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Research article

The generalized inverse eigenvalue problem of Hamiltonian matrices and its approximation

  • Received: 31 January 2021 Accepted: 23 June 2021 Published: 01 July 2021
  • MSC : 15A24, 65F18

  • Let J=[0InIn0]R2n×2n. A matrix AR2n×2n is said to be Hamiltonian if (AJ)=AJ. In this paper, we first consider the following generalized inverse eigenvalue problem (GIEP): Given a pair of matrices (Λ,X) in the form Λ=diag{λ1,,λp}Cp×p and X=[x1,,xp]C2n×p, where diagonal elements of Λ are all distinct with rank(X)=p, and both Λ and X are closed under complex conjugation in the sense that λ2i=ˉλ2i1C, x2i=ˉx2i1C2n for i=1,,l, and λjR, xjR2n for j=2l+1,,p. Find Hamiltonian matrices A and B such that AXΛ=BX. Then, we consider the associated optimal approximation problem (OAP): Given ˜A,˜BR2n×2n. Find (ˆA,ˆB)SE such that ˆA˜A2+ˆB˜B2=min(A,B)SE(A˜A2+B˜B2), where SE is the solution set of Problem GIEP. By using the QR-decomposition, we deduce the representation of the general solution of Problem GIEP. Also, we obtain the unique optimal approximation solution (ˆA,ˆB) of Problem OAP.

    Citation: Lina Liu, Huiting Zhang, Yinlan Chen. The generalized inverse eigenvalue problem of Hamiltonian matrices and its approximation[J]. AIMS Mathematics, 2021, 6(9): 9886-9898. doi: 10.3934/math.2021574

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  • Let J=[0InIn0]R2n×2n. A matrix AR2n×2n is said to be Hamiltonian if (AJ)=AJ. In this paper, we first consider the following generalized inverse eigenvalue problem (GIEP): Given a pair of matrices (Λ,X) in the form Λ=diag{λ1,,λp}Cp×p and X=[x1,,xp]C2n×p, where diagonal elements of Λ are all distinct with rank(X)=p, and both Λ and X are closed under complex conjugation in the sense that λ2i=ˉλ2i1C, x2i=ˉx2i1C2n for i=1,,l, and λjR, xjR2n for j=2l+1,,p. Find Hamiltonian matrices A and B such that AXΛ=BX. Then, we consider the associated optimal approximation problem (OAP): Given ˜A,˜BR2n×2n. Find (ˆA,ˆB)SE such that ˆA˜A2+ˆB˜B2=min(A,B)SE(A˜A2+B˜B2), where SE is the solution set of Problem GIEP. By using the QR-decomposition, we deduce the representation of the general solution of Problem GIEP. Also, we obtain the unique optimal approximation solution (ˆA,ˆB) of Problem OAP.



    Throughout this paper, Cm×n, Rm×n, ORn×n and SRn×n stand for the sets of all m×n complex matrices, all m×n real matrices, all n×n orthogonal matrices and all n×n real-valued symmetric matrices, respectively. The symbol A and tr(A) stand for the transpose and the trace of a matrix A, respectively. In represents the identity matrix of size n, and HR2n×2n represents the set of all 2n×2n Hamiltonian matrices, that is, HR2n×2n={A|(AJ)=AJ, AR2n×2n}, where J=[0InIn0]R2n×2n.

    Hamiltonian matrices are widely applied in Hamiltonian systems of differential equations [1,2], optimal quadratic linear control [3] and H optimization [4], etc. For example, Hamiltonian matrix elements from a symmetric wave function are necessary to study the structure of deuterated molecules [5]. Also, the eigenvalue problems for Hamiltonian and skew-Hamiltonian matrices appear frequently in scientific and engineering applications. Such as to compute the conformal parameterization via a constrained energy minimization problem in the field of digital geometry processing [6,7], quantum mechanical problems with time reversal symmetry [8,9], the study of closed shell Hartree-Fock wave functions in response theory [10,11] and total least squares problems with symmetric constraints [12].

    Inverse eigenvalue problems emerge from many application areas, and have been studied by many scholars [13,14,15,16,17,18]. Generalized inverse eigenvalue problems are concerned in structural dynamics [19,20,21], parameter identification [22] and pole assignment [23], etc. Recently, Zhao and Zhang [24] derived the solvability conditions for the inverse eigenvalue problem of normal skew J-Hamiltonian matrices by the Moore-Penrose generalized inverse and the generalized singular value decomposition. Zhang and Yuan [25] solved the generalized inverse eigenvalue problems of Hermitian and J-Hamiltonian/skew-Hamiltonian matrices by applying the singular value decomposition and the spectral decomposition. However, the problem of OAP cannot be considered due to the complexity of the expression of the general solution. Very recently, Yuan and Chen [26] solved the inverse eigenvalue problem and the optimal approximation problem for Hamiltonian matrices by using the generalized singular value decomposition. Nevertheless, the generalized inverse eigenvalue problem of Hamiltonian matrices seems rarely to be discussed in the literatures, which motivates us to study such kind of inverse problem and the associated approximation problem. That is, in this paper, we will consider the following generalized inverse eigenvalue problem and the associated optimal approximation problem, which is a generalization of the problems discussed in [26].

    Problem GIEP. Given a pair of matrices (Λ,X) in the form Λ=diag{λ1,,λp}Cp×p, and X=[x1,,xp]C2n×p, where diagonal elements of Λ are all distinct, X is of full column rank p, and both Λ and X are closed under complex conjugation in the sense that λ2i=ˉλ2i1C, x2i=ˉx2i1C2n for i=1,,l, and λjR, xjR2n for j=2l+1,,p. Find A,BHR2n×2n such that

    AXΛ=BX. (1.1)

    Problem OAP. Given ˜A,˜BR2n×2n. Find (ˆA,ˆB)SE such that

    ˆA˜A2+ˆB˜B2=min(A,B)SE(A˜A2+B˜B2), (1.2)

    where is the Frobenius norm and SE is the solution set of Problem GIEP.

    By using the QR-decomposition, the representation of the general solution of Problem GIEP is deduced and the unique optimal approximation solution (ˆA,ˆB) of Problem OAP is obtained. Finally, two numerical examples are presented to illustrate the efficiency of the results.

    Define Tp as

    Tp=diag{12[1i1i],,12[1i1i],Ip2l}Cp×p,

    where i=1. It is easy to verify that Tp is a unitary matrix, that is, ˉTpTp=Ip. With this matrix, we have

    ˜Λ=ˉTpΛTp=diag{[α1β1β1α1],,[α2l1β2l1β2l1α2l1],λ2l+1,,λp} (2.1)
    \begin{equation} \tilde{X} = XT_{p} = [\sqrt{2}{\bf{y}}_{1},\sqrt{2}{\bf{z}}_{1},\cdots,\sqrt{2}{\bf{y}}_{2l-1}, \sqrt{2}{\bf{z}}_{2l-1},{\bf{x}}_{2l+1},\cdots,{\bf{x}}_{p}]\in \mathbb{R}^{2n\times p}, \end{equation} (2.2)

    where \alpha_{i} and \beta_{i} are the real part and imaginary part of the complex number \lambda_{i} , and {\bf{y}}_{i} and {\bf{z}}_{i} are, respectively, the real part and imaginary part of the complex vector {\bf{x}}_{i} for i = 1, 3, \cdots, 2l-1. Then, Eq (1.1) can be equivalently written as

    \begin{equation} A\tilde{X}\tilde{\Lambda} = B\tilde{X}, \end{equation} (2.3)

    clearly, Eq (2.3) is equivalent to

    \begin{equation} AJJ^{\top}\tilde{X}\tilde{\Lambda} = {B}JJ^{\top}\tilde{X}. \end{equation} (2.4)

    Since rank (X) = rank (\tilde{X}) = p , the QR-decomposition of J^{\top}\tilde{X} is of the form

    \begin{equation} J^{\top}\tilde{X} = Q\left [\begin{matrix} R\\0\end{matrix}\right] = [Q_{1},Q_{2}]\left [\begin{matrix} R\\0\end{matrix}\right], \end{equation} (2.5)

    where Q = [Q_{1}, Q_{2}]\in \mathbb{OR}^{2n\times 2n} with Q_{1}\in \mathbb{R}^{2n\times p} , and R\in \mathbb{R}^{p\times p} is nonsingular. Partition the parameter matrices Q^{\top}AJQ and Q^{\top}BJQ into blocks:

    \begin{equation} \begin{array}{cc} Q^{\top}AJQ = \left[\begin{array}{cc} A_{11} & A_{12} \\ A_{12}^{\top} & A_{22} \\ \end{array}\right] & \begin{array}{c} p \\ 2n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ \ 2n-p \\ \end{array}& \\ \end{array}, \ \ \ \begin{array}{cc} Q^{\top}BJQ = \left[\begin{array}{cc} B_{11} & B_{12} \\ B_{12}^{\top} & B_{22} \\ \end{array}\right] & \begin{array}{c} p \\ 2n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ \ 2n-p \\ \end{array} & \\ \end{array}, \end{equation} (2.6)

    where A_{11}, \ A_{22}, \ B_{11} and B_{22} are real-valued symmetric matrices. By (2.5) and (2.6) , Eq (2.4) is equivalent to

    \begin{equation} \left[ \begin{matrix} A_{11} & A_{12} \\ A_{12}^{\top} & A_{22} \\ \end{matrix} \right] \left[ \begin{matrix} R\tilde{\Lambda} \\ 0 \end{matrix} \right] = \left[ \begin{matrix} B_{11} & B_{12} \\ B_{12}^{\top} & B_{22} \\ \end{matrix}\right]\left[ \begin{matrix} R \\ 0 \end{matrix} \right]. \end{equation} (2.7)

    Then, it follows from Eq (2.7) that

    \begin{equation} A_{11}R\tilde{\Lambda} = B_{11}R, \end{equation} (2.8)
    \begin{equation} A_{12}^{\top}R\tilde{\Lambda} = B_{12}^{\top}R. \end{equation} (2.9)

    By Eq (2.8) , B_{11} is a symmetric matrix implies that

    \begin{equation} R^{\top}A_{11}R\tilde{\Lambda} = \tilde{\Lambda}^{\top}R^{\top}A_{11}R. \end{equation} (2.10)

    Write

    \begin{equation} C = R^{\top}A_{11}R, \end{equation} (2.11)

    then Eq (2.10) can be written as

    \begin{equation} C\tilde{\Lambda} = \tilde{\Lambda}^{\top}C, \ \: \mathrm{s.\ t.}\ \: C = C^{\top}. \end{equation} (2.12)

    By direct calculation, we have

    \begin{equation} C = {\rm{diag}}\left\{\left[\begin{matrix} a_{1} & b_{1}\\ b_{1} & -a_{1} \end{matrix}\right],\cdots,\left[\begin{matrix} a_{2l-1} & b_{2l-1}\\ b_{2l-1} & -a_{2l-1} \end{matrix}\right],c_{2l+1},\cdots,c_{p} \right\}\in \mathbb{R}^{p\times p}, \end{equation} (2.13)

    where a_{2i-1}, b_{2i-1}, i = 1, \cdots, l, and c_{j}, j = 2l+1, \cdots, p, are arbitrary real numbers. Thus

    \begin{equation} A_{11} = R^{-\top}CR^{-1}. \end{equation} (2.14)

    Combining (2.8) with (2.11) , we find that

    \begin{equation} B_{11} = R^{-\top}C\tilde{\Lambda}R^{-1}. \end{equation} (2.15)

    By Eq (2.9), we have

    \begin{equation} B_{12} = R^{-\top}\tilde{\Lambda}^{\top}R^{\top}A_{12}, \end{equation} (2.16)

    where A_{12}\in \mathbb{R}^{p\times (2n-p)} is an arbitrary matrix.

    Summing up above discussion, we can obtain the following result.

    Theorem 2.1. Suppose that \Lambda = \mathit{{\rm{diag}}}\{\lambda_1, \cdots, \lambda_{p}\}\in \mathbb{C}^{p\times p}, X = [{\bf{x}}_{1}, \cdots, {\bf{x}}_{p}]\in \mathbb{C}^{2n\times p} , where diagonal elements of \Lambda are all distinct, X is of full column rank p , and both \Lambda and X are closed under complex conjugation. Let the real matrices \tilde{\Lambda} and \tilde{X} be given by (2.1) and (2.2) and the QR-decomposition of J^{\top}\tilde{X} be given by (2.5) . Then the general solution of Problem GIEP can be expressed as

    \begin{equation} \mathbb{S_{E}} = \left\{(A,B)\left | \left. \begin{array}{c} A = Q\left[\begin{array}{cc} R^{-\top}CR^{-1} & A_{12}\\ A_{12}^{\top} & A_{22} \end{array} \right]Q^{\top}J^{\top},\\ B = Q\left [\begin{array}{cc} R^{-\top}C\tilde{\Lambda}R^{-1} & R^{-\top}\tilde{\Lambda}^{\top}R^{\top}A_{12}\\ A_{12}^{\top}R\tilde{\Lambda}R^{-1} & B_{22} \end{array} \right]Q^{\top}J^{\top} \end{array} \right. \right.\right\}, \end{equation} (2.17)

    where A_{12}\in \mathbb{R}^{p\times (2n-p)} , A_{22}\in \mathbb{SR}^{(2n-p)\times (2n-p)} and B_{22}\in \mathbb{SR}^{(2n-p)\times (2n-p)} are arbitrary matrices, and C is given by (2.13) .

    According to (2.17), we know that the solution set \mathbb{S_E} is always nonempty and \mathbb{S_E} is a closed convex subset, which implies that Problem OAP has a unique solution (\hat{A}, \hat{B})\in \mathbb{S_E} by the optimal approximation theorem (see Ref. [27]). For the given matrices \tilde{A}, \tilde{B}\in \mathbb{R}^{2n\times 2n}, write

    \begin{array}{*{20}{c}} {{Q^ \top }\tilde AJQ = \left[ {\begin{array}{*{20}{c}} {{{\tilde A}_{11}}}&{{{\tilde A}_{12}}}\\ {{{\tilde A}_{21}}}&{{{\tilde A}_{22}}} \end{array}} \right]}&{\begin{array}{*{20}{c}} p\\ {2n - p} \end{array}}\\ {\begin{array}{*{20}{c}} {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;p}&{\;\;2n - p} \end{array}}&{} \end{array},\begin{array}{*{20}{c}} {{Q^ \top }\tilde BJQ = \left[ {\begin{array}{*{20}{c}} {{{\tilde B}_{11}}}&{{{\tilde B}_{12}}}\\ {{{\tilde B}_{21}}}&{{{\tilde B}_{22}}} \end{array}} \right]}&{\begin{array}{*{20}{c}} p\\ {2n - p} \end{array}}\\ {\begin{array}{*{20}{c}} {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;p}&{\;\;2n - p} \end{array}}&{} \end{array}, (3.1)

    then

    \begin{array}{lllllll} & \|A-\tilde{A}\|^2+\|B -\tilde{B}\|^2\\ = & \left\|Q\left[\begin{array}{cc} R^{-\top}CR^{-1} & A_{12}\\ A_{12}^{\top} & A_{22} \end{array} \right]Q^{\top}J^{\top} - \tilde{A}\right\|^{2} + \left\|Q\left [\begin{array}{cc} R^{-\top}C\tilde{\Lambda}R^{-1} & R^{-\top}\tilde{\Lambda}^{\top}R^{\top}A_{12}\\ A_{12}^{\top}R\tilde{\Lambda}R^{-1} & B_{22} \end{array} \right]Q^{\top}J^{\top} - \tilde{B}\right\|^{2} \\ = & \|R^{-\top}CR^{-1}- \tilde{A}_{11} \|^2+\|A_{12}- \tilde{A}_{12}\|^2+ \|A_{12}^{\top}-\tilde{A}_{21}\|^2+\|A_{22}-\tilde{A}_{22}\|^2 \\ +& \|R^{-\top}C\tilde{\Lambda}R^{-1} -\tilde{B}_{11}\|^2+\|R^{-\top}\tilde{\Lambda}^{\top}R^{\top}A_{12}-\tilde{B}_{12}\|^2+ \| A_{12}^{\top}R\tilde{\Lambda}R^{-1}- \tilde{B}_{21}\|^2+\|B_{22}-\tilde{B}_{22}\|^2. \end{array}

    Therefore, \|A-\tilde{A}\|^2+\|B -\tilde{B}\|^2 = min if and only if

    \begin{equation} f(C) = \|R^{-\top}CR^{-1}- \tilde{A}_{11}\|^2+\|R^{-\top}C\tilde{\Lambda}R^{-1}-\tilde{B}_{11}\|^2 = \min,\\ \end{equation} (3.2)
    \begin{equation} \|A_{12}-\tilde{A}_{12}\|^2+\|R^{-\top}\tilde{\Lambda}^{\top}R^{\top}A_{12} -\tilde{B}_{12}\|^2 +\|A_{12}^{\top}-\tilde{A}_{21}\|^2 + \|A_{12}^{\top}R\tilde{\Lambda}R^{-1}-\tilde{B}_{21}\|^2 = \min, \end{equation} (3.3)
    \begin{equation} \|A_{22}-\tilde{A}_{22}\|^2 = \min,\\ \end{equation} (3.4)
    \begin{equation} \|B_{22}-\tilde{B}_{22}\|^2 = \min. \end{equation} (3.5)

    Let

    \begin{equation} R^{-1} = \left[\begin{array}{c} R_{1}\\ R_{2} \\ \end{array}\right],\ \end{equation} (3.6)

    where

    R_{1} = \left[\begin{array}{c} R_{1,1}\\ \vdots\\ R_{1,2l-1} \\ \end{array}\right],\ R_{2} = \left[\begin{array}{c} R_{2,2l+1}\\ \vdots\\ R_{2,p} \\ \end{array}\right],

    and R_{1, 2i-1}\in \mathbb{R}^{2\times p}, R_{2, j}\in \mathbb{R}^{1\times p} \ (i = 1, \cdots, l, j = 2l+1, \cdots, p). Furthermore, let

    \begin{equation} \left\{ \begin{array}{ll} D_{2i-1} = R_{1,2i-1}^{\top}F_{1}R_{1,2i-1},\\ D_{2i} = R_{1,2i-1}^{\top}F_{2}R_{1,2i-1},\\ D_{j} = R_{2,j}^{\top}R_{2,j},\\ E_{2i-1} = R_{1,2i-1}^{\top}F_{1}\tilde{\Lambda}_{2i-1}R_{1,2i-1},\\ E_{2i} = R_{1,2i-1}^{\top}F_{2}\tilde{\Lambda}_{2i-1}R_{1,2i-1},\\ E_{j} = R_{2,j}^{\top}\lambda_{j}R_{2,j},\\ i = 1,\cdots,l,j = 2l+1,\cdots,p, \end{array} \right. \end{equation} (3.7)

    where

    F_{1} = \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \\ \end{array}\right],\ F_{2} = \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array}\right].

    Then the relation of (3.2) is equivalent to

    \begin{array}{lll} & f(C) = f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p}) \\ = & \|a_{1}D_{1}+b_{1}D_{2}+\cdots+a_{2l-1}D_{2l-1}+b_{2l-1}D_{2l}+c_{2l+1}D_{2l+1}+\cdots+c_{p}D_{p}-\tilde{A}_{11}\|^{2}\\ +& \|a_{1}E_{1}+b_{1}E_{2}+\cdots+a_{2l-1}E_{2l-1}+b_{2l-1}E_{2l}+c_{2l+1}E_{2l+1}+\cdots+c_{p}E_{p}-\tilde{B}_{11}\|^{2} = \min, \end{array}

    that is,

    \begin{array}{ll} & f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p}) \\ = & \rm{tr}[(a_{1}D_{1}^{\top}+b_{1}D_{2}^{\top}+\cdots+a_{2l-1}D_{2l-1}^{\top}+b_{2l-1}D_{2l}^{\top}+c_{2l+1}D_{2l+1}^{\top}+\cdots+c_{p}D_{p}^{\top}-\tilde{A}_{11}^{\top})\\ &(a_{1}D_{1}+b_{1}D_{2}+\cdots+a_{2l-1}D_{2l-1}+b_{2l-1}D_{2l}+c_{2l+1}D_{2l+1}+\cdots+c_{p}D_{p}-\tilde{A}_{11})\\ +&(a_{1}E_{1}^{\top}+b_{1}E_{2}^{\top}+\cdots+a_{2l-1}E_{2l-1}^{\top}+b_{2l-1}E_{2l}^{\top}+c_{2l+1}E_{2l+1}^{\top}+\cdots+c_{p}E_{p}^{\top}-\tilde{B}_{11}^{\top})\\ &(a_{1}E_{1}+b_{1}E_{2}+\cdots+a_{2l-1}E_{2l-1}+b_{2l-1}E_{2l}+c_{2l+1}E_{2l+1}+\cdots+c_{p}E_{p}-\tilde{B}_{11})]\\ = &a_{1}^{2}g_{1,1}+2a_{1}b_{1}g_{1,2}+\cdots+2a_{1}a_{2l-1}g_{1,2l-1}+2a_{1}b_{2l-1}g_{1,2l}+2a_{1}c_{2l+1}g_{1,2l+1}+\cdots\\ +&2a_{1}c_{p}g_{1,p}-2a_{1}h_{1}\\ +&b_{1}^{2}g_{2,2}+\cdots+2a_{2l-1}b_{1}g_{2,2l-1}+2b_{1}b_{2l-1}g_{2,2l}+2b_{1}c_{2l+1}g_{2,2l+1}+\cdots+2b_{1}c_{p}g_{2,p}\\-&2b_{1}h_{2}\\ +&\cdots,\cdots \\ +&a_{2l-1}^{2}g_{2l-1,2l-1}+2a_{2l-1}b_{2l-1}g_{2l-1,2l}+2a_{2l-1}c_{2l+1}g_{2l-1,2l+1}+\cdots+2a_{2l-1}c_{p}g_{2l-1,p}\\ -&2a_{2l-1}h_{2l-1}\\ +&b_{2l-1}^{2}g_{2l,2l}+2b_{2l-1}c_{2l+1}g_{2l,2l+1}+\cdots+2b_{2l-1}c_{p}g_{2l,p}-2b_{2l-1}h_{2l}\\ +&c_{2l+1}^{2}g_{2l+1,2l+1}+\cdots+2c_{2l+1}c_{p}g_{2l+1,p}-2c_{2l+1}h_{2l+1}\\ +&\cdots,\cdots\\ +&c_{p}^{2}g_{p,p}-2c_{p}h_{p}+e, \end{array}

    where g_{m, n} = \rm{tr}(D_{m}^{\top}D_{n})+\rm{tr}(E_{m}^{\top}E_{n}), h_{m} = \rm{tr}(D_{m}^{\top}\tilde{A}_{11})+\rm{tr}(E_{m}^{\top}\tilde{B}_{11}), e = \rm{tr}(\tilde{A}_{11}^{\top}\tilde{A}_{11})+\rm{tr}(\tilde{B}_{11}^{\top}\tilde{B}_{11}), m, n = 1, \cdots, p .

    Consequently,

    \begin{array}{ll} \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial a_{1}}& = 2a_{1}g_{1,1}+2b_{1}g_{1,2}+\cdots+2a_{2l-1}g_{1,2l-1}+2b_{2l-1}g_{1,2l}+2c_{2l+1}g_{1,2l+1}\\&+ \cdots+2c_{p}g_{1,p}-2h_{1},\\ \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial b_{1}}& = 2a_{1}g_{2,1}+2b_{1}g_{2,2}+\cdots+2a_{2l-1}g_{2,2l-1}+2b_{2l-1}g_{2,2l}+2c_{2l+1}g_{2,2l+1}\\&+ \cdots+2c_{p}g_{2,p}-2h_{2},\\ \cdots,\cdots,\\ \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial a_{2l-1}}& = 2a_{1}g_{2l-1,1}+2b_{1}g_{2l-1,2}+\cdots+2a_{2l-1}g_{2l-1,2l-1}+2b_{2l-1}g_{2l-1,2l}\\&+2c_{2l+1}g_{2l-1,2l+1}+\cdots +2c_{p}g_{2l-1,p}-2h_{2l-1},\\ \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial b_{2l-1}}& = 2a_{1}g_{2l,1}+2b_{1}g_{2l,2}+\cdots+2a_{2l-1}g_{2l,2l-1}+2b_{2l-1}g_{2l,2l} +2c_{2l+1}g_{2l,2l+1}\\&+\cdots +2c_{p}g_{2l,p}-2h_{2l},\\ \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial c_{2l+1}}& = 2a_{1}g_{2l+1,1}+2b_{1}g_{2l+1,2}+\cdots+2a_{2l-1}g_{2l+1,2l-1}+2b_{2l-1}g_{2l+1,2l}\\ &+2c_{2l+1}g_{2l+1,2l+1}+\cdots +2c_{p}g_{2l+1,p}-2h_{2l+1},\\ \cdots,\cdots,\\ \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial c_{p}}& = 2a_{1}g_{p,1}+2b_{1}g_{p,2}+\cdots+2a_{2l-1}g_{p,2l-1}+2b_{2l-1}g_{p,2l} +2c_{2l+1}g_{p,2l+1}\\&+\cdots +2c_{p}g_{p,p}-2h_{p}.\\ \end{array}

    Clearly, f(a_{1}, b_{1}, \cdots, a_{2l-1}, b_{2l-1}, c_{2l+1}, \cdots, c_{p}) = min if and only if

    \frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial a_{1}} = 0,\cdots,\frac{\partial f(a_{1},b_{1},\cdots,a_{2l-1},b_{2l-1},c_{2l+1},\cdots,c_{p})}{\partial c_{p}} = 0.

    Therefore,

    \begin{equation} \begin{split} &a_{1}g_{1,1}+b_{1}g_{1,2}+\cdots+a_{2l-1}g_{1,2l-1}+b_{2l-1}g_{1,2l}+c_{2l+1}g_{1,2l+1}+\cdots+c_{p}g_{1,p} = h_{1},\\ &a_{1}g_{2,1}+b_{1}g_{2,2}+\cdots+a_{2l-1}g_{2,2l-1}+b_{2l-1}g_{2,2l} +c_{2l+1}g_{2,2l+1}+\cdots+c_{p}g_{2,p} = h_{2},\\ &\cdots,\cdots,\\ &a_{1}g_{2l-1,1}+b_{1}g_{2l-1,2}+\cdots+a_{2l-1}g_{2l-1,2l-1}+b_{2l-1}g_{2l-1,2l}+c_{2l+1}g_{2l-1,2l+1}+\cdots+c_{p}g_{2l-1,p} = h_{2l-1},\\ &a_{1}g_{2l,1}+b_{1}g_{2l,2}+\cdots+a_{2l-1}g_{2l,2l-1}+b_{2l-1}g_{2l,2l} +c_{2l+1}g_{2l,2l+1}+\cdots+c_{p}g_{2l,p} = h_{2l},\\ &a_{1}g_{2l+1,1}+b_{1}g_{2l+1,2}+\cdots+a_{2l-1}g_{2l+1,2l-1}+b_{2l-1}g_{2l+1,2l}+c_{2l+1}g_{2l+1,2l+1}+\cdots+c_{p}g_{2l+1,p} = h_{2l+1},\\ &\cdots,\cdots,\\ &a_{1}g_{p,1}+b_{1}g_{p,2}+\cdots+a_{2l-1}g_{p,2l-1}+b_{2l-1}g_{p,2l} +c_{2l+1}g_{p,2l+1}+\cdots+c_{p}g_{p,p} = h_{p}. \end{split} \end{equation} (3.8)

    If let

    G = \left[ \begin{array}{ccccccccc} g_{1,1}& g_{1,2}& \cdots& g_{1,2l-1}& g_{1,2l}& g_{1,2l+1}& \cdots& g_{1,p}& \\ g_{2,1}& g_{2,2}& \cdots& g_{2,2l-1}& g_{2,2l}& g_{2,2l+1}& \cdots& g_{2,p}& \\ \vdots& \vdots& & \vdots& \vdots& \vdots& & \vdots&\\ g_{2l-1,1}& g_{2l-1,2}& \cdots& g_{2l-1,2l-1}& g_{2l-1,2l}& g_{2l-1,2l+1}& \cdots& g_{2l-1,p}& \\ g_{2l,1}& g_{2l,2}& \cdots& g_{2l,2l-1}& g_{2l,2l}& g_{2l,2l+1}& \cdots& g_{2l,p}& \\ g_{2l+1,1}& g_{2l+1,2}& \cdots& g_{2l+1,2l-1}& g_{2l+1,2l}& g_{2l+1,2l+1}& \cdots& g_{2l+1,p}& \\ \vdots& \vdots& & \vdots& \vdots& \vdots& & \vdots&\\ g_{p,1}& g_{p,2}& \cdots& g_{p,2l-1}& g_{p,2l}& g_{p,2l+1}& \cdots& g_{p,p}& \\ \end{array} \right],
    T = \left[ \begin{array}{c} a_{1}\\ b_{1}\\ \vdots\\ a_{2l-1}\\ b_{2l-1}\\ c_{2l+1}\\ \vdots\\ c_{p}\\ \end{array} \right],\ H = \left[ \begin{array}{c} h_{1}\\ h_{2}\\ \vdots\\ h_{2l-1}\\ h_{2l}\\ h_{2l+1}\\ \vdots\\ h_{p}\\ \end{array} \right],

    where G is symmetric matrix. Then Eq (3.8) is equivalent to

    \begin{equation} GT = H, \end{equation} (3.9)

    and the solution of Eq (3.9) is

    \begin{equation} T = G^{-1}H. \end{equation} (3.10)

    Substituting (3.10) into (2.13) , we can obtain C explicitly. Similarly, Eq (3.3) is equivalent to

    \begin{array}{ll} f(A_{12})& = tr[(A_{12}^{\top}-\tilde{A}_{12}^{\top})(A_{12}-\tilde{A}_{12})]+tr[(A_{12}^{\top}P -\tilde{B}_{12}^{\top})(P^{\top}A_{12} -\tilde{B}_{12})]\\ &+tr[(A_{12}-\tilde{A}_{21}^{\top})(A_{12}^{\top}-\tilde{A}_{21})]+tr[(P^{\top}A_{12}-\tilde{B}_{21}^{\top})(A_{12}^{\top}P-\tilde{B}_{21})]. \end{array}

    Thus,

    \frac{\partial f(A_{12})}{\partial A_{12}} = 2A_{12}-2\tilde{A}_{12}+2PP^{\top}A_{12}-2P\tilde{B}_{12}+2A_{12}-2\tilde{A}_{21}^{\top}+2PP^{\top}A_{12}-2P\tilde{B}_{21}^{\top},

    setting \frac{\partial f(A_{12})}{\partial A_{12}} = 0, we obtain

    \begin{equation} A_{12} = \frac{1}{2}(I_{p}+PP^{\top})^{-1}(\tilde{A}_{12}+P\tilde{B}_{12}+\tilde{A}_{21}^{\top}+P\tilde{B}_{21}^{\top}), \end{equation} (3.11)

    where P = R\tilde{\Lambda}R^{-1}. A_{22}, B_{22} are symmetric matrices implies that the relations of (3.4) and (3.5) are equivalent to

    \|A_{22}-\tilde{A}_{22}\|^2 = \left\|A_{22} - \frac{1}{2}(\tilde{A}_{22}+\tilde{A}_{22}^{\top})\right\|^2+\left\|\frac{1}{2}(\tilde{A}_{22}-\tilde{A}_{22}^{\top})\right\|^2,
    \|B_{22} -\tilde{B}_{22}\|^2 = \left\|B_{22} - \frac{1}{2}(\tilde{B}_{22}+\tilde{B}_{22}^{\top})\right\|^2+\left \| \frac{1}{2}(\tilde{B}_{22}-\tilde{B}_{22}^{\top})\right\|^2,

    therefore, we have

    \begin{equation} A_{22} = \frac{1}{2}(\tilde{A}_{22}+\tilde{A}_{22}^{\top}),\ B_{22} = \frac{1}{2}(\tilde{B}_{22}+\tilde{B}_{22}^{\top}). \end{equation} (3.12)

    Theorem 3.1. Given \tilde{A}, \tilde{B}\in \mathbb{R}^{2n\times 2n}, then the Problem OAP has a unique solution and the unique solution of Problem OAP is

    \begin{equation} \hat{A} = Q\left[\begin{array}{cc} R^{-T}CR^{-1} & A_{12}\\ A_{12}^{\top} & {A}_{22} \end{array} \right]Q^{\top}J^{\top},\ \hat{B} = Q\left [\begin{array}{cc} R^{-T}C\tilde{\Lambda}R^{-1} & R^{-T}\tilde{\Lambda}^{\top}R^{\top}A_{12}\\ A_{12}^{\top}R\tilde{\Lambda}R^{-1} & {B}_{22} \end{array} \right]Q^{\top}J^{\top}, \end{equation} (3.13)

    where A_{12} and \ A_{22}, \ B_{22} are given by (3.11) and (3.12), and a_{1}, b_{1}, \cdots, a_{2l-1}, b_{2l-1}, c_{2l+1}, \cdots, c_{p} are given by (3.10) , respectively.

    According to Theorems 2.1 and 3.1, we have the following algorithm for solving Problem OAP.

    Algorithm 4.1.

    1). Input \Lambda, \ X, \ J, \ \tilde{A}, \ \tilde{B}.

    2). Compute real-valued matrices \tilde{\Lambda}, \ \tilde{X} by (2.1) and (2.2) , respectively.

    3). Compute the QR-decomposition of the matrix J^{\top}\tilde{X} by (2.5) .

    4). Compute \tilde{A}_{ij} , \tilde{B}_{ij} by (3.1), i, j = 1, 2 .

    5). Compute R^{-1} by (3.6) to form R_{1}, R_{2} .

    6). Compute D_{2i-1}, D_{2i}, D_{j} , E_{2i-1}, E_{2i} and E_{j}\ (i = 1, \cdots, l, j = 2l+1, \cdots, p ) by (3.7).

    7). Compute g_{m, n} = \mbox{tr}(D_{m}^{\top}D_{n})+\mbox{tr}(E_{m}^{\top}E_{n}) and h_{m} = \mbox{tr}(D_{m}^{\top}\tilde{A}_{11})+\mbox{tr}(E_{m}^{\top}\tilde{B}_{11}) ( m, n = 1, \cdots, p ).

    8). Compute a_{1}, b_{1}, \cdots, a_{2l-1}, b_{2l-1}, c_{2l+1}, \cdots, c_{p} by (3.10) .

    9). Compute A_{12} by (3.11).

    10). Compute A_{22} and B_{22} by (3.12) .

    11). Compute \hat{A} and \hat{B} by (3.13).

    Remark 4.1. After statistics, we find that the amount of computations required by Algorithm 1 is about p^5+p^4+\frac{19}{3}p^3+14np^2+64n^3 flops.

    Example 4.1. Let n = 5, \ p = 5 , and the matrices \Lambda, \ X, \ \tilde{A} and \tilde{B} be given by

    \Lambda = \mbox{diag}\left\{-0.2218 + 2.0231i,\ -0.2218 - 2.0231i,\ -0.1617 + 0.5721i,\ -0.1617 - 0.5721i,\ 2.7670 \right\},
    X = \left[ \begin{array}{rrrrrrrrrr} -0.6377 - 0.1444i& -0.6377 + 0.1444i& -0.0405 + 0.3341i& -0.0405 - 0.3341i& -1.0000\\ 0.2678 - 0.0983i& 0.2678 + 0.0983i& -0.2615 - 0.3973i& -0.2615 + 0.3973i& 0.1151\\ 0.4260 + 0.5740i& 0.4260 - 0.5740i& 0.1348 - 0.2946i& 0.1348 + 0.2946i& 0.0278\\ -0.2032 - 0.0489i& -0.2032 + 0.0489i& -0.5238 + 0.4762i& -0.5238 - 0.4762i& 0.0357\\ 0.2111 - 0.1510i& 0.2111 + 0.1510i& 0.7982 - 0.1668i& 0.7982 + 0.1668i& 0.6793\\ -0.5233 + 0.1151i& -0.5233 - 0.1151i& -0.3033 - 0.4132i& -0.3033 + 0.4132i& -0.1091\\ 0.4820 - 0.2541i& 0.4820 + 0.2541i& 0.1398 - 0.0715i& 0.1398 + 0.0715i& 0.8550\\ 0.3183 - 0.3431i& 0.3183 + 0.3431i& -0.2716 - 0.3411i& -0.2716 + 0.3411i& 0.6436\\ 0.1376 + 0.3007i& 0.1376 - 0.3007i& -0.1391 + 0.0981i& -0.1391 - 0.0981i& 0.0883\\ -0.0672 + 0.0189i& -0.0672 - 0.0189i& 0.3143 + 0.1462i& 0.3143 - 0.1462i& -0.2828\\ \end{array} \right],
    \tilde{A} = \left[ \begin{array}{rrrrrrrrrr} 1.8351& 3.0635& 9.3900& 1.9476& 9.7975& 1.1742& 7.3033& 6.2406& 2.6187& 9.0372\\ 3.6848& 5.0851& 8.7594& 2.2592& 4.3887& 2.9668& 4.8861& 6.7914& 3.3536& 8.9092\\ 6.2562& 5.1077& 5.5016& 1.7071& 1.1112& 3.1878& 5.7853& 3.9552& 6.7973& 3.3416\\ 7.8023& 8.1763& 6.2248& 2.2766& 2.5806& 4.2417& 2.3728& 3.6744& 1.3655& 6.9875\\ 0.8113& 7.9483& 5.8704& 4.3570& 4.0872& 5.0786& 4.5885& 9.8798& 7.2123& 1.9781\\ 9.2939& 6.4432& 2.0774& 3.1110& 5.9490& 0.8552& 9.6309& 0.3774& 1.0676& 0.3054\\ 7.7571& 3.7861& 3.0125& 9.2338& 2.6221& 2.6248& 5.4681& 8.8517& 6.5376& 7.4407\\ 4.8679& 8.1158& 4.7092& 4.3021& 6.0284& 8.0101& 5.2114& 9.1329& 4.9417& 5.0002\\ 4.3586& 5.3283& 2.3049& 1.8482& 7.1122& 0.2922& 2.3159& 7.9618& 7.7905& 4.7992\\ 4.4678& 3.5073& 8.4431& 9.0488& 2.2175& 9.2885& 4.8890& 0.9871& 7.1504& 9.0472\\ \end{array} \right],
    \tilde{B} = \left[ \begin{array}{rrrrrrrrrr} 6.0987& 1.6793& 0.9673& 4.5380& 3.9926& 1.0622& 4.2284& 6.6653& 3.6892& 1.2061\\ 6.1767& 9.7868& 8.1815& 4.3239& 5.2688& 3.7241& 5.4787& 1.7813& 4.6073& 5.8951\\ 8.5944& 7.1269& 8.1755& 8.2531& 4.1680& 1.9812& 9.4274& 1.2801& 9.8164& 2.2619\\ 8.0549& 5.0047& 7.2244& 0.8347& 6.5686& 4.8969& 4.1774& 9.9908& 1.5640& 3.8462\\ 5.7672& 4.7109& 1.4987& 1.3317& 6.2797& 3.3949& 9.8305& 1.7112& 8.5552& 5.8299\\ 1.8292& 0.5962& 6.5961& 1.7339& 2.9198& 9.5163& 3.0145& 0.3260& 6.4476& 2.5181\\ 2.3993& 6.8197& 5.1859& 3.9094& 4.3165& 9.2033& 7.0110& 5.6120& 3.7627& 2.9044\\ 8.8651& 0.4243& 9.7297& 8.3138& 0.1549& 0.5268& 6.6634& 8.8187& 1.9092& 6.1709\\ 0.2867& 0.7145& 6.4899& 8.0336& 9.8406& 7.3786& 5.3913& 6.6918& 4.2825& 2.6528\\ 4.8990& 5.2165& 8.0033& 0.6047& 1.6717& 2.6912& 6.9811& 1.9043& 4.8202& 8.2438\\ \end{array} \right].

    By applying Algorithm 4.1 , we can obtain the unique solution (\hat{A}, \hat{B}) of Problem OAP as follows:

    \hat{A} = \left[ \begin{array}{rrrrrrrrrr} 3.0180& -0.4225& -0.1555& 0.8402& 0.9125& 4.0071& 3.8840& 3.2121& 4.0178& 6.8173\\ -0.3558& -1.7726& -0.3149& 2.4300& -2.4421& 3.8840& 1.9734& 5.0988& 2.4334& 6.2185\\ 1.8766& -3.0189& -0.9503& -1.1732& -1.8790& 3.2121& 5.0988& 2.6896& 6.6750& 5.4058\\ 1.9510& -0.1294& -0.2763& -0.7668& -1.9008& 4.0178& 2.4334& 6.6750& 2.9195& 5.2316\\ 3.0211& 0.2273& -0.6075& -0.9217& -2.7818& 6.8173& 6.2185& 5.4058& 5.2316& 3.9665\\ 6.3731& 6.9288& 5.5322& 4.4397& 5.4356& -3.0180& 0.3558& -1.8766& -1.9510& -3.0211\\ 6.9288& 4.5231& 6.2946& 7.2049& 3.6851& 0.4225& 1.7726& 3.0189& 0.1294& -0.2273\\ 5.5322& 6.2946& 4.8453& 3.7706& 4.4618& 0.1555& 0.3149& 0.9503& 0.2763& 0.6075\\ 4.4397& 7.2049& 3.7706& 2.8249& 6.6239& -0.8402& -2.4300& 1.1732& 0.7668& 0.9217\\ 5.4356& 3.6851& 4.4618& 6.6239& 2.8468& -0.9125& 2.4421& 1.8790& 1.9008& 2.7818\\ \end{array} \right],
    \hat{B} = \left[ \begin{array}{rrrrrrrrrr} 0.6709& -1.2381& -0.7557& -1.1218& -1.9835& -1.1000& 3.7817& 3.8263& 0.6376& 3.5484\\ 1.3451& 0.2456& -0.0257& -0.2862& 0.2915& 3.7817& 5.3559& 5.3224& 3.6947& 8.6240\\ 3.0095& 1.2171& -1.0933& 0.9515& 2.1292& 3.8263& 5.3224& 1.2147& 6.8013& 2.4324\\ 2.0994& -1.8754& 3.1477& -0.1903& 0.6793& 0.6376& 3.6947& 6.8013& 3.3691& 7.2508\\ 1.4047& 0.3032& -1.6547& -2.5697& 1.2202& 3.5484& 8.6240& 2.4324& 7.2508& 6.0558\\ 2.7999& 2.6538& 5.9368& 4.0530& 1.3808& -0.6709& -1.3451& -3.0095& -2.0994& -1.4047\\ 2.6538& 5.3390& 2.3523& 2.6326& 4.1208& 1.2381& -0.2456& -1.2171& 1.8754& -0.3032\\ 5.9368& 2.3523& 8.4193& 6.9480& 4.5598& 0.7557& 0.0257& 1.0933& -3.1477& 1.6547\\ 4.0530& 2.6326& 6.9480& 8.0838& 5.2373& 1.1218& 0.2862& -0.9515& 0.1903& 2.5697\\ 1.3808& 4.1208& 4.5598& 5.2373& 3.4028& 1.9835& -0.2915& -2.1292& -0.6793& -1.2202\\ \end{array} \right],

    and

    \|\hat{A}X \Lambda -\hat{B}X\| = 2.0806 \times10^{-14},\

    which implies that \hat{A} X \Lambda = \hat{B}X reproduces the desired eigenvalues and eigenvectors.

    Example 4.2. We consider an inverse problem for the spectral conformal parameterization (see Refs.[6,7]). Let n = 5, \ p = 4 , and the matrices \Lambda, \ X, \ \tilde{B} and \tilde{L}_{C} be given by

    \Lambda = \mbox{diag}\left\{ -0.0822,\ -0.0250,\ 0,\ 0.0757 \right\}\triangleq\mbox{diag}\{\lambda_1,\lambda_2,\lambda_3,\lambda_{4}\},
    X = \left[ \begin{array}{rrrrrrrrrr} 0.5760& 0.2684& 0.0000& 0.1340\\ -0.1510& -0.0673& 0.0000& -1.0000\\ -0.3890& -0.0610& 0.0000& -0.1246\\ -0.0359& -0.1401& 0.0000& 0.9907\\ 0.3409& 1.0000& 1.0000& 0.0366\\ 0.0000& 0.0000& 0.0000& 0.0000\\ -0.5672& 0.0665& 0.0000& 0.3072\\ 1.0000& -0.1288& 0.0000& -0.1395\\ -0.4328& 0.0623& 0.0000& -0.1677\\ 0.0650& 0.2282& 0.2000& -0.1508\\ \end{array} \right]\triangleq\mbox{diag}\{f_1,\ f_2,\ f_3,\ f_{4}\},
    \tilde{B} = \left[ \begin{array}{rrrrrrrrrr} 0.7000& 0.5483& 0.5369& 0.6919& 0.6032& 0& -0.8083& 0.0306& -0.7715& 0.0234\\ 0.5483& 1.6617& 1.3425& 0.6037& 1.4860& 0.8083& 0& 0.4237& -0.5357& -0.9292\\ 0.5369& 1.3425& 1.5075& 0.9112& 1.0372& -0.0306& -0.4237& 0& 0.4304& 0.0488\\ 0.6919& 0.6037& 0.9112& 1.5583& 0.9459& 0.7715& 0.5357& -0.4304& 0& -0.0255\\ 0.6032& 1.4860& 1.0372& 0.9459& 0.6742& -0.0234& 0.9292& -0.0488& 0.0255& 0\\ 0& 0.8083& -0.0306& 0.7715& -0.0234& 0.7000& 0.5483& 0.5369& 0.6919& 0.6032\\ -0.8083& 0& -0.4237& 0.5357& 0.9292& 0.5483& 1.6617& 1.3425& 0.6037& 1.4860\\ 0.0306& 0.4237& 0& -0.4304& -0.0488& 0.5369& 1.3425& 1.5075& 0.9112& 1.0372\\ -0.7715& -0.5357& 0.4304& 0& 0.0255 & 0.6919& 0.6037& 0.9112& 1.5583& 0.9459\\ 0.0234& -0.9292& 0.0488& -0.0255& 0& 0.6032& 1.4860& 1.0372& 0.9459& 0.6742\\ \end{array} \right],
    \tilde{L}_{C} = \left[ \begin{array}{rrrrrrrrrr} 1.7000& 2.5483& 3.5369& 4.6919& -9.3968& 0& -1.6166& 0.0612& -1.5430& 0.0468\\ 2.5483& 4.6617& 5.3425& -9.3963& 2.4860& 1.6166& 0& 0.8474& -1.0714& -1.8584\\ 3.5369& 5.3425& -8.4925& 1.9112& 3.0372& -0.0612& -0.8474& 0& 0.8608& 0.0976\\ 4.6919& -9.3963& 1.9112& 3.5583& 3.9459& 1.5430& 1.0714& -0.8608& 0& -0.0510\\ -9.3968& 2.4860& 3.0372& 3.9459& 4.6742& -0.0468& 1.8584& -0.0976& 0.0510& 0\\ 0& 1.6166& -0.0612& 1.5430& -0.0468& 1.7000& 2.5483& 3.5369& 4.6919& -9.3968\\ -1.6166& 0& -0.8474& 1.0714& 1.8584& 2.5483& 4.6617& 5.3425& -9.3963& 2.4860\\ 0.0612& 0.8474& 0& -0.8608& -0.0976& 3.5369& 5.3425& -8.4925& 1.9112& 3.0372\\ -1.5430& -1.0714& 0.8608& 0& 0.0510& 4.6919& -9.3963& 1.9112& 3.5583& 3.9459\\ 0.0468& -1.8584& 0.0976& -0.0510& 0& -9.3968& 2.4860& 3.0372& 3.9459& 4.6742\\ \end{array} \right].

    By calculating, we can obtain the unique solution (\hat{B}, \hat{L}_{C}) of Problem OAP as follows:

    \hat{B} = \left[ \begin{array}{rrrrrrrrrr} 1.0866& -0.0847& -0.0520 & 0.0501 & 0.1093 & 0.0000& -0.0330 & 0.0833 & -0.0503& 0.0134\\ -0.1591 & 1.7380& -0.2522& -0.3467& -0.2414 & 0.0330 & -0.0000 & -0.0191 & 0.0087 & 0.0234\\ -0.1345& -0.2315 & 1.6597 & -0.2656 & -0.2222& -0.0833 & 0.0191& 0.0000 & 0.0206 & -0.0230\\ -0.0049 & -0.3583 & -0.2945 & 1.6496 & -0.3472 & 0.0503 & -0.0087& -0.0206 & -0.0000 & -0.1716\\\ 0.2724 & -0.0519& -0.1132& -0.1159 & 0.8731 & -0.0134 & -0.0234 & 0.0230& 0.1716& 0.0000\\ 0.0000 & 0.0864& -0.2014 & 0.1221& -0.0239 & 1.0866 & -0.1591 & -0.1345 & -0.0049 & 0.2724\\ -0.0864 & -0.0000& 0.0415 & 0.0244 & 0.1288 & -0.0847 & 1.7380 & -0.2315 & -0.3583& -0.0519\\ 0.2014 & -0.0415 & 0.0000& -0.1586 & -0.1553 & -0.0520 & -0.2522 & 1.6597 & -0.2945& -0.1132\\ -0.1221 & -0.0244 & 0.1586 & -0.0000 & -0.0000 & 0.0501 & -0.3467 & -0.2656 & 1.6496 & -0.1159\\ 0.0239 & -0.1288 & 0.1553 & 0.0000 & 0.0000 & 0.1093 & -0.2414& -0.2222 & -0.3472 & 0.8731\\ \end{array} \right],
    \hat{L}_{C} = \left[ \begin{array}{rrrrrrrrrr} 3.0749& 3.0182& 2.9855& 3.3985& -0.0221& -0.0000 & -0.5788& 0.0130 & 1.0166& 0.1107\\ 0.8672 & 0.7497& 1.1429& 0.2011& 0.4437& 0.5788& 0.0000& -0.0656& -0.3528& -2.2185\\ 0.7590& 0.8360& 0.5193& 0.6574& 0.1300& -0.0130& 0.0656& 0.0000& -0.0313 & -0.6502\\ 0.1065& 0.3474& -0.4752 & 0.5081& -0.2263 & -1.0166 & 0.3528 & 0.0313 & 0.0000 & 1.1314\\ -0.1139& 0.6946& -0.6294& -0.1520 & 0.0000& -0.1107& 2.2185& 0.6502 & -1.1314 & 0.0000\\ 0.0000& -0.3435 & 0.7467& -0.4158& 0.0228& 3.0749 & 0.8672& 0.7590 & 0.1065 & -0.1139\\ 0.3435& 0.0000 & 0.9259 & 0.1743& -0.1389& 3.0182& 0.7497 & 0.8360 & 0.3474 & 0.6946\\ -0.7467& -0.9259& -0.0000& -1.3216& 0.1259& 2.9855& 1.1429& 0.5193 & -0.4752 & -0.6294\\ 0.4158& -0.1743 & 1.3216& -0.0000& 0.0304& 3.3985& 0.2011& 0.6574 & 0.5081 & -0.1520\\ -0.0228 & 0.1389 & -0.1259 & -0.0304& -0.0000 & -0.0221& 0.4437 & 0.1300 & -0.2263 & 0.0000\\ \end{array} \right].

    Furthermore, we can obtain the following numerical results: Therefore, the new model \hat{B} X \Lambda = \hat{L}_{C}X reproduces the desired eigenvalues and eigenvectors.

    Table 4.1.  Residuals of the eigenpairs (\lambda_{i},f_{i}) .
    (\lambda_{i},f_{i}) (\lambda_{1},f_{1}) (\lambda_{2},f_{2}) (\lambda_{3},f_{3}) (\lambda_{4},f_{4})
    \|\lambda_{i}\hat{B}f_{i}-\hat{L}_{C}f_{i}\| 1.1762 \times10^{-15} 1.1322 \times10^{-15} 1.0372 \times10^{-15} 9.7141 \times10^{-16}

     | Show Table
    DownLoad: CSV

    The authors would like to express their gratitude to the anonymous referees for their valuable suggestions and comments for the revision of this manuscript.

    The authors declare no conflict of interest.



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