On the inverse eigenvalue problem for periodic Jacobi matrices

ABSTRACT

The problem of reconstructing a matrix with a specific structure from a partial or total spectral data is known as inverse eigenvalue problem which arises in a variety of applications. In this paper, we study a partially described inverse eigenvalue problem of periodic Jacobi matrices and prove some spectral properties of such matrices. The problem involves the reconstruction of the matrix by one eigenvalue of each of its leading principal submatrices and one eigenvector of the required matrix and one more additional piece of information. The conditions for solvability of the problem are presented and finally an algorithm and some numerical results are given.

Keywords:

• Inverse eigenvalue problem
• leading principal minors
• periodic Jacobi matrix
• characteristic polynomial
• eigenpair

2010 Mathematics Subject Classifications:

• 65F18
• 65F15

1. Introduction

An inverse eigenvalue problem (IEP) concerns the reconstruction of a matrix with a special structure from prescribed spectral data. By structure, we mean the pattern of entries that are either zero or nonzero. There are many different types of IEPs and the level of their difficulty depends on the structure of the matrices which are to be reconstructed and the available eigen information. M.T. Chu in [1], gave an elaborate characterization of IEPs. Inverse eigenvalue problems have many practical applications such as control theory, mechanical system simulation, geophysics, molecular spectroscopy, structural analysis, mass spring vibrations, circuit theory and graph theory [1–5].

A periodic Jacobi matrix is the following symmetric matrix: (1) $A_n=\left[\begin{array}{cccccccc}{a}_1& b_1& 0& 0& 0& 0& \cdots & b_n\\ b_1& a_2& b_2& 0& 0& 0& \cdots & 0\\ 0& b_2& a_3& b_3& 0& 0& \cdots & 0\\ 0& 0& b_3& a_4& b_4& 0& \cdots & 0\\ 0& 0& 0& b_4& a_5& b_5& \cdots & 0\\ 0& 0& 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& 0& 0& \cdots & b_{n-2}& a_{n-1}& b_{n-1}\\ b_n& 0& 0& 0& \cdots & 0& b_{n-1}& a_n\end{array}\right],$(1)

where $a_i$s and $b_i$s are real numbers and $b_i\ne 0$ and $a_i$s may be zero or nonzero, $i=1,2,\dots ,n$.

Periodic Jacobi matrices have many practical applications such as periodic Toda lattices [6] and continued fractions [7]. The application of this matrix also appears in Quantum and Schrodinger equation which is studied by Damanik et al. [8].

Spectral properties of periodic Jacobi matrix were first investigated by Ferguson [9] and further studies made by Boley and Golub [10,11]. They gave numerical algorithm for constructing periodic Jacobi matrix using the given eigendata and one more piece of data which is the parameter β, defined as product of nonzero off-diagonal entries. Ying-Hong Xu in 2006 [12] investigated an IEP of this matrix and also proved some of its spectral properties.

In this paper, we solve the IEP stated below, denoted by IEPPJ, consisting in constructing a matrix of form $A_n$ with partial eigendata. This problem was originally introduced by Peng [4] for acyclic bordered-diagonal matrices and then Sharma in 2016 [13] studied this problem for symmetric tridiagonal matrices with nonzero off-diagonal entries.

IEPPJ: Given n real numbers $\mathrm{\Lambda }=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\}$, a vector $X_n=(x_1,x_2,\dots ,x_n{)}^{\mathrm{T}}$ and a real nonzero number c, find an $n\times n$ periodic Jacobi matrix $A_n$ such that ${\lambda }_i$ is an eigenvalue of $A_i$, $({\lambda }_n,X_n)$ is an eigenpair of $A_n$ and $b_n=c$.

The paper is organized as follows: Section 2 describes the preliminary concepts and the notations used in this paper. Section 3 deals with the analysis of the IEPPJ and the main results. In Section 4, we present some numerical examples to illustrate the solutions of IEPPJ. In Section 5 we conclude the paper.

2. Preliminaries

Throughout the paper, let $A_j$ be the jth leading principal submatrix of the matrix $A_n$. Let ${\phi }_j\left(\lambda \right)=\mathit{d}\mathit{e}\mathit{t}(\lambda I_j-A_j)$, $j=1,2,\dots ,n$, be the characteristic polynomial of the jth leading principal minor of $A_n$, where $I_j$ is the identity matrix of order j. Let us define (2) $B_j=\left[\begin{array}{cccccc}{a}_2& b_2& 0& 0& \cdots & 0\\ b_2& a_3& b_3& 0& \cdots & 0\\ 0& b_3& a_4& b_4& \cdots & 0\\ 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& \cdots & b_{j-2}& a_{j-1}& b_{j-1}\\ 0& 0& \cdots & 0& b_{j-1}& a_j\end{array}\right],j=2,3,\dots ,n,$(2) hence the matrix $B_j$ of order $(j-1)\times (j-1)$, is the matrix $A_j$ after removing the first row and column. Thus as an example, the matrix $B_2$ is a $1\times 1$ matrix with entry $a_2$. Let ${\chi }_j\left(\lambda \right)=\mathrm{d}\mathrm{e}\mathrm{t}(\lambda I_{j-1}-B_j)$, $j=2,\dots ,n-1$, and finally by a Jacobi matrix $J_n$ we mean a periodic Jacobi matrix $A_n$ so that $b_n=0$.

In what follows, we give the necessary lemmas that are used in the paper.

Lemma 2.1

[13]

For a Jacobi matrix $J_n,$ no two successive principal minors of $J_n$ have a common eigenvalue.

Lemma 2.2

[14]

Let T be the following symmetric Jacobi matrix and assume that $b_i\ne 0$ for $i=1,\dots ,n-1:$ $T=\left[\begin{array}{cccccccc}{a}_1& b_1& 0& 0& 0& 0& \cdots & 0\\ b_1& a_2& b_2& 0& 0& 0& \cdots & 0\\ 0& b_2& a_3& b_3& 0& 0& \cdots & 0\\ 0& 0& b_3& a_4& b_4& 0& \cdots & 0\\ 0& 0& 0& b_4& a_5& b_5& \cdots & 0\\ 0& 0& 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& 0& 0& \cdots & b_{n-2}& a_{n-1}& b_{n-1}\\ 0& 0& 0& 0& \cdots & 0& b_{n-1}& a_n\end{array}\right],$ and let $T_{i,j}$ denotes the entry at the ith row and jth column. Then (3) $(T{)}_{i,j}^{-1}=(-1{)}^{i+j}{b}_i\cdots b_{j-1}\frac{d_{j+1}\cdots d_n}{{\delta }_i\cdots {\delta }_n}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}i\le j,$(3) with the convention that empty product equals 1 such that $\begin{array}{rl}& d_n=a_n,d_i=a_i-\frac{b_i^2}{d_{i+1}},i=n-1,\dots ,1,\\ & {\delta }_1=a_1,{\delta }_i=a_i-\frac{b_{i-1}^2}{{\delta }_{i-1}},i=2,\dots ,n-1.\end{array}$

Because inverse of a symmetric matrix is symmetric, thus in Lemma 2.2, $(T{)}_{i,j}^{-1}=(T{)}_{j,i}^{-1}$.

Lemma 2.3

Let A and B be the following matrices: $A=\left[\begin{array}{cccccc}{a}_1& b_1& 0& 0& \cdots & 0\\ b_1& a_2& b_2& 0& \cdots & 0\\ 0& b_2& a_3& b_3& \cdots & 0\\ 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& \cdots & b_{j-2}& a_{j-1}& b_{j-1}\\ 0& 0& \cdots & 0& b_{j-1}& a_j\end{array}\right],B=\left[\begin{array}{cccccc}1& 0& 0& 0& \cdots & 0\\ 0& a_2& b_2& 0& \cdots & 0\\ 0& b_2& a_3& b_3& \cdots & 0\\ 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& \cdots & b_{j-2}& a_{j-1}& b_{j-1}\\ 0& 0& \cdots & 0& b_{j-1}& a_j\end{array}\right],$ then matrix $E=A^{-1}B$ has the following form: $E=\left[\begin{array}{cccccc}\ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 1& 0& \cdots & 0\\ \ast & \ast & 0& 1& \cdots & 0\\ ⋮& ⋮& 0& 0& \ddots & 0\\ \ast & \ast & 0& 0& \cdots & 1\end{array}\right],$ such that entries shown by $\ast$ represent real value numbers.

Proof.

We can represent the matrix B as follows: $B=A-D,$ where $D=\left[\begin{array}{cccccc}{a}_1-1& b_1& 0& 0& \cdots & 0\\ b_1& 0& 0& 0& \cdots & 0\\ 0& 0& 0& 0& \cdots & 0\\ 0& 0& 0& 0& \cdots & 0\\ ⋮& ⋮& & \ddots & \ddots & \\ 0& 0& 0& 0& \cdots & 0\end{array}\right],$ then $E=A^{-1}B=I-A^{-1}D.$ Since D has nonzero entries only at the first two columns, we have: $A^{-1}D=\left[\begin{array}{cccccc}\ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ ⋮& ⋮& & \ddots & \ddots & \\ \ast & \ast & 0& 0& \cdots & 0\end{array}\right],$ and so $\begin{array}{rl}E& =\left[\begin{array}{cccccc}1& 0& 0& 0& \cdots & 0\\ 0& 1& 0& 0& \cdots & 0\\ 0& 0& 1& 0& \cdots & 0\\ 0& 0& 0& 1& \cdots & 0\\ ⋮& ⋮& & \ddots & \ddots & \\ 0& 0& 0& 0& \cdots & 1\end{array}\right]-\left[\begin{array}{cccccc}\ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ ⋮& ⋮& & \ddots & \ddots & \\ \ast & \ast & 0& 0& \cdots & 0\end{array}\right]\\ & =\left[\begin{array}{cccccc}\ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 0& 0& \cdots & 0\\ \ast & \ast & 1& 0& \cdots & 0\\ \ast & \ast & 0& 1& \cdots & 0\\ ⋮& ⋮& 0& 0& \ddots & 0\\ \ast & \ast & 0& 0& \cdots & 1\end{array}\right].\end{array}$

3. Solution of IEPPJ

In this section, we first obtain the recurrence relation of characteristic polynomials of the matrix $A_n$ in Lemma 3.1. Then in Lemma 3.2, we show how to obtain the component $x_i$ of eigenvector $X_n$ from entries $A_{i-1}$, $x_1$ and $b_n{x}_n$. This information enables us to compute the unknown entry $b_{i-1}$. In Lemma 3.4, we investigate the condition under which the component $x_i\in X_n$ becomes zero. We close this section by giving the solution to IEPPJ in Theorem 3.5 and deriving an algorithm.

Lemma 3.1

The sequence $\left\{{\phi }_j\right(\lambda )=\mathit{d}\mathit{e}\mathit{t}(\lambda I_j-A_j){\}}_{j=1}^n$ of characteristic polynomials of $A_j$ satisfies the following recurrence relations:

1. ${\phi }_1\left(\lambda \right)=\lambda -a_1$.

2. ${\phi }_j\left(\lambda \right)=(\lambda -a_j){\phi }_{j-1}\left(\lambda \right)-b_{j-1}^2{\phi }_{j-2}\left(\lambda \right),2\le j\le n-1$.

3. ${\phi }_n\left(\lambda \right)=(\lambda -a_n){\phi }_{n-1}\left(\lambda \right)-b_{n-1}^2{\phi }_{n-2}\left(\lambda \right)-2\prod _{i=1}^n{b}_i-b_n^2\left(\mathrm{d}\mathrm{e}\mathrm{t}\right(\lambda I_{n-2}-B_{n-1}\left)\right)$.

To write the recurrences we define ${\phi }_0\left(\lambda \right)=1$.

Proof.

The recurrence relations can be verified by expanding the determinant.

The relation 2 of the Lemma 3.1 expresses the recurrence relation of characteristic polynomial of a Jacobi matrix.

Since $({\lambda }_n,X_n)$ is an eigenpair of $A_n$, so $A_n{X}_n={\lambda }_n{X}_n,$ we get the following relation (4) $b_{i-1}{x}_{i-1}+a_i{x}_i+b_i{x}_{i+1}={\lambda }_n{x}_i\mathrm{f}\mathrm{o}\mathrm{r}i=1,2,\dots ,n,$(4) with the following conventions $x_n=x_0,x_{n+1}=x_1,\mathrm{a}\mathrm{n}\mathrm{d}{b}_0=b_n.$ In the following lemma, we show that every component $x_i$ of eigenvector $X_n$, $i=2,3,\dots ,n-1$, is a linear combination of $x_1$ and $b_n{x}_n$.

Lemma 3.2

If $Y=(y_1,y_2,\dots ,y_n{)}^{\mathrm{T}}$ and $(\lambda ,Y)$ is an eigenpair of $A_n$, then $|y_1|+|y_n|>0$ and entries of Y are given by: (5) $y_j=\frac{{\phi }_{j-1}\left(\lambda \right)y_1-{\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-1}{b}_i},j=2,\dots ,n-1\mathrm{a}\mathrm{n}\mathrm{d}{\chi }_1\left(\lambda \right)=1.$(5)

Proof.

We prove this by induction on the rows of matrix $A_n$. For the base cases we have: $\begin{array}{rl}& a_1{y}_1+b_1{y}_2+b_n{y}_n=\lambda y_1\\ & \Rightarrow y_2=\frac{(\lambda -a_1)y_1-b_n{y}_n}{b_1}=\frac{{\phi }_1\left(\lambda \right)y_1-{\chi }_1\left(\lambda \right)b_n{y}_n}{b_1},\end{array}$ and $\begin{array}{rl}& b_1{y}_1+a_2{y}_2+b_2{y}_3=\lambda y_2\\ & \Rightarrow y_3=\frac{(\lambda -a_2)y_2-b_1{y}_1}{b_2}=\frac{(\lambda -a_2)b_1{y}_2-b_1^2{y}_1}{b_1{b}_2}\\ & =\frac{(\lambda -a_2)\left[\right(\lambda -a_1)y_1-b_n{y}_n]-b_1^2{y}_1}{b_1{b}_2}\\ & =\frac{\left(\right(\lambda -a_2\left)\right(\lambda -a_1)-b_1^2)y_1-(\lambda -a_2)b_n{y}_n}{b_1{b}_2}\\ & =\frac{{\phi }_2\left(\lambda \right)y_1-{\chi }_2\left(\lambda \right)b_n{y}_n}{b_1{b}_2}.\end{array}$ Now suppose the lemma holds for $y_2,y_3,\dots ,y_j$, we prove it for $y_{j+1}$. From $A_{n}Y=\lambda Y$ we obtain: $\begin{array}{rl}{y}_{j+1}& =\frac{(\lambda -a_j)y_j-b_{j-1}{y}_{j-1}}{b_j}\\ & =\frac{(\lambda -a_j)}{b_j}\left(\frac{{\phi }_{j-1}\left(\lambda \right)y_1-{\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-1}{b}_i}\right)-\frac{b_{j-1}}{b_j}\left(\frac{{\phi }_{j-2}\left(\lambda \right)y_1-{\chi }_{j-2}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-2}{b}_i}\right)\\ & =\frac{(\lambda -a_j){\phi }_{j-1}\left(\lambda \right)y_1-(\lambda -a_j){\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^j{b}_i}-\frac{b_{j-1}^2{\phi }_{j-2}\left(\lambda \right)y_1-b_{j-1}^2{\chi }_{j-2}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^j{b}_i}\\ & =\frac{(\lambda -a_j){\phi }_{j-1}\left(\lambda \right)-b_{j-1}^2{\phi }_{j-2}\left(\lambda \right)}{\prod _{i=1}^j{b}_i}{y}_1-\frac{(\lambda -a_j){\chi }_{j-1}\left(\lambda \right)-b_{j-1}^2{\chi }_{j-2}\left(\lambda \right)}{\prod _{i=1}^j{b}_i}{b}_n{y}_n\\ & =\frac{{\phi }_j\left(\lambda \right)y_1-{\chi }_j\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^j{b}_i}.\end{array}$ Since Y is an eigenvector thus $Y\ne 0$. If $y_1=y_n=0$, then from (5(5) $y_j=\frac{{\phi }_{j-1}\left(\lambda \right)y_1-{\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-1}{b}_i},j=2,\dots ,n-1\mathrm{a}\mathrm{n}\mathrm{d}{\chi }_1\left(\lambda \right)=1.$(5) ), all other entries of Y become zero, hence $|y_1|+|y_n|>0$.

Corollary 3.3

Let $x_n\ne 0,$ if $x_{j+1}=0$ then the matrix ${\lambda }_n{I}_j-A_j$ is invertible for $j=1,2,\dots ,n-2$.

Proof.

We prove this by contradiction. By taking $\lambda ={\lambda }_n$ and $Y=X$ in Lemma 3.2, we derive that if $x_{j+1}=0$ then ${\phi }_j\left({\lambda }_n\right)x_1={\chi }_j\left({\lambda }_n\right)b_n{x}_n$. Suppose the matrix ${\lambda }_n{I}_j-A_j$ is not invertible, thus ${\phi }_j\left({\lambda }_n\right)=det({\lambda }_n{I}_j-A_j)=0$ then ${\lambda }_j$ is an eigenvalue of $A_j$. It implies that ${\chi }_j\left({\lambda }_n\right)b_n{x}_n=0$. Since $b_n{x}_n\ne 0$, hence ${\chi }_j\left({\lambda }_n\right)=det({\lambda }_n{I}_{j-1}-B_j)=0$ then ${\lambda }_n$ is an eigenvalue of $B_j$. Since $B_j$ is a principal submatrix of the Jacobi matrix $A_j$ by Lemma 2.1, we have a contradiction therefore the matrix ${\lambda }_n{I}_j-A_j$ is invertible.

The result of Corollary 3.3 is independent of the value of $x_1$. If $x_1=0$, then ${\chi }_j\left({\lambda }_n\right)=0$ which in turn by Lemma 2.1 implies that the determinant of ${\lambda }_n{I}_j-A_j$ is nonzero thus it is invertible and if $x_1\ne 0$, the equation ${\phi }_j\left({\lambda }_n\right)x_1={\chi }_j\left(\lambda \right)b_n{x}_n$ is true if either ${\phi }_j\left({\lambda }_n\right)={\chi }_j\left({\lambda }_n\right)=0$ or ${\phi }_j\left({\lambda }_n\right)\ne 0,{\chi }_j\left({\lambda }_n\right)\ne 0$. By Lemma 2.1 the first one is impossible. Thus in this case ${\lambda }_n{I}_j-A_j$ is also invertible.

In the following lemma, we show the conditions that make an entry $x_j$ of eigenvector $X_n=(x_1,\dots ,x_n{)}^{\mathrm{T}}$, $j=2,\dots ,n-1$, becomes zero.

Lemma 3.4

Let $X_n=(x_1,x_2,\dots ,x_n{)}^{\mathrm{T}},({\lambda }_n,X_n)$ be an eigenpair of $A_n$ and $x_n\ne 0$. The entry $x_j=0,j=2,3,\dots ,n-1,$ if $\frac{x_1}{b_n{x}_n}=e_{1,1}{e}_{2,2}-e_{1,2}{e}_{2,1},$ such that using notations of Lemma 2.2 for $T={\lambda }_n{I}_{j-1}-A_{j-1},$ the $e_{1,1},e_{1,2},e_{2,1}$ and $e_{2,2}$ are as follow: $\begin{array}{rl}{e}_{1,1}& =\frac{d_2\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{1,2}& =b_1({\lambda }_n-a_2)\frac{d_3\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{2,1}& =b_1\frac{d_3\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{2,2}& =({\lambda }_n-a_2)\left(\frac{d_3\cdots d_{j-1}}{{\delta }_2\cdots {\delta }_{j-1}}\right)-b_2^2\left(\frac{d_4\cdots d_{j-1}}{{\delta }_2\cdots {\delta }_{j-1}}\right).\end{array}$

Proof.

By (5(5) $y_j=\frac{{\phi }_{j-1}\left(\lambda \right)y_1-{\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-1}{b}_i},j=2,\dots ,n-1\mathrm{a}\mathrm{n}\mathrm{d}{\chi }_1\left(\lambda \right)=1.$(5) ), $x_j=0$ implies that ${\phi }_{j-1}\left({\lambda }_n\right)x_1={\chi }_{j-1}\left({\lambda }_n\right)b_n{x}_n$. We can write this equation as follows: $\mathrm{d}\mathrm{e}\mathrm{t}({\lambda }_n{I}_{j-1}-A_{j-1})x_1=\mathrm{d}\mathrm{e}\mathrm{t}({\lambda }_n{I}_{j-2}-B_{j-1})b_n{x}_n=\mathrm{d}\mathrm{e}\mathrm{t}({\lambda }_n{I}_{j-1}-C_{j-1})b_n{x}_n,$ such that $B_{j-1}$ is the matrix defined in (2(2) $B_j=\left[\begin{array}{cccccc}{a}_2& b_2& 0& 0& \cdots & 0\\ b_2& a_3& b_3& 0& \cdots & 0\\ 0& b_3& a_4& b_4& \cdots & 0\\ 0& 0& \ddots & \ddots & \ddots & 0\\ 0& 0& \cdots & b_{j-2}& a_{j-1}& b_{j-1}\\ 0& 0& \cdots & 0& b_{j-1}& a_j\end{array}\right],j=2,3,\dots ,n,$(2) ) and $C_{j-1}=\left[\begin{array}{cc}{\lambda }_n-1& 0\\ 0& B_{j-1}\end{array}\right]$. From Corollary 3.3, the matrix ${\lambda }_n{I}_{j-1}-A_{j-1}$ is invertible hence by multiplying the $\mathrm{d}\mathrm{e}\mathrm{t}\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1})$ from the left side to both sides of the equation, we get the following: (6) $x_1=\mathrm{d}\mathrm{e}\mathrm{t}\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}({\lambda }_n{I}_{j-1}-C_{j-1}))b_n{x}_n.$(6) The matrix $({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}({\lambda }_n{I}_{j-1}-C_{j-1})$ has the form of matrix E given in Lemma 2.3. Let $e_{i,j}$ denotes its entry at the ith row and jth column, determinant of this matrix by expansion on the first row is $e_{1,1}{e}_{2,2}-e_{1,2}{e}_{2,1}$, hence it implies $\frac{x_1}{b_n{x}_n}=e_{1,1}{e}_{2,2}-e_{1,2}{e}_{2,1}.$ Each $e_{i,j},i,j=1,2$, is computed as follows: $\begin{array}{rl}{e}_{1,1}& =\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}{)}_{1,1},\\ e_{1,2}& =({\lambda }_n-a_2)\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}{)}_{1,2},\\ e_{2,1}& =\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}{)}_{2,1},\\ e_{2,2}& =({\lambda }_n-a_2)\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}{)}_{2,2}-b_2\left(\right({\lambda }_n{I}_{j-1}-A_{j-1}{)}^{-1}{)}_{2,3},\end{array}$ which by Lemma 2.2 are: $\begin{array}{rl}{e}_{1,1}& =\frac{d_2\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{1,2}& =b_1({\lambda }_n-a_2)\frac{d_3\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{2,1}& =b_1\frac{d_3\cdots d_{j-1}}{{\delta }_1\cdots {\delta }_{j-1}},\\ e_{2,2}& =({\lambda }_n-a_2)\left(\frac{d_3\cdots d_{j-1}}{{\delta }_2\cdots {\delta }_{j-1}}\right)-b_2^2\left(\frac{d_4\cdots d_{j-1}}{{\delta }_2\cdots {\delta }_{j-1}}\right),\end{array}$ and this completes the proof.

In the following theorem, the solution to the IEPPJ and the conditions under which the problem has a unique solution are given.

Theorem 3.5

The IEPPJ has a unique solution given by the following relations (7) $\begin{array}{rl}{a}_1& ={\lambda }_1,b_n=c,\\ a_j& ={\lambda }_j-\frac{b_{j-1}^2{\phi }_{j-2}\left({\lambda }_j\right)}{{\phi }_{j-1}\left({\lambda }_j\right)},j=2,3,\dots ,n-1,\\ b_{j-1}& =\frac{{\phi }_{j-1}\left({\lambda }_n\right)x_1-{\chi }_{j-1}\left({\lambda }_n\right)b_n{x}_n}{x_j\prod _{i=1}^{j-2}{b}_i},j=2,3,\dots ,n,\\ a_n& =\frac{{\lambda }_n{\phi }_{n-1}\left({\lambda }_n\right)-b_{n-1}^2{\phi }_{n-2}\left({\lambda }_n\right)-2\prod _{i=1}^n{b}_i-b_n^2\mathrm{d}\mathrm{e}\mathrm{t}({\lambda }_n{I}_{n-2}-B_{n-1})}{{\phi }_{n-1}\left({\lambda }_n\right)},\end{array}$(7) if for $j=2,3,\dots ,n,$ $x_j\ne 0,$ ${\lambda }_j\ne {\lambda }_{j-1}$ and ${\lambda }_n$ not to be an eigenvalue of $A_{n-1}$.

Proof.

Let $x_j\ne 0$ for all $j=2,\dots ,n$. As per conditions of IEPPJ, ${\lambda }_j$ is an eigenvalue of $A_j$ for all $j=1,2,\dots ,n$. Thus ${\phi }_1\left({\lambda }_1\right)=0\Rightarrow {\lambda }_1=a_1$.

By Lemma 3.1, the relation ${\phi }_j\left({\lambda }_j\right)=0$ for $j=2,\dots ,n-1$ implies $({\lambda }_j-a_j){\phi }_{j-1}\left({\lambda }_j\right)-b_{j-1}^2{\phi }_{j-2}\left({\lambda }_j\right)=0\Rightarrow a_j={\lambda }_j-\frac{b_{j-1}^2{\phi }_{j-2}\left({\lambda }_j\right)}{{\phi }_{j-1}\left({\lambda }_j\right)},$ and from Lemma 2.1, the denominator of this expression is nonzero. By applying ${\phi }_n\left({\lambda }_n\right)=0$, it implies $a_n=\frac{{\lambda }_n{\phi }_{n-1}\left({\lambda }_n\right)-b_{n-1}^2{\phi }_{n-2}\left({\lambda }_n\right)-2\prod _{i=1}^n{b}_i-b_n^2\mathit{d}\mathit{e}\mathit{t}(\lambda I_{n-2}-B_{n-1})}{{\phi }_{n-1}\left({\lambda }_n\right)},$ by the assumption of this theorem, ${\lambda }_n$ is not an eigenvalue of $A_{n-1}$ hence ${\phi }_{n-1}\left({\lambda }_n\right)\ne 0$. By (5(5) $y_j=\frac{{\phi }_{j-1}\left(\lambda \right)y_1-{\chi }_{j-1}\left(\lambda \right)b_n{y}_n}{\prod _{i=1}^{j-1}{b}_i},j=2,\dots ,n-1\mathrm{a}\mathrm{n}\mathrm{d}{\chi }_1\left(\lambda \right)=1.$(5) ), for $j=2,3,\dots ,n-1$, we obtain $b_{j-1}=\frac{{\phi }_{j-1}\left({\lambda }_n\right)x_1-{\chi }_{j-1}\left({\lambda }_n\right)b_n{x}_n}{x_j\prod _{i=1}^{j-2}{b}_i}.$ From (4(4) $b_{i-1}{x}_{i-1}+a_i{x}_i+b_i{x}_{i+1}={\lambda }_n{x}_i\mathrm{f}\mathrm{o}\mathrm{r}i=1,2,\dots ,n,$(4) ), we get $b_{n-2}{x}_{n-2}+(a_{n-1}-{\lambda }_n)x_{n-1}+b_{n-1}{x}_n=0,$ and by a proof similar to the proof of Lemma 3.2 it can be verified that (8) $x_n=\frac{{\phi }_{n-1}\left({\lambda }_n\right)x_1-{\chi }_{n-1}\left({\lambda }_n\right)b_n{x}_n}{\prod _{j=1}^{n-1}{b}_j},$(8) which implies $b_{n-1}=\frac{{\phi }_{n-1}\left({\lambda }_n\right)x_1-{\chi }_{n-1}\left({\lambda }_n\right)b_n{x}_n}{x_n\prod _{j=1}^{n-2}{b}_j}.$ Since $x_j\ne 0,j=2,3,\dots ,n$, it follows ${\phi }_{j-1}\left({\lambda }_n\right)x_1-{\chi }_{j-1}\left({\lambda }_n\right)b_n{x}_n\ne 0$, hence the expression for $b_{j-1},j=2,3,\dots ,n$, is valid and $b_{j-1}\ne 0$.

It is necessary to enforce the condition ${\lambda }_{j-1}\ne {\lambda }_j$, $j=2,3,\dots ,n$, because the leading principal minors of a periodic Jacobi matrix form a Jacobi matrix and by Lemma 2.1, no two successive principal minors of a Jacobi matrix have common eigenvalues, therefore there is no $\lambda \in \mathbb{R}$ that will be a root of ${\phi }_{j-1}\left(\lambda \right)$ and ${\phi }_j\left(\lambda \right)$.

The Algorithm 1 solves the IEPPJ on the inputs $\mathrm{\Lambda }=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\}$, $X_n=\{x_1,x_2,\dots ,x_n\}$ and a real nonzero value c.

Remark

The relations ${\phi }_j\left(\lambda \right),{\chi }_j\left(\lambda \right)$ in Algorithm 1, compute determinant of a symmetric tridiagonal matrix. Ed S.Coakley et al. in [15] presented an algorithm to compute this determinant in $O(n\log n)$, therefore the time complexity of the Algorithm 1 is $O(n^2\log n)$.

4. Numerical examples

In this section, some examples are given to demonstrate the correctness of the method. The numerical computations are performed by using MATLAB R-2014a.

Example 4.1

Given 7 real numbers ${\lambda }_1=3,{\lambda }_2=5,{\lambda }_3=-4,{\lambda }_4=1,{\lambda }_5=13,{\lambda }_6=6,{\lambda }_7=5$ and a real vector $X_7=(5,3,1,9,7,10,-6{)}^{\mathrm{T}}$, find a periodic Jacobi matrix $A_7$ such that ${\lambda }_j$ is an eigenvalue of $A_j$ for each $j=1,2,\dots ,7$ and $({\lambda }_7,X_7)$ is an eigenpair of $A_7$ and $c=6$.

Solution: Using Theorem 3.5, we obtain the following matrix as the solution: $A_7=\left[\begin{array}{ccccccc}3.0000& 15.3333& 0& 0& 0& 0& 6.0000\\ 15.3333& -112.5556& 276.0000& 0& 0& 0& 0\\ 0& 276.0000& -539.9116& -31.4543& 0& 0& 0\\ 0& 0& -31.4543& -3.6822& 15.6563& 0& 0\\ 0& 0& 0& 15.6563& 1.7523& -11.8173& 0\\ 0& 0& 0& 0& -11.8173& 13.2405& -0.0526\\ 6.0000& 0& 0& 0& 0& -0.0526& 9.9123\end{array}\right].$ The eigenvalues of all of the leading principal submatrices are $\begin{array}{rl}& \sigma \left(A_1\right)=\left\{\mathbf{\text{3.0000}}\right\}\\ & \sigma \left(A_2\right)=\{-114.5556,\mathbf{\text{5.0000}}\}\\ & \sigma \left(A_3\right)=\{-675.3485,\mathbf{-}\mathbf{\text{4.0000}},29.8814\}\\ & \sigma \left(A_4\right)=\{-676.5339,-11.6891,\mathbf{\text{1.0000}},34.0736\}\\ & \sigma \left(A_5\right)=\{-676.5345,-19.6720,-3.1896,\mathbf{\text{13.0000}},34.9991\}\\ & \sigma \left(A_6\right)=\{-676.5345,-21.0304,-4.3346,\mathbf{\text{6.0000}},22.4873,35.2557\}\\ & \sigma \left(A_7\right)=\{-676.5345,-21.0636,-5.8089,\mathbf{\text{5.0000}},12.1182,22.6154,35.4293\}\end{array}$ To verify $A_7{X}_7={\lambda }_7{X}_7$, we compute both terms: $A_7{X}_7=(25.0000,15.0000,5.0000,45.0000,35.0000,50.0000,-30.0000{)}^{\mathrm{T}},$ and ${\lambda }_7{X}_7=(25.0000,15.0000,5.0000,45.0000,35.0000,50.0000,-30.0000{)}^{\mathrm{T}}.$

Example 4.2

The solution to the Example 4.1 for c = −2 and $X_7=(0,8,2,-5,1,4,7{)}^{\mathrm{T}}$ is: $A_7=\left[\begin{array}{ccccccc}3.0000& 1.7500& 0& 0& 0& 0& -2.0000\\ 1.7500& 3.4688& 6.1250& 0& 0& 0& 0\\ 0& 6.1250& 1.3356& 8.3342& 0& 0& 0\\ 0& 0& 8.3342& -0.7504& -45.4206& 0& 0\\ 0& 0& 0& -45.4206& -434.6772& 53.1435& 0\\ 0& 0& 0& 0& 53.1435& -5.8521& -1.3908\\ -2.0000& 0& 0& 0& 0& -1.3908& 5.7947\end{array}\right].$ The eigenvalues of all of the leading principal submatrices are $\begin{array}{rl}& \sigma \left(A_1\right)=\left\{\mathbf{\text{3.0000}}\right\}\\ & \sigma \left(A_2\right)=\{1.4688,\mathbf{\text{5.0000}}\}\\ & \sigma \left(A_3\right)=\{-\mathbf{\text{4.0000}},2.8771,8.9272\}\\ & \sigma \left(A_4\right)=\{-9.5241,\mathbf{\text{1.0000}},3.8896,11.6883\}\\ & \sigma \left(A_5\right)=\{-439.3822,-7.8691,1.8719,4.7560,\mathbf{\text{13.0000}}\}\\ & \sigma \left(A_6\right)=\{-445.7354,-8.9183,-1.2984,2.6047,\mathbf{\text{6.0000}},13.8720\}\\ & \sigma \left(A_7\right)=\{-445.7355,-8.9278,-1.5992,2.1255,\mathbf{\text{5.0000}},7.5767,13.8796\}\end{array}$ To verify $A_7{X}_7={\lambda }_7{X}_7$ we compute both terms: $A_7{X}_7=(0,40.0000,10.0000,-25.0000,5.0000,20.0000,35.0000{)}^{\mathrm{T}},$ and ${\lambda }_7{X}_7=(0,40.0000,10.0000,-25.0000,5.0000,20.0000,35.0000{)}^{\mathrm{T}}.$

5. Conclusions

In this work, we have considered a partially described inverse eigenvalue problem for periodic Jacobi matrices. The problem involves the reconstruction of this matrix by one eigenvalue of each of the leading principal submatrices and one eigenpair of the required matrix and one additional piece of information which is the entry $b_n$. The recurrence relation of the leading principal minors and the relation of obtaining the component $x_i$ of the given eigenvector $X_n$ from the entries of leading principal minor $A_{i-1}$, are central in obtaining the solution. To the best of our knowledge, this is the first work studied this problem for a cyclic matrix. In other acyclic matrices such as Jacobi and bordered-diagonal matrices [4,13], the relation $x_i\in X_n$ is a multiple of $x_1$, while in this paper, due to the matrix structure, this relation is linear combination in $x_1$ and $b_n{x}_n$ that created distinct proofs from these papers. It also omits the condition $x_1\ne 0$ which is required in mentioned works. Finally the necessary conditions under which the problem has a unique solution and an algorithm are derived. Some illustrative examples were also given to test the efficiency of the algorithm. As a future work, it would be interesting to study application of this matrix in control theory. Ayatollahi [16] studied an application of Jacobi matrices in control theory, her study is equivalent to an inverse problem of Jacobi matrices. The case of periodic Jacobi matrices is under investigation.

Disclosure statement

No potential conflict of interest was reported by the authors.