$.
\end{theorem}

\begin{proof}
 Using Lemma \ref{lem3.1}, suppose
$\|\mathbb{G}\|=\text{max}\langle\mathbb{G}u,u\rangle$ for
$\|u\|=1,~u\in~\mathbb{C}^n$. Since
$u\longmapsto\langle\mathbb{G}u,u\rangle$ is continuous on the
compact sphere $\|u\|=1$, thus it attains its maximum on the
sphere, i.e., there exists an $x_0$ with $\|x_0\|=1$ such that
$\text{max}\langle\mathbb{G}u,u\rangle=\langle\mathbb{G}x_0,x_0\rangle$,
for $\|u\|=1$. Hence there is a sequence $\{u_m\}$ in
$\mathbb{C}^n$ such that $\|u_m\|=1$ and
$\langle\mathbb{G}u_m,u_m\rangle\to \|\mathbb{G}\|$, as
$m\to \infty$. Therefore, using Ascoli theorem there is a
subsequence of $\{\mathbb{G}u_m\}$, call it $\{\mathbb{G}u_m\}$
also, which is convergent to a vector $v_0$ in $\mathbb{C}^n$;
that is,
\begin{equation}\label{e30}
\|\mathbb{G}u_m-v_0\|\to 0,\quad (m\to \infty).
\end{equation}
Let $\mu_0=\|\mathbb{G}\|$. Now we prove that $v_0$ is eigenvector
of $\mathbb{G}$ corresponding to eigenvalue of $\mu_0$. Clearly
$\|\mathbb{G}u_m\|\to \|v_0\|$ and we have
\begin{equation}\label{e31}
\|\mathbb{G}u_m-\mu_0u_m\|^2=\|\mathbb{G}u_m\|^2
+\mu_0^2\|u_m\|^2-2\mu_0\langle\mathbb{G}u_m,u_m\rangle
\end{equation}
and the right side tends to $\|v_0\|^2-\mu_0^2$.
It follows that $\|v_0\|^2\geq \mu_0^2>0$, hence $v_0$ is not
zero. From (\ref{e31}) it follows that since
$\|\mathbb{G}u_m\|^2\leq \mu_0^2$,
$$
0\leq\|\mathbb{G}u_m-\mu_0u_m\|^2\leq2\mu_0^2
-2\mu_0\langle\mathbb{G}u_m,u_m\rangle
$$
which tends to zero as $m\to\infty$. Thus
\begin{equation}\label{e32}
\|\mathbb{G}u_m-\mu_0u_m\|\to 0
\end{equation}
On the other hand by using the Triangle inequality we have
$$
0\leq\|\mathbb{G}v_0-\mu_0v_0\|\leq\|\mathbb{G}v_0
-\mathbb{G}(\mathbb{G}u_m)\|+\|
\mathbb{G}(\mathbb{G}u_m)-\mu_0\mathbb{G}u_m\|
+\|\mu_0\mathbb{G}u_m-\mu_0v_0\|
$$
Combining (\ref{e30}), (\ref{e32}), and the inequality
$\|\mathbb{G}u\|\leq\|\mathbb{G}\|\|u\|$, the last inequality
shows that $\|\mathbb{G}v_0-\mu_0v_0\|=0$, which proves that
$\mathbb{G}v_0=\mu_0v_0$. If the other case of Lemma \ref{lem3.1} holds;
i.e., $-\|\mathbb{G}\|=\text{min}\langle\mathbb{G}u,u\rangle$, the
proof is similar.
\end{proof}

\subsection*{Conclusion}
In both discrete and continuous Sturm-Liouville problems Green's
functions and Weyl m-functions have similar expressions in terms
of spectral data. In both cases Weyl m-function uniquely
determines the Sturm-Liouville operators. Rayleigh-Ritz Theorem
may be proved by using the concept of Green's function in discrete
case to characterize the eigenvalues and eigenvectors of a given
Hermitian matrix as follows. Let $\mu_0=\|\mathbb{G}\|$ and let
$v_0$ be the corresponding eigenvector. Define
$\chi_0=v_0/\|v_0\|$. The eigenvector $\chi_0$ is said to be
\emph{normalized}. Let
\begin{equation}\label{e33}
G_1(i,j)=G(i,j)-\mu_0\chi_0(i)\bar{\chi}_0(j),
\end{equation}
and similar to $\mathbb{G}$ the operator
$\mathbb{G}_1$ is defined by
\begin{equation}\label{e34}
\mathbb{G}_1u_i=\sum_{j=1}^nG_1(i,j)u_j, \quad i=1,\ldots,n.
\end{equation}
Then $\mathbb{G}_1$ has the same properties as $\mathbb{G}$, in
particular $\mu_1=\|\mathbb{G}_1\|$ is an eigenvalue of
$\mathbb{G}_1$ and there is a corresponding eigenfunction
$\varphi_1$. Let $\chi_1=v_1/\|v_1\|$. We can easily verify that
$\langle\mathbb{G}_1u,\chi_0\rangle=0$ for all $u\in\mathbb{C}^n$.
This follows that $\chi_1$ is orthogonal to $\chi_0$. Therefore,
\begin{equation}\label{e35}
\mathbb{G}\chi_1=\mathbb{G}_1\chi_1=\mu_1\chi_1.
\end{equation}
Moreover, we have
\begin{equation}\label{e36}
|\mu_0|\geq|\langle\mathbb{G}\chi_1,\chi_1\rangle|=|\mu_1|\|\chi_1\|^2=|\mu_1|.
\end{equation}
Letting
\begin{equation}\label{e37}
G_2(i,j)=G_1(i,j)-\mu_1\chi_1(i)\bar{\chi}_1(j),
\end{equation}
and proceeding as above, the existence of $\chi_2$ and $\mu_2$ is
established with $|\mu_2|\leq|\mu_1|,$ and $\chi_2$ orthogonal to
$\chi_1$ and $\chi_0$. In this way establish an orthonormal basis
of eigenvectors and all eigenvalues of $A^{-1}$, hence the
eigenvalues of $A$.

\subsection*{Acknowledgments} The author would like to thank the
anonymous referee for careful reading of the original manuscript
and for his/her valuable comments. The author also
thanks the Sahand University of Technology for the financial support.

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\end{thebibliography}

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