# Completing a2×2 Block Matrix of Real Quaternions with a Partial Specified Inverse

Authors: Yong Lin; Qing-Wen Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper considers a completion problem of a nonsingular2×2 block matrix over the real quaternion algebra ℍ: Let m1,m2,n1,n2 be nonnegative integers, m1+m2=n1+n2=n>0, and A12∈ℍm1×n2,A21∈ℍm2×n1,A22∈ℍm2×n2,B11∈ℍn1×m1 be given. We determine necessary and sufficient conditions so that there exists a variant block entry matrix A11∈ℍm1×n1 such that A=(A11A12A21A22)∈ℍn×n is nonsingular, and B11 is the upper left block of a partitioning of A-1. The general expression for A11 is also obtained. Finally, a numerical example is presented to verify the theoretical findings.


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## Body

## 1. Introduction

The problem of completing a block-partitioned matrix of a specified type with some of its blocks given has been studied by many authors. Fiedler and Markham [1] considered the following completion problem over the real number field ℝ. Suppose m1,m2,n1,n2 are nonnegative integers, m1+m2=n1+n2=n>0,A11∈ℝm1×n1,A12∈ℝm1×n2,A21∈ℝm2×n1, and B22∈ℝn2×m2. Determine a matrix A22∈ℝm2×n2 such that
(1)A=(A11A12A21A22)
is nonsingular and B22 is the lower right block of a partitioning of A-1. This problem has the form of
(2)(A11A12A21?)-1=(???B22),
and the solution and the expression for A22 were obtained in [1]. Dai [2] considered this form of completion problems with symmetric and symmetric positive definite matrices over ℝ.Some other particular forms for2×2 block matrices over ℝ have also been examined (see, e.g., [3]), such as
(3)(A11A12A21?)-1=(B11???),(A11???)-1=(???B22),(A11??A22)-1=(?B12B21?).The real quaternion matrices play a role in computer science, quantum physics, and so on (e.g., [4–6]). Quaternion matrices are receiving much attention as witnessed recently (e.g., [7–9]). Motivated by the work of [1, 10] and keeping such applications of quaternion matrices in view, in this paper we consider the following completion problem over the real quaternion algebra:
(4)ℍ={k2a0+a1i+a2j+a3k∣i2=j2=k2=ijk=-1anda0,a1,a2,a3∈ℝ∣}.Problem 1.
Supposem1,m2,n1,n2 are nonnegative integers, m1+m2=n1+n2=n>0, and A12∈ℍm1×n2,A21∈ℍm2×n1,A22∈Rm2×n2,B11∈ℍn1×m1. Find a matrix A11∈ℍm1×n1 such that
(5)A=(A11A12A21A22)∈ℍn×n
is nonsingular, and B11 is the upper left block of a partitioning of A-1. That is
(6)(?A12A21A22)-1=(B11???),
where ℍm×n denotes the set of all m×n matrices over ℍ and A-1 denotes the inverse matrix of A.Throughout, over the real quaternion algebraℍ, we denote the identity matrix with the appropriate size by I, the transpose of A by AT, the rank of A by r(A), the conjugate transpose of A by A*=(A¯)T, a reflexive inverse of a matrix A over ℍ by A+ which satisfies simultaneously AA+A=A and A+AA+=A+. Moreover, LA=I-A+A,RA=I-AA+, where A+ is an arbitrary but fixed reflexive inverse of A. Clearly, LA and RA are idempotent, and each is a reflexive inverse of itself. ℛ(A) denotes the right column space of the matrix A.The rest of this paper is organized as follows. In Section2, we establish some necessary and sufficient conditions to solve Problem 1 over ℍ, and the general expression for A11 is also obtained. In Section 3, we present a numerical example to illustrate the developed theory.


## 2. Main Results

In this section, we begin with the following lemmas.Lemma 1 (singular-value decomposition [9]).
LetA∈ℍm×n be of rank r. Then there exist unitary quaternion matrices U∈ℍm×m and V∈ℍn×n such that
(7)UAV=(Dr000),
where Dr=diag(d1,…,dr) and the dj's are the positive singular values of A.Letℍcn denote the collection of column vectors with n components of quaternions and A be an m×n quaternion matrix. Then the solutions of Ax=0 form a subspace of ℍcn of dimension n(A). We have the following lemma.Lemma 2.
Let(8)(A11A12A21A22)
be a partitioning of a nonsingular matrix A∈ℍn×n, and let
(9)(B11B12B21B22)
be the corresponding (i.e., transpose) partitioning of A-1. Then n(A11)=n(B22).Proof.
It is readily seen that(10)(B22B21B12B11),(A22A21A12A11)
are inverse to each other, so we may suppose that n(A11)<n(B22).
Ifn(B22)=0, necessarily n(A11)=0 and we are finished. Let n(B22)=c>0, then there exists a matrix F with c right linearly independent columns, such that B22F=0. Then, using
(11)A11B12+A12B22=0,
we have
(12)A11B12F=0.
From(13)A21B12+A22B22=I,
we have
(14)A21B12F=F.
It follows that the rank r(B12F)≥c. In view of (12), this implies
(15)n(A11)≥r(B12F)≥c=n(B22).
Thus(16)n(A11)=n(B22).Lemma 3 (see [10]).
LetA∈ℍm×n, B∈ℍp×q, D∈ℍm×q be known and X∈ℍn×p unknown. Then the matrix equation
(17)AXB=D
is consistent if and only if
(18)AA+DB+B=D.
In that case, the general solution is
(19)X=A+DB++LAY1+Y2RB,
where Y1, Y2 are any matrices with compatible dimensions over ℍ.By Lemma1, let the singular value decomposition of the matrix A22 and B11 in Problem 1 be
(20)A22=Q(Λ000)R*,(21)B11=U(Σ000)V*,
where Λ=diag(λ1,λ2,…,λs) is a positive diagonal matrix, λi≠0(i=1,…,s) are the singular values of A22, s=r(A22), Σ=diag(σ1,σ2,…,σr) is a positive diagonal matrix, σi≠0(i=1,…,r) are the singular values of B11 and r=r(B11).Q
=
(
Q
1
Q
2
)
∈
ℍ
m
2
×
m
2, R=(R1R2)∈ℍn2×n2, U=(U1U2)∈ℍn1×n1, V=(V1V2)∈ℍm1×m1 are unitary quaternion matrices, where Q1∈ℍm2×s, R1∈ℍn2×s, U1∈ℍn1×r,  and  V1∈ℍm1×r.Theorem 4.
Problem1 has a solution if and only if the following conditions are satisfied:(a)
r(A12A22)=n2,(b)
n2-r(A22)=m1-r(B11),  that is n2-s=m1-r,(c)
ℛ(A21B11)⊂ℛ(A22),(d)
ℛ(A12*B11*)⊂ℛ(A22*).
In that case, the general solution has the form of(22)A11=B11++A12R(Λ-1Q1*A21U1Σ0H-(V2*A12R2)-1)×V*B11++Y-YB11B11+,
where H is an arbitrary matrix in ℍ(n2-s)×r and Y is an arbitrary matrix in ℍm1×n1.Proof.
If there exists anm1×n1 matrix A11 such that A is nonsingular and B11 is the corresponding block of A-1, then (a) is satisfied. From AB=BA=I, we have that
(23)A21B11+A22B21=0,B11A12+B12A22=0,
so that (c) and (d) are satisfied.
By (11), we have
(24)r(A22)+n(A22)=n2,r(B11)+n(B11)=m1.
From Lemma 2, Notice that (A11A12A21A22) is the corresponding partitioning of B-1, we have
(25)n(B11)=n(A22),
implying that (b) is satisfied.
Conversely, from (c), we know that there exists a matrixK∈ℍn2×m1 such that
(26)A21B11=A22K.
Let
(27)B21=-K.
From (20), (21), and (26), we have
(28)A21U(Σ000)V*=Q(Λ000)R*K.
It follows that(29)Q*A21U(Σ000)V*V=Q*Q(Λ000)R*KV.
This implies that(30)(Q1*A21U1Q1*A21U2Q2*A21U1Q2*A21U2)(Σ000)=(Λ000)(R1*KV1R1*KV2R2*KV1R2*KV2).
Comparing corresponding blocks in (30), we obtain
(31)Q2*A21U1=0.
LetR*KV=K^. From (29), (30), we have
(32)K^=(Λ-1Q1*A21U1Σ0HK22),H∈ℍ(n2-s)×r,K22∈ℍ(n2-s)×(m1-r).
In the same way, from (d), we can obtain(33)V1*A12R2=0.
Notice that (A12A22) in (a) is a full column rank matrix. By (20), (21), and (33), we have
(34)(0Q*V*0)(A12A22)R=(Λ000V1*A12R1V1*A12R2V2*A12R1V2*A12R2),
so that
(35)n2=r(A12A22)=r((0Q*V*0)(A12A22)R)=r(Λ000V1*A12R1V1*A12R2V2*A12R1V2*A12R2)=r(Λ)+r(V2*A12R2)=s+r(V2*A12R2).
It follows from (b) and (35) that V2TA12R2 is a full column rank matrix, so it is nonsingular.
FromAB=I, we have the following matrix equation:
(36)A11B11+A12B21=I,
that is
(37)A11B11=I-A12B21,I∈ℍm1×m1,
where B11, A12 were given, B21=-K (from (27)). By Lemma 3, the matrix equation (37) has a solution if and only if
(38)(I-A12B21)B11+B11=I-A12B21.
By (21), (27), (32), and (33), we have that (38) is equivalent to:
(39)(I+A12K)V(Σ-1000)U*U(Σ000)V*=I+A12K.
We simplify the equation above. The left hand side reduces to (I+A12K)V1V1* and so we have
(40)A12KV1V1*-A12K=I-V1V1*.
So,
(41)A12RK^V*V1V1*-A12RK^V*=(V1V2)(V1*V2*)-V1V1*.
This implies that
(42)A12RK^(V1*V1V2*V1)V1*-A12RK^(V1*V2*)=V2V2*,
so that
(43)A12RK^(I0)V1*-A12RK^(V1*V2*)=V2V2*.
So,
(44)-A12RK^(0V2*)=V2V2*,
and hence,
(45)-(A12R1A12R2)(Λ-1Q1*A21U1Σ0HK22)(0V2*)=V2V2*.
Finally, we obtain
(46)A12R2K22V2*=-V2V2*.
Multiplying both sides of (46) by V* from the left, considering (33) and the fact that V2*A12R2 is nonsingular, we have
(47)K22=-(V2*A12R2)-1.
From Lemma 3, (38), (47), Problem 1 has a solution and the general solution is
(48)A11=B11++A12R(Λ-1Q1*A21U1Σ0H-(V2*A12R2)-1)×V*B11++Y-YB11B11+,
where H is an arbitrary matrix in ℍ(n2-s)×r and Y is an arbitrary matrix in ℍm1×n1.


## 3. An Example

In this section, we give a numerical example to illustrate the theoretical results.Example 5.
Consider Problem1 with the parameter matrices as follows:
(49)A12=(2+j12k-k1+12j),A21=(32+12i-12j-12k12j+12k32+12i),A22=(2i2jk),B11=(1ijk).
It is easy to show that(c), (d) are satisfied, and that
(50)n2=r(A12A22)=2,n2-r(A22)=m1-r(B11)=0,
so (a), (b) are satisfied too. Therefore, we have
(51)B11+=(12-12j-12i-12k),A22=Q(Λ000)R*,B11=U(Σ000)V*,
where
(52)Q=12(1ijk),Λ=(22002),R=(1001),U=12(1ijk),Σ=(2002),V=(1001).
We also have
(53)Q1=12(1ijk),R1=(1001),U1=12(1ijk),V1=(1001).
By Theorem4, for an arbitrary matrices Y∈ℍ2×2, we have(54)A11=B11++A12R(Λ-1Q1*A21U1Σ)V*B11++Y-YB11B11+=(32+14j+14k34+14i-32j12-i+14j-14k14-34i-12j-k),
it follows that
(55)A=(32+14j+14k34+14i-32j2+j12k12-i+14j-14k14-34i-12j-k-k1+12j32+12i-12j-12k2i12j+12k32+12i2jk),A-1=(1i-1-1jk0-1-103412-34j-1-112-i12-12i-12j-k).
The results verify the theoretical findings of Theorem 4.



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*Source: 271978-2013-04-10.xml*