Generalized inverse spectral problem for pseudo-Jacobi matrices with mixed eigendata

ABSTRACT

In this paper, we investigate a generalized inverse eigenproblem for pseudo-Jacobi matrices with mixed eigendata. These matrices appear in non-Hermitian Quantum Mechanics and extend the well-known concept of Jacobi matrices. It is shown that a unique pseudo-Jacobi matrix may be recovered from certain prescribed mixed eigendata, i.e. its leading principal submatrix, two distinct real eigenvalues, and part of the corresponding eigenvectors. An algorithm is provided for the reconstruction of such a matrix, and illustrative numerical experiments are presented to test the algorithm. The recurrence relation involving leading principal minors is crucial for the problem solution.

KEYWORDS:

• Generalized inverse eigenvalue problem
• pseudo-Jacobi matrix
• tridiagonal matrix
• non-Hermitian quantum mechanics
• partial spectral data

AMS CLASSIFICATIONS:

• 15A18
• 15A29
• 65F15
• 65F18

1. Introduction

In this paper, we shall be concerned with the set of real tridiagonal matrices $T=\left[\begin{array}{ccccccccc}{\alpha }_1& {\beta }_1& & & & & & & \\ {\gamma }_1& \ddots & \ddots & & & & & & \\ & \ddots & \ddots & {\beta }_{r-1}& & & & & \\ & & {\gamma }_{r-1}& {\alpha }_r& {\beta }_r& & & & \\ & & & {\gamma }_r& {\alpha }_{r+1}& {\beta }_{r+1}& & & \\ & & & & {\gamma }_{r+1}& {\alpha }_{r+2}& {\beta }_{r+2}& & \\ & & & & & {\gamma }_{r+2}& \ddots & \ddots & \\ & & & & & & \ddots & \ddots & {\beta }_{n-1}\\ & & & & & & & {\gamma }_{n-1}& {\alpha }_n\end{array}\right]$ with ${\beta }_i$, $i=1,\dots ,n-1$ positive and satisfying ${\gamma }_r=-{\beta }_r$, ${\gamma }_i={\beta }_i$, $i\ne r$ and 0<r<n. This set is denoted by $\mathcal{J}(n,r,\mathit{\beta })$, and its elements are known as pseudo-Jacobi matrices (see the definition in [1]). If the values of ${\gamma }_i$ and ${\beta }_i$, $i=1,2,\dots ,n-1$ in the above matrix T are equal and positive respectively, then T is changed into a Jacobi matrix. These matrices attracted the attention of scientists as they appear in different areas of science, such as mechanics and theoretical physics (e.g. see [1–4] and the references therein). In particular, the literature on inverse eigenvalue problems for Jacobi matrices is very rich. Much of the basic theory on existence and uniqueness of solution of these problems is due to Hochstadt [5] and Hald [6], and numerical algorithms for reconstructing Jacobi matrices from prescribed spectral data were proposed by Boley and Golub [7], Ferguson [8], etc. For other results, we refer the readers to [9–15]. Inverse eigenproblems have many applications in vibrating systems [12], classical moment problems [2], etc. The study of inverse eigenproblem for pseudo-Jacobi matrices also deserves attention since these matrices arise in connection with the discretization and truncation of the Schrödinger equation in non-Hermitian quantum mechanics [1] (see also [3,16,17]). In addition, generalized inverse eigenproblems on the reconstruction of a Jacobi matrix and a symmetric tridiagonal matrix from its leading principal submatrix, two distinct real eigenvalues, and part of the corresponding eigenvectors, were presented, respectively in [18,19]. In this article, we will consider the generalized inverse eigenproblem with mixed eigendata in the pseudo-Jacobi case.

Consider the eigenvalue problem $J_{n}X=\lambda C_{n}X,0\ne X\in {\mathbb{R}}^n,$ with $J_n\in \mathcal{J}(n,r,\mathit{\beta })$ and $C_n$ being the non-Hermitian matrix (1) $C_n=\left[\begin{array}{cccccccc}{a}_1& -b_1& & & & & & \\ -b_1& \ddots & \ddots & & & & & \\ & \ddots & \ddots & -b_{r-1}& & & & \\ & & -b_{r-1}& a_r& -b_r& & & \\ & & & b_r& a_{r+1}& -b_{r+1}& & \\ & & & & -b_{r+1}& \ddots & \ddots & \\ & & & & & \ddots & \ddots & -b_{n-1}\\ & & & & & & -b_{n-1}& a_n\end{array}\right],$(1) where the pair $(\lambda ,X)$ is said to be an eigenpair of $(J_n,C_n).$ The goal of this work is to solve the generalized inverse eigenproblem stated below, denoted by the acronym of generalized inverse eigenproblem for pseudo-Jacobi matrices (GIEPPJ), consisting in reconstructing a matrix in $\mathcal{J}(n,r,\mathit{\beta })$ from the knowledge of partial spectral data.

The GIEPPJ statement is as follows.

GIEPPJ: Given $C_n$ as in (1(1) $C_n=\left[\begin{array}{cccccccc}{a}_1& -b_1& & & & & & \\ -b_1& \ddots & \ddots & & & & & \\ & \ddots & \ddots & -b_{r-1}& & & & \\ & & -b_{r-1}& a_r& -b_r& & & \\ & & & b_r& a_{r+1}& -b_{r+1}& & \\ & & & & -b_{r+1}& \ddots & \ddots & \\ & & & & & \ddots & \ddots & -b_{n-1}\\ & & & & & & -b_{n-1}& a_n\end{array}\right],$(1) ), an $r\times r$ Jacobi matrix $J_r$, real distinct numbers λ and μ, real vectors $X_2=(x_{r+1},x_{r+2},\dots ,x_n{)}^{\mathrm{T}}$, $Y_2=(y_{r+1},y_{r+2},\dots ,y_n{)}^{\mathrm{T}}$, where 0<r<n, find nonzero real vectors $X_1=(x_1,x_2,\dots ,x_r{)}^{\mathrm{T}}$, $Y_1=(y_1,y_2,\dots ,y_r{)}^{\mathrm{T}}$ and a pseudo-Jacobi matrix $J_n\in \mathcal{J}(n,r,\mathit{\beta })$ such that $J_r$ is its $r\times r$ leading principal submatrix, and $(\lambda ,X)$ and $(\mu ,Y)$ are eigenpairs of $(J_n,C_n)$, i.e. $J_{n}X=\lambda C_{n}X$ and $J_{n}Y=\mu C_{n}Y$, where $X=(X_1^{\mathrm{T}},X_2^{\mathrm{T}}{)}^{\mathrm{T}}$ and $Y=(Y_1^{\mathrm{T}},Y_2^{\mathrm{T}}{)}^{\mathrm{T}}$.

The rest of this paper is organized as follows. In Section 2, the eigenvectors X and Y of the matrix pair $(J_n,C_n)$ are characterized. In Section 3, a necessary and sufficient condition for the existence of a unique solution for the GIEPPJ is presented. In Section 4, an algorithm is proposed to reconstruct the pseudo-Jacobi matrix in $\mathcal{J}(n,r,\mathit{\beta })$, and examples are given to illustrate the algorithm. Finally, in Section 5 some conclusions are presented.

2. Eigenvectors of the matrix pair $(J_n,C_n)$

Throughout this paper, let $\sigma (J_n,C_n)$ denote the set of eigenvalues of the matrix pair $(J_n,C_n)$, and the determination of these pairs is the direct generalized spectral problem. Let $f_i\left(\lambda \right)=\det (J_i-\lambda C_i)$ be the ith leading principal minor of $J_n-\lambda C_n$, where $J_i$ and $C_i$ are, respectively, the $i\times i$ leading principal submatrix of matrices $J_n$ and $C_n$ for $i=1,2,\dots ,n$. Assume that $f_{-1}\left(\lambda \right)=0$ and $f_0\left(\lambda \right)=1$. Let us define: (2) $\begin{array}{rl}{\phi }_i(\lambda ,\mu )& :=\det \left[\begin{array}{cc}{f}_{i-1}\left(\lambda \right)& f_i\left(\lambda \right)\\ f_{i-1}\left(\mu \right)& f_i\left(\mu \right)\end{array}\right],i=1,2,\dots ,r,\end{array}$(2) (3) $\begin{array}{rl}{t}_j& :=\det \left[\begin{array}{cc}{x}_{j+1}& y_{j+1}\\ x_j& y_j\end{array}\right],j=r+1,r+2,\dots ,n-1,\end{array}$(3) (4) $\begin{array}{rl}{d}_i& :=\sum _{j=i+1}^n{a}_j{x}_j{y}_j,i=r+1,r+2,\dots ,n-1,\end{array}$(4) (5) $\begin{array}{rl}{g}_i& :=\sum _{j=i+1}^{n-1}{b}_j(x_j{y}_{j+1}+x_{j+1}{y}_j),i=r+1,r+2,\dots ,n-2\mathrm{with}{g}_{n-1}=0,\end{array}$(5) where λ, μ, $X_2=(x_{r+1},x_{r+2},\dots ,x_n{)}^{\mathrm{T}}$ and $Y_2=(y_{r+1},y_{r+2},\dots ,y_n{)}^{\mathrm{T}}$ are as in the GIEPPJ.

The following three-term recurrence relation holds: (6) $\begin{array}{rl}{f}_j\left(\lambda \right)& =({\alpha }_j-\lambda a_j)f_{j-1}\left(\lambda \right)-({\beta }_{j-1}+\lambda b_{j-1}{)}^2{f}_{j-2}(\lambda ),j=1,\dots ,r,\\ f_{r+1}\left(\lambda \right)& =({\alpha }_{r+1}-\lambda a_{r+1})f_r\left(\lambda \right)+({\beta }_r+\lambda b_r{)}^2{f}_{r-1}(\lambda ),\\ f_k\left(\lambda \right)& =({\alpha }_k-\lambda a_k)f_{k-1}\left(\lambda \right)-({\beta }_{k-1}+\lambda b_{k-1}{)}^2{f}_{k-2}(\lambda ),k=r+2,\dots ,n.\end{array}$(6)

Before stating our main result we present preliminary lemmas.

Lemma 2.1

Let $(\lambda ,X)$ be an eigenpair of the matrix pair $(J_n,C_n),$ where $X=(x_1,x_2,$ $\dots ,x_n{)}^{\mathrm{T}}$. If ${\beta }_i+\lambda b_i\ne 0$ for $i=1,2,\dots ,n-1,$ then

1. $x_1{x}_n\ne 0;$

2. two consecutive components of X cannot simultaneously vanish;

3. $x_{\ell }=\left((-1{)}^{\ell -1}{x}_1{f}_{\ell -1}(\lambda )\right)/\left(\prod _{i=1}^{\ell -1}({\beta }_i+\lambda b_i)\right)$, $\ell =2,\dots ,n$.

Proof.

(a), (b) Assume that $(\lambda ,X)$ is an eigenpair of $(J_n,C_n)$, $(J_n-\lambda C_n)X=\mathbf{0},X\ne \mathbf{0},$ and write this matrix equation as the system: (7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) where $x_0$ and $x_{n+1}$ are interpreted as zero.

If ${\beta }_i+\lambda b_i\ne 0$ for $i=1,2,\dots ,n-1$, then by the first equation in (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ) we conclude that $x_1=0$ implies $x_2=0$. By the second equation, taking $x_1=x_2=0$ implies $x_3=0$, and successively, considering the jth equation (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ), we find that $x_{j-1}=x_j=0$ implies $x_{j+1}=0$. So, up to the $(n-1)$th equation in (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ), $x_{n-2}=x_{n-1}=0$ implies that $x_n=0$. Conversely, $x_n=0$ implies $x_{n-1}=\cdots =x_2=x_1=0$, and all the components of X must vanish, contradicting that $X\ne \mathbf{0}$. Then (a) and (b) hold.

(c) Writing ${\chi }_1=x_1,{\chi }_{\ell }=(-1{)}^{\ell -1}{x}_{\ell }\prod _{i=1}^{\ell -1}({\beta }_i+\lambda b_i),\ell =2,\dots ,n+1,$ and multiplying the first, second and third equations in (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ), respectively, by $\begin{array}{rl}& (-1{)}^{j-1}\prod _{i=1}^{j-1}({\beta }_i+\lambda b_i),1\le j\le r,\\ & (-1{)}^r\prod _{i=1}^r({\beta }_i+\lambda b_i)\\ & \mathrm{and}(-1{)}^{k-1}\prod _{i=1}^{k-1}({\beta }_i+\lambda b_i),r+2\le k\le n,\end{array}$ we obtain (8) $\begin{array}{rl}{\chi }_{j+1}& =({\alpha }_j-\lambda a_j){\chi }_j-({\beta }_{j-1}+\lambda b_{j-1}{)}^2{\chi }_{j-1},j=1,\dots ,r,\\ {\chi }_{r+2}& =({\alpha }_{r+1}-\lambda a_{r+1}){\chi }_{r+1}+({\beta }_r+\lambda b_r{)}^2{\chi }_r,\\ {\chi }_{k+1}& =({\alpha }_k-\lambda a_k){\chi }_k-({\beta }_{k-1}+\lambda b_{k-1}{)}^2{\chi }_{k-1},k=r+2,\dots ,n,\end{array}$(8) where ${\chi }_0={\chi }_{n+1}=0$.

Comparing (8(8) $\begin{array}{rl}{\chi }_{j+1}& =({\alpha }_j-\lambda a_j){\chi }_j-({\beta }_{j-1}+\lambda b_{j-1}{)}^2{\chi }_{j-1},j=1,\dots ,r,\\ {\chi }_{r+2}& =({\alpha }_{r+1}-\lambda a_{r+1}){\chi }_{r+1}+({\beta }_r+\lambda b_r{)}^2{\chi }_r,\\ {\chi }_{k+1}& =({\alpha }_k-\lambda a_k){\chi }_k-({\beta }_{k-1}+\lambda b_{k-1}{)}^2{\chi }_{k-1},k=r+2,\dots ,n,\end{array}$(8) ) with (6(6) $\begin{array}{rl}{f}_j\left(\lambda \right)& =({\alpha }_j-\lambda a_j)f_{j-1}\left(\lambda \right)-({\beta }_{j-1}+\lambda b_{j-1}{)}^2{f}_{j-2}(\lambda ),j=1,\dots ,r,\\ f_{r+1}\left(\lambda \right)& =({\alpha }_{r+1}-\lambda a_{r+1})f_r\left(\lambda \right)+({\beta }_r+\lambda b_r{)}^2{f}_{r-1}(\lambda ),\\ f_k\left(\lambda \right)& =({\alpha }_k-\lambda a_k)f_{k-1}\left(\lambda \right)-({\beta }_{k-1}+\lambda b_{k-1}{)}^2{f}_{k-2}(\lambda ),k=r+2,\dots ,n.\end{array}$(6) ), we conclude that ${\chi }_{\ell }=K\cdot f_{\ell -1}\left(\lambda \right)$, $\ell =1,2,\dots ,n+1$, where K is a nonzero constant. If $\ell =1$, then $K=x_1$. Thus, we get $(-1{)}^{\ell -1}{x}_{\ell }\prod _{i=1}^{\ell -1}({\beta }_i+\lambda b_i)=x_1{f}_{\ell -1}(\lambda ),$

Therefore, if ${\beta }_i+\lambda b_i\ne 0$ for any $i=1,2,\dots ,\ell -1$, $x_{\ell }=\frac{(-1{)}^{\ell -1}{x}_1{f}_{\ell -1}(\lambda )}{\prod _{i=1}^{\ell -1}({\beta }_i+\lambda b_i)}$ for any $\ell =2,\dots ,n$.

An immediate consequence of Lemma 2.1 is the following.

Lemma 2.2

Let $(\lambda ,X)$ be an eigenpair of $(J_n,C_n)$. Assume that $X_1=(x_1,x_2,\dots ,x_r{)}^{\mathrm{T}}$ $(0<r<n)$ is a nonzero vector. If ${\beta }_i+\lambda b_i\ne 0$ for $i=1,2,\dots ,r-1,$ then $x_1\ne 0$ and $x_j=\frac{(-1{)}^{j-1}{x}_1{f}_{j-1}(\lambda )}{\prod _{i=1}^{j-1}({\beta }_i+\lambda b_i)},j=2,\dots ,r.$

Lemma 2.3

Under the assumptions in the GIEPPJ and Lemma 2.2, and assuming that ${\beta }_r+\lambda b_r\ne 0,$ we have:

1. If $\lambda \notin \sigma (J_r,C_r),$ then $x_{r+1}\ne 0$ and $\begin{array}{rl}{X}_1& =\frac{(-1{)}^r({\beta }_r+\lambda b_r)x_{r+1}}{f_r\left(\lambda \right)}\left(\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i),-f_1\left(\lambda \right)\prod _{i=2}^{r-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^j{f}_j(\lambda )\right.\\ & \times {\left.\prod _{i=j+1}^{r-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^{r-2}{f}_{r-2}(\lambda \left)\right({\beta }_{r-1}+\lambda b_{r-1}),(-1{)}^{r-1}{f}_{r-1}\left(\lambda \right)\right)}^{\mathrm{T}};\end{array}$

2. If $\lambda \in \sigma (J_r,C_r),$ then $x_{r+1}=0$ and $\begin{array}{rl}{X}_1& =\frac{x_1}{\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i)}\left(\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i),-f_1\left(\lambda \right)\prod _{i=2}^{r-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^j{f}_j(\lambda )\right.\\ & \times {\left.\prod _{i=j+1}^{r-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^{r-2}{f}_{r-2}(\lambda \left)\right({\beta }_{r-1}+\lambda b_{r-1}),(-1{)}^{r-1}{f}_{r-1}\left(\lambda \right)\right)}^{\mathrm{T}},\end{array}$ where $x_1$ is an arbitrary nonzero real number.

Proof.

(1) Since $\lambda \notin \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)\ne 0$. By Lemma 2.2 we have $x_1\ne 0$. If ${\beta }_i+\lambda b_i\ne 0$ for $i=1,2,\dots ,r$, from the proof of Lemma 2.1 we conclude that $x_{r+1}=\left((-1{)}^r{x}_1{f}_r(\lambda )\right)/\left(\prod _{i=1}^r({\beta }_i+\lambda b_i)\right)$. Therefore, $x_{r+1}\ne 0$ and $x_1=\frac{(-1{)}^r{x}_{r+1}}{f_r\left(\lambda \right)}\prod _{i=1}^r({\beta }_i+\lambda b_i).$

From the first equation $({\alpha }_1-\lambda a_1)x_1+({\beta }_1+\lambda b_1)x_2=0$ in (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ), as ${\beta }_1+\lambda b_1\ne 0$, we obtain by using the value of $x_1$ $x_2=-\frac{{\alpha }_1-\lambda a_1}{{\beta }_1+\lambda b_1}{x}_1=(-1{)}^{r+1}\frac{x_{r+1}{f}_1\left(\lambda \right)}{f_r\left(\lambda \right)}\prod _{i=2}^r({\beta }_i+\lambda b_i).$

The proof proceeds by induction. We assume $x_j=(-1{)}^{r+j-1}\frac{x_{r+1}{f}_{j-1}\left(\lambda \right)}{f_r\left(\lambda \right)}\prod _{i=j}^r({\beta }_i+\lambda b_i),j=2,\dots ,l-1,l\le r,$ and we show that $x_l=(-1{)}^{r+l-1}\frac{x_{r+1}{f}_{l-1}\left(\lambda \right)}{f_r\left(\lambda \right)}\prod _{i=l}^r({\beta }_i+\lambda b_i),l\le r.$

Using the recurrence relation $({\beta }_{l-2}+\lambda b_{l-2})x_{l-2}+({\alpha }_{l-1}-\lambda a_{l-1})x_{l-1}+({\beta }_{l-1}+\lambda b_{l-1})x_l=0$ in (7(7) $\begin{array}{rl}({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\alpha }_j-\lambda a_j)x_j+({\beta }_j+\lambda b_j)x_{j+1}& =0,j=1,\dots ,r,\\ -({\beta }_r+\lambda b_r)x_r+({\alpha }_{r+1}-\lambda a_{r+1})x_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =0,\\ ({\beta }_{k-1}+\lambda b_{k-1})x_{k-1}+({\alpha }_k-\lambda a_k)x_k+({\beta }_k+\lambda b_k)x_{k+1}& =0,k=r+2,\dots ,n,\end{array}$(7) ), as ${\beta }_{l-1}+\lambda b_{l-1}\ne 0$, we obtain $\begin{array}{rl}{x}_l& =-\frac{{\beta }_{l-2}+\lambda b_{l-2}}{{\beta }_{l-1}+\lambda b_{l-1}}{x}_{l-2}-\frac{{\alpha }_{l-1}-\lambda a_{l-1}}{{\beta }_{l-1}+\lambda b_{l-1}}{x}_{l-1}\\ & =(-1{)}^{r+l-1}\frac{x_{r+1}}{f_r\left(\lambda \right)}\left[-f_{l-3}({\beta }_{l-2}+\lambda b_{l-2}{)}^2+({\alpha }_{l-1}-\lambda a_{l-1}\left)f_{l-2}\right(\lambda )\right]\prod _{i=l}^r({\beta }_i+\lambda b_i)\\ & =(-1{)}^{r+l-1}\frac{x_{r+1}{f}_{l-1}\left(\lambda \right)}{f_r\left(\lambda \right)}\prod _{i=l}^r({\beta }_i+\lambda b_i).\end{array}$

Hence (1) holds.

(2) If $\lambda \in \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)=0$. If ${\beta }_i+\lambda b_i\ne 0$ for $i=1,2,\dots ,r$, from the proof of Lemma 2.1 we get $x_{r+1}=\left((-1{)}^r{x}_1{f}_r(\lambda )\right)/\left(\prod _{i=1}^r({\beta }_i+\lambda b_i)\right)=0$, and we easily obtain $x_l=(-1{)}^{l-1}\frac{x_1{f}_{l-1}\left(\lambda \right)}{\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i)}\prod _{i=l}^{r-1}({\beta }_i+\lambda b_i),l\le r.$

Since $x_1\ne 0$, by Lemma 2.2, we may consider $x_1$ as an arbitrary nonzero real number, and the result follows.

Lemma 2.4

Under the hypothesis of Lemma 2.3, we have: $\begin{array}{rl}& \left(\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i),-f_1\left(\lambda \right)\prod _{i=2}^{r-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^{r-2}{f}_{r-2}(\lambda )\right.\\ & \times \left.\phantom{\prod _{i=2}^{r-1}}({\beta }_{r-1}+\lambda b_{r-1}),(-1{)}^{r-1}{f}_{r-1}(\lambda )\right)C_r\cdot \\ & \left(\prod _{i=1}^{r-1}({\beta }_i+\mu b_i),-f_1\left(\mu \right)\prod _{i=2}^{r-1}({\beta }_i+\mu b_i),\dots ,(-1{)}^{r-2}{f}_{r-2}(\mu )\right.\\ & \times {\left.\phantom{\prod _{i=2}^{r-1}}({\beta }_{r-1}+\mu b_{r-1}),(-1{)}^{r-1}{f}_{r-1}(\mu )\right)}^{\mathrm{T}}\\ & =\frac{{\phi }_r(\lambda ,\mu )}{\lambda -\mu }.\end{array}$

Proof.

For r=1, the statement is trivial. For r=2, we get $\begin{array}{rl}& \left({\beta }_1+\lambda b_1,-f_1\left(\lambda \right)\right)C_2{\left({\beta }_1+\mu b_1,-f_1\left(\mu \right)\right)}^{\mathrm{T}}\\ & =a_1{\beta }_1^2+2b_1{\alpha }_1{\beta }_1+{\alpha }_1{b}_1^2(\lambda +\mu )-\lambda \mu a_1{b}_1^2+a_2{f}_1\left(\lambda \right)f_1\left(\mu \right)\\ & =\frac{{\beta }_1^2\left(f_1\right(\mu )-f_1(\lambda \left)\right)+2{\beta }_1{b}_1\left(\lambda f_1\right(\mu )-\mu f_1(\lambda \left)\right)+b_1^2\left({\lambda }^2{f}_1\right(\mu )-{\mu }^2{f}_1(\lambda \left)\right)}{\lambda -\mu }\\ & +\frac{f_1\left(\mu \right)f_1\left(\lambda \right)[{\alpha }_2-\mu a_2-({\alpha }_2-\lambda a_2\left)\right]}{\lambda -\mu }\\ & =\frac{f_1\left(\mu \right)\left[\right({\beta }_1+\lambda b_1{)}^2-({\alpha }_2-\lambda a_2)f_1\left(\lambda \right)]-f_1(\lambda \left)\right[({\beta }_1+\mu b_1{)}^2-({\alpha }_2-\mu a_2\left)f_1\right(\mu \left)\right]}{\lambda -\mu }\\ & =\frac{-f_1\left(\mu \right)f_2\left(\lambda \right)+f_1\left(\lambda \right)f_2\left(\mu \right)}{\lambda -\mu }=\frac{{\phi }_2(\lambda ,\mu )}{\lambda -\mu }.\end{array}$

The proof follows by induction. We assume the validity of the statement for r=p, $2\le p\le n-2$, i.e. $\begin{array}{rl}& \left(\prod _{i=1}^{p-1}({\beta }_i+\lambda b_i),-f_1\left(\lambda \right)\prod _{i=2}^{p-1}({\beta }_i+\lambda b_i),\dots ,(-1{)}^{p-2}{f}_{p-2}(\lambda )\right.\\ & \left.\times \phantom{\prod _{i=2}^{p-1}}({\beta }_{p-1}+\lambda b_{p-1}),(-1{)}^{p-1}{f}_{p-1}(\lambda )\right)C_p\cdot \\ & \left(\prod _{i=1}^{p-1}({\beta }_i+\mu b_i),-f_1\left(\mu \right)\prod _{i=2}^{p-1}({\beta }_i+\mu b_i),\dots ,(-1{)}^{p-2}{f}_{p-2}(\mu )\right.\\ & \times {\left.\phantom{\prod _{i=2}^{p-1}}({\beta }_{p-1}+\mu b_{p-1}),(-1{)}^{p-1}{f}_{p-1}(\mu )\right)}^{\mathrm{T}}\\ & =\frac{{\phi }_p(\lambda ,\mu )}{\lambda -\mu }.\end{array}$

By considering the partition $C_{p+1}=\left[\begin{array}{cc}{C}_p& -b_p{\mathbf{e}}_p\\ -b_p{\mathbf{e}}_p^{\mathrm{T}}& a_{p+1}\end{array}\right],$ where ${\mathbf{e}}_p=(0,0,\dots ,0,1{)}^{\mathrm{T}}\in {\mathbb{R}}^p$, we obtain $\begin{array}{rl}& \left(\prod _{i=1}^p({\beta }_i+\lambda b_i),-f_1\left(\lambda \right)\prod _{i=2}^p({\beta }_i+\lambda b_i),\dots ,(-1{)}^{p-1}{f}_{p-1}(\lambda \left)\right({\beta }_p+\lambda b_p),(-1{)}^p{f}_p\left(\lambda \right)\right)C_{p+1}\cdot \\ & {\left(\prod _{i=1}^p({\beta }_i+\mu b_i),-f_1\left(\mu \right)\prod _{i=2}^p({\beta }_i+\mu b_i),\dots ,(-1{)}^{p-1}{f}_{p-1}(\mu \left)\right({\beta }_p+\mu b_p),(-1{)}^p{f}_p\left(\mu \right)\right)}^{\mathrm{T}}\\ & =({\beta }_p+\lambda b_p)({\beta }_p+\mu b_p)\frac{{\phi }_p(\lambda ,\mu )}{\lambda -\mu }+b_p{f}_p\left(\lambda \right)f_{p-1}\left(\mu \right)({\beta }_p+\mu b_p)\\ & +b_p{f}_{p-1}\left(\lambda \right)f_p\left(\mu \right)({\beta }_p+\lambda b_p)+a_{p+1}{f}_p\left(\lambda \right)f_p\left(\mu \right)\\ & =\frac{1}{\lambda -\mu }\left[({\beta }_p+\lambda b_p)({\beta }_p+\mu b_p)\left(f_{p-1}\right(\lambda \left)f_p\right(\mu )-f_p(\lambda \left)f_{p-1}\right(\mu \left)\right)\right.\\ & +b_p(\lambda -\mu )f_p\left(\lambda \right)f_{p-1}\left(\mu \right)({\beta }_p+\mu b_p)+b_p(\lambda -\mu )f_{p-1}\left(\lambda \right)f_p\left(\mu \right)({\beta }_p+\lambda b_p)\\ & \left.+a_{p+1}(\lambda -\mu )f_p\left(\lambda \right)f_p\left(\mu \right)\right]\\ & =\frac{1}{\lambda -\mu }\left[f_{p-1}\left(\lambda \right)f_p\left(\mu \right)({\beta }_p+\lambda b_p{)}^2-f_p(\lambda \left)f_{p-1}\right(\mu \left)\right({\beta }_p+\mu b_p{)}^2\right.\\ & \left.+a_{p+1}(\lambda -\mu )f_p\left(\lambda \right)f_p\left(\mu \right)\right]\\ & =\frac{1}{\lambda -\mu }\left\{f_p\left(\lambda \right)\left[({\alpha }_{p+1}-\mu a_{p+1})f_p\left(\mu \right)-({\beta }_p+\mu b_p{)}^2{f}_{p-1}(\mu )\right]\right.\\ & \left.-f_p\left(\mu \right)\left[({\alpha }_{p+1}-\lambda a_{p+1})f_p\left(\lambda \right)-({\beta }_p+\lambda b_p{)}^2{f}_{p-1}(\lambda )\right]\right\}\\ & =\frac{f_p\left(\lambda \right)f_{p+1}\left(\mu \right)-f_p\left(\mu \right)f_{p+1}\left(\lambda \right)}{\lambda -\mu }=\frac{{\phi }_{p+1}(\lambda ,\mu )}{\lambda -\mu }.\end{array}$

This shows the validity of the statement for r = p+1, and so the statement is true for $1\le r\le n-1$.

Lemma 2.5

Under the hypothesis of Lemma 2.3, the following holds:

1. If $\lambda ,\mu \notin \sigma (J_r,C_r),$ then $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{x_{r+1}{y}_{r+1}({\beta }_r+\lambda b_r)({\beta }_r+\mu b_r){\phi }_r(\lambda ,\mu )}{(\lambda -\mu )f_r\left(\lambda \right)f_r\left(\mu \right)};$

2. If $\lambda \in \sigma (J_r,C_r)$ and $\mu \notin \sigma (J_r,C_r),$ then $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{(-1{)}^r{x}_1{y}_{r+1}({\beta }_r+\mu b_r\left)f_{r-1}\right(\lambda )}{(\lambda -\mu )\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i)};$

3. If $\lambda \notin \sigma (J_r,C_r)$ and $\mu \in \sigma (J_r,C_r),$ then $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{(-1{)}^{r+1}{y}_1{x}_{r+1}({\beta }_r+\lambda b_r\left)f_{r-1}\right(\mu )}{(\lambda -\mu )\prod _{i=1}^{r-1}({\beta }_i+\mu b_i)};$

4. If $\lambda ,\mu \in \sigma (J_r,C_r),$ then $X_1^{\mathrm{T}}{C}_r{Y}_1=0,$

where $x_1$ and $y_1$ are arbitrary nonzero real numbers.

Proof.

(1) If $\lambda ,\mu \notin \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)\ne 0$ and $f_r\left(\mu \right)\ne 0$. By Lemmas 2.3 and 2.4, we obtain $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{x_{r+1}{y}_{r+1}({\beta }_r+\lambda b_r)({\beta }_r+\mu b_r){\phi }_r(\lambda ,\mu )}{(\lambda -\mu )f_r\left(\lambda \right)f_r\left(\mu \right)}.$

(2) If $\lambda \in \sigma (J_r,C_r)$ and $\mu \notin \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)=0$ and $f_r\left(\mu \right)\ne 0$. Thus ${\phi }_r(\lambda ,\mu )=f_{r-1}\left(\lambda \right)f_r\left(\mu \right)$. By Lemmas 2.3 and 2.4, we have $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{(-1{)}^r{x}_1{y}_{r+1}({\beta }_r+\mu b_r\left)f_{r-1}\right(\lambda )}{(\lambda -\mu )\prod _{i=1}^{r-1}({\beta }_i+\lambda b_i)},$ where $x_1$ is an arbitrary nonzero real number.

(3) If $\lambda \notin \sigma (J_r,C_r)$ and $\mu \in \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)\ne 0$, $f_r\left(\mu \right)=0$ and ${\phi }_r(\lambda ,\mu )=-f_r\left(\lambda \right)f_{r-1}\left(\mu \right)$. Thus, by Lemmas 2.3 and 2.4 we get $X_1^{\mathrm{T}}{C}_r{Y}_1=\frac{(-1{)}^{r+1}{y}_1{x}_{r+1}({\beta }_r+\lambda b_r\left)f_{r-1}\right(\mu )}{(\lambda -\mu )\prod _{i=1}^{r-1}({\beta }_i+\mu b_i)},$ where $y_1$ is an arbitrary nonzero real number.

(4) If $\lambda ,\mu \in \sigma (J_r,C_r)$, then $f_r\left(\lambda \right)=f_r\left(\mu \right)=0$ and ${\phi }_r(\lambda ,\mu )=0$. Therefore, by Lemmas 2.3 and 2.4 we find, $X_1^{\mathrm{T}}{C}_r{Y}_1=0,$ and the proof is complete.

3. Solution of the GIEPPJ

Assume that $(\lambda ,X)$ and $(\mu ,Y)$ are eigenpairs of $(J_n,C_n)$. Thus, (9) $\begin{array}{r}\begin{array}{rl}& (J_n-\lambda C_n)X=\mathbf{0},\\ & (J_n-\mu C_n)Y=\mathbf{0},\\ & Y^{\mathrm{T}}(J_n-\lambda C_n)X=0,\\ & X^{\mathrm{T}}(J_n-\mu C_n)Y=0.\end{array}\end{array}$(9) Let $(J_{r+1,n}$, $C_{r+1,n})$ be the $(n-r)\times (n-r)$ trailing principal submatrix of $(J_n$, $C_n)$ in lines $r+1,\dots ,n$, and consider the partition $J_n-\lambda C_n=\left[\begin{array}{cc}{J}_r-\lambda C_r& ({\beta }_r+\lambda b_r){\mathbf{e}}_r{\mathit{\delta }}_1^{\mathrm{T}}\\ -({\beta }_r+\lambda b_r){\mathit{\delta }}_1{\mathbf{e}}_r^{\mathrm{T}}& J_{r+1,n}-\lambda C_{r+1,n}\end{array}\right],$ where ${\mathbf{e}}_r=(0,0,\dots ,0,1{)}^{\mathrm{T}}\in {\mathbb{R}}^r$ and ${\mathit{\delta }}_1=(1,0,\dots ,0{)}^{\mathrm{T}}\in {\mathbb{R}}^{n-r}$. The system of matrix equations (9(9) $\begin{array}{r}\begin{array}{rl}& (J_n-\lambda C_n)X=\mathbf{0},\\ & (J_n-\mu C_n)Y=\mathbf{0},\\ & Y^{\mathrm{T}}(J_n-\lambda C_n)X=0,\\ & X^{\mathrm{T}}(J_n-\mu C_n)Y=0.\end{array}\end{array}$(9) ) can be equivalently expressed as follows: (10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) (11) $\begin{array}{rl}& (J_{r+1,n}-\lambda C_{r+1,n})X_2=({\beta }_r+\lambda b_r)x_r{\mathit{\delta }}_1,\end{array}$(11) (12) $\begin{array}{rl}& (J_r-\mu C_r)Y_1=-({\beta }_r+\mu b_r)y_{r+1}{\mathbf{e}}_r,\end{array}$(12) (13) $\begin{array}{rl}& (J_{r+1,n}-\mu C_{r+1,n})Y_2=({\beta }_r+\mu b_r)y_r{\mathit{\delta }}_1,\end{array}$(13) (14) $\begin{array}{rl}& Y_1^{\mathrm{T}}(J_r-\lambda C_r)X_1+Y_2^{\mathrm{T}}(J_{r+1,n}-\lambda C_{r+1,n})X_2+({\beta }_r+\lambda b_r)(y_r{x}_{r+1}-y_{r+1}{x}_r)=0,\end{array}$(14) (15) $\begin{array}{rl}& X_1^{\mathrm{T}}(J_r-\mu C_r)Y_1+X_2^{\mathrm{T}}(J_{r+1,n}-\mu C_{r+1,n})Y_2+({\beta }_r+\mu b_r)(x_r{y}_{r+1}-x_{r+1}{y}_r)=0.\end{array}$(15)

We intend to find a necessary and sufficient condition for the existence of a unique solution of the GIEPPJ. The solution of the GIEPPJ is unique if and only if there exists a unique positive number ${\beta }_r$, unique nonzero real vectors $X_1=(x_1,x_2,\dots ,x_r{)}^{\mathrm{T}}$, $Y_1=(y_1,y_2,\dots ,y_r{)}^{\mathrm{T}}$ with $x_1{y}_1\ne 0$, and a unique Jacobi matrix $J_{r+1,n}$ satisfying Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) )–(15(15) $\begin{array}{rl}& X_1^{\mathrm{T}}(J_r-\mu C_r)Y_1+X_2^{\mathrm{T}}(J_{r+1,n}-\mu C_{r+1,n})Y_2+({\beta }_r+\mu b_r)(x_r{y}_{r+1}-x_{r+1}{y}_r)=0.\end{array}$(15) ).

Theorem 3.1

There exists a unique triplet $(X_1,Y_1,J_n),$ with $J_n\in \mathcal{J}(n,r,\mathit{\beta }),$ solving the GIEPPJ if and only if the following conditions are satisfied:

1. ${\beta }_i+\lambda b_i\ne 0$ and ${\beta }_i+\mu b_i\ne 0$ for any $i=1,2,\dots ,r-1;$

2. $\lambda ,\mu \notin \sigma (J_r,C_r);$

3. One of the following holds: $\begin{array}{rl}& \mathrm{\Delta }=0,\mathcal{A}<0,\mathcal{B}>0,(-\mathcal{A}+\lambda b_r)(-\mathcal{A}+\mu b_r)\ne 0;\\ & \mathrm{\Delta }>0,\mathcal{B}<0,\left(-\mathcal{A}+\frac{1}{2}\sqrt{\mathrm{\Delta }}+\lambda b_r\right)\left(-\mathcal{A}+\frac{1}{2}\sqrt{\mathrm{\Delta }}+\mu b_r\right)\ne 0;\\ & \mathrm{\Delta }>0,\mathcal{A}<0,\mathcal{B}=0,(-2\mathcal{A}+\lambda b_r)(-2\mathcal{A}+\mu b_r)\ne 0;\end{array}$ where $\begin{array}{rl}\mathrm{\Delta }& =\frac{4(\lambda -\mu )f_r\left(\lambda \right)f_r\left(\mu \right)\left[(\lambda -\mu )f_{r-1}\left(\lambda \right)f_{r-1}\left(\mu \right)b_r^2{x}_{r+1}{y}_{r+1}+X_2^{\mathrm{T}}{C}_{r+1,n}{Y}_2{\phi }_r(\lambda ,\mu )\right]}{x_{r+1}{y}_{r+1}{\phi }_r^2(\lambda ,\mu )},\\ \mathcal{A}& =\frac{\lambda f_{r-1}\left(\lambda \right)f_r\left(\mu \right)-\mu f_{r-1}\left(\mu \right)f_r\left(\lambda \right)}{{\phi }_r(\lambda ,\mu )}{b}_r,\\ \mathcal{B}& =\frac{{\lambda }^2{f}_{r-1}\left(\lambda \right)f_r\left(\mu \right)-{\mu }^2{f}_{r-1}\left(\mu \right)f_r\left(\lambda \right)}{{\phi }_r(\lambda ,\mu )}{b}_r^2-\frac{(\lambda -\mu )f_r\left(\lambda \right)f_r\left(\mu \right)X_2^{\mathrm{T}}{C}_{r+1,n}{Y}_2}{x_{r+1}{y}_{r+1}{\phi }_r(\lambda ,\mu )};\end{array}$

4. for any $i=r+1,r+2,\dots ,n-1,$ $\frac{b_i}{t_i}(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+\frac{\lambda -\mu }{t_i}(g_i-d_i)>0,$ where the parameters $t_i,$ $d_i$ and $g_i$ are defined in (3(3) $\begin{array}{rl}{t}_j& :=\det \left[\begin{array}{cc}{x}_{j+1}& y_{j+1}\\ x_j& y_j\end{array}\right],j=r+1,r+2,\dots ,n-1,\end{array}$(3) )–(5(5) $\begin{array}{rl}{g}_i& :=\sum _{j=i+1}^{n-1}{b}_j(x_j{y}_{j+1}+x_{j+1}{y}_j),i=r+1,r+2,\dots ,n-2\mathrm{with}{g}_{n-1}=0,\end{array}$(5) ).

Proof.

Firstly, we prove the sufficiency. Under the condition (1), the results in Lemmas 2.2–2.5 hold. Under the condition (2), then $f_r\left(\lambda \right)\ne 0$ and $f_r\left(\mu \right)\ne 0$, and this imply that Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) ) and (12(12) $\begin{array}{rl}& (J_r-\mu C_r)Y_1=-({\beta }_r+\mu b_r)y_{r+1}{\mathbf{e}}_r,\end{array}$(12) ) have unique solutions $X_1$ and $Y_1$. From the first part of Lemma 2.3, we obtain the unique expressions of $X_1$ and $Y_1$. Meanwhile, $x_{r+1}\ne 0$ and $y_{r+1}\ne 0$.

As $Y_1^{\mathrm{T}}{J}_r{X}_1=X_1^{\mathrm{T}}{J}_r{Y}_1,Y_1^{\mathrm{T}}{C}_r{X}_1=X_1^{\mathrm{T}}{C}_r{Y}_1,Y_2^{\mathrm{T}}{J}_{r+1,n}{X}_2=X_2^{\mathrm{T}}{J}_{r+1,n}{Y}_2$ and $Y_2^{\mathrm{T}}{C}_{r+1,n}{X}_2=X_2^{\mathrm{T}}{C}_{r+1,n}{Y}_2,$ subtracting (15(15) $\begin{array}{rl}& X_1^{\mathrm{T}}(J_r-\mu C_r)Y_1+X_2^{\mathrm{T}}(J_{r+1,n}-\mu C_{r+1,n})Y_2+({\beta }_r+\mu b_r)(x_r{y}_{r+1}-x_{r+1}{y}_r)=0.\end{array}$(15) ) from (14(14) $\begin{array}{rl}& Y_1^{\mathrm{T}}(J_r-\lambda C_r)X_1+Y_2^{\mathrm{T}}(J_{r+1,n}-\lambda C_{r+1,n})X_2+({\beta }_r+\lambda b_r)(y_r{x}_{r+1}-y_{r+1}{x}_r)=0,\end{array}$(14) ) yields $(\lambda -\mu )(X_1^{\mathrm{T}}{C}_r{Y}_1+X_2^{\mathrm{T}}{C}_{r+1,n}{Y}_2)-\left[\right(\lambda +\mu )b_r+2{\beta }_r](x_{r+1}{y}_r-x_r{y}_{r+1})=0.$

Since the condition (3) holds, then ${\phi }_r(\lambda ,\mu )\ne 0$. By substituting in the above equation the unique values of $x_r$ and $y_r$ provided by the first part of Lemma 2.3, and the unique value of $X_1^{\mathrm{T}}{C}_r{Y}_1$ in the first part of Lemma 2.5, we obtain the following quadratic equation (16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) in the unknown ${\beta }_r$, where $\mathcal{A}$ and $\mathcal{B}$ are given in the condition (3). Let Δ be the discriminant of Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ).

If $\mathrm{\Delta }=0$, $\mathcal{A}<0$, $\mathcal{B}>0$, then Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ) has a double positive root. In this case, ${\beta }_r=-\mathcal{A}$.

If $\mathrm{\Delta }>0$, $\mathcal{B}<0$, then Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ) has a positive and a negative root, and we consider the positive root ${\beta }_r=-\mathcal{A}+\frac{1}{2}\sqrt{\mathrm{\Delta }}$.

If $\mathrm{\Delta }>0$, $\mathcal{A}<0$, $\mathcal{B}=0$, then Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ) has a positive and a zero root, and we take the positive root ${\beta }_r=-2\mathcal{A}$.

Therefore, the condition (3) ensures that Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ) has a unique positive root. Since ${\beta }_r+\lambda b_r\ne 0$ and ${\beta }_r+\mu b_r\ne 0$, the solutions of Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) ) and (12(12) $\begin{array}{rl}& (J_r-\mu C_r)Y_1=-({\beta }_r+\mu b_r)y_{r+1}{\mathbf{e}}_r,\end{array}$(12) ) are both nonzero, that is $X_1\ne \mathbf{0}$ and $Y_1\ne \mathbf{0}$ in the first part of Lemma 2.3.

Now, by Equation (11(11) $\begin{array}{rl}& (J_{r+1,n}-\lambda C_{r+1,n})X_2=({\beta }_r+\lambda b_r)x_r{\mathit{\delta }}_1,\end{array}$(11) ) we obtain (17) $\begin{array}{rl}{\alpha }_{r+1}{x}_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =\lambda a_{r+1}{x}_{r+1}+({\beta }_r+\lambda b_r)x_r,\end{array}$(17) (18) $\begin{array}{rl}{\alpha }_j{x}_j+({\beta }_j+\lambda b_j)x_{j+1}& =\lambda a_j{x}_j-({\beta }_{j-1}+\lambda b_{j-1})x_{j-1},j=r+2,\dots ,n-1,\end{array}$(18) (19) $\begin{array}{rl}{\alpha }_n{x}_n& =\lambda a_n{x}_n-({\beta }_{n-1}+\lambda b_{n-1})x_{n-1}.\end{array}$(19) Similarly, by Equation (13(13) $\begin{array}{rl}& (J_{r+1,n}-\mu C_{r+1,n})Y_2=({\beta }_r+\mu b_r)y_r{\mathit{\delta }}_1,\end{array}$(13) ) we find (20) $\begin{array}{rl}{\alpha }_{r+1}{y}_{r+1}+({\beta }_{r+1}+\mu b_{r+1})y_{r+2}& =\mu a_{r+1}{y}_{r+1}+({\beta }_r+\mu b_r)y_r,\end{array}$(20) (21) $\begin{array}{rl}{\alpha }_j{y}_j+({\beta }_j+\mu b_j)y_{j+1}& =\mu a_j{y}_j-({\beta }_{j-1}+\mu b_{j-1})y_{j-1},j=r+2,\dots ,n-1,\end{array}$(21) (22) $\begin{array}{rl}{\alpha }_n{y}_n& =\mu a_n{y}_n-({\beta }_{n-1}+\mu b_{n-1})y_{n-1}.\end{array}$(22) Eliminating ${\alpha }_j$ from Equations (18(18) $\begin{array}{rl}{\alpha }_j{x}_j+({\beta }_j+\lambda b_j)x_{j+1}& =\lambda a_j{x}_j-({\beta }_{j-1}+\lambda b_{j-1})x_{j-1},j=r+2,\dots ,n-1,\end{array}$(18) ) and (21(21) $\begin{array}{rl}{\alpha }_j{y}_j+({\beta }_j+\mu b_j)y_{j+1}& =\mu a_j{y}_j-({\beta }_{j-1}+\mu b_{j-1})y_{j-1},j=r+2,\dots ,n-1,\end{array}$(21) ) yields (23) ${\beta }_j{t}_j+b_j(\lambda x_{j+1}{y}_j-\mu x_j{y}_{j+1})=(\lambda -\mu )a_j{x}_j{y}_j+{\beta }_{j-1}{t}_{j-1}-b_{j-1}(\lambda x_{j-1}{y}_j-\mu x_j{y}_{j-1})$(23) for $j=r+2,\dots ,n-1$. Next, by eliminating ${\alpha }_n$ from Equations (19(19) $\begin{array}{rl}{\alpha }_n{x}_n& =\lambda a_n{x}_n-({\beta }_{n-1}+\lambda b_{n-1})x_{n-1}.\end{array}$(19) ) and (22(22) $\begin{array}{rl}{\alpha }_n{y}_n& =\mu a_n{y}_n-({\beta }_{n-1}+\mu b_{n-1})y_{n-1}.\end{array}$(22) ), we find (24) $-{\beta }_{n-1}{t}_{n-1}+b_{n-1}(\lambda x_{n-1}{y}_n-\mu x_n{y}_{n-1})=(\lambda -\mu )a_n{x}_n{y}_n.$(24) Hence summing Equation (23(23) ${\beta }_j{t}_j+b_j(\lambda x_{j+1}{y}_j-\mu x_j{y}_{j+1})=(\lambda -\mu )a_j{x}_j{y}_j+{\beta }_{j-1}{t}_{j-1}-b_{j-1}(\lambda x_{j-1}{y}_j-\mu x_j{y}_{j-1})$(23) ) over $j=i+1,i+2,\dots ,n-1$ $(r+1\le i\le n-2)$, we get (25) $\begin{array}{rl}& {\beta }_{n-1}{t}_{n-1}+b_i(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+b_{n-1}(\lambda x_n{y}_{n-1}-\mu x_{n-1}{y}_n)\\ & +(\lambda -\mu )\sum _{j=i+1}^{n-2}{b}_j(x_j{y}_{j+1}+x_{j+1}{y}_j)=(\lambda -\mu )\sum _{j=i+1}^{n-1}{a}_j{x}_j{y}_j+{\beta }_i{t}_i.\end{array}$(25) Further, summing Equations (24(24) $-{\beta }_{n-1}{t}_{n-1}+b_{n-1}(\lambda x_{n-1}{y}_n-\mu x_n{y}_{n-1})=(\lambda -\mu )a_n{x}_n{y}_n.$(24) ) and (25(25) $\begin{array}{rl}& {\beta }_{n-1}{t}_{n-1}+b_i(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+b_{n-1}(\lambda x_n{y}_{n-1}-\mu x_{n-1}{y}_n)\\ & +(\lambda -\mu )\sum _{j=i+1}^{n-2}{b}_j(x_j{y}_{j+1}+x_{j+1}{y}_j)=(\lambda -\mu )\sum _{j=i+1}^{n-1}{a}_j{x}_j{y}_j+{\beta }_i{t}_i.\end{array}$(25) ) yields (26) ${\beta }_i{t}_i=b_i(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+(\lambda -\mu )(g_i-d_i),i=r+1,r+2,\dots ,n-1.$(26) Equation (26(26) ${\beta }_i{t}_i=b_i(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+(\lambda -\mu )(g_i-d_i),i=r+1,r+2,\dots ,n-1.$(26) ) and the condition (4) imply that ${\beta }_i>0$ exists uniquely and can be expressed as (27) ${\beta }_i=\frac{b_i}{t_i}(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+\frac{\lambda -\mu }{t_i}(g_i-d_i),i=r+1,r+2,\dots ,n-1.$(27) Since $x_{r+1}\ne 0$ and $y_{r+1}\ne 0$, from Equations (17(17) $\begin{array}{rl}{\alpha }_{r+1}{x}_{r+1}+({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}& =\lambda a_{r+1}{x}_{r+1}+({\beta }_r+\lambda b_r)x_r,\end{array}$(17) ) and (20(20) $\begin{array}{rl}{\alpha }_{r+1}{y}_{r+1}+({\beta }_{r+1}+\mu b_{r+1})y_{r+2}& =\mu a_{r+1}{y}_{r+1}+({\beta }_r+\mu b_r)y_r,\end{array}$(20) ) we have (28) $\begin{array}{rl}{\alpha }_{r+1}& =\lambda a_{r+1}+\frac{({\beta }_r+\lambda b_r)x_r-({\beta }_{r+1}+\lambda b_{r+1})x_{r+2}}{x_{r+1}},\\ {\alpha }_{r+1}& =\mu a_{r+1}+\frac{({\beta }_r+\mu b_r)y_r-({\beta }_{r+1}+\mu b_{r+1})y_{r+2}}{y_{r+1}}.\end{array}$(28) The above values of ${\alpha }_{r+1}$ are equal, and ${\alpha }_{r+1}$ are uniquely determined. As $t_j\ne 0$ $(j=r+1,r+2,\dots ,n-1)$, then $x_j$ and $y_j$ $(j=r+2,\dots ,n)$ cannot be simultaneously zero. This implies that ${\alpha }_j$ is unique for $j=r+2,\dots ,n$. Thus, from Equations (18(18) $\begin{array}{rl}{\alpha }_j{x}_j+({\beta }_j+\lambda b_j)x_{j+1}& =\lambda a_j{x}_j-({\beta }_{j-1}+\lambda b_{j-1})x_{j-1},j=r+2,\dots ,n-1,\end{array}$(18) ) and (21(21) $\begin{array}{rl}{\alpha }_j{y}_j+({\beta }_j+\mu b_j)y_{j+1}& =\mu a_j{y}_j-({\beta }_{j-1}+\mu b_{j-1})y_{j-1},j=r+2,\dots ,n-1,\end{array}$(21) ) we get (29) ${\alpha }_j=\left\{\begin{array}{ll}\lambda a_j-\frac{({\beta }_{j-1}+\lambda b_{j-1})x_{j-1}+({\beta }_j+\lambda b_j)x_{j+1}}{x_j},& x_j\ne 0,\\ \mu a_j-\frac{({\beta }_{j-1}+\mu b_{j-1})y_{j-1}+({\beta }_j+\mu b_j)y_{j+1}}{y_j},& y_j\ne 0.\end{array}\right.$(29) From Equations (19(19) $\begin{array}{rl}{\alpha }_n{x}_n& =\lambda a_n{x}_n-({\beta }_{n-1}+\lambda b_{n-1})x_{n-1}.\end{array}$(19) ) and (22(22) $\begin{array}{rl}{\alpha }_n{y}_n& =\mu a_n{y}_n-({\beta }_{n-1}+\mu b_{n-1})y_{n-1}.\end{array}$(22) ), ${\alpha }_n$ can be expressed as (30) ${\alpha }_n=\left\{\begin{array}{ll}\lambda a_n-\frac{({\beta }_{n-1}+\lambda b_{n-1})x_{n-1}}{x_n},& x_n\ne 0,\\ \mu a_n-\frac{({\beta }_{n-1}+\mu b_{n-1})y_{n-1}}{y_n},& y_n\ne 0.\end{array}\right.$(30) In addition, if $x_j{y}_j\ne 0$ $(j=r+2,\dots ,n)$, then the above two values of ${\alpha }_j$ are equal, and all the ${\alpha }_j$ are unique. Thus, the positive number ${\beta }_r$ and the truncated matrix $J_{r+1,n}$ are unique.

Now we prove the necessity. Assume the GIEPPJ has a unique solution for the triplet $(X_1,Y_1,J_n)$, where $J_n\in \mathcal{J}(n,r,\mathit{\beta })$, i.e. Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) )–(15(15) $\begin{array}{rl}& X_1^{\mathrm{T}}(J_r-\mu C_r)Y_1+X_2^{\mathrm{T}}(J_{r+1,n}-\mu C_{r+1,n})Y_2+({\beta }_r+\mu b_r)(x_r{y}_{r+1}-x_{r+1}{y}_r)=0.\end{array}$(15) ) have unique solutions. Because Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) ) and (12(12) $\begin{array}{rl}& (J_r-\mu C_r)Y_1=-({\beta }_r+\mu b_r)y_{r+1}{\mathbf{e}}_r,\end{array}$(12) ) have unique solutions, then $f_r\left(\lambda \right)\ne 0$ and $f_r\left(\mu \right)\ne 0$, i.e. $\lambda ,\mu \notin \sigma (J_r,C_r)$. Since $X_1\ne \mathbf{0}$ and $Y_1\ne \mathbf{0}$, then $({\beta }_r+\lambda b_r)x_{r+1}\ne 0$ and $({\beta }_r+\mu b_r)y_{r+1}\ne 0$, otherwise, from Equations (10(10) $\begin{array}{rl}& (J_r-\lambda C_r)X_1=-({\beta }_r+\lambda b_r)x_{r+1}{\mathbf{e}}_r,\end{array}$(10) ) and (12(12) $\begin{array}{rl}& (J_r-\mu C_r)Y_1=-({\beta }_r+\mu b_r)y_{r+1}{\mathbf{e}}_r,\end{array}$(12) ) we would obtain $X_1=Y_1=\mathbf{0}$, a contradiction. Moreover, $x_{r+1}{y}_{r+1}\ne 0$ implies that $x_1{y}_1\ne 0$, because $x_{r+1}=\frac{(-1{)}^r{x}_1{f}_r(\lambda )}{\prod _{i=1}^r({\beta }_i+\lambda b_i)}\mathrm{and}{y}_{r+1}=\frac{(-1{)}^r{y}_1{f}_r(\mu )}{\prod _{i=1}^r({\beta }_i+\mu b_i)}$ from the first part of Lemma 2.3. Hence there exist ${\beta }_i+\lambda b_i\ne 0$ and ${\beta }_i+\mu b_i\ne 0$ for $i=1,2,\dots ,r$. Since ${\beta }_r>0$ exists uniquely and Equation (16(16) ${\beta }_r^2+2\mathcal{A}{\beta }_r+\mathcal{B}=0$(16) ) can be obtained from Equations (14(14) $\begin{array}{rl}& Y_1^{\mathrm{T}}(J_r-\lambda C_r)X_1+Y_2^{\mathrm{T}}(J_{r+1,n}-\lambda C_{r+1,n})X_2+({\beta }_r+\lambda b_r)(y_r{x}_{r+1}-y_{r+1}{x}_r)=0,\end{array}$(14) ) and (15(15) $\begin{array}{rl}& X_1^{\mathrm{T}}(J_r-\mu C_r)Y_1+X_2^{\mathrm{T}}(J_{r+1,n}-\mu C_{r+1,n})Y_2+({\beta }_r+\mu b_r)(x_r{y}_{r+1}-x_{r+1}{y}_r)=0.\end{array}$(15) ), and the first part of Lemmas 2.3 and 2.5, then the condition (3) holds. Finally, the condition (4) is satisfied because Equation (26(26) ${\beta }_i{t}_i=b_i(\lambda x_i{y}_{i+1}-\mu x_{i+1}{y}_i)+(\lambda -\mu )(g_i-d_i),i=r+1,r+2,\dots ,n-1.$(26) ) holds for all $i=r+1,r+2,\dots ,n-1$ and ${\beta }_i>0$ exists uniquely for any $i=r+1,r+2,\dots ,n-1$.

4. Numerical algorithm and experiments

Based on Theorem 3.1 we present the following algorithm to solve the GIEPPJ.

By using MATLAB R2013a with a machine precision of around ${10}^{-16}$, we give the following example to illustrate Algorithm 1.

Example 4.1

Let $\lambda =-8.0276$, $\mu =1.9579$, $\begin{array}{rl}{C}_7& =\left[\begin{array}{ccccccc}3.0000& -0.8571& 0& 0& 0& 0& 0\\ -0.8571& 2.7143& -3.8571& 0& 0& 0& 0\\ 0& -3.8571& 2.4286& -6.8571& 0& 0& 0\\ 0& 0& -6.8571& 2.1429& -9.8571& 0& 0\\ 0& 0& 0& -9.8571& 1.8571& -12.8571& 0\\ 0& 0& 0& 0& 12.8571& 1.5714& -15.8571\\ 0& 0& 0& 0& 0& -15.8571& 3.0000\end{array}\right],\\ J_5& =\left[\begin{array}{ccccc}5.5714& 4.5714& 0& 0& 0\\ 4.5714& 5.4286& 5.1429& 0& 0\\ 0& 5.1429& 5.2857& 5.7143& 0\\ 0& 0& 5.7143& 5.1429& 6.2857\\ 0& 0& 0& 6.2857& 5.0000\end{array}\right],X_2=\left[\begin{array}{c}0.1417\\ 0.5644\end{array}\right],\\ Y_2& =\left[\begin{array}{c}0.0010\\ -0.3075\end{array}\right].\end{array}$

By using Algorithm 1, the four conditions in Theorem 3.1 hold. Then we get $\begin{array}{rl}{X}_1& =[-0.0744,-0.9550,-1.0000,-0.0025,0.6765{]}^{\mathrm{T}},\\ Y_1& =[-1.0000,-0.0484,0.4927,0.0184,-0.3693{]}^{\mathrm{T}}\end{array}$ and $J_7=\left[\begin{array}{ccccccc}5.5714& 4.5714& 0& 0& 0& 0& 0\\ 4.5714& 5.4286& 5.1429& 0& 0& 0& 0\\ 0& 5.1429& 5.2857& 5.7143& 0& 0& 0\\ 0& 0& 5.7143& 5.1429& 6.2857& 0& 0\\ 0& 0& 0& 6.2857& 5.0000& 6.8571& 0\\ 0& 0& 0& 0& -6.8571& 4.8571& 7.4286\\ 0& 0& 0& 0& 0& 7.4286& 6.0000\end{array}\right].$

Moreover, let $X=(X_1^{\mathrm{T}},X_2^{\mathrm{T}}{)}^{\mathrm{T}}$ and $Y=(Y_1^{\mathrm{T}},Y_2^{\mathrm{T}}{)}^{\mathrm{T}}$. We can reconfirm that $\det (J_7-\lambda C_7)=1.1822\times {10}^{-4}$, $\det (J_7-\mu C_7)=6.6448\times {10}^{-9}$, $\parallel J_{7}X-\lambda C_{7}X{\parallel }_2=1.5441\times {10}^{-14}$ and $\parallel J_{7}Y-\mu C_{7}Y{\parallel }_2=1.7334\times {10}^{-14}$. That is to say $(\lambda ,X)$ and $(\mu ,Y)$ are the eigenpairs of $(J_7,C_7)$.

In the following, we present an example to test Algorithm 1 in a different way.

Example 4.2

Given the matrix $C_n$ as in (1(1) $C_n=\left[\begin{array}{cccccccc}{a}_1& -b_1& & & & & & \\ -b_1& \ddots & \ddots & & & & & \\ & \ddots & \ddots & -b_{r-1}& & & & \\ & & -b_{r-1}& a_r& -b_r& & & \\ & & & b_r& a_{r+1}& -b_{r+1}& & \\ & & & & -b_{r+1}& \ddots & \ddots & \\ & & & & & \ddots & \ddots & -b_{n-1}\\ & & & & & & -b_{n-1}& a_n\end{array}\right],$(1) ) with entries $\begin{array}{rl}{b}_i& =0,\\ a_i& =3+\frac{2-2i}{n},i=1,2,\dots ,n-1,\\ a_n& =3.\end{array}$

Let $J_r$ be the rth leading principal matrix of ${\tilde{J}}_n\in \mathcal{J}(n,r,\stackrel{\mathbf{~}}{\mathit{\beta }})$ with entries $\begin{array}{rl}{\tilde{\beta }}_i& =4+\frac{4i}{n},\\ {\tilde{\alpha }}_i& =6-\frac{1}{n}(i+2),i=1,2,\dots ,n-1,\\ {\tilde{\alpha }}_n& =6.\end{array}$

Assume λ and μ are the first two real and distinct generalized eigenvalues of the matrix pair $({\tilde{J}}_n,C_n)$, $\tilde{X}$ and $\tilde{Y}$ are their corresponding generalized eigenvectors. Take the partition $\tilde{X}=({\tilde{X}}_1^{\mathrm{T}},X_2^{\mathrm{T}}{)}^{\mathrm{T}}\mathrm{and}\tilde{Y}=({\tilde{Y}}_1^{\mathrm{T}},Y_2^{\mathrm{T}}{)}^{\mathrm{T}},$ where ${\tilde{X}}_1,{\tilde{Y}}_1\in {\mathbb{R}}^r$. By using Algorithm 1, we obtain the pseudo-Jacobi matrix $J_n\in \mathcal{J}(n,r,\mathit{\beta })$ and real eigenvectors $X_1$ and $Y_1$. Let $X=(X_1^{\mathrm{T}},X_2^{\mathrm{T}}{)}^{\mathrm{T}}$ and $Y=(Y_1^{\mathrm{T}},Y_2^{\mathrm{T}}{)}^{\mathrm{T}}$. In Table 1, we show that $(\lambda ,X)$ and $(\mu ,Y)$ are eigenpairs of $(J_n,C_n)$. On the other hand, $X_1$ and $Y_1$ are the unique nonzero real vectors that can be obtained as well as $J_n$ is the unique recovered pseudo-Jacobi matrix within the machine precision. Furthermore, if the value of r is larger, the error $\parallel J_n-{\tilde{J}}_n{\parallel }_F$ will be smaller. In addition, the larger the value of n is, the larger the error $\parallel J_n-{\tilde{J}}_n{\parallel }_F$ will be. Thus, Algorithm 1 is feasible and effective to solve the GIEPPJ for a modest order n.

Table 1. Numerical results of Example 4.2.

n	r	$\parallel X_1-{\tilde{X}}_1{\parallel }_2$	$\parallel Y_1-{\tilde{Y}}_1{\parallel }_2$	$\parallel J_{n}X-\lambda C_{n}X{\parallel }_2$	$\parallel J_{n}Y-\mu C_{n}Y{\parallel }_2$	$\parallel J_n-{\tilde{J}}_n{\parallel }_F$
10	5	2.4921e−15	5.6494e−15	2.2377e−16	2.3604e−15	3.0859e−13
	8	9.5243e−15	1.4804e−14	2.1756e−15	3.8089e−15	7.5464e−14
20	10	2.7673e−14	8.5526e−13	3.3308e−15	4.2930e−14	2.6677e−10
	15	4.6134e−15	3.7625e−15	4.6132e−16	4.9763e−15	1.8023e−13
30	15	2.0888e−13	2.1721e−11	4.3512e−15	1.4433e−11	1.7724e−07
	20	6.0255e−14	1.0732e−12	1.9863e−15	6.1537e−15	2.3616e−10
40	25	2.2695e−13	1.4160e−11	3.6893e−15	7.7265e−13	4.8966e−08
	30	5.1089e−15	6.1997e−14	1.9868e−15	1.4889e−14	1.8808e−11
50	35	7.6030e−14	3.4969e−12	2.2350e−15	8.4302e−13	6.1181e−09
	40	8.9247e−15	1.6774e−14	8.9566e−16	1.1685e−14	1.1573e−11
60	40	3.2497e−12	3.7061e−10	2.0353e−15	1.0266e−11	2.0745e−05
	50	2.5871e−14	1.7225e−13	6.9560e−18	4.7107e−15	3.7397e−11
70	45	1.5545e−09	3.6859e−07	3.9725e−15	9.6494e−10	1.0918
	50	6.9044e−13	6.1867e−11	2.6762e−15	8.4085e−14	1.7100e−06

5. Conclusions

In this paper, we have considered the generalized inverse eigenvalue problem for the class of pseudo-Jacobi matrices $\mathcal{J}(n,r,\mathit{\beta })$. The three-term recurrence relation of the leading principal minors of the matrix pencil $J_n-\lambda C_n$ is central in obtaining the solution. We analysed the properties of the eigenvectors of the matrix pair $(J_n,C_n)$, and we obtained a necessary and sufficient condition for the solvability of the $\mathbf{G}\mathbf{I}\mathbf{E}\mathbf{P}\mathbf{P}\mathbf{J}$. The given conditions ensure that the $\mathbf{G}\mathbf{I}\mathbf{E}\mathbf{P}\mathbf{P}\mathbf{J}$ has a unique solution. The obtained pseudo-Jacobi matrix in $\mathcal{J}(n,r,\mathit{\beta })$ was reconstructed from the prescribed mixed spectral data, namely, the given matrices $C_n$, $J_r$, two distinct real eigenvalues and their corresponding partial eigenvectors. Finally, a numerical algorithm to solve the $\mathbf{G}\mathbf{I}\mathbf{E}\mathbf{P}\mathbf{P}\mathbf{J}$ was provided. Some illustrative examples were also given to test the efficiency of the algorithm. The results in this paper extend the previous results obtained by Ghanbari and Parvizpour [18], and Sen and Sharma [19] into the non-self-adjoint case. For the class of pseudo-Jacobi matrices in $\mathcal{J}(n,r,\mathit{\beta })$ with ${\gamma }_r=-{\beta }_r$, ${\gamma }_{r+1}=-{\beta }_{r+1}$, ${\gamma }_i={\beta }_i$, $i\ne r,r+1$ and 0<r<n−1, the GIEPPJ is an open problem under investigation.

Acknowledgements

The authors are very grateful to the anonymous referees for careful reading of the manuscript and valuable comments and suggestions which helped to improve the presentation of this paper.

Disclosure statement

No potential conflict of interest was reported by the authors.

Additional information

Funding

This work was supported by National Natural Science Foundation of China [No. 11471122], and in part by Science and Technology Commission of Shanghai Municipality [No. 18dz2271000]. The first author was also partially supported by Natural Science Foundation of Shandong Province [No. ZR2017MA050]. The second author acknowledges financial support from the Centre for Mathematics of University of Coimbra (funded by the Portuguese Government through FCT/MEC and by European RDF through Partnership Agreement PT2020).