Citation: Jiao Xu, Hairui Zhang, Lina Liu, Huiting Zhang, Yongxin Yuan. A unified treatment for the restricted solutions of the matrix equation AXB=C[J]. AIMS Mathematics, 2020, 5(6): 6594-6608. doi: 10.3934/math.2020424
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Throughout this paper, the complex m×n matrix space is denoted by Cm×n. The conjugate transpose, the Moore-Penrose inverse, the range space and the null space of a complex matrix A∈Cm×n are denoted by AH, A+, R(A) and N(A), respectively. In denotes the n×n identity matrix. PL stands for the orthogonal projector on the subspace L⊂Cn. Furthermore, for a matrix A∈Cm×n, EA and FA stand for two orthogonal projectors: EA=Im−AA+=PN(AH), FA=In−A+A=PN(A).
A number of papers have been published for solving linear matrix equations. For example, Chen et al. [1] proposed LSQR iterative method to solve common symmetric solutions of matrix equations AXB=E and CXD=F. Zak and Toutounian [2] studied the matrix equation of AXB=C with nonsymmetric coefficient matrices by using nested splitting conjugate gradient (NSCG) iteration method. By applying a Hermitian and skew-Hermitian splitting (HSS) iteration method, Wang et al. [3] computed the solution of the matrix equation AXB=C. Tian et al. [4] obtained the solution of the matrix equation AXB=C by applying the Jacobi and Gauss-Seidel-type iteration methods. Liu et al. [5] solved the matrix equation AXB=C by employing stationary splitting iterative methods. In addition, some scholars studied matrix equations by direct methods. Yuan and Dai [6] obtained generalized reflexive solutions of the matrix equation AXB=D and the optimal approximation solution by using the generalized singular value decomposition. Zhang et al. [7] provided the explicit expression of the minimal norm least squares Hermitian solution of the complex matrix equation AXB+CXD=E by using the structure of the real representations of complex matrices and the Moore-Penrose inverse. By means of the definitions of the rank and inertias of matrices, Song and Yu [8] obtained the existence conditions and expressions of the nonnegative (positive) definite and the Re-nonnegative (Re-positive) definite solutions to the matrix equations AXAH=C and BXBH=D.
In this paper, we will focus on the restricted solutions to the following well-known linear matrix equation
AXB=C, | (1) |
where A∈Cm×n,B∈Cn×p and C∈Cm×p. We observe that the structured matrix, such as Hermitian matrix, skew-Hermitian matrix, Re-nonnegative definite matrix and Re-positive definite matrix, is of important applications in structural dynamics, numerical analysis, stability and robust stability analysis of control theory and so on [9,10,11,12,13,14]. Conditions for the existence of Hermitian solutions to Eq (1) were studied in [15,16,17]. A solvability criterion for the existence of Re-nonnegative definite solutions of Eq (1) by using generalized inner inverses was investigated by Cvetković-Ilić [19]. Recently, a direct method for solving Eq (1) by using the generalized inverses of matrices and orthogonal projectors was proposed by Yuan and Zuo [21]. In addition, the Re-nonnegative definite and Re-positive definite solutions to some special cases of Eq (1) were considered by Wu [22], Wu and Cain [23] and Groß [24]. In [25], necessary and sufficient conditions for the existence of common Re-nonnegative definite and Re-positive definite solutions to the matrix equations AX=C,XB=D were discussed by virtue of the extremal ranks of matrix polynomials.
In this paper, necessary and sufficient conditions for the existence of Hermitian (skew-Hermitian), Re-nonnegative (Re-positive) definite, and Re-nonnegative (Re-positive) definite least-rank solutions to Eq (1) are deduced by using the Moore-Penrose inverse of matrices, and the explicit representations of the general solutions are given when the solvability conditions are satisfied. Compared with the approaches proposed in [18,19,20,21], the coefficient matrices of Eq (1) have no any constraints and the method in this paper is straightforward and easy to implement.
Definition 1. A matrix A∈Cn×n is said to be Re-nonnegative definite (Re-nnd) if H(A):=12(A+AH) is Hermitian nonnegative definite (i.e., H(A)≥0), and A is said to be Re-positive definite (Re-pd) if H(A) is Hermitian positive definite (i.e., H(A)>0). The set of all Re-nnd (Re-pd) matrices in Cn×n is denoted by RNDn×n (RPDn×n).
Lemma 1. [26] Let A∈Cm×n,B∈Cn×p and C∈Cm×p. Then the matrix equation AXB=C is solvable if and only if AA+CB+B=C. In this case, the general solution can be written in the following parametric form
X=A+CB++FAL1+L2EB, |
where L1,L2∈Cn×n are arbitrary matrices.
Lemma 2. [27,28] Let B1∈Cl×q,D1∈Cl×l. Then the matrix equation
YBH1±B1YH=D1, |
has a solution Y∈Cl×q if and only if
D1=±DH1, EB1D1EB1=0. |
In which case, the general solution is
Y=12D1(B+1)H+12EB1D1(B+1)H+2V−VB+1B1∓B1VH(B+1)H−EB1VB+1B1, |
where V∈Cl×q is an arbitrary matrix.
Lemma 3. [29] Let A∈Cm×n and B∈Cm×p. Then the Moore-Penrose inverse of the matrix [A,B] is
[A,B]+=[(I+TTH)−1(A+−A+BC+)C++TH(I+TTH)−1(A+−A+BC+)], |
where C=(I−AA+)B and T=A+B(I−C+C).
Lemma 4. [30] Suppose that a Hermitian matrix M is partitioned as
M=[M11M12MH12M22], |
where M11, M22 are square. Then
(i). M is Hermitian nonnegative definite if and only if
M11≥0, M11M+11M12=M12, M22−MH12M+11M12≜H2≥0. |
In which case, M can be expressed as
M=[M11M11H1HH1M11H2+HH1M11H1], |
where H1 is an arbitrary matrix and H2 is an arbitrary Hermitian nonnegative definite matrix.
(ii). M is Hermitian positive definite if and only if
M11>0, M22−MH12M−111M12≜H3>0. |
In the case, M can be expressed as
M=[M11M12MH12H3+MH12M−111M12], |
where H3 is an arbitrary Hermitian positive definite matrix.
Lemma 5. [31] Let
M=[CAB0], N=[0In], S=[0In], N1=NFM, S1=EMS. |
Then the general least-rank solution to Eq.(1) can be written as
X=−NM+S+N1R1+R2S1, |
where R1∈C(p+n)×n,R2∈Cn×(m+n) are arbitrary matrices.
Theorem 1. Eq (1) has a Hermitian solution X if and only if
AA+CB+B=C,PT(A+CB+−(A+CB+)H)PT=0, | (2) |
where T=R(AH)∩R(B). In which case, the general Hermitian solution of Eq (1) is
X=A+CB++FAL1+L2EB, | (3) |
where
L1=−P1D1+12P1D1WH1+2VH1−2FAZH1+P1(V1FA−V2EB)+FAZH1WH1, | (4) |
L2=D1QH1−12W1D1QH1+2V2+2Z1EB+(FAVH1−EBVH2)QH1−W1Z1EB, | (5) |
D1=A+CB+−(A+CB+)H, C1=−(I−FAF+A)EB, T1=−F+AEB(I−C+1C1),P1=(I+T1TH1)−1(F+A+F+AEBC+1), Q1=C+1+TH1P1,W1=FAP1−EBQ1, Z1=V1P1+V2Q1, |
and V1,V2∈Cn×n are arbitrary matrices.
Proof. By Lemma 1, if the first condition of (2) holds, then the general solution of Eq (1) is given by (3). Now, we will find L1 and L2 such that AXB=C has a Hermitian solution, that is,
A+CB++FAL1+L2EB=(A+CB+)H+LH1FA+EBLH2. | (6) |
Clearly, Eq (6) can be equivalently written as
X1AH1−A1XH1=D1, | (7) |
where A1=[FA,−EB], X1=[LH1,L2], D1=A+CB+−(A+CB+)H.
By Lemma 2, Eq (7) has a solution X1 if and only if
D1=−DH1, EA1D1EA1=0. | (8) |
The first condition of (8) is obviously satisfied. And note that
EA1=PN(AH1)=PN(FA)∩N(EB)=PR(AH)∩R(B). |
Thus, the second condition of (8) is equivalent to
PTD1PT=0, |
where T=R(AH)∩R(B), which is the second condition of (2). In which case, the general solution of Eq (7) is
X1=D1(A+1)H−12A1A+1D1(A+1)H+2V−2VA+1A1+A1VH(A+1)H+A1A+1VA+1A1, | (9) |
where V=[V1,V2] is an arbitrary matrix. By Lemma 3, we have
[FA,−EB]+=[(I+T1TH1)−1(F+A+F+AEBC+1)C+1+TH1(I+T1TH1)−1(F+A+F+AEBC+1)], | (10) |
where C1=−(I−FAF+A)EB and T1=−F+AEB(I−C+1C1). Upon substituting (10) into (9), we can get (4) and (5).
Corollary 1. Eq (1) has a skew-Hermitian solution X if and only if
AA+CB+B=C,PT(A+CB++(A+CB+)H)PT=0, | (11) |
where T=R(AH)∩R(B). In which case, the general skew-Hermitian solution of Eq (1) is
X=A+CB++FAL3+L4EB, | (12) |
where
L3=P2D2−12P2D2WH2+2VH3−2FAZH2−P2(V3FA+V4EB)+FAZH2WH2, | (13) |
L4=D2QH2−12W2D2QH2+2V4−2Z2EB−(FAVH3+EBVH4)QH2+W2Z2EB, | (14) |
D2=−A+CB+−(A+CB+)H, C2=(I−FAF+A)EB, T2=F+AEB(I−C+2C2),P2=(I+T2TH2)−1(F+A−F+AEBC+2), Q2=C+2+TH2P2,W2=FAP2+EBQ2, Z2=V3P2+V4Q2, |
and V3, V4∈Cn×n are arbitrary matrices.
Theorem 2. Let A∈Cm×n, B∈Cn×p, C∈Cm×p and T=R(AH)∩R(B). Assume that the spectral decomposition of PT is
PT=U[Is000]UH, | (15) |
where U=[U1,U2]∈Cn×n is a unitary matrix and s=dim(T). Then
(a). Eq (1) has a Re-nnd solution if and only if
AA+CB+B=C, UH1A+CB+U1∈RNDs×s. | (16) |
In which case, the general Re-nnd solution of (1) is
X=A+CB++FAJ1+J2EB, | (17) |
where
J1=P3D3−12P3D3WH3+2VH5−2FAZH3−P3(V5FA+V6EB)+FAZH3WH3, | (18) |
J2=D3QH3−12W3D3QH3+2V6−2Z3EB−(FAVH5+EBVH6)QH3+W3Z3EB, | (19) |
D3=K−A+CB+−(A+CB+)H, C3=(I−FAF+A)EB,T3=F+AEB(I−C+3C3), P3=(I+T3TH3)−1(F+A−F+AEBC+3), Q3=C+3+TH3P3,W3=FAP3+EBQ3, Z3=V5P3+V6Q3, K11=UH1(A+CB++(A+CB+)H)U1,K=U[K11K11H1HH1K11H2+HH1K11H1]UH, |
V5,V6∈Cn×n,H1∈Cs×(n−s) are arbitrary matrices, and H2∈C(n−s)×(n−s) is an arbitrary Hermitian nonnegative definite matrix.
(b). Eq (1) has a Re-pd solution if and only if
AA+CB+B=C, UH1A+CB+U1∈RPDs×s. | (20) |
In which case, the general Re-pd solution of (1) is
X=A+CB++FAJ1+J2EB, | (21) |
where
K=U[K11K12KH12H3+KH12K−111K12]UH, |
J1,J2,D3,C3,T3,P3,Q3,W3,Z3 and K11 are given by (18) and (19), K12∈Cs×(n−s) is an arbitrary matrix and H3∈C(n−s)×(n−s) is an arbitrary Hermitian positive definite matrix.
Proof. By Lemma 1, if the first condition of (16) holds, then the general solution of Eq (1) is given by (17). Now, we will find J1 and J2 such that AXB=C has a Re-nnd (Re-pd) solution, that is, we will choose suitable matrices J1 and J2 such that
A+CB++(A+CB+)H+FAJ1+JH1FA+J2EB+EBJH2≜K≥0 (K>0). | (22) |
Clearly, Eq (22) can be equivalently written as
X3AH3+A3XH3=D3, | (23) |
where A3=[FA,EB], X3=[JH1,J2], D3=K−A+CB+−(A+CB+)H.
By Lemma 2, Eq (23) has a solution X1 if and only if
D3=DH3, EA3D3EA3=0. | (24) |
The first condition of (24) is obviously satisfied. And note that
EA3=PN(AH3)=PN(FA)∩N(EB)=PR(AH)∩R(B). |
Then the second condition of (24) is equivalent to
PTD3PT=0, |
where T=R(AH)∩R(B). By (15), we can obtain
[Is000]UHKU[Is000]=[Is000]UH(A+CB++(A+CB+)H)U[Is000], | (25) |
Let
UHKU=[K11K12KH12K22], | (26) |
where U=[U1,U2]. By (25), we can obtain
K11=UH1(A+CB++(A+CB+)H)U1, | (27) |
it is easily known that K≥0 (K>0) if and only if UHKU≥0 (UHKU>0). And X is Re-nnd (Re-pd) if and only if K≥0 (K>0). Thus, by (26) and (27), we can get
K≥0⟺K11=UH1(A+CB++(A+CB+)H)U1≥0,K>0⟺K11=UH1(A+CB++(A+CB+)H)U1>0, |
equivalently,
K≥0⟺UH1A+CB+U1∈RNDs×s,K>0⟺UH1A+CB+U1∈RPDs×s, |
which are the second conditions of (16) and (20). In which case, by Lemma 4,
K≥0⟺K=U[K11K11H1HH1K11H2+HH1K11H1]UH, |
K>0⟺K=U[K11K12KH12H3+KH12K−111K12]UH, |
where H1∈Cs×(n−s) is an arbitrary matrix, H2∈C(n−s)×(n−s) is an arbitrary Hermitian nonnegative definite matrix and H3∈C(n−s)×(n−s) is an arbitrary Hermitian positive definite matrix. And the general solution of Eq (23) is
X3=D3(A+3)H−12A3A+3D3(A+3)H+2V−2VA+3A3−A3VH(A+3)H+A3A+3VA+3A3, | (28) |
where V=[V5,V6] is an arbitrary matrix. By Lemma 3, we have
[FA,EB]+=[(I+T3TH3)−1(F+A−F+AEBC+3)C+3+TH3(I+T3TH3)−1(F+A−F+AEBC+3)], | (29) |
where C3=(I−FAF+A)EB and T3=F+AEB(I−C+3C3). Upon substituting (29) into (28), we can get (18) and (19).
Theorem 3. Let A∈Cm×n,B∈Cn×p, C∈Cm×p and ˜T=N(NH1)∩N(S1). Assume that the spectral decomposition of P˜T is
P˜T=˜U[Ik000]˜UH, | (30) |
where ˜U=[˜U1,˜U2]∈Cn×n is a unitary matrix and k=dim(˜T), and M,N,S,N1,S1 are given by Lemma 5. Then
(a). Eq (1) has a Re-nnd least-rank solution if and only if
AA+CB+B=C, ˜UH1(−NM+S)˜U1∈RNDk×k. | (31) |
In which case, the general Re-nnd least-rank solution of Eq (1) is
X=−NM+S+N1R1+R2S1, | (32) |
where
R1=P4D4−12P4D4WH4+2VH7−2NH1ZH4−P4(V7NH1+V8S1)+NH1ZH4WH4, | (33) |
R2=D4QH4−12W4D4QH4+2V8−2Z4SH1−(N1VH7+SH1VH8)QH4+W4Z4SH1, | (34) |
D4=˜K+NM+S+(NM+S)H, C4=(I−N1N+1)SH1, T4=N+1SH1(I−C+4C4),P4=(I+T4TH4)−1(N+1−N+1SH1C+4), Q4=C+4+TH4P4,W4=N1P4+SH1Q4, Z4=V7P4+V8Q4, ˜K11=˜UH1(−NM+S−(NM+S)H)˜U1,˜K=˜U[˜K11˜K11˜H1˜HH1˜K11˜H2+˜HH1˜K11˜H1]˜UH, |
V7∈Cn×(p+n), V8∈Cn×(m+n), ˜H1∈Ck×(n−k) are arbitrary matrices, and ˜H2∈C(n−k)×(n−k) is an arbitrary Hermitian nonnegative definite matrix.
(b). Eq (1) has a Re-pd least-rank solution if and only if
AA+CB+B=C, ˜UH1(−NM+S)˜U1∈RPDk×k. | (35) |
In which case, the general Re-nnd least-rank solution of Eq (1) is
X=−NM+S+N1R1+R2S1, | (36) |
where
˜K=˜U[˜K11˜K12˜KH12˜H3+˜KH12˜K−111˜K12]˜UH, |
R1,R2,D4,C4,T4,P4,Q4,W4,Z4 and ˜K11 are given by (33) and (34), ˜K12∈Ck×(n−k) is an arbitrary matrix and ˜H3∈C(n−k)×(n−k) is an arbitrary Hermitian positive definite matrix.
Proof. By Lemmas 1 and 5, if the first condition of (31) holds, then the least-rank solution of Eq (1) is given by (32). Now, we will find R1 and R2 such that AXB=C has a Re-nnd (Re-pd) least-rank solution, that is, we will choose suitable matrices R1 and R2 such that
−NM+S−(NM+S)H+N1R1+RH1NH1+R2S1+SH1RH2≜˜K≥0 (˜K>0). | (37) |
Clearly, Eq (37) can be equivalently written as
X4AH4+A4XH4=D4, | (38) |
where A4=[N1,SH1], X4=[RH1,R2], D4=˜K+NM+S+(NM+S)H.
By Lemma 2, Eq (38) has a solution X4 if and only if
D4=DH4, EA4D4EA4=0. | (39) |
The first condition of (39) is obviously satisfied. And note that
EA4=PN(AH4)=PN(NH1)∩N(S1). |
Thus, the second condition of (39) is equivalent to
P˜TD4P˜T=0, |
where ˜T=N(NH1)∩N(S1). By (30), we can obtain
[Ik000]˜UH˜K˜U[Ik000]=[Ik000]˜UH(−NM+S−(NM+S)H)˜U[Ik000], | (40) |
Let
˜UH˜K˜U=[˜K11˜K12˜KH12˜K22], | (41) |
where ˜U=[˜U1,˜U2]. By (40), we can obtain
˜K11=˜UH1(−NM+S−(NM+S)H)˜U1, | (42) |
it is easily known that ˜K≥0 (˜K>0) if and only if ˜UH˜K˜U≥0 (˜UH˜K˜U>0). And X is Re-nnd (Re-pd) least-rank solution if and only if ˜K≥0 (˜K>0). Thus, by (41) and (42), we can get
˜K≥0⟺˜K11=˜UH1(−NM+S−(NM+S)H)˜U1≥0,˜K>0⟺˜K11=˜UH1(−NM+S−(NM+S)H)˜U1>0, |
equivalently,
˜K≥0⟺˜UH1(−NM+S)˜U1∈RNDk×k,˜K>0⟺˜UH1(−NM+S)˜U1∈RPDk×k, |
which are the second conditions of (31) and (35). In which case, by Lemma 4,
˜K≥0⟺˜K=˜U[˜K11˜K11˜H1˜HH1˜K11˜H2+HH1˜K11˜H1]˜UH, |
˜K>0⟺˜K=˜U[˜K11˜K12˜KH12˜H3+˜KH12˜K−111˜K12]˜UH, |
where ˜H1∈Ck×(n−k) is an arbitrary matrix, ˜H2∈C(n−k)×(n−k) is an arbitrary Hermitian nonnegative definite matrix and ˜H3∈C(n−k)×(n−k) is an arbitrary Hermitian positive definite matrix. And the general solution of Eq (38) is
X4=D4(A+4)H−12A4A+4D4(A+4)H+2V−2VA+4A4−A4VH(A+4)H+A4A+4VA+4A4, | (43) |
where V=[V7,V8] is an arbitrary matrix. By Lemma 3, we have
[N1,SH1]+=[(I+T4TH4)−1(N+1−N+1SH1C+4)C+4+TH4(I+T4TH4)−1(N+1−N+1SH1C+4)], | (44) |
where C4=(I−N1N+1)SH1 and T4=N+1SH1(I−C+4C4). Upon substituting (44) into (43), we can get (33) and (34).
The following example comes from [9].
Example 1. Consider a 7-DOF system modelled analytically with the first three measured modal data given by
Λ=diag(3.5498, 101.1533, 392.8443), X=[0.55850.4751−0.4241−0.0841−0.23530.28380.3094−0.17170.2512−0.0800−0.16460.08520.0996−0.3562−0.0508−0.05530.0404−0.21050.0084−0.1788−0.4113]. |
and the corrected symmetric mass matrix M and symmetric stiffness matrix K should satisfy the orthogonality conditions, that is,
X⊤MX=I3,X⊤KX=Λ. |
Since
‖X⊤(X⊤)+X+X−I3‖=1.5442×10−15,‖PT((X⊤)+X+−((X⊤)+X+)⊤)PT‖=0, |
which means that the conditions of (2) are satisfied. Choose V1=0,V2=0. Then, by the equation of (3), we can obtain a corrected mass matrix given by
M=[1.1968−0.10730.8201−0.16780.1977−0.2079−0.1439−0.10730.30570.62920.09980.2953−0.2547−0.20950.82010.62922.27480.13471.1398−0.7290−0.3249−0.16780.09980.13470.09980.34270.01770.28040.19770.29531.13980.34271.87750.03701.4878−0.2079−0.2547−0.72900.01770.03700.37420.6461−0.1439−0.2095−0.32490.28041.48780.64612.1999], |
and
‖X⊤MX−I3‖=1.5016×10−15, |
which implies that M is a symmetric solution of X⊤MX=I3.
Since
‖X⊤(X⊤)+ΛX+X−Λ‖=4.0942×10−13,‖PT((X⊤)+ΛX+−((X⊤)+ΛX+)⊤)PT‖=1.5051×10−14, |
which means that the conditions of (2) are satisfied. Choose V1=0,V2=0. Then, by the equation of (3), we obtain a corrected stiffness matrix given by
K=[50.0364−47.8369−66.10521.462121.655862.090444.8915−47.836993.9748169.400741.942349.9653−78.9100−244.1315−66.1052169.4007176.9957−3.6625−103.7349−173.8782−169.81701.462141.9423−3.662525.7528−2.6730−189.498698.311021.655849.9653−103.7349−2.673095.3473−51.6395446.806262.0904−78.9100−173.8782−189.4986−51.639556.3200448.690044.8915−244.1315−169.817098.3110446.8062448.6900394.1690], |
and
‖X⊤KX−Λ‖=3.5473×10−13, |
which implies that K is a symmetric solution of X⊤KX=Λ.
Example 2. Given matrices
A=[7.94829.79751.36526.61445.82792.25959.56842.71450.11762.84414.23505.79815.22592.52338.93904.69225.15517.60378.80148.75741.99140.64783.33955.29821.72967.37312.98729.88334.32916.4053],B=[1.93433.78378.21633.41193.70416.82228.60016.44915.34087.02743.02768.53668.17977.27115.46575.41675.93566.60233.09294.44881.50874.96553.41978.38506.94576.97908.99772.89735.68076.2131],C=[745.63171194.55431060.3913995.63791010.8200535.5044831.5304791.3676684.6897711.7632845.43241338.10651123.73801077.06151096.2601629.07681006.4762928.6566804.5134824.7220868.41581299.0559995.07861012.70461054.9264]. |
Since ‖AA+CB+B−C‖=9.3907×10−12, and the eigenvalues of K11 are
Λ1=diag{0.3561,11.0230,6.8922,5,3274}, |
which means that the conditions of (16) are satisfied. Choose V,H1 and H2 as
V=[I6,I6], H1=[0.52110.67910.23160.39550.48890.36740.62410.9880], H2=[6000]. |
Then, by the equation of (17), we get a solution of Eq (1):
X=[7.9305−1.8515−1.13640.56150.54111.2825−1.72282.34252.30291.46770.4975−0.49200.85821.68172.52851.83251.0895−0.80953.0514−0.8250−0.79223.37411.52491.6918−6.17873.85492.5294−2.69452.45680.6923−0.19681.06000.32140.6383−0.31253.2860] |
with corresponding residual
‖AXB−C‖=6.4110×10−12. |
Furthermore, it can be computed that the eigenvalues of X+XH are
Λ2=diag{0, 0.3555, 3.9363, 7.0892, 11.1802, 21.2758}, |
which implies that X is a Re-nnd solution of Eq (1).
Example 3. Let the matrices A and B be the same as those in Example 2 and the matrix C be given by
C=[1841.23232726.84152555.29262172.76882365.39391281.07881957.59321843.72521614.62111699.39851826.13253077.61412600.81682539.01812467.35581583.35432422.31692153.66251910.01562035.21232065.87612950.77832362.00602197.71292389.1173]. |
Since ‖AA+CB+B−C‖=1.7944×10−11, and the eigenvalues of K11 are
Λ1=diag{25.3998, 20.3854, 19.4050, 19.0160}, |
which means that the conditions of (20) are satisfied. If select V,K12 and H3 as
V=[I6,I6], K12=[0.52110.67910.23160.39550.48890.36740.62410.9880], H3=[6008]. |
Then, by the equation of (21), we can achieve a Re-pd solution of Eq (1):
X=[7.07011.25000.86771.76852.7735−1.46891.01529.90320.1535−0.4383−0.05711.22970.80630.487410.0878−0.59562.02351.14262.16620.0136−0.83648.29371.87212.58871.66270.72791.28871.21985.66820.4925−1.60281.17371.80042.67180.27298.1489], |
and the eigenvalues of X+XH are
Λ2=diag{5.9780, 7.9395, 19.1387, 19.4286, 20.4585, 25.4004}. |
Furthermore, it can be computed that
‖AXB−C‖=1.7808×10−11. |
In this paper, we mainly consider some special solutions of Eq (1). By imposing some constraints on the expression X=A+CB++FAL1+L2EB, we succeed in obtaining a set of necessary and sufficient conditions for the existence of the Hermitian, skew-Hermitian, Re-nonnegative definite, Re-positive definite, Re-nonnegative definite least-rank and Re-positive definite least-rank solutions of Eq (1), respectively. Moreover, we give the explicit expressions for these special solutions, when the consistent conditions are satisfied.
The authors wish to give special thanks to the editor and the anonymous reviewers for their helpful comments and suggestions which have improved the presentation of the paper.
The authors declare no conflict of interest.
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