Reconstructing real symmetric matrices from eigenvalues of finite dimensional perturbations

Abstract

An algorithm for the reconstruction of a symmetric matrix from the eigenvalues of finite dimensional perturbations is given. The sufficient condition of determining the finite symmetric matrices is derived.

Keywords:

• Reconstruction
• symmetric matrix
• eigenvalue
• perturbation

2010 Mathematics Subject Classification:

• 05C50

1. Introduction

Let $A_0$ be an $n\times n$ real valued symmetric matrix. Let $\{{\lambda }_i^{(0)}{\}}_{i=1}^n$ be the spectrum of $A_0$. Consider the inverse eigenvalue problem of the real valued symmetric matrices A which are finite dimensional perturbations of $A_0$, expressed by (1) $A=A_0+\sum _{i=1}^p{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,p\in {\mathbb{Z}}^{+}$(1) where ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_p$ are unit column vectors in ${\mathbb{R}}^n$ and $a_1,a_2,\dots ,a_p$ are p real numbers. Let $\{{\lambda }_i^{(p)}{\}}_{i=1}^n$ be the spectrum of A. Let $\{{\lambda }_i^{(j)}{\}}_{i=1}^n$ be the spectrum of (2) $A_j=A_0+\sum _{i=1}^j{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T$(2) for $j=1,2,\dots ,p-1$. We intend to reconstruct the matrices A from the spectra $\{{\lambda }_i^{(j)}{\}}_{i=1}^n$,$j=1,2,\dots ,p$.

There are some results about inverse eigenvalue problem for real valued symmetric matrices (see, for example, [1–8] and references therein). Our main motivation comes from the inverse two spectra problem for Jacobi matrices in [4,5]. In [4], Nylen and Uhlig concerned the reconstruction of an $n\times n$ Jacobi matrix T from eigenvalues of $T+E_{11}$ where $E_{jj}$ signifies the matrix with a one in the j, j position and zero everywhere else. In [5], they concerned the reconstruction of an $n\times n$ Jacobi matrix T from eigenvalues of $(I+m{E}_{jj},T+k{E}_{jj})$ which are the roots of the equation $det(t(I+m{E}_{jj})-(T+k{E}_{jj}))=0$. From different perspectives, we study the inverse eigenvalue problem for real valued symmetric matrices and give a resolution of the problems in [4,5]. In the paper, we consider finite dimensional perturbations of real valued symmetric matrices while the analysis in [4,5] was concerntrated on rank one perteubations of Jacobi matrices.

The main result of this paper asserts that given the matrix $A_0$ and p sequences of real numbers $\{{\lambda }_i^{(j)}{\}}_{i=1}^n,j=1,2,\dots ,p$ satisfying the interlacing properties (3) ${\lambda }_i^{(j)}\le {\lambda }_i^{(j+1)}\le {\lambda }_{i+1}^{(j)},i=1,2,\dots ,n-1$(3) or (4) ${\lambda }_i^{(j+1)}\le {\lambda }_i^{(j)}\le {\lambda }_{i+1}^{(j+1)},i=1,2,\dots ,n-1.$(4) Then there exist finite $n\times n$ real valued symmetric matrices A such that the spectrum of A is $\{{\lambda }_i^{(p)}{\}}_{i=1}^n$ and the spectrum of $A_j$ is $\{{\lambda }_i^{(j)}{\}}_{i=1}^n$ for $j=1,2,\dots ,p-1$. The proof of this result is achieved by giving an explicit algorithm for reconstructing all such matrices A.

In Section 3, we apply our results to the inverse eigenvalue problem of Jacobi matrices and resolve the problem in [4,5].

2. The main theorem

To state our main result, we give two propositions and a preliminary theorem.

Proposition 2.1

Each $n\times n$ real valued symmetric matrix $A^{\mathrm{\prime }}$ can be expressed by (5) $A^{\mathrm{\prime }}=\sum _{i=1}^n{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,$(5) where ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_n$ are unit column vectors in ${\mathbb{R}}^n$ and $a_1,a_2,\dots ,a_n$ are n real numbers.

Proof.

By Theorem 2.9 of Chapter 7 in [9], there exists an $n\times n$ real valued invertible matrix X that (6) $A^{\mathrm{\prime }}=X^{T}BX,$(6) where $B=diag(b_1,b_2,\dots ,b_n)$. Let the vector ${\overrightarrow{x}}_i$ be the ith row vector of matrix X. By (6(6) $A^{\mathrm{\prime }}=X^{T}BX,$(6) ), we get $A^{\mathrm{\prime }}=b_1{\overrightarrow{x}}_1^T{\overrightarrow{x}}_1+b_2{\overrightarrow{x}}_2^T{\overrightarrow{x}}_2+\cdots +b_n{\overrightarrow{x}}_n^T{\overrightarrow{x}}_n.$ Let $a_i=b_i\Vert {\overrightarrow{x}}_i{\Vert }^2,{\overrightarrow{w}}_i=\frac{{\overrightarrow{x}}_i^T}{\Vert {\overrightarrow{x}}_i\Vert }$ for $i=1,2,\dots ,n$, then we have (5(5) $A^{\mathrm{\prime }}=\sum _{i=1}^n{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,$(5) ).

Corollary 2.2

Let $A^{\mathrm{\prime }}$ be an $n\times n$ real valued symmetric matrix with rank r, then $A^{\mathrm{\prime }}$ can be expressed by (5(5) $A^{\mathrm{\prime }}=\sum _{i=1}^n{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,$(5) ). If ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_n$ are linearly independent, then the number of nonzero real numbers $a_1,a_2,\dots ,a_n$ is r.

Proof.

According to Proposition 2.1, $A^{\mathrm{\prime }}$ can be expressed by (5(5) $A^{\mathrm{\prime }}=\sum _{i=1}^n{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,$(5) ). Let $X=[{\overrightarrow{w}}_1{\overrightarrow{w}}_2\cdots {\overrightarrow{w}}_n{]}^T$ and $B=diag(a_1,a_2,\dots ,a_n)$. Then (7) $A^{\mathrm{\prime }}=X^{T}BX.$(7) If ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_n$ are linearly independent, then the matrix X is invertible. By Property 2.9 of Chapter 4 in [9], the rank of matrix B is r. That means the number of nonzero real numbers $a_1,a_2,\dots ,a_n$ is r. The proof is complete.

Proposition 2.3

Let $A_0$ be a real valued symmetric matrix, $\overrightarrow{w}$ be a unit column vector in ${\mathbb{R}}^n$ and a be a real number. Assume that two sequences of real numbers $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^n$ arranged in the increasing order, represent the spectra of $A_0$ and (8) $A=A_0+a\overrightarrow{w}{\overrightarrow{w}}^T,$(8) respectively. Then we have (9) ${\lambda }_1\le {\mu }_1\le {\lambda }_2\le {\mu }_2\le \cdots \le {\lambda }_n\le {\mu }_n$(9) or (10) ${\mu }_1\le {\lambda }_1\le {\mu }_2\le {\lambda }_2\le \cdots \le {\mu }_n\le {\lambda }_n.$(10)

Proof.

We firstly find the characteristic polynomial of A. Let $\overrightarrow{x}$ be a nontrivial solution of (11) $Ax=\lambda x,x\in {\mathbb{R}}^n.$(11) We can find the vectors set $\{{\overrightarrow{y}}_i{\}}_{i=1}^n$ such that ${\overrightarrow{y}}_i$ is the eigenvector of $A_0$ corresponding to the eigenvalue ${\lambda }_i$ and $\{{\overrightarrow{y}}_i{\}}_{i=1}^n$ forms a normalised orthonormal basis in ${\mathbb{R}}^n$. Then the vectors $\overrightarrow{x}$ and $\overrightarrow{w}$ can be expressed by  (12) $\overrightarrow{x}=\sum _{i=1}^n{x}_i{\overrightarrow{y}}_i,\overrightarrow{w}=\sum _{i=1}^n{w}_i{\overrightarrow{y}}_i,$(12) where $\{x_i{\}}_{i=1}^n,\{w_i{\}}_{i=1}^n$ are real. Substituting (12(12) $\overrightarrow{x}=\sum _{i=1}^n{x}_i{\overrightarrow{y}}_i,\overrightarrow{w}=\sum _{i=1}^n{w}_i{\overrightarrow{y}}_i,$(12) ) into (11(11) $Ax=\lambda x,x\in {\mathbb{R}}^n.$(11) ) yields $A_0\sum _{i=1}^n{x}_i{\overrightarrow{y}}_i+a\sum _{i=1}^n{w}_i{\overrightarrow{y}}_i\sum _{i=1}^n{w}_i{\overrightarrow{y}}_i^T\sum _{i=1}^n{x}_i{\overrightarrow{y}}_i=\lambda \sum _{i=1}^n{x}_i{\overrightarrow{y}}_i.$ Since eigenvectors ${\overrightarrow{y}}_1,{\overrightarrow{y}}_2,\dots ,{\overrightarrow{y}}_n$ are orthogonal, then $\sum _{i=1}^n(\lambda x_i-{\lambda }_i{x}_i-a{w}_i\sum _{j=1}^n{x}_j{w}_j){\overrightarrow{y}}_i=0,$ that is (13) $\begin{array}{rl}& (\lambda -{\lambda }_1-a{w}_1^2)x_1-a{w}_1\sum _{j\ne 1,j=1}^n{w}_j{x}_j=0,\\ & (\lambda -{\lambda }_2-a{w}_2^2)x_2-a{w}_2\sum _{j\ne 2,j=1}^n{w}_j{x}_j=0,\\ & ⋮\\ & (\lambda -{\lambda }_n-a{w}_n^2)x_n-a{w}_n\sum _{j\ne n,j=1}^n{w}_j{x}_j=0.\end{array}$(13) Then (11(11) $Ax=\lambda x,x\in {\mathbb{R}}^n.$(11) ) has a nontrivial solution if and only if the determinant of the coefficients of the system (13(13) $\begin{array}{rl}& (\lambda -{\lambda }_1-a{w}_1^2)x_1-a{w}_1\sum _{j\ne 1,j=1}^n{w}_j{x}_j=0,\\ & (\lambda -{\lambda }_2-a{w}_2^2)x_2-a{w}_2\sum _{j\ne 2,j=1}^n{w}_j{x}_j=0,\\ & ⋮\\ & (\lambda -{\lambda }_n-a{w}_n^2)x_n-a{w}_n\sum _{j\ne n,j=1}^n{w}_j{x}_j=0.\end{array}$(13) ) is zero. Denote (14) $Q\left(\lambda \right)=\left|\begin{array}{cccc}\lambda -{\lambda }_1-a{w}_1^2& -a{w}_1{w}_2& \cdots & -a{w}_1{w}_n\\ -a{w}_1{w}_2& \lambda -{\lambda }_2-a{w}_2^2& \cdots & -a{w}_2{w}_n\\ ⋮& ⋮& \ddots & ⋮\\ -a{w}_1{w}_n& -a{w}_2{w}_n& \cdots & \lambda -{\lambda }_n-a{w}_n^2\end{array}\right|,$(14) then $Q(\lambda )$ is the characteristic polynomial of matrix A and the zeros set of $Q(\lambda )$ is $\{{\mu }_i{\}}_{i=1}^n$. So $Q(\lambda )$ can also be expressed by $Q\left(\lambda \right)=\prod _{i=1}^n(\lambda -{\mu }_i).$ Let $P(\lambda )=\prod _{i=1}^n(\lambda -{\lambda }_i)$. By (14(14) $Q\left(\lambda \right)=\left|\begin{array}{cccc}\lambda -{\lambda }_1-a{w}_1^2& -a{w}_1{w}_2& \cdots & -a{w}_1{w}_n\\ -a{w}_1{w}_2& \lambda -{\lambda }_2-a{w}_2^2& \cdots & -a{w}_2{w}_n\\ ⋮& ⋮& \ddots & ⋮\\ -a{w}_1{w}_n& -a{w}_2{w}_n& \cdots & \lambda -{\lambda }_n-a{w}_n^2\end{array}\right|,$(14) ), $Q\left(\lambda \right)=P(\lambda )-a\sum _{i=1}^n{w}_i^2\prod _{j=1,j\ne i}^n(\lambda -{\lambda }_j).$ Hence (15) $\frac{Q\left(\lambda \right)}{P\left(\lambda \right)}=1-a\sum _{i=1}^n{w}_i^2\frac{1}{\lambda -{\lambda }_i}$(15) for $\lambda \in \mathbb{C}\mathrm{\setminus }\{{\lambda }_i{\}}_{i=1}^n$. We will prove the conclusion in two cases:

1. If a is a positive real number, by the result of the classically known fact [1], the coefficients of the partial fraction decomposition of $Q(\lambda )/P(\lambda )$ are nonpositive if and only if the roots of $Q(\lambda )$ interlace those of $P(\lambda )$. Thus (9(9) ${\lambda }_1\le {\mu }_1\le {\lambda }_2\le {\mu }_2\le \cdots \le {\lambda }_n\le {\mu }_n$(9) ) holds.

2. If a is a negative real number, then the coefficients of the partial fraction decomposition of $Q(\lambda )/P(\lambda )$ are nonnegative. By [1], (10(10) ${\mu }_1\le {\lambda }_1\le {\mu }_2\le {\lambda }_2\le \cdots \le {\mu }_n\le {\lambda }_n.$(10) ) holds. It finishs the proof.

Corollary 2.4

Let $A_0$ be a real valued symmetric matrix, a be a real number and $\overrightarrow{w}$ be a column vector in ${\mathbb{R}}^n$. Assume that two sequences of real numbers $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^n$ arranged in the increasing order, represent the spectra of $A_0$ and A definded by (8(8) $A=A_0+a\overrightarrow{w}{\overrightarrow{w}}^T,$(8) ) respectively. If $\overrightarrow{w}$ is the eigenvector of $A_0$ corresponding to ${\lambda }_j,j\in \{1,2,\dots ,n\},$ then we have ${\left\{{\mu }_i\right\}}_{i=1}^n={\left\{{\lambda }_i\right\}}_{i=1}^{j-1}\cup {\left\{{\lambda }_i\right\}}_{i=j+1}^n\cup \{{\lambda }_j+a\}.$

Proof.

Let ${\overrightarrow{y}}_i$ be the eigenvector of $A_0$ corresponding to ${\lambda }_i,i=1,2,\dots ,n,i\ne j$. Noting that $A\overrightarrow{w}=(A_0+a\overrightarrow{w}{\overrightarrow{w}}^T)\overrightarrow{w}=({\lambda }_j+a)\overrightarrow{w}$ and $A{\overrightarrow{y}}_i=(A_0+a\overrightarrow{w}{\overrightarrow{w}}^T){\overrightarrow{y}}_i={\lambda }_i{\overrightarrow{y}}_i,i\ne j,$ the proof is complete.

Theorem 2.5

Let $A_0$ be an $n\times n$ real valued symmetric matrix and denote its spectrum by $\{{\lambda }_i{\}}_{i=1}^n$. Suppose that $\{{\mu }_i{\}}_{i=1}^n$ is a sequence of real numbers satisfying (9(9) ${\lambda }_1\le {\mu }_1\le {\lambda }_2\le {\mu }_2\le \cdots \le {\lambda }_n\le {\mu }_n$(9) ) or (10(10) ${\mu }_1\le {\lambda }_1\le {\mu }_2\le {\lambda }_2\le \cdots \le {\mu }_n\le {\lambda }_n.$(10) ). Then there exists a unit column vector $\overrightarrow{w}$ in ${\mathbb{R}}^n$ and a real number a such that the spectrum of A defined by (8(8) $A=A_0+a\overrightarrow{w}{\overrightarrow{w}}^T,$(8) ) is $\{{\mu }_i{\}}_{i=1}^n$.

Let $S\left(\lambda \right)=\prod _{i=1}^n(\lambda -{\mu }_i),P\left(\lambda \right)=\prod _{i=1}^n(\lambda -{\lambda }_i).$ Denote by $R(\lambda )$ the greatest common divisor of $S(\lambda )$ and $P(\lambda )$. Let $S\left(\lambda \right)=R\left(\lambda \right)s\left(\lambda \right),P\left(\lambda \right)=R\left(\lambda \right)p\left(\lambda \right),$ where $s\left(\lambda \right)=\prod _{i=1}^k(\lambda -s_i),p\left(\lambda \right)=\prod _{i=1}^k(\lambda -p_i).$ It is possible to arrange the roots of $s(\lambda )$ and $p(\lambda )$ such that (16) $p_1<s_1<p_2<s_2<\cdots <p_k<s_k$(16) or (17) $s_1<p_1<s_2<p_2<\cdots <s_k<p_k.$(17) Moreover the number of A is $2^k$ at most.

Proof

Proof of Theorem 2.5

We consider two cases.

(i) Suppose that (9(9) ${\lambda }_1\le {\mu }_1\le {\lambda }_2\le {\mu }_2\le \cdots \le {\lambda }_n\le {\mu }_n$(9) ) holds. Denote by I the following subset of $\{1,2,\dots ,n\}$:  $I=\{i|{\lambda }_i=p_j,1\le j\le k\}.$ By (16(16) $p_1<s_1<p_2<s_2<\cdots <p_k<s_k$(16) ), $s(\lambda )/p(\lambda )$ is a Herglotz function. From the Herglotz representation theorem in [3, P4 ], $s(\lambda )/p(\lambda )$ can be expressed by (18) $\frac{s\left(\lambda \right)}{p\left(\lambda \right)}=1-\sum _{i\in I}\frac{b_i}{\lambda -{\lambda }_i},$(18) where $b_i=s({\nu }_i)/\dot{p}({\nu }_i)>0,\dot{p}(\lambda )=\mathrm{d}p(\lambda )/\mathrm{d}\lambda$. We can find the vectors set $\{{\overrightarrow{y}}_i{\}}_{i=1}^n$ such that ${\overrightarrow{y}}_i$ is the eigenvector of $A_0$ corresponding to the eigenvalue ${\lambda }_i$ and $\{{\overrightarrow{y}}_i{\}}_{i=1}^n$ forms a normalised orthonormal basis in ${\mathbb{R}}^n$. Let (19) $a=\sqrt{\sum _{i\in I}|b_i|}$(19) and (20) $\overrightarrow{w}=\sum _{i\in I}{w}_i{\overrightarrow{y}}_i,$(20) where $w_i^2=|b_i|/a$. By (18(18) $\frac{s\left(\lambda \right)}{p\left(\lambda \right)}=1-\sum _{i\in I}\frac{b_i}{\lambda -{\lambda }_i},$(18) ), (21) $\frac{s\left(\lambda \right)}{p\left(\lambda \right)}=1-a\sum _{i\in I}{w}_i^2\frac{1}{\lambda -{\lambda }_i}.$(21) Next we claim that the spectrum of the matrix A defined by (8(8) $A=A_0+a\overrightarrow{w}{\overrightarrow{w}}^T,$(8) ) is  $\{{\mu }_i{\}}_{i=1}^n$. In fact, define $Q\left(\lambda \right)=P\left(\lambda \right)(1-a\sum _{i\in I}{w}_i^2\frac{1}{\lambda -{\lambda }_i}).$ According to (15(15) $\frac{Q\left(\lambda \right)}{P\left(\lambda \right)}=1-a\sum _{i=1}^n{w}_i^2\frac{1}{\lambda -{\lambda }_i}$(15) ), it follows that $Q(\lambda )$ is the characteristic polynomial of A. Comparing with (21(21) $\frac{s\left(\lambda \right)}{p\left(\lambda \right)}=1-a\sum _{i\in I}{w}_i^2\frac{1}{\lambda -{\lambda }_i}.$(21) ), we obtain $Q\left(\lambda \right)=P\left(\lambda \right)\frac{s\left(\lambda \right)}{p(\lambda )},$ i.e. $Q\left(\lambda \right)=S\left(\lambda \right).$ This proves that the spectrum of A is $\{{\mu }_i{\}}_{i=1}^n$. Moreover by (20(20) $\overrightarrow{w}=\sum _{i\in I}{w}_i{\overrightarrow{y}}_i,$(20) ) the number of the vectors $\overrightarrow{w}$ is $2^k$ at most.

(ii) Suppose that (10(10) ${\mu }_1\le {\lambda }_1\le {\mu }_2\le {\lambda }_2\le \cdots \le {\mu }_n\le {\lambda }_n.$(10) ) holds, then we have (18(18) $\frac{s\left(\lambda \right)}{p\left(\lambda \right)}=1-\sum _{i\in I}\frac{b_i}{\lambda -{\lambda }_i},$(18) ) with $b_i=s({\nu }_i)/\dot{p}({\nu }_i)<0,i\in I$. Let the vector $\overrightarrow{w}$ and the real number a be defined by (20(20) $\overrightarrow{w}=\sum _{i\in I}{w}_i{\overrightarrow{y}}_i,$(20) ) and (19(19) $a=\sqrt{\sum _{i\in I}|b_i|}$(19) ). By the proof of case (i), we see that the spectrum of A is $\{{\mu }_i{\}}_{i=1}^n$ and the number of the vectors $\overrightarrow{w}$ is $2^k$ at most. Then the proof is complete.

Now we state our main result.

Theorem 2.6

Let $A_0$ be an $n\times n$ real valued symmetric matrix and denote its spectrum by $\{{\lambda }_i^{(0)}{\}}_{i=1}^n$. Suppose that $\{{\lambda }_i^{(j)}{\}}_{i=1}^n,j=1,2,\dots ,p$ are p sequences of real numbers satisfying (3(3) ${\lambda }_i^{(j)}\le {\lambda }_i^{(j+1)}\le {\lambda }_{i+1}^{(j)},i=1,2,\dots ,n-1$(3) ) or (4(4) ${\lambda }_i^{(j+1)}\le {\lambda }_i^{(j)}\le {\lambda }_{i+1}^{(j+1)},i=1,2,\dots ,n-1.$(4) ). Then there exist finitely many matrices A defined by (1(1) $A=A_0+\sum _{i=1}^p{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T,p\in {\mathbb{Z}}^{+}$(1) ) such that the spectrum of A is $\{{\lambda }_i^{(p)}{\}}_{i=1}^n$ and the spectrum of $A_j$ defined by (2(2) $A_j=A_0+\sum _{i=1}^j{a}_i{\overrightarrow{w}}_i{\overrightarrow{w}}_i^T$(2) ) is $\{{\lambda }_i^{(j)}{\}}_{i=1}^n,j=1,2,\dots ,p-1$.

Proof.

We prove our theorem by the induction. Denote by $\sigma (A_j)$ the spectrum of $A_j$,$j=1,2,\dots ,p-1$.

For k = 1, by Theorem 2.5 there exists the vector ${\overrightarrow{w}}_1$ and the real number $a_1$ such that $\sigma (A_1)=\{{\lambda }_i^{(1)}{\}}_{i=1}^n$ and the number of $A_1$ is finite.

For k>1, if there exist k−1 vectors ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_{k-1}$ and k−1 real numbers $a_1,a_2,\dots ,a_{k-1}$ such that $\sigma (A_j)=\{{\lambda }_i^{(j)}{\}}_{i=1}^n$ for $j=1,2,\dots ,k-1$, then by Theorem 2.5 there exists the vector ${\overrightarrow{w}}_k$ and the real number $a_k$ such that $\sigma (A_k)=\{{\lambda }_i^{(k)}{\}}_{i=1}^n$ and the number of $A_k$ is finite. That is, there exists k vectors ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_k$ and k real numbers $a_1,a_2,\dots ,a_k$ such that $\sigma (A_j)=\{{\lambda }_i^{(j)}{\}}_{i=1}^n$ for $j=1,2,\dots ,k$ and the number of $A_k$ is finite.

From above, there exist finitely many matrices A such that the eigenvalues sets of A is $\{{\lambda }_i^{(p)}{\}}_{i=1}^n$ and the spectrum of $A_j$ is $\{{\lambda }_i^{(j)}{\}}_{i=1}^n$ for $j=1,2,\dots ,p-1$. The proof is ended.

Remark 2.7

By Proposition 2.1, we know that all matrices A reconstructed in Theorem 2.6 have been found do not satisfying any other relations except the interlacing property (3(3) ${\lambda }_i^{(j)}\le {\lambda }_i^{(j+1)}\le {\lambda }_{i+1}^{(j)},i=1,2,\dots ,n-1$(3) ) or (4(4) ${\lambda }_i^{(j+1)}\le {\lambda }_i^{(j)}\le {\lambda }_{i+1}^{(j+1)},i=1,2,\dots ,n-1.$(4) ).

3. Application

We can apply our results to the inverse eigenvalue problem of Jacobi matrices defined by (22) $A_0=\left[\begin{array}{ccccc}{\alpha }_1& {\beta }_1& & & \\ {\beta }_1& {\alpha }_2& {\beta }_2& & \\ & {\beta }_2& \ddots & \ddots & \\ & & \ddots & {\alpha }_{n-1}& {\beta }_{n-1}\\ & & & {\beta }_{n-1}& {\alpha }_n\end{array}\right],$(22) where ${\alpha }_i\in {\mathbb{R}}^{+},{\beta }_j\in {\mathbb{R}}^{-}$. Let $\overrightarrow{w}$ be a unit column vector in ${\mathbb{R}}^n$ and a be a positive real number. Let $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^n$ be the spectra of $A_0$ and $A=A_0+a\overrightarrow{w}\overrightarrow{w}{}^T$ represently. Then the following inverse eigenvalue problem is stated: reconstruct the Jacobi matrix $A_0$ and the positive real number a from the spectra $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^n$ and the vector $\overrightarrow{w}$.

We first mention some preliminaries which will be needed subsequently. Let ${\mathrm{\Lambda }}_0=\left[\begin{array}{cccc}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_n\end{array}\right],$ where $\{{\lambda }_i{\}}_{i=1}^n$ is the spectrum of $A_0$. By Theorem 2.5, there exists the column vector ${\overrightarrow{w}}_0=(w_i{)}_{i=1}^n$, where (23) $w_i^2=\frac{\prod _{j=1}^n({\lambda }_i-{\mu }_j)}{\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j)},$(23) such that the spectrum of $({\mathrm{\Lambda }}_0+{\overrightarrow{w}}_0{\overrightarrow{w}}_0{}^T)$ is $\{{\mu }_i{\}}_{i=1}^n$.

For fundamental aspects of spectral theory for Jacobi matrices, we refer to [10]. Let $\overrightarrow{\phi }(\lambda )$ be a nontrivial solution of (24) $A_{0}x=\lambda x,x\in {\mathbb{R}}^n.$(24) Noting that $\overrightarrow{\phi }(\lambda )$ can be written as $\overrightarrow{\phi }(\lambda )=(P_1(\lambda ),P_2(\lambda ),\dots ,P_n(\lambda ){)}^T$ where $P_i(\lambda )$ is a real valued function of λ. By (24(24) $A_{0}x=\lambda x,x\in {\mathbb{R}}^n.$(24) ), we get $\begin{array}{rl}& P_1(\lambda ){\alpha }_1+{\beta }_1{P}_2(\lambda )=\lambda P_1(\lambda ),\\ & {\beta }_{i-1}{P}_{i-1}(\lambda )+{\alpha }_i{P}_i(\lambda )+{\beta }_i{P}_{i+1}(\lambda )=\lambda P_i(\lambda ),i=2,3,\dots ,n-1,\\ & {\beta }_{n-1}{P}_{n-1}(\lambda )+{\alpha }_n{P}_n(\lambda )=\lambda P_n(\lambda ).\end{array}$ For $i\le j$, the matrix $A_{ij}$ denotes the $(j-i+1)\times (j-i+1)$ submatrix of A contained in the rows and columns of A indexed by $i,i+1,\dots ,j$. Then (25) $P_i(\lambda )=P_1(\lambda ){\theta }_{1,i-1}{A}_{1,i-1}(\lambda ),i=2,3,\dots ,n$(25) where $A_{ij}(\lambda )$ is the determinant of $(\lambda E-A_{ij})$ and ${\theta }_{ij}=(\prod _{k=i}^j{\beta }_k{)}^{-1}$.

Let (26) $V=(\overrightarrow{\phi }({\lambda }_1),\overrightarrow{\phi }({\lambda }_2),\dots ,\overrightarrow{\phi }({\lambda }_n))$(26) and $\Vert \overrightarrow{\phi }({\lambda }_i)\Vert =1$. Then V is an orthogonal matrix and $A_0=V{\mathrm{\Lambda }}_0{V}^T$. If  (27) $V{\overrightarrow{w}}_0=\sqrt{a}\overrightarrow{w},$(27) then the matrix $({\mathrm{\Lambda }}_0+{\overrightarrow{w}}_0{\overrightarrow{w}}_0{}^T)$ is similar to $(A_0+a\overrightarrow{w}{\overrightarrow{w}}^T)$ and the spectrum of$(A_0+a\overrightarrow{w}{\overrightarrow{w}}^T)$ is $\{{\mu }_i{\}}_{i=1}^n$.

Let $\{{\overrightarrow{e}}_i{\}}_{i=1}^n$ be the standard basis in ${\mathbb{R}}^n$. Next we will treat two cases:

1. $\overrightarrow{w}={\overrightarrow{e}}_n$;

2. $\overrightarrow{w}={\overrightarrow{e}}_l,0<l<n$.

Case (i). Let $\{{\lambda }_i{\}}_{i=1}^n$ and $\{{\mu }_i{\}}_{i=1}^n$ be two sequences of real numbers satisfying (28) ${\lambda }_1<{\mu }_1<{\lambda }_2<{\mu }_2<\cdots <{\lambda }_n<{\mu }_n.$(28) Then we can reconstruct the unique Jacobi matrix $A_0$ and the positive real number a such that the spectrum of $A_0$ is $\{{\lambda }_i{\}}_{i=1}^n$ and the spectrum of A is $\{{\mu }_i{\}}_{i=1}^n$, where $A=A_0+a{\overrightarrow{e}}_n{\overrightarrow{e}}_n^T$.

Calculate $w_i,i=1,2,\dots ,n$ by (23(23) $w_i^2=\frac{\prod _{j=1}^n({\lambda }_i-{\mu }_j)}{\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j)},$(23) ). By Theorem 2 in [5], there exists a uniqe Jacobi matrix $A_0$. Suppose that the orthogonal matrix V defined by (26(26) $V=(\overrightarrow{\phi }({\lambda }_1),\overrightarrow{\phi }({\lambda }_2),\dots ,\overrightarrow{\phi }({\lambda }_n))$(26) ) satisfies (27(27) $V{\overrightarrow{w}}_0=\sqrt{a}\overrightarrow{w},$(27) ) i.e. (29) $\begin{array}{rl}& \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_n{P}_n({\lambda }_i)=\sqrt{a}.\end{array}$(29) Due to $V^{T}V=I$, it follows that $\sqrt{a}{V}^T{\overrightarrow{e}}_n={\overrightarrow{w}}_0$, i.e. (30) $\begin{array}{rl}& \sqrt{a}{P}_n({\lambda }_1)=w_1,\\ & \sqrt{a}{P}_n({\lambda }_2)=w_2,\\ & ⋮\\ & \sqrt{a}{P}_n({\lambda }_n)=w_n.\end{array}$(30) According to (25(25) $P_i(\lambda )=P_1(\lambda ){\theta }_{1,i-1}{A}_{1,i-1}(\lambda ),i=2,3,\dots ,n$(25) ), (31) $\begin{array}{rl}& P_1(\lambda )=P_n(\lambda )\frac{1}{{\theta }_{1,n-1}{A}_{1,n-1}(\lambda )},\\ & P_i(\lambda )=P_n(\lambda )\frac{{\theta }_{1,i-1}{A}_{1,i-1}(\lambda )}{{\theta }_{1,n-1}{A}_{1,n-1}(\lambda )},i=2,3,\dots ,n.\end{array}$(31) Putting (30(30) $\begin{array}{rl}& \sqrt{a}{P}_n({\lambda }_1)=w_1,\\ & \sqrt{a}{P}_n({\lambda }_2)=w_2,\\ & ⋮\\ & \sqrt{a}{P}_n({\lambda }_n)=w_n.\end{array}$(30) ) and (31(31) $\begin{array}{rl}& P_1(\lambda )=P_n(\lambda )\frac{1}{{\theta }_{1,n-1}{A}_{1,n-1}(\lambda )},\\ & P_i(\lambda )=P_n(\lambda )\frac{{\theta }_{1,i-1}{A}_{1,i-1}(\lambda )}{{\theta }_{1,n-1}{A}_{1,n-1}(\lambda )},i=2,3,\dots ,n.\end{array}$(31) ) into (29(29) $\begin{array}{rl}& \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_n{P}_n({\lambda }_i)=\sqrt{a}.\end{array}$(29) ), we get  $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{A_{1,j-1}({\lambda }_i)}{A_{1,n-1}({\lambda }_i)}=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_i^2=a,\end{array}$ that is (32) $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,n-1}({\lambda }_i)}=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{n-1}}{A_{1,n-1}({\lambda }_i)}=a.\end{array}$(32) By (28(28) ${\lambda }_1<{\mu }_1<{\lambda }_2<{\mu }_2<\cdots <{\lambda }_n<{\mu }_n.$(28) ), (32(32) $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,n-1}({\lambda }_i)}=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{n-1}}{A_{1,n-1}({\lambda }_i)}=a.\end{array}$(32) ) has a unique solution (33) $\begin{array}{rl}& a=\sum _{i=1}^n{w}_i^2,\\ & A_{1,n-1}({\lambda }_i)=\frac{w_i^2}{\sum _{j=1}^n{w}_j^2}\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j),i=1,2,\dots ,n.\end{array}$(33) Let $A_{1,n-1}(\lambda )=p_{n-1}{\lambda }^{n-1}+p_{n-2}{\lambda }^{n-2}+\cdots +p_0.$ Therefore, it is possible to determine $A_{1,n-1}(\lambda )$ by solving the Vandermonde system (34) $\left[\begin{array}{ccccc}{\lambda }_1^{n-1}& {\lambda }_1^{n-2}& \cdots & {\lambda }_1& 1\\ {\lambda }_2^{n-1}& {\lambda }_2^{n-2}& \cdots & {\lambda }_2& 1\\ ⋮& ⋮& \cdots & ⋮& ⋮\\ {\lambda }_n^{n-1}& {\lambda }_n^{n-2}& \cdots & {\lambda }_n& 1\end{array}\right]\left[\begin{array}{c}{p}_{n-1}\\ p_{n-2}\\ ⋮\\ p_0\end{array}\right]=\left[\begin{array}{c}{A}_{1,n-1}({\lambda }_1)\\ A_{1,n-1}({\lambda }_2)\\ ⋮\\ A_{1,n-1}({\lambda }_n)\end{array}\right].$(34) By $A_{1,n-1}({\lambda }_i)A_{1,n-1}({\lambda }_{i+1})<0$ and $A_{1,n-1}(\lambda )$ is a continuous function in $\mathbb{R}$, there is at least one zero in $({\lambda }_i,{\lambda }_{i+1})$. However, $A_{1,n-1}(\lambda )$ has only n−1 zeros. Hence it necessarily follows that the zeros set of $A_{1,n-1}(\lambda )$ interlaces $\{{\lambda }_i{\}}_{i=1}^n$. By Theorem 1 and Remark in [4, p414], we can exactly reconstruct the uniqe matrix $A_0$ from $\{{\lambda }_i{\}}_{i=1}^n$ and the zeros of $A_{1,n-1}(\lambda )$.

Example 3.1

Find all $3\times 3$ Jacobi matrices $A_0$ and the positive number a such that the spectrum of$A_0$ is $\{1,3,5\}$ and the spectrum of A is $\{2,4,6\}$ where $A=A_0+a{\overrightarrow{e}}_3{\overrightarrow{e}}_3^T$ and ${\overrightarrow{e}}_3^T=(0,0,1)$.

The explicit algorithm is:

1. Calculate $w_1,w_2,w_3$ by (23(23) $w_i^2=\frac{\prod _{j=1}^n({\lambda }_i-{\mu }_j)}{\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j)},$(23) );

2. Let $A_{1,2}(\lambda )=p_2{\lambda }^2+p_1\lambda +p_0$;

3. Calculate $A_{1,2}(1),A_{1,2}(3),A_{1,2}(5)$ and a by (33(33) $\begin{array}{rl}& a=\sum _{i=1}^n{w}_i^2,\\ & A_{1,n-1}({\lambda }_i)=\frac{w_i^2}{\sum _{j=1}^n{w}_j^2}\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j),i=1,2,\dots ,n.\end{array}$(33) );

4. Calculate $p_0,p_1,p_2$ by (34(34) $\left[\begin{array}{ccccc}{\lambda }_1^{n-1}& {\lambda }_1^{n-2}& \cdots & {\lambda }_1& 1\\ {\lambda }_2^{n-1}& {\lambda }_2^{n-2}& \cdots & {\lambda }_2& 1\\ ⋮& ⋮& \cdots & ⋮& ⋮\\ {\lambda }_n^{n-1}& {\lambda }_n^{n-2}& \cdots & {\lambda }_n& 1\end{array}\right]\left[\begin{array}{c}{p}_{n-1}\\ p_{n-2}\\ ⋮\\ p_0\end{array}\right]=\left[\begin{array}{c}{A}_{1,n-1}({\lambda }_1)\\ A_{1,n-1}({\lambda }_2)\\ ⋮\\ A_{1,n-1}({\lambda }_n)\end{array}\right].$(34) );

5. Calculate the zeros of $A_{1,2}(\lambda )$, denoted by $\{t_1,t_2\}$;

6. Reconstruct the uniqe matrix $A_0$ from $\{1,3,5\}$ and $\{t_1,t_2\}$ by $\frac{(\lambda -1)(\lambda -3)(\lambda -5)}{(\lambda -t_1)(\lambda -t_2)}=\lambda -{\alpha }_3-\frac{{\beta }_2^2}{\lambda -{\alpha }_2-{\beta }_1^2/(\lambda -{\alpha }_1)}.$

It yiels the unique matrix $A_0$, i.e. $\left[\begin{array}{ccc}3.499994164804373& -1.118031999993422& 0\\ -1.118031999993422& 3.500001835195626& -1.414217255927815\\ 0& -1.414217255927815& 2.000004000000000\end{array}\right].$ This computation was performed using MATLAB.

Case (ii). Let $\{{\lambda }_i{\}}_{i=1}^n$ and $\{{\mu }_i{\}}_{i=1}^n$ be two sequences of real numbers satisfying (28(28) ${\lambda }_1<{\mu }_1<{\lambda }_2<{\mu }_2<\cdots <{\lambda }_n<{\mu }_n.$(28) ). Then we can reconstruct the Jacobi matrix $A_0$ and a positive real number a such that the spectrum of $A_0$ is $\{{\lambda }_i{\}}_{i=1}^n$ and the spectrum of A is $\{{\mu }_i{\}}_{i=1}^n$, where $A=A_0+a{\overrightarrow{e}}_l{\overrightarrow{e}}_l^T,0<l<n$.

Calculate $w_i,i=1,2,\dots ,n$ by (23(23) $w_i^2=\frac{\prod _{j=1}^n({\lambda }_i-{\mu }_j)}{\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j)},$(23) ). By Theorem 2 in [5], there exist finitely many $(\genfrac{}{}{0ex}{}{n-1}{l-1})$ Jacobi matrices $A_0$. Suppose that the orthogonal matrix V defined by (26(26) $V=(\overrightarrow{\phi }({\lambda }_1),\overrightarrow{\phi }({\lambda }_2),\dots ,\overrightarrow{\phi }({\lambda }_n))$(26) ), satisfies (27(27) $V{\overrightarrow{w}}_0=\sqrt{a}\overrightarrow{w},$(27) ) i.e. (35) $\begin{array}{rl}& \sum _{i=1}^n{w}_i{P}_1({\lambda }_i)=0,\\ & \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=2,3,\dots ,l-1,\\ & \sum _{i=1}^n{w}_i{P}_l({\lambda }_i)=\sqrt{a},\\ & \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=l+1,j+2,\dots ,n.\end{array}$(35) By $V^{T}V=I$, it follows that $\sqrt{a}{V}^T{\overrightarrow{e}}_l={\overrightarrow{w}}_0$ i.e. (36) $\begin{array}{rl}& \sqrt{a}{P}_l({\lambda }_1)=w_1,\\ & \sqrt{a}{P}_l({\lambda }_2)=w_2,\\ & ⋮\\ & \sqrt{a}{P}_l({\lambda }_n)=w_n.\end{array}$(36) According to (25(25) $P_i(\lambda )=P_1(\lambda ){\theta }_{1,i-1}{A}_{1,i-1}(\lambda ),i=2,3,\dots ,n$(25) ), (37) $\begin{array}{rl}& P_1(\lambda )=P_l(\lambda )\frac{1}{{\theta }_{1,l-1}{A}_{1,l-1}(\lambda )},\\ & P_i(\lambda )=P_l(\lambda )\frac{{\theta }_{1,i-1}{A}_{1,i-1}(\lambda )}{{\theta }_{1,l-1}{A}_{1,l-1}(\lambda )},i=2,3,\dots ,n.\end{array}$(37) Putting (36(36) $\begin{array}{rl}& \sqrt{a}{P}_l({\lambda }_1)=w_1,\\ & \sqrt{a}{P}_l({\lambda }_2)=w_2,\\ & ⋮\\ & \sqrt{a}{P}_l({\lambda }_n)=w_n.\end{array}$(36) ) and (37(37) $\begin{array}{rl}& P_1(\lambda )=P_l(\lambda )\frac{1}{{\theta }_{1,l-1}{A}_{1,l-1}(\lambda )},\\ & P_i(\lambda )=P_l(\lambda )\frac{{\theta }_{1,i-1}{A}_{1,i-1}(\lambda )}{{\theta }_{1,l-1}{A}_{1,l-1}(\lambda )},i=2,3,\dots ,n.\end{array}$(37) ) into (35(35) $\begin{array}{rl}& \sum _{i=1}^n{w}_i{P}_1({\lambda }_i)=0,\\ & \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=2,3,\dots ,l-1,\\ & \sum _{i=1}^n{w}_i{P}_l({\lambda }_i)=\sqrt{a},\\ & \sum _{i=1}^n{w}_i{P}_j({\lambda }_i)=0,j=l+1,j+2,\dots ,n.\end{array}$(35) ), we have $\begin{array}{rl}& \sum _{i=1}^n{w}_j^2\frac{1}{A_{1,l-1}({\lambda }_i)}=0,\\ & \sum _{i=1}^n{w}_i^2\frac{A_{1,j-1}({\lambda }_i)}{A_{1,l-1}({\lambda }_i)}=0,j=2,3,\dots ,l-1,\\ & \sum _{i=1}^n{w}_i^2=a,\\ & \sum _{i=1}^n{w}_i^2\frac{A_{1,j-1}({\lambda }_i)}{A_{1,l-1}({\lambda }_i)}=0,j=l+1,l+2,\dots ,n,\end{array}$ that is (38) $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,l-1}({\lambda }_i)}=0,j=1,2,\dots ,l-1,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{l-1}}{A_{1,l-1}({\lambda }_i)}=a,\\ & \sum _{i=1}^n{w}_i^2\frac{A_{1,l}({\lambda }_i){\lambda }_i^{j-l-1}}{A_{1,l-1}({\lambda }_i)}=0,j=l+1,l+2,\dots ,n.\end{array}$(38) Combining with $A_{1,l}({\lambda }_i)A_{l+1,n}({\lambda }_i)-{\beta }_l^2=0$, we obtain $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,l-1}({\lambda }_i)}=0,j=1,2,\dots ,l-1,\\ & \sum _{i=1}^n{w}_i^2=a,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-l-1}}{A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)}=0,j=l+1,l+2,\dots ,n.\end{array}$ Then, it is easily shown that (39) $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)}=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{n-1}}{A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)}=a.\end{array}$(39) By (28(28) ${\lambda }_1<{\mu }_1<{\lambda }_2<{\mu }_2<\cdots <{\lambda }_n<{\mu }_n.$(28) ), (39(39) $\begin{array}{rl}& \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{j-1}}{A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)}=0,j=1,2,\dots ,n-1,\\ & \sum _{i=1}^n{w}_i^2\frac{{\lambda }_i^{n-1}}{A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)}=a.\end{array}$(39) ) has a unique solution (40) $\begin{array}{rl}& a=\sum _{i=1}^n{w}_i^2,\\ & A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)=\frac{w_i^2}{\sum _{j=1}^n{w}_j^2}\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j),i=1,2,\dots ,n.\end{array}$(40) Let $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )=p_{n-1}{\lambda }^{n-1}+p_{n-2}{\lambda }^{n-2}+\cdots +p_0.$ Therefore, it is possible to determine $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )$ by solving the Vandermonde system (41) $\left[\begin{array}{ccccc}{\lambda }_1^{n-1}& {\lambda }_1^{n-2}& \cdots & {\lambda }_1& 1\\ {\lambda }_2^{n-1}& {\lambda }_2^{n-2}& \cdots & {\lambda }_2& 1\\ ⋮& ⋮& \cdots & ⋮& ⋮\\ {\lambda }_n^{n-1}& {\lambda }_n^{n-2}& \cdots & {\lambda }_n& 1\end{array}\right]\left[\begin{array}{c}{p}_{n-1}\\ p_{n-2}\\ ⋮\\ p_0\end{array}\right]=\left[\begin{array}{c}{A}_{1,l-1}({\lambda }_1)A_{l+1,n}({\lambda }_1)\\ A_{1,l-1}({\lambda }_2)A_{l+1,n}({\lambda }_2)\\ ⋮\\ A_{1,l-1}({\lambda }_n)A_{l+1,n}({\lambda }_n)\end{array}\right].$(41) By $A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)A_{1,l-1}({\lambda }_{i+1})A_{l+1,n}({\lambda }_{i+1})<0$ and $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )$ is a continuous function in $\mathbb{R}$, there is at least one zero in $({\lambda }_i,{\lambda }_{i+1})$. However, $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )$ has only n−1 zeros. Hence it necessarily follows that the zeros set of $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )$ interlaces $\{{\lambda }_i{\}}_{i=1}^n$. By Theorem 3 in [4], we can exactly reconstruct the finitely many $(\genfrac{}{}{0ex}{}{n-1}{l-1})$ matrices ${\tilde{A}}_0=A_0$ from $\{{\lambda }_i{\}}_{i=1}^n$ and the zeros of $A_{1,l-1}(\lambda )A_{l+1,n}(\lambda )$.

Example 3.2

Find all $3\times 3$ Jacobi matrices $A_0$ and the positive number a such that the spectrum of$A_0$ is $\{1,3,5\}$ and the spectrum of A is $\{2,4,6\}$ where $A=A_0+a{\overrightarrow{e}}_2{\overrightarrow{e}}_2^T$ and ${\overrightarrow{e}}_2^T=(0,1,0)$.

The explicit algorithm is:

1. Calculate $w_1,w_2,w_3$ by (23(23) $w_i^2=\frac{\prod _{j=1}^n({\lambda }_i-{\mu }_j)}{\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j)},$(23) );

2. Let $A_{1,1}(\lambda )A_{3,3}(\lambda )=p_2{\lambda }^2+p_1\lambda +p_0$;

3. Calculate $A_{1,1}(1)A_{3,3}(1),A_{1,1}(3)A_{3,3}(3),A_{1,1}(5)A_{3,3}(5)$ and a by (40(40) $\begin{array}{rl}& a=\sum _{i=1}^n{w}_i^2,\\ & A_{1,l-1}({\lambda }_i)A_{l+1,n}({\lambda }_i)=\frac{w_i^2}{\sum _{j=1}^n{w}_j^2}\prod _{j\ne i,j=1}^n({\lambda }_i-{\lambda }_j),i=1,2,\dots ,n.\end{array}$(40) );

4. Calculate $p_0,p_1,p_2$ by (41(41) $\left[\begin{array}{ccccc}{\lambda }_1^{n-1}& {\lambda }_1^{n-2}& \cdots & {\lambda }_1& 1\\ {\lambda }_2^{n-1}& {\lambda }_2^{n-2}& \cdots & {\lambda }_2& 1\\ ⋮& ⋮& \cdots & ⋮& ⋮\\ {\lambda }_n^{n-1}& {\lambda }_n^{n-2}& \cdots & {\lambda }_n& 1\end{array}\right]\left[\begin{array}{c}{p}_{n-1}\\ p_{n-2}\\ ⋮\\ p_0\end{array}\right]=\left[\begin{array}{c}{A}_{1,l-1}({\lambda }_1)A_{l+1,n}({\lambda }_1)\\ A_{1,l-1}({\lambda }_2)A_{l+1,n}({\lambda }_2)\\ ⋮\\ A_{1,l-1}({\lambda }_n)A_{l+1,n}({\lambda }_n)\end{array}\right].$(41) );

5. Calculate the zeros of $A_{1,1}(\lambda )A_{3,3}(\lambda )$, denoted by $\{t_1,t_2\}$;

6. Reconstruct the matrices $A_0$ from $\{1,3,5\}$ and $\{t_1,t_2\}$ by $\frac{(\lambda -1)(\lambda -3)(\lambda -5)}{(\lambda -t_1)(\lambda -t_2)}=\lambda -{\alpha }_2-\frac{{\beta }_1^2}{\lambda -{\alpha }_1}-\frac{{\beta }_3^2}{\lambda -{\alpha }_3}.$

It shows two matrices $A_0$, i.e. $\begin{array}{rl}& \left[\begin{array}{ccc}3.864028955680140& -1.344732872875984& 0\\ -1.344732872875984& 2.000004000000000& -0.437840093608048\\ 0& -0.437840093608048& 0.066203959516563\end{array}\right],\\ & \left[\begin{array}{ccc}0.066203959516563& -0.437840093608048& 0\\ -0.437840093608048& 2.000004000000000& -1.344732872875984\\ 0& -1.344732872875984& 3.864028955680140\end{array}\right].\end{array}$ This computation was performed using MATLAB.

Disclosure statement

No potential conflict of interest was reported by the author(s).