Fan-Liang Li

# Left and right inverse eigenpairs problem with a submatrix constraint for the generalized centrosymmetric matrix

De Gruyter | Published online: June 18, 2020

# Abstract

Left and right inverse eigenpairs problem is a special inverse eigenvalue problem. There are many meaningful results about this problem. However, few authors have considered the left and right inverse eigenpairs problem with a submatrix constraint. In this article, we will consider the left and right inverse eigenpairs problem with the leading principal submatrix constraint for the generalized centrosymmetric matrix and its optimal approximation problem. Combining the special properties of left and right eigenpairs and the generalized singular value decomposition, we derive the solvability conditions of the problem and its general solutions. With the invariance of the Frobenius norm under orthogonal transformations, we obtain the unique solution of optimal approximation problem. We present an algorithm and numerical experiment to give the optimal approximation solution. Our results extend and unify many results for left and right inverse eigenpairs problem and the inverse eigenvalue problem of centrosymmetric matrices with a submatrix constraint.

MSC 2010: 65F18; 15A24

## 1 Introduction

Throughout this article we use some notations as follows. Let C n×m be the set of all n × m complex matrices, R n×m be the set of all n × m real matrices, C n = C n×1, R n = R n×1, R denote the set of all real numbers, OR n×n denote the set of all n × n orthogonal matrices, R(A), A T , r(A), tr(A) and A + be the column space, the transpose, rank, trace and the Moore–Penrose generalized inverse of a matrix A, respectively. I n is the identity matrix of size n. Let e i be the ith column of I n , and set J n = (e n ,…,e 1). For A, BR n×m , 〈A, B〉 = tr(B T A) denotes the inner product of matrices A and B. The induced matrix norm is called the Frobenius norm, i.e. | | A | | = A , A 1 / 2 = ( tr ( A T A ) ) 1 / 2 , then R n×m is a Hilbert inner product space.

Generally, the left and right inverse eigenpairs problem is as follows: given partial left and right eigenpairs (eigenvalue and corresponding eigenvector) (γ j , y j ), j = 1,…,l; (λ i , x i ), i = 1,…,h, and a special n × m matrix set S, to find AS such that

(1.1) { A x i = λ i x i , i = 1, , h , y j T A = γ j y j T , j = 1, , l ,
where hn and ln. If X = ( x 1,…, x h ), Λ = diag( λ 1,…, λ h ), Y = ( y 1,…, y l ), Γ = diag( γ 1,…, γ l ), then ( 1.1) is equivalent to
(1.2) { A X = X Λ , Y T A = Γ Y T .
This problem, which mainly arises in perturbation analysis of matrix eigenvalue and in recursive matters, has some practical applications in economic and scientific computation fields [ 1, 2, 3].

Many important results have been achieved on this problem associated with many kinds of matrix sets. Li et al. [4,5,6,7,8,9] have solved the left and right inverse eigenpairs problems for skew-centrosymmetric matrices, generalized centrosymmetric matrices, κ-persymmetric matrices, symmetrizable matrices, orthogonal matrices and κ-Hermitian matrices by using the special properties of eigenpairs of matrix. Zhang and Xie [10], Ouyang [11], Liang and Dia [12] and Yin and Huang [13] have, respectively, solved the left and right inverse eigenvalue problems for real matrices, semipositive subdefinite matrices, generalized reflexive and anti-reflexive matrices and (R,S) symmetric matrices with the special structure of matrix.

Arav et al. [2] and Loewy and Mehrmann [3] studied the recursive inverse eigenvalue problem which arises in the Leontief economic model. Namely, given eigenvalue λ i of A i , in which A i is the ith leading principal submatrix of A, corresponding left eigenvector y i and right eigenvector x i of λ i , construct a matrix AC n×m such that

{ A i x i = λ i x i , y i T A i = λ i y i T , i = 1 , , n .
This recursive inverse eigenvalue problem is a special case of the left and right inverse eigenvalue problem with the leading principal submatrix constraint. Few authors have considered the left and right inverse eigenpairs problem with a submatrix constraint. In this article, we will consider the left and right inverse eigenpairs problem with the leading principal submatrix constraint for the generalized centrosymmetric matrix, which has not been discussed.

## Definition 1

Let κ be a real fixed product of disjoint transpositions and J be the associated permutation matrix. A = (a ij ) ∈ R n×m , if a ij = a κ(i)κ(j) (or a ij = −a κ(i)κ(j)), then A is called the generalized centrosymmetric matrix (or generalized centro-skewsymmetric matrix), and GCSR n×m (or GCSSR n×n ) denote the set of all generalized centrosymmetric matrices (or the set of all generalized centro-skewsymmetric matrices).

From Definition 1, it is easy to derive the following conclusions.

1. (1)

J T = J and J 2 = I n . Real matrices and centrosymmetric matrices are the special cases of generalized centrosymmetric matrices with κ(i) = i and κ(i) = ni + 1 or J = I n and J = J n , respectively.

2. (2)

A ∈ GCSR n×n if and only if A = JAJ and A ∈ GCSSR n×n if and only if A = −JAJ.

3. (3)

R n×n = GCSR n×n ⊕ GCSSR n×n , where the notation V 1V 2 stands for the orthogonal direct sum of linear subspaces V 1 and V 2.

Centrosymmetry, persymmetry and symmetry are three important symmetric structures of a square n × n matrix and have profound applications, such as engineering, statistics and so on [14,15,16]. There are many meaningful results about the inverse problem and the inverse eigenvalue problem of centrosymmetric matrices with a submatrix constraint. Peng et al. [17] and Bai [18] discussed the inverse problem and the inverse eigenvalue problem of centrosymmetric matrices with a principal submatrix constraint, respectively. Zhao et al. [19] studied least squares solutions to AX = B for symmetric centrosymmetric matrices under a central principal submatrix constraint and the optimal approximation. The matrix inverse problem (or inverse eigenvalue problem) with a submatrix constraint is also called the matrix extension problem. Since de Boor and Golub [20] first put forward and considered the Jacobi matrix extension problem in 1978, many authors have studied the matrix extension problem and a series of meaningful results have been achieved [17,18,19,21,22,23,24,25,26].

Assume (λ i ,x i ), i = 1,…,m, be right eigenpairs of A; (μ i ,y i ), j = 1,…,h, be left eigenpairs of A. Let X = (x 1,…,x m ) ∈ C n×m , Λ = diag(λ 1,…,λ m ) ∈ C m×m ; Y = (y 1,…,y h ) ∈ C n×h , Γ = diag(μ 1,…,μ h ) ∈ C h×h . The problems studied in this article can be described as follows.

## Problem I.

Given X = (x 1,…,x m ) ∈ C n×m , Y = (y 1,…,y h ) ∈ C n×h , Λ = diag(λ 1,…,λ m ) ∈ C m×m , Γ = diag(μ 1,…,μ h ) ∈ C h×h , A 0R p×p , mn, hn, pn, find A ∈ GCSR n×n such that

{ A X = X Λ , Y T A = Γ Y T , A [ 1 : p ] = A 0 ,
where A[1: p] denotes the p × p leading principal submatrix.

## Problem II.

Given A* ∈ R n×n , find A ˆ S E such that

A A ˆ = min A S E A A ,
where S E is the solution set of Problem I.

This article is organized as follows. In Section 2, we first study the special properties of eigenpairs and the structure of generalized centrosymmetric matrices. Then, we provide the solvability conditions for and the general solutions of Problem I. In Section 3, we first attest the existence and uniqueness theorem of Problem II and then present the unique approximation solution with the orthogonal invariance of the Frobenius norm. Finally, we provide an algorithm to compute the unique approximation solution. Some conclusions are provided in Section 4.

## Definition 2

Let xC n . If Jx = x (or Jx = −x), then x is called the generalized symmetric (or generalized skew-symmetric) vector. Denote the set of all generalized symmetric (or generalized skew-symmetric) vectors by GC n (or GSC n ).

Denote

P 1 = 1 2 ( I n + J ) , P 2 = 1 2 ( I n J ) .
Let ( u 1, u 2,…, u nr ) and ( u nr+1, u nr+2,…, u n ) are the orthonormal bases for R( P 1) and R( P 2), respectively, and are denoted as K 1 = ( u 1, u 2,…, u nr ), K 2 = ( u nr+1, u nr+2,…, u n ) and K = ( K 1, K 2). Combining Definitions 1 and 2, it is easy to derive the following equalities.
(2.1) P 1 = K 1 K 1 T , P 2 = K 2 K 2 T .
(2.2) J = K 1 K 1 T K 2 K 2 T = K ( I n r 0 0 I r ) K T .
Combining conclusion (2) of Definition 1, ( 2.1) and ( 2.2), it is easy to derive the following lemma.

## Lemma 1

AGCSR n×n if and only if

A = K ( A 11 0 0 A 22 ) K T ,
where A 11 = K 1 T A K 1 R ( n r ) × ( n r ) , A 22 = K 2 T A K 2 R r × r .

If J = J n and n = 2k, then

K = 1 2 ( I k I k J k J k ) .

If J = J n , and n = 2k + 1, then

K = 1 2 ( I k 0 I k 0 2 0 J k 0 J k ) .
Similarly, we have the following splitting of centrosymmetric matrices into smaller submatrices using K.

## Lemma 2

[27] (1) If ACSR 2k×2k , then A can be written as

A = ( B C J k J k C J k B J k ) = K ( B + C 0 0 B C ) K T , B , C R k × k .

(2) If ACSR (2k+1)×(2k+1) , then A can be written as

A = ( B p C J k q T d q T J k J k C J k p J k B J k ) = K ( B + C 2 p 0 2 q T d 0 0 0 B C ) K T , B , C R k × k , p , q R k , d R ,
where CSR n×n denotes the set of all n × n centrosymmetric matrices, k = [ n 2 ] denotes the largest integer number that is not greater than n 2 . In fact, Lemma 2 is a special result of Lemma 1 with J = J n .

For a real matrix AR n×m , its complex right eigenpairs are conjugate pairs. That is, if a + b 1 and x + 1 y are one of its right eigenpairs, then a b 1 and x 1 y are one of its right eigenpairs. This implies Ax = axby and Ay = bx + ay, i.e.,

A ( x , y ) = ( x , y ) ( a b b a ) .
Therefore, in Problem I, we can assume that XR n×m and
Λ = diag ( λ ¯ 1 , , λ ¯ m ¯ ) R m × m ,
where λ ¯ i , i = 1 , , m ¯ are real numbers or 2 × 2 real matrices, m ¯ m . m ¯ = m holds if and only if all right eigenvalues of A are real numbers. We can also prove that the complex left eigenpairs of A are conjugate pairs. Hence, in Problem I, we can also assume that YR n×h and
Γ = diag ( μ ¯ 1 , , μ ¯ h ¯ ) R h × h ,
where μ ¯ i , i = 1 , , h ¯ are real numbers or 2 × 2 real matrices, h ¯ h . h ¯ = h holds if and only if all left eigenvalues of A are real numbers.

Let A ∈ GCSR n×n , if Ax = λx, where λ is a number, xC n , and x ≠ 0, then we have

J A J J x = λ J x , A J x = λ J x .
Hence, we have A( x ± Jx) = λ( x ± Jx). It is obvious that x + Jx and xJx is a generalized symmetric vector and a generalized skew-symmetric vector, respectively. If a + b 1 and x + 1 y are one of its right eigenpairs, then we have
(2.3) A ( x , y ) = ( x , y ) ( a b b a ) .
According to conclusion (2) of Definition 1, we have
(2.4) A J ( x , y ) = J ( x , y ) ( a b b a ) .
Combining ( 2.3) and ( 2.4) implies
A [ ( x , y ) ± J ( x , y ) ] = [ ( x , y ) ± J ( x , y ) ] ( a b b a ) .
It is easy to see that the columns of ( x; y) + J( x; y) (or ( x; y) − J( x; y)) is a generalized symmetric vector (or generalized skew-symmetric vector). Hence, the right eigenvectors of the generalized centrosymmetric matrix can be expressed as generalized symmetric vectors or generalized skew-symmetric vectors. It is clear that the left eigenvectors of the generalized centrosymmetric matrix have the same properties as the right ones. According to the aforementioned analysis, in Problem I, we may assume as follows:
(2.5) X = ( X 1 , X 2 ) R n × m , X 1 = J X 1 R n × m 1 , X 2 = J X 2 R n × ( m m 1 ) , Y = ( Y 1 , Y 2 ) R n × h , Y 1 = J Y 1 R n × h 1 , Y 2 = J Y 2 R n × ( h h 1 ) ,
Λ = ( Λ 1 0 0 Λ 2 ) , Λ 1 = diag ( λ ¯ 1 , , λ ¯ m ¯ 1 ) R m 1 × m 1 , Λ 2 = diag ( λ ¯ m ¯ 1 + 1 , , λ ¯ m ¯ ) R ( m m 1 ) × ( m m 1 ) ,
Γ = ( Γ 1 0 0 Γ 2 ) , Γ 1 = diag ( μ ¯ 1 , , μ ¯ h ¯ 1 ) R h 1 × h 1 , Γ 2 = diag ( μ ¯ h ¯ 1 + 1 , , μ ¯ h ¯ ) R ( h h 1 ) × ( h h 1 ) .

## Lemma 3

[6] If X, Λ, Y, Γ are given by (2.5), then (AX = XΛ,Y T A = ΓY T ) has a solution in GCSR n×n if and only if

(2.6) Y 1 T X 1 Λ 1 = Γ 1 Y 1 T X 1 , X 1 Λ 1 = X 1 Λ 1 X 1 + X 1 , Y 1 Γ 1 = Y 1 Γ 1 Y 1 + Y 1 ,
(2.7) Y 2 T X 2 Λ 2 = Γ 2 Y 2 T X 2 , X 2 Λ 2 = X 2 Λ 2 X 2 + X 2 , Y 2 Γ 2 = Y 2 Γ 2 Y 2 + Y 2 .

Moreover, its general solution can be expressed as

(2.8) A = A 10 + E F G , F G C S R n × n ,
where
(2.9) A 10 = X 1 Λ 1 X 1 + + ( Y 1 T ) + Γ 1 Y 1 T ( I n X 1 X 1 + ) + X 2 Λ 2 X 2 + + ( Y 2 T ) + Γ 2 Y 2 T ( I n X 2 X 2 + ) G C S R n × n ,
(2.10) E = I n Y 1 Y 1 + Y 2 Y 2 + G C S R n × n , G = I n X 1 X 1 + X 2 X 2 + G C S R n × n .

Combining Lemmas 1 and 3, it is easy to prove that A in (2.8) can be expressed as

(2.11) A = K ( A 110 + E 1 F 1 G 1 0 0 A 210 + E 2 F 2 G 2 ) K T ,
where A 110, A 210 are denoted by A 10, E 1, E 2 are denoted by E, G 1, G 2 are denoted by G, and A 110, E 1, G 1R (nk)×(nk), A 210, E 2, G 2R k×k , for any F 1R (nk)×(nk), F 2R k×k .

Denote

(2.12) ( I p , 0 ) K 1 E 1 = E ¯ 1 , ( I p , 0 ) K 2 E 2 = E ¯ 2 , G 1 K 1 T ( I p , 0 ) T = G ¯ 1 , G 2 K 2 T ( I p , 0 ) T = G ¯ 2 , A 0 ( I p , 0 ) K 1 A 110 K 1 T ( I p , 0 ) T ( I p , 0 ) K 2 A 210 K 2 T ( I p , 0 ) T = A ¯ 0 .
Combining ( 2.11) and ( 2.12), A[1: p] = A 0 if and only if the following equation holds.
(2.13) E ¯ 1 F 1 G ¯ 1 + E ¯ 2 F 2 G ¯ 2 = A ¯ 0 .

Suppose that the generalized singular value decomposition (GSVD) of matrix pairs ( E ¯ 1 T , E ¯ 2 T ) is as follows:

(2.14) E ¯ 1 T = Q 1 Σ 1 S T , E ¯ 2 T = Q 2 Σ 2 S T ,
where Q 1OR (nk)×(nk), Q 2OR k×k , SR p×p is nonsingular, and
(2.15) Σ 1 = ( I r 1 0 0 0 0 Θ 1 0 0 0 0 0 0 ) , Σ 2 = ( 0 0 0 0 0 Θ 2 0 0 0 0 I s 1 r 1 t 1 0 ) r 1 t 1 s 1 r 1 t 1 p s 1 r 1 t 1 s 1 r 1 t 1 p s 1
with s 1 = r ( E ¯ 1 , E ¯ 2 ) T , r 1 = s 1 r ( E ¯ 2 T ) , t 1 = r ( E ¯ 1 T ) + r ( E ¯ 2 T ) s 1 , Θ 1 = diag ( γ 1 , , γ t 1 ) , Θ 2 = diag ( δ 1 , , δ t 1 ) , with 1 γ t 1 γ 1 > 0 , 0 < δ 1 δ t 1 1 , γ i 2 + δ i 2 = 1 , i = 1 , , t 1 .

Suppose that the GSVD of matrix pairs ( G ¯ 1 , G ¯ 2 ) is

(2.16) G ¯ 1 = P 1 Σ 3 W T , G ¯ 2 = P 2 Σ 4 W T ,
where P 1OR (nk)×(nk), P 2OR k×k , WR p×p is nonsingular, and
(2.17) Σ 3 = ( I r 1 0 0 0 0 Θ 3 0 0 0 0 0 0 ) , Σ 4 = ( 0 0 0 0 0 Θ 4 0 0 0 0 I s 2 r 2 t 2 0 ) r 2 t 2 s 2 r 2 t 2 p s 2 r 2 t 2 s 2 r 2 t 2 p s 2
with s 2 = r ( G ¯ 1 G ¯ 2 ) , r 2 = s 2 r ( G ¯ 2 ) , t 2 = r ( G ¯ 1 ) + r ( G ¯ 2 ) s 2 , Θ 3 = diag ( α 1 , , α t 2 ) , Θ 4 = diag ( β 1 , , β t 2 ) , with 1 α t 2 α 1 > 0 , 0 < β 1 β t 2 1 , α i 2 + β i 2 = 1 , i = 1 , , t 2 , where r 1, t 1, s 1r 1t 1, ps 1, r 2, t 2, s 2r 2t 2, ps 2 denote the number of columns of the sub-block-matrix of Σ 1, Σ 2, Σ 3 and Σ 4.

Combining (2.14) and (2.16) implies that (2.13) can be written as

(2.18) Σ 1 T Q 1 T F 1 P 1 Σ 3 + Σ 2 T Q 2 T F 2 P 2 Σ 4 = S 1 A ¯ 0 W T .
Partition Q 1 T F 1 P 1 , Q 2 T F 2 P 2 , S 1 A ¯ 0 W T into the following form:
(2.19) Q 1 T F 1 P 1 = ( F 111 F 112 F 113 F 121 F 122 F 123 F 131 F 132 F 133 ) , Q 2 T F 2 P 2 = ( F 211 F 212 F 213 F 221 F 222 F 223 F 231 F 232 F 233 ) ,
(2.20) S 1 A ¯ 0 W T = ( A 011 A 012 A 013 A 014 A 021 A 022 A 023 A 024 A 031 A 032 A 033 A 034 A 041 A 042 A 043 A 044 ) .
Combining ( 2.19) and ( 2.20) implies that ( 2.18) can be written as
(2.21) ( F 111 F 112 Θ 3 0 0 Θ 1 F 121 Θ 1 F 122 Θ 3 + Θ 2 F 222 Θ 4 Θ 2 F 223 0 0 F 232 Θ 4 F 233 0 0 0 0 0 ) = ( A 011 A 012 A 013 A 014 A 021 A 022 A 023 A 024 A 031 A 032 A 033 A 034 A 041 A 042 A 043 A 044 ) .
Combining Lemma 3 and ( 2.11)–( 2.21) derives the following theorem.

## Theorem 1

If X, Λ, Y, Γ are given by (2.5) and given A 0R p×p , then Problem I has a solution in GCSR n×n if and only if (2.6), (2.7) and the following equations hold:

(2.22) A 013 = 0 , A 014 = 0 , A 024 = 0 , A 031 = 0 , A 034 = 0 , A 041 = 0 , A 042 = 0 , A 043 = 0 , A 044 = 0 .
Moreover, the general solution is
(2.23) A = K ( A 110 + E 1 F 1 G 1 0 0 A 210 + E 2 F 2 G 2 ) K T ,
where A 110, E 1, G 1, A 210, E 2, G 2 are denoted by ( 2.11) , and
(2.24) F 1 = Q 1 ( A 011 A 012 Θ 3 1 F 113 Θ 1 1 A 021 Θ 1 1 ( A 022 Θ 2 F 222 Θ 4 ) Θ 3 1 F 123 F 131 F 132 F 133 ) P 1 T ,
(2.25) F 2 = Q 2 ( F 211 F 212 F 213 F 221 F 222 Θ 2 1 A 023 F 231 A 032 Θ 4 1 A 033 ) P 2 T
where F 113, F 123, F 131, F 133, F 211, F 212, F 213, F 221, F 222 and F 231 are the arbitrary matrices.

## 3 An expression of the solution of Problem II

From (2.23), it is easy to prove that the solution set S E of Problem I is a nonempty closed convex set if Problem I has a solution in GCSR n×n . We claim that for any given A* ∈ R n×n , there exists a unique optimal approximation for Problem II.

Combining (2.8)–(2.11) and Lemma 1, (2.23) can be written as

(3.1) A = A 10 + E F G , F = K ( F 1 0 0 F 2 ) K T ,
where E and G are denoted by ( 2.10), F 1 and F 2 are denoted by ( 2.24) and ( 2.25), respectively.

According to conclusion (3) of Definition 1, for any A* ∈ R n×n , there exist only one A 1 G C S R n × n and only one A 2 G C S S R n × n which satisfy

(3.2) A = A 1 + A 2 ,
where
(3.3) A 1 = 1 2 ( A + J A J ) , A 2 = 1 2 ( A J A J ) .
According to Lemma 1, A 1 can be written as
(3.4) A 1 = K ( A 11 0 0 A 22 ) K T ,
where A 11 R ( n k ) × ( n k ) , A 22 R n × n are given by A 1 . Denote
(3.5) Q 1 T A 11 P 1 = ( A 111 A 112 A 113 A 121 A 122 A 123 A 131 A 132 A 133 ) , Q 2 T A 22 P 2 = ( A 211 A 212 A 213 A 221 A 222 A 223 A 231 A 232 A 233 ) .

## Theorem 2

Given X, Y, Λ, Γ according to (2.5) and A 0. If they satisfy the conditions of Theorem 1, and given A* ∈ R n×n , then Problem II has the unique solution A ˆ . Moreover, A ˆ can be expressed as

(3.6) A ˆ = A 10 + E F ˆ G ,
where A 10, E, G are denoted by ( 2.9) and ( 2.10) with
(3.7) F ˆ = K ( F ˆ 1 0 0 F ˆ 2 ) K T ,
where
(3.8) F ˆ 1 = Q 1 ( A 011 A 012 Θ 3 1 A 113 Θ 1 1 A 021 Θ 1 1 ( A 022 Θ 2 A 222 Θ 4 ) Θ 3 1 A 123 A 131 A 132 A 133 ) P 1 T ,
(3.9) F ˆ 2 = Q 2 ( A 211 A 212 A 213 A 221 A 222 Θ 2 1 A 023 A 231 A 032 Θ 4 1 A 033 ) P 2 T .

## Proof

Combining Theorem 1 and (3.2), for any AS E , we have

A A 2 = A 1 A 2 + A 2 2 = A 1 A 10 E F G 2 + A 2 2 .
According to ( 2.10), it is easy to prove that E, F are orthogonal projection matrices. Hence, there exist orthogonal projection matrices E ¯ , G ¯ which satisfy
(3.10) E ¯ + E = I n , E ¯ E = 0 ; G ¯ + G = I n , G ¯ G = 0 .
From this, we have
A A 2 = ( E ¯ + E ) A 1 A 10 E F G 2 + A 2 2 = E ¯ A 1 A 10 2 + E A 1 A 10 E F G 2 + A 2 2 = E ¯ A 1 A 10 2 + E A 1 A 10 ( G + G ¯ ) E F G 2 + A 2 2 = E ¯ A 1 A 10 2 + E A 1 A 10 G E F G 2 + E A 1 A 10 G ¯ 2 + A 2 2 .
This implies that
min for any A S E A A min F denoted by ( 3.1 ) E ( A 1 A 10 ) G E F G .
According to ( 2.9) and ( 2.10), it is easy to prove EA 10 G = 0. Hence, we have
min for any A S E A A min F denoted by ( 3.1 ) E ( E A 1 G E F G ) .
It is clear that if F ˆ = A 1 + E ¯ F ¯ G ¯ , for any F ¯ G C S R n × n , then
(3.11) E A 1 G E F ˆ G = 0 .
Combining Lemma 1, ( 2.11) and ( 3.10), we have
E ¯ = K ( I n k E 1 0 0 I k E 2 ) K T , G ¯ = K ( I n k G 1 0 0 I k G 2 ) K T ,
where E 1, E 2, G 1, G 2 are denoted by ( 2.11).
F ¯ = K ( F ¯ 1 0 0 F ¯ 2 ) K T , F ¯ 1 R ( n k ) × ( n k ) , F ¯ 2 R k × k .

Denote

Q 1 T ( I n k E 1 ) F ¯ 1 ( I n k G 1 ) P 1 = ( F ¯ 111 F ¯ 112 F ¯ 113 F ¯ 121 F ¯ 122 F ¯ 123 F ¯ 131 F ¯ 132 F ¯ 133 ) , Q 2 T ( I k E 2 ) F ¯ 2 ( I k G 2 ) P 2 = ( F ¯ 211 F ¯ 212 F ¯ 213 F ¯ 221 F ¯ 222 F ¯ 223 F ¯ 231 F ¯ 232 F ¯ 233 ) .
Combining ( 3.1) and ( 3.5), we have
(3.12) ( A 011 A 012 Θ 3 1 F 113 Θ 1 1 A 021 Θ 1 1 ( A 022 Θ 2 F 222 Θ 4 ) Θ 3 1 F 123 F 131 F 132 F 133 ) = ( A 111 A 112 A 113 A 121 A 122 A 123 A 131 A 132 A 133 ) + ( F ¯ 111 F ¯ 112 F ¯ 113 F ¯ 121 F ¯ 122 F ¯ 123 F ¯ 131 F ¯ 132 F ¯ 133 ) ,
(3.13) ( F 211 F 212 F 213 F 221 F 222 Θ 2 1 A 023 F 231 A 032 Θ 4 1 A 033 ) = ( A 211 A 212 A 213 A 221 A 222 A 223 A 231 A 232 A 233 ) + ( F ¯ 211 F ¯ 212 F ¯ 213 F ¯ 221 F ¯ 222 F ¯ 223 F ¯ 231 F ¯ 232 F ¯ 233 ) .
( 3.12) and ( 3.13) imply that if F ¯ satisfies the following conditions, then we can also obtain ( 3.11).
(3.14) F ¯ 111 = A 011 A 111 , F ¯ 112 = A 012 Θ 3 1 A 112 , F ¯ 113 = 0 , F ¯ 121 = Θ 1 1 A 021 A 121 , F ¯ 122 = Θ 1 1 ( A 022 Θ 2 F 222 Θ 4 ) Θ 3 1 A 122 , F ¯ 123 = 0 , F ¯ 131 = 0 , F ¯ 132 = 0 , F ¯ 133 = 0 . F ¯ 211 = 0 , F ¯ 212 = 0 , F ¯ 213 = 0 , F ¯ 221 = 0 , F ¯ 222 = F 222 A 222 , F ¯ 223 = Θ 2 1 A 023 A 223 , F ¯ 231 = 0 , F ¯ 232 = A 032 Θ 4 1 A 232 , F ¯ 233 = A 033 A 233 .
According to ( 3.14), we have
(3.15) F 113 = A 113 , F 123 = A 123 , F 131 = A 131 , F 132 = A 132 , F 133 = A 133 , F 211 = A 211 , F 212 = A 212 , F 213 = A 213 , F 221 = A 221 , F 222 = A 222 , F 231 = A 231 .
( 3.15) gives the results of Theorem 2.□

Algorithm

1. Input A*, A 0, and input X, Y, Λ, Γ according to (2.5).

2. Compute Y 1 T X 1 Λ 1 , Γ 1 Y 1 T X 1 , X 1 Λ 1 , X 1 Λ 1 X 1 + X 1 , Y 1 Γ 1 , Y 1 Γ 1 Y 1 + Y 1 , Y 2 T X 2 Λ 2 , Γ 2 Y 2 T X 2 , X 2 Λ 2 , X 2 Λ 2 X 2 + X 2 , Y 2 Γ 2 , Y 2 Γ 2 Y 2 + Y 2 . If (2.6) and (2.7) hold, then go to 3; otherwise stop.

3. Compute A 10, E, G according to (2.9) and (2.10), and compute A 110, A 210, E 1, E 2, G 1, G 2 according to (2.11).

4. Compute E ¯ 1 , E ¯ 2 , G ¯ 1 , G ¯ 2 , A ¯ 0 according to (2.12).

5. Compute the GSVD of matrix pairs ( E ¯ 1 T , E ¯ 2 T ) and ( G ¯ 1 , G ¯ 2 ) , respectively.

6. Partition S 1 A ¯ 0 W T according to (2.20). If (2.22) holds, go to 7; otherwise stop.

7. Compute A 1 according to (3.3).

8. Compute A 11 , A 22 according to (3.4), and partition Q 1 T A 11 P 1 , Q 2 T A 22 P 2 according to (3.5).

9. Compute F ˆ 1 and F ˆ 2 according to (3.8) and (3.9), respectively.

10. Compute F ˆ according to (3.7).

11. Calculate A ˆ according to (3.6).

Example (n = 10, h = 6, l = 2, p = 3).

Give J and choose a random matrix A in GCSR10×10 as follows.

J = ( 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ) ,
A = ( 0.9129 0.3401 0.3762 0.1782 0.8459 0.2490 0.3496 0.4997 0.8081 0.2734 0.4842 0.8901 0.4871 0.6404 0.3066 0.4359 0.7860 0.2541 0.7008 0.4527 0.4296 0.6955 0.6290 0.3674 0.7706 0.3658 0.6734 0.7211 0.2705 0.9054 0.3787 0.5113 0.2782 0.6369 0.3789 0.6256 0.7456 0.3834 0.6683 0.5673 0.9355 0.4189 0.4208 0.5631 0.7648 0.3268 0.6183 0.2866 0.4675 0.3522 0.4675 0.3522 0.2866 0.6183 0.3268 0.7648 0.5631 0.4208 0.9355 0.4189 0.6683 0.5673 0.3834 0.7456 0.6256 0.3789 0.6369 0.2782 0.3787 0.5113 0.2705 0.9054 0.7211 0.6734 0.3658 0.7706 0.3674 0.6290 0.4296 0.6955 0.8081 0.2734 0.4997 0.3496 0.2490 0.8459 0.1782 0.3762 0.9129 0.3401 0.7008 0.4527 0.2541 0.2541 0.4359 0.3066 0.6404 0.4871 0.4842 0.8901 ) .

Compute the eigenvalues and the right eigenvectors of A, choose partial eigenpairs of A and obtain X 1, X 2, Λ 1, Λ 2 according to (2.5).

X 1 = ( 0.2852 0.2264 0.2136 0.3242 0.0454 0.0978 0.3553 0.5749 0 0.3076 0.0458 0.1363 0.3044 0.2003 0.0148 0.3044 0.2003 0.0148 0.3076 0.0458 0.1363 0.3553 0.5749 0 0.2852 0.2264 0.2136 0.3242 0.0454 0.0978 ) , X 2 = ( 0.3432 0.1016 0.0780 0.0483 0.5103 0 0.3278 0.4345 0.1443 0.2424 0.0393 0.0108 0.4623 0.1051 0.0290 0.4623 0.1051 0.0290 0.2424 0.0393 0.0108 0.3278 0.4345 0.1443 0.3432 0.1016 0.0780 0.0483 0.5103 0 ) , Λ 1 = ( 5.2477 0 0 0 0.7218 0.2883 0 0.2883 0.7128 ) , Λ 2 = ( 0.9026 0 0 0 0.2113 0.0948 0 0.0948 0.2113 ) .
Compute the eigenvalues and the right eigenvectors of A T , choose partial eigenpairs of A T and obtain Y 1, Y 2, Γ 1, Γ 2 according to ( 2.5).
Y 1 = ( 0.1898 0.4155 0.1347 0.0559 0.5197 0.5197 0.0559 0.1347 0.1898 0.4155 ) , Y 2 = ( 0.3865 0.0860 0.2930 0.4379 0.2560 0.2560 0.4379 0.2930 0.3865 0.0860 ) , Γ 1 = ( 0.2847 ) , Γ 2 = ( 0.4610 ) .
It is clear that ( 2.6) and ( 2.7) hold. Input A 0 is
A 0 = ( 0.9129 0.3401 0.3762 0.4842 0.8901 0.4871 0.4296 0.6955 0.6290 ) .
We can also prove that ( 2.22) holds. For a given matrix
A = ( 0.4326 0.1867 0.2944 0.3999 1.6041 1.0106 0.0000 0.5689 0.6232 0.3899 1.6656 0.7258 1.3362 0.6900 0.2573 0.6145 0.3179 0.2556 0.7990 0.08808 0.1253 0.5883 0.7143 0.8156 1.0565 0.5077 1.0950 0.3775 0.9409 0.6355 0.2877 2.1832 1.6236 0.7119 1.4151 1.6924 1.8740 0.2959 0.9921 0.5596 1.1465 0.1364 0.6918 1.2902 0.8051 0.5913 0.4282 1.4751 0.2120 0.4437 1.1909 0.1139 0.8580 0.6686 0.5287 0.6436 0.8956 0.2340 0.2379 0.9499 1.1892 1.0668 1.2540 1.1908 0.2193 0.3803 0.7310 0.1184 1.0078 0.7812 0.0376 0.0593 1.5937 1.2025 0.9219 1.0091 0.5779 0.3148 0.7420 0.5690 0.3273 0.0956 1.4410 0.0198 2.1707 0.0195 0.0403 1.4435 1.0823 0.8217 0.1746 0.8323 0.5711 0.1567 0.0592 0.0482 0.6771 0.3510 0.1315 0.2656 ) ,
by Algorithm, the unique solution of Problem II is
A ˆ = ( 0.9266 0.3432 0.3968 0.0394 0.6860 0.3670 0.6008 0.4813 0.8063 0.2759 0.4930 0.8920 0.4941 0.9728 0.5025 0.2942 0.4168 0.2558 0.6723 0.4423 0.4356 0.6968 0.6337 0.3239 0.7364 0.3935 0.7209 0.7153 0.2671 0.9055 0.1859 0.7142 0.3284 0.3009 0.9766 1.0656 0.0199 0.39999 0.4142 0.7545 0.9811 0.4541 0.4843 0.5740 0.7820 0.4512 0.4709 0.2349 0.3621 0.3539 0.3621 0.3539 0.2349 0.4709 0.4512 0.7820 0.5740 0.4843 0.9811 0.4541 0.4142 0.7545 0.3999 0.0199 0.0656 0.9766 0.3009 0.3284 0.1859 0.7142 0.2671 0.9055 0.7153 0.7209 0.3935 0.7364 0.3239 0.6337 0.4356 0.6968 0.8063 0.2759 0.4813 0.6008 0.3670 0.6860 0.0394 0.3868 0.9266 0.3432 0.6723 0.4423 0.2558 0.4168 0.2942 0.5025 0.9728 0.4941 0.4930 0.8920 ) .

## 4 Conclusion

In this article, we have obtained the necessary and sufficient conditions and associated general solutions of Problem I (Theorem 1). For given matrix A* ∈ R n×n , the unique optimal approximation solution of Problem II has been derived (Theorem 2). Our results extend and unify many results for left and right inverse eigenpairs problem, the inverse problem and the inverse eigenvalue problem of centrosymmetric matrices with a submatrix constraint, which is the first motivation of this work. For instance, in Problem I, if Y = 0, then this problem becomes Problem I in [17]; in Problem I, if p = 0, this problem becomes Problem I in [4,5,6,7,8,9,10,11,12,13].

The left and right eigenpairs of a real matrix are not all real eigenpairs, and its complex eigenpairs are all conjugate pairs. Hence, the supposition for Problem I in [4,5,6,7,10,11] is not suitable. In this article, we derive the suitable supposition for Problem I (X, Y, Λ, Γ are given by (2.5)), which is another motivation of this work.

# Acknowledgements

This research was supported by the Natural Science Foundation of Hunan Province (2015JJ4090) and by Scientific Fund of Hunan Provincial Education Department of China (Grant no 13C1139). The authors are grateful to the anonymous reviewer and Dr. Justyna Zuk for their valuable comments and careful reading of the original manuscript of this article.

Conflicts of interest: The authors declare that there is no conflict of interests regarding the publication of this paper.

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