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}diag{˜Λ1,,˜Λ2l1,λ2l+1,,λp}Rp×p, (2.1)
    ˜X=XTp=[2y1,2z1,,2y2l1,2z2l1,x2l+1,,xp]R2n×p, (2.2)

    where αi and βi are the real part and imaginary part of the complex number λi, and yi and zi are, respectively, the real part and imaginary part of the complex vector xi for i=1,3,,2l1. Then, Eq (1.1) can be equivalently written as

    A˜X˜Λ=B˜X, (2.3)

    clearly, Eq (2.3) is equivalent to

    AJJ˜X˜Λ=BJJ˜X. (2.4)

    Since rank(X) = rank(˜X) = p, the QR-decomposition of J˜X is of the form

    J˜X=Q[R0]=[Q1,Q2][R0], (2.5)

    where Q=[Q1,Q2]OR2n×2n with Q1R2n×p, and RRp×p is nonsingular. Partition the parameter matrices QAJQ and QBJQ into blocks:

    QAJQ=[A11A12A12A22]p2np                       p  2np,   QBJQ=[B11B12B12B22]p2np                       p  2np, (2.6)

    where A11, A22, B11 and B22 are real-valued symmetric matrices. By (2.5) and (2.6), Eq (2.4) is equivalent to

    [A11A12A12A22][R˜Λ0]=[B11B12B12B22][R0]. (2.7)

    Then, it follows from Eq (2.7) that

    A11R˜Λ=B11R, (2.8)
    A12R˜Λ=B12R. (2.9)

    By Eq (2.8), B11 is a symmetric matrix implies that

    RA11R˜Λ=˜ΛRA11R. (2.10)

    Write

    C=RA11R, (2.11)

    then Eq (2.10) can be written as

    C˜Λ=˜ΛC, s. t. C=C. (2.12)

    By direct calculation, we have

    C=diag{[a1b1b1a1],,[a2l1b2l1b2l1a2l1],c2l+1,,cp}Rp×p, (2.13)

    where a2i1, b2i1, i=1,,l, and cj, j=2l+1,,p, are arbitrary real numbers. Thus

    A11=RCR1. (2.14)

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

    B11=RC˜ΛR1. (2.15)

    By Eq (2.9), we have

    B12=R˜ΛRA12, (2.16)

    where A12Rp×(2np) is an arbitrary matrix.

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

    Theorem 2.1. Suppose that Λ=diag{λ1,,λp}Cp×p, 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. Let the real matrices ˜Λ and ˜X be given by (2.1) and (2.2) and the QR-decomposition of J˜X be given by (2.5). Then the general solution of Problem GIEP can be expressed as

    SE={(A,B)|A=Q[RCR1A12A12A22]QJ,B=Q[RC˜ΛR1R˜ΛRA12A12R˜ΛR1B22]QJ}, (2.17)

    where A12Rp×(2np), A22SR(2np)×(2np) and B22SR(2np)×(2np) are arbitrary matrices, and C is given by (2.13).

    According to (2.17), we know that the solution set SE is always nonempty and SE is a closed convex subset, which implies that Problem OAP has a unique solution (ˆA,ˆB)SE by the optimal approximation theorem (see Ref. [27]). For the given matrices ˜A,˜BR2n×2n, write

    Q˜AJQ=[˜A11˜A12˜A21˜A22]p2npp2np,Q˜BJQ=[˜B11˜B12˜B21˜B22]p2npp2np, (3.1)

    then

    A˜A2+B˜B2=Q[RCR1A12A12A22]QJ˜A2+Q[RC˜ΛR1R˜ΛRA12A12R˜ΛR1B22]QJ˜B2=RCR1˜A112+A12˜A122+A12˜A212+A22˜A222+RC˜ΛR1˜B112+R˜ΛRA12˜B122+A12R˜ΛR1˜B212+B22˜B222.

    Therefore, A˜A2+B˜B2 = min if and only if

    f(C)=RCR1˜A112+RC˜ΛR1˜B112=min, (3.2)
    A12˜A122+R˜ΛRA12˜B122+A12˜A212+A12R˜ΛR1˜B212=min, (3.3)
    A22˜A222=min, (3.4)
    B22˜B222=min. (3.5)

    Let

    R1=[R1R2],  (3.6)

    where

    R1=[R1,1R1,2l1], R2=[R2,2l+1R2,p],

    and R1,2i1R2×p, R2,jR1×p (i=1,,l,j=2l+1,,p). Furthermore, let

    {D2i1=R1,2i1F1R1,2i1,D2i=R1,2i1F2R1,2i1,Dj=R2,jR2,j,E2i1=R1,2i1F1˜Λ2i1R1,2i1,E2i=R1,2i1F2˜Λ2i1R1,2i1,Ej=R2,jλjR2,j,i=1,,l,j=2l+1,,p, (3.7)

    where

    F1=[1001], F2=[0110].

    Then the relation of (3.2) is equivalent to

    f(C)=f(a1,b1,,a2l1,b2l1,c2l+1,,cp)=a1D1+b1D2++a2l1D2l1+b2l1D2l+c2l+1D2l+1++cpDp˜A112+a1E1+b1E2++a2l1E2l1+b2l1E2l+c2l+1E2l+1++cpEp˜B112=min,

    that is,

    f(a1,b1,,a2l1,b2l1,c2l+1,,cp)=tr[(a1D1+b1D2++a2l1D2l1+b2l1D2l+c2l+1D2l+1++cpDp˜A11)(a1D1+b1D2++a2l1D2l1+b2l1D2l+c2l+1D2l+1++cpDp˜A11)+(a1E1+b1E2++a2l1E2l1+b2l1E2l+c2l+1E2l+1++cpEp˜B11)(a1E1+b1E2++a2l1E2l1+b2l1E2l+c2l+1E2l+1++cpEp˜B11)]=a21g1,1+2a1b1g1,2++2a1a2l1g1,2l1+2a1b2l1g1,2l+2a1c2l+1g1,2l+1++2a1cpg1,p2a1h1+b21g2,2++2a2l1b1g2,2l1+2b1b2l1g2,2l+2b1c2l+1g2,2l+1++2b1cpg2,p2b1h2+,+a22l1g2l1,2l1+2a2l1b2l1g2l1,2l+2a2l1c2l+1g2l1,2l+1++2a2l1cpg2l1,p2a2l1h2l1+b22l1g2l,2l+2b2l1c2l+1g2l,2l+1++2b2l1cpg2l,p2b2l1h2l+c22l+1g2l+1,2l+1++2c2l+1cpg2l+1,p2c2l+1h2l+1+,+c2pgp,p2cphp+e,

    where gm,n=tr(DmDn)+tr(EmEn), hm=tr(Dm˜A11)+tr(Em˜B11), e=tr(˜A11˜A11)+tr(˜B11˜B11), m,n=1,,p.

    Consequently,

    f(a1,b1,,a2l1,b2l1,c2l+1,,cp)a1=2a1g1,1+2b1g1,2++2a2l1g1,2l1+2b2l1g1,2l+2c2l+1g1,2l+1++2cpg1,p2h1,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)b1=2a1g2,1+2b1g2,2++2a2l1g2,2l1+2b2l1g2,2l+2c2l+1g2,2l+1++2cpg2,p2h2,,,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)a2l1=2a1g2l1,1+2b1g2l1,2++2a2l1g2l1,2l1+2b2l1g2l1,2l+2c2l+1g2l1,2l+1++2cpg2l1,p2h2l1,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)b2l1=2a1g2l,1+2b1g2l,2++2a2l1g2l,2l1+2b2l1g2l,2l+2c2l+1g2l,2l+1++2cpg2l,p2h2l,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)c2l+1=2a1g2l+1,1+2b1g2l+1,2++2a2l1g2l+1,2l1+2b2l1g2l+1,2l+2c2l+1g2l+1,2l+1++2cpg2l+1,p2h2l+1,,,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)cp=2a1gp,1+2b1gp,2++2a2l1gp,2l1+2b2l1gp,2l+2c2l+1gp,2l+1++2cpgp,p2hp.

    Clearly, f(a1,b1,,a2l1,b2l1,c2l+1,,cp) = min if and only if

    f(a1,b1,,a2l1,b2l1,c2l+1,,cp)a1=0,,f(a1,b1,,a2l1,b2l1,c2l+1,,cp)cp=0.

    Therefore,

    a1g1,1+b1g1,2++a2l1g1,2l1+b2l1g1,2l+c2l+1g1,2l+1++cpg1,p=h1,a1g2,1+b1g2,2++a2l1g2,2l1+b2l1g2,2l+c2l+1g2,2l+1++cpg2,p=h2,,,a1g2l1,1+b1g2l1,2++a2l1g2l1,2l1+b2l1g2l1,2l+c2l+1g2l1,2l+1++cpg2l1,p=h2l1,a1g2l,1+b1g2l,2++a2l1g2l,2l1+b2l1g2l,2l+c2l+1g2l,2l+1++cpg2l,p=h2l,a1g2l+1,1+b1g2l+1,2++a2l1g2l+1,2l1+b2l1g2l+1,2l+c2l+1g2l+1,2l+1++cpg2l+1,p=h2l+1,,,a1gp,1+b1gp,2++a2l1gp,2l1+b2l1gp,2l+c2l+1gp,2l+1++cpgp,p=hp. (3.8)

    If let

    G=[g1,1g1,2g1,2l1g1,2lg1,2l+1g1,pg2,1g2,2g2,2l1g2,2lg2,2l+1g2,pg2l1,1g2l1,2g2l1,2l1g2l1,2lg2l1,2l+1g2l1,pg2l,1g2l,2g2l,2l1g2l,2lg2l,2l+1g2l,pg2l+1,1g2l+1,2g2l+1,2l1g2l+1,2lg2l+1,2l+1g2l+1,pgp,1gp,2gp,2l1gp,2lgp,2l+1gp,p],
    T=[a1b1a2l1b2l1c2l+1cp], H=[h1h2h2l1h2lh2l+1hp],

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

    GT=H, (3.9)

    and the solution of Eq (3.9) is

    T=G1H. (3.10)

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

    f(A12)=tr[(A12˜A12)(A12˜A12)]+tr[(A12P˜B12)(PA12˜B12)]+tr[(A12˜A21)(A12˜A21)]+tr[(PA12˜B21)(A12P˜B21)].

    Thus,

    f(A12)A12=2A122˜A12+2PPA122P˜B12+2A122˜A21+2PPA122P˜B21,

    setting f(A12)A12=0, we obtain

    A12=12(Ip+PP)1(˜A12+P˜B12+˜A21+P˜B21), (3.11)

    where P=R˜ΛR1. A22, B22 are symmetric matrices implies that the relations of (3.4) and (3.5) are equivalent to

    A22˜A222=A2212(˜A22+˜A22)2+12(˜A22˜A22)2,
    B22˜B222=B2212(˜B22+˜B22)2+12(˜B22˜B22)2,

    therefore, we have

    A22=12(˜A22+˜A22), B22=12(˜B22+˜B22). (3.12)

    Theorem 3.1. Given ˜A,˜BR2n×2n, then the Problem OAP has a unique solution and the unique solution of Problem OAP is

    ˆA=Q[RTCR1A12A12A22]QJ, ˆB=Q[RTC˜ΛR1RT˜ΛRA12A12R˜ΛR1B22]QJ, (3.13)

    where A12 and  A22, B22 are given by (3.11) and (3.12), and a1,b1,,a2l1,b2l1,c2l+1,,cp 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 Λ, X, J, ˜A, ˜B.

    2). Compute real-valued matrices ˜Λ, ˜X by (2.1) and (2.2), respectively.

    3). Compute the QR-decomposition of the matrix J˜X by (2.5).

    4). Compute ˜Aij, ˜Bij by (3.1), i,j=1,2.

    5). Compute R1 by (3.6) to form R1,R2.

    6). Compute D2i1,D2i,Dj, E2i1,E2i and Ej (i=1,,l,j=2l+1,,p) by (3.7).

    7). Compute gm,n=tr(DmDn)+tr(EmEn) and hm=tr(Dm˜A11)+tr(Em˜B11) (m,n=1,,p).

    8). Compute a1,b1,,a2l1,b2l1,c2l+1,,cp by (3.10).

    9). Compute A12 by (3.11).

    10). Compute A22 and B22 by (3.12).

    11). Compute ˆA and ˆB by (3.13).

    Remark 4.1. After statistics, we find that the amount of computations required by Algorithm 1 is about p5+p4+193p3+14np2+64n3 flops.

    Example 4.1. Let n=5, p=5, and the matrices Λ, X, ˜A and ˜B be given by

    Λ=diag{0.2218+2.0231i, 0.22182.0231i, 0.1617+0.5721i, 0.16170.5721i, 2.7670},
    X=[0.63770.1444i0.6377+0.1444i0.0405+0.3341i0.04050.3341i1.00000.26780.0983i0.2678+0.0983i0.26150.3973i0.2615+0.3973i0.11510.4260+0.5740i0.42600.5740i0.13480.2946i0.1348+0.2946i0.02780.20320.0489i0.2032+0.0489i0.5238+0.4762i0.52380.4762i0.03570.21110.1510i0.2111+0.1510i0.79820.1668i0.7982+0.1668i0.67930.5233+0.1151i0.52330.1151i0.30330.4132i0.3033+0.4132i0.10910.48200.2541i0.4820+0.2541i0.13980.0715i0.1398+0.0715i0.85500.31830.3431i0.3183+0.3431i0.27160.3411i0.2716+0.3411i0.64360.1376+0.3007i0.13760.3007i0.1391+0.0981i0.13910.0981i0.08830.0672+0.0189i0.06720.0189i0.3143+0.1462i0.31430.1462i0.2828],
    ˜A=[1.83513.06359.39001.94769.79751.17427.30336.24062.61879.03723.68485.08518.75942.25924.38872.96684.88616.79143.35368.90926.25625.10775.50161.70711.11123.18785.78533.95526.79733.34167.80238.17636.22482.27662.58064.24172.37283.67441.36556.98750.81137.94835.87044.35704.08725.07864.58859.87987.21231.97819.29396.44322.07743.11105.94900.85529.63090.37741.06760.30547.75713.78613.01259.23382.62212.62485.46818.85176.53767.44074.86798.11584.70924.30216.02848.01015.21149.13294.94175.00024.35865.32832.30491.84827.11220.29222.31597.96187.79054.79924.46783.50738.44319.04882.21759.28854.88900.98717.15049.0472],
    ˜B=[6.09871.67930.96734.53803.99261.06224.22846.66533.68921.20616.17679.78688.18154.32395.26883.72415.47871.78134.60735.89518.59447.12698.17558.25314.16801.98129.42741.28019.81642.26198.05495.00477.22440.83476.56864.89694.17749.99081.56403.84625.76724.71091.49871.33176.27973.39499.83051.71128.55525.82991.82920.59626.59611.73392.91989.51633.01450.32606.44762.51812.39936.81975.18593.90944.31659.20337.01105.61203.76272.90448.86510.42439.72978.31380.15490.52686.66348.81871.90926.17090.28670.71456.48998.03369.84067.37865.39136.69184.28252.65284.89905.21658.00330.60471.67172.69126.98111.90434.82028.2438].

    By applying Algorithm 4.1, we can obtain the unique solution (ˆA,ˆB) of Problem OAP as follows:

    ˆA=[3.01800.42250.15550.84020.91254.00713.88403.21214.01786.81730.35581.77260.31492.43002.44213.88401.97345.09882.43346.21851.87663.01890.95031.17321.87903.21215.09882.68966.67505.40581.95100.12940.27630.76681.90084.01782.43346.67502.91955.23163.02110.22730.60750.92172.78186.81736.21855.40585.23163.96656.37316.92885.53224.43975.43563.01800.35581.87661.95103.02116.92884.52316.29467.20493.68510.42251.77263.01890.12940.22735.53226.29464.84533.77064.46180.15550.31490.95030.27630.60754.43977.20493.77062.82496.62390.84022.43001.17320.76680.92175.43563.68514.46186.62392.84680.91252.44211.87901.90082.7818],
    ˆB=[0.67091.23810.75571.12181.98351.10003.78173.82630.63763.54841.34510.24560.02570.28620.29153.78175.35595.32243.69478.62403.00951.21711.09330.95152.12923.82635.32241.21476.80132.43242.09941.87543.14770.19030.67930.63763.69476.80133.36917.25081.40470.30321.65472.56971.22023.54848.62402.43247.25086.05582.79992.65385.93684.05301.38080.67091.34513.00952.09941.40472.65385.33902.35232.63264.12081.23810.24561.21711.87540.30325.93682.35238.41936.94804.55980.75570.02571.09333.14771.65474.05302.63266.94808.08385.23731.12180.28620.95150.19032.56971.38084.12084.55985.23733.40281.98350.29152.12920.67931.2202],

    and

    ˆAXΛˆBX=2.0806×1014, 

    which implies that ˆAXΛ=ˆBX 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 Λ, X, ˜B and ˜LC be given by

    Λ=diag{0.0822, 0.0250, 0, 0.0757}diag{λ1,λ2,λ3,λ4},
    X=[0.57600.26840.00000.13400.15100.06730.00001.00000.38900.06100.00000.12460.03590.14010.00000.99070.34091.00001.00000.03660.00000.00000.00000.00000.56720.06650.00000.30721.00000.12880.00000.13950.43280.06230.00000.16770.06500.22820.20000.1508]diag{f1, f2, f3, f4},
    ˜B=[0.70000.54830.53690.69190.603200.80830.03060.77150.02340.54831.66171.34250.60371.48600.808300.42370.53570.92920.53691.34251.50750.91121.03720.03060.423700.43040.04880.69190.60370.91121.55830.94590.77150.53570.430400.02550.60321.48601.03720.94590.67420.02340.92920.04880.0255000.80830.03060.77150.02340.70000.54830.53690.69190.60320.808300.42370.53570.92920.54831.66171.34250.60371.48600.03060.423700.43040.04880.53691.34251.50750.91121.03720.77150.53570.430400.02550.69190.60370.91121.55830.94590.02340.92920.04880.025500.60321.48601.03720.94590.6742],
    ˜LC=[1.70002.54833.53694.69199.396801.61660.06121.54300.04682.54834.66175.34259.39632.48601.616600.84741.07141.85843.53695.34258.49251.91123.03720.06120.847400.86080.09764.69199.39631.91123.55833.94591.54301.07140.860800.05109.39682.48603.03723.94594.67420.04681.85840.09760.0510001.61660.06121.54300.04681.70002.54833.53694.69199.39681.616600.84741.07141.85842.54834.66175.34259.39632.48600.06120.847400.86080.09763.53695.34258.49251.91123.03721.54301.07140.860800.05104.69199.39631.91123.55833.94590.04681.85840.09760.051009.39682.48603.03723.94594.6742].

    By calculating, we can obtain the unique solution (ˆB,ˆLC) of Problem OAP as follows:

    ˆB=[1.08660.08470.05200.05010.10930.00000.03300.08330.05030.01340.15911.73800.25220.34670.24140.03300.00000.01910.00870.02340.13450.23151.65970.26560.22220.08330.01910.00000.02060.02300.00490.35830.29451.64960.34720.05030.00870.02060.00000.1716 0.27240.05190.11320.11590.87310.01340.02340.02300.17160.00000.00000.08640.20140.12210.02391.08660.15910.13450.00490.27240.08640.00000.04150.02440.12880.08471.73800.23150.35830.05190.20140.04150.00000.15860.15530.05200.25221.65970.29450.11320.12210.02440.15860.00000.00000.05010.34670.26561.64960.11590.02390.12880.15530.00000.00000.10930.24140.22220.34720.8731],
    ˆLC=[3.07493.01822.98553.39850.02210.00000.57880.01301.01660.11070.86720.74971.14290.20110.44370.57880.00000.06560.35282.21850.75900.83600.51930.65740.13000.01300.06560.00000.03130.65020.10650.34740.47520.50810.22631.01660.35280.03130.00001.13140.11390.69460.62940.15200.00000.11072.21850.65021.13140.00000.00000.34350.74670.41580.02283.07490.86720.75900.10650.11390.34350.00000.92590.17430.13893.01820.74970.83600.34740.69460.74670.92590.00001.32160.12592.98551.14290.51930.47520.62940.41580.17431.32160.00000.03043.39850.20110.65740.50810.15200.02280.13890.12590.03040.00000.02210.44370.13000.22630.0000].

    Furthermore, we can obtain the following numerical results: Therefore, the new model ˆBXΛ=ˆLCX reproduces the desired eigenvalues and eigenvectors.

    Table 4.1.  Residuals of the eigenpairs (λi,fi).
    (λi,fi) (λ1,f1) (λ2,f2) (λ3,f3) (λ4,f4)
    λiˆBfiˆLCfi 1.1762×1015 1.1322×1015 1.0372×1015 9.7141×1016

     | 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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