A note on the group inverse of some 2 × 2 block matrices over skew fields

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Abstract

Let K be a skew field and Km×n be the set of all m × n matrices over K. In this paper, we give the necessary and sufficient conditions to the existence and the representations of the group inverse for block matrix AX+YBAB0 under some conditions, where A, Y  Km×n, X, B  Kn×m.

Introduction

It is well known that the representations for the Drazin (group) inverse of block matrices are very important not only in matrix theory, but also in singular differential and difference equations, probability statistical, numerical analysis, game theory, econometrics, control theory and so on (see [1], [2], [3], [4], [5], [6], [19], [20]). In 1979, Campbell and Meyer proposed an open problem in [20], that is to find the explicit representation for the Drazin (group) inverse of the block matrixABCD, where A and D are square matrices. Until now, this problem has not been solved completely. However, under some conditions, there have been some results about this problem (see [7], [8], [9], [10], [11], [23], [24], [25]). In 1983, on the background of second-order systems of differential equations, Campbell in [4] proposed an open problem to give an explicit representation for the Drazin inverse of a 2 × 2 anti-triangular block matrix CAB0, where C is a square matrix and 0 is a square zero matrix. And this open problem has not been solved yet. Furthermore, the Drazin (group) inverse of this kind of block matrix has some applications in other fields. For example, it can be applied to solve constrained optimization problems (KKT Linear Systems), and also can be used in finding the solution of differential equations (see [20], [21], [22]). About this open problem, there have been some results under some certain conditions (see [10], [11], [12], [13], [14], [15], [16], [17], [18], [23], [24], [25]).

Specially, in papers [15], [16], [18], [24], the existence and the representations of the group inverse for the following block matrices are researched:

  • (i)

    M=0AB0 (see [24]);

  • (ii)

    M=AAB0,A=A2 (see [15]);

  • (iii)

    M=AAB0,rank(B)rank(A) (see [16]);

  • (iv)

    M=AkBlAB0, where k and l are positive integers (see [18]);

  • (v)

    M=c1A+c2BAB0, where non-zero elements c1, c2 are in the center of K and A, B exist (see [18]).

Inspired by the above results, we mainly consider a class of 2 × 2 anti-triangular block matrices with the formM=AX+YBAB0where A, Y  Km×n, X, B  Kn×m. Clearly, this form include all cases in the above results (i)–(v).

For a matrix A  Kn×n, the matrix X  Kn×n satisfyingAXA=A,XAX=X,AX=XAis called the group inverse of A and is denoted by X = A. By [1], A exists if and only if rank(A) = rank(A2). If A exists, then it is unique. Sincerank(M2)=rank(AX+YB)2+AB(AX+YB)AB(AX+YB)BA=rankAXYB+ABAXABYBBA,by the existence of group inverse, we get M exists if and only ifrankAXYB+ABAXABYBBA=rank(A)+rank(B).In this paper, we give the necessary and sufficient conditions to the existence and the representations of the group inverse for block matrixAX+YBAB0 when A, B, X, Y satisfy one of the following conditions:

  • (1)

    A, B, X, Y  Kn×n, XA = AX and X is invertible, A exists;

  • (2)

    Y = 0, A  Km×n, X, B  Kn×m and rank(B)  rank(A);

  • (3)

    Y = 0 and replace X with XB, A  Km×n, B  Kn×m, X  Kn×n.

We mainly show three theorems as following:

Theorem 1.1

Suppose M=XA+YBAB0K2n×2n, where A, B, X, Y  Kn×n, A exists, XA = AX and X is invertible, P = TB, then

  • (i)

    M exists iff rank(B) = rank(BP);

  • (ii)

    if M exists, then P exists and M=M1M2M3M4,

    whereM1=PAπ+PπAX-1-PπAZBPAπ,M2=-PAAX-1+PπAX-2+PπAZBPAAX-1,M3=-BPAX-1+B(P)2Aπ+BPAZBPAπ,M4=-BPAX-2-B(P)2AAX-1-BPAZBPAAX-1,T=AπY-X-1AA,Z=X-2+X-1Y.

Theorem 1.2

Suppose M=AXAB0K(m+n)×(m+n),AKm×n,X,BKn×m and rank(B)  rank(A), then

  • (i)

    M exists iff rank(A) = rank(B) = rank(AB) = rank(BA);

  • (ii)

    if M exists, then M=M1M2M3M4, whereM1=(AB)AX-(AB)AXA(BA)B,M2=(AB)A,M3=(BA)B-B(AB)AXA(BA)X+B(AB)AX(AB)AXA(BA)B,M4=-B(AB)AXA(BA).

Theorem 1.3

Suppose M=AXBAB0K(m+n)×(m+n),AKm×n,BKn×m,XKn×n, then

  • (i)

    M exists iff rank(A) = rank(B) = rank(AB) = rank(BA);

  • (ii)

    if M exists, then M=0A(BA)B(AB)-B(AB)AXB(AB)A.

The above three theorems generalize the main results in [15], [16], [18], [24], respectively. In fact, in Theorem1.1, if B exists, X = c1In, Y = c2In, and 0  c1, c2  C(K), it is just the result in Theorem 3.2 of [18]. In order to show this, we need to prove equations S = BBPB and BSB = BP, where S = BP.

Suppose rank(B) = r, since B exists, there exist invertible matrices Q  Kn×n and B1  Kr×r such thatB=QB1000Q-1andB=QB1-1000Q-1.LetT=QT1T2T3T4Q-1,thenP=QB1-1T1-100Q-1,S=QB1-1T1-1B1-1000Q-1.By direct computation, the results hold. In Theorem1.1, let X = In, Y = 0, A2 = A, we also get Theorem 3.1 in [15]; in Theorem1.2, let m = n, X = In, we have Theorem3.1 in [16]; in Theorem1.3, let m = n, X = Ak−1Bl−1, we get Theorem 3.5 in [18] and if X = 0, then we get Theorem 2 in [24]; if we replace X with XB, and apply equation BA(BA)B = B, the formulas of Theorem1.3(ii) is obtained, (the proof of BA(BA)B = B in Lemma2.2). In a word, we not only extend the main results in [15], [16], [18] and [24], but also find the unity of the formulas about Theorem 3.1 in [16] and Theorem 3.5 in [18].

In this paper, we mainly use the knowledge of range and null space, this can avoid complicated decomposition. We end this section by some notations. Let K be any skew field, Km×n and C(K) = {c : c  K, ck = kc, for all k  K} be the set of all m × n matrices over K and the center of K, respectively. For A  Km×n, rank(A), R(A) = {Ax : x  Kn} and N(A) = {x  Kn : Ax = 0} are called the rank, the range and the null space of A, respectively. Aπ denotes I  AA.

Section snippets

Some lemmas

In order to prove the main results, we give some lemmas.

Lemma 2.1

Let A  Km×n, B  Kn×m, (AB) and (BA) exist, then the following equalities hold:

  • (a)

    (AB) = A((BA))2B;

  • (b)

    (BA) = B((AB))2A;

  • (c)

    (AB)A = A(BA);

  • (d)

    (BA)B = B(AB);

  • (e)

    (AB)AB = A(BA)B;

  • (f)

    (BA)BA = B(AB)A.

Proof

The proof is similar to the proof of lemma in paper [12]. 

Lemma 2.2

Let A  Km×n, B  Kn×m, and rank(A) = rank(B) = rank(AB) = rank(BA), then (AB) and (BA) exist, and the following equalities hold:

  • (g)

    A(BA)BA = (AB)ABA = A;

  • (h)

    B(AB)AB = (BA)BAB = B.

Proof

From [13] (or [18]), we know that AB and BA are

The proof of Theorem1.1

Proof

  • (i)

    From (1.2), we see thatrank(A)+rank(B)=rankXAYB+ABXA2BYBBA=rank0XA2BPBA(asA2(AB)=AB,A2(AYB)=AYB)=rank0XA2BP0(as(BA)A2=BA)=rank(BP)+rank(A2).Sincerank(A)=rank(A2),then M exists if and only if rank(B) = rank(BP).

  • (ii)

    By rank(B) = rank(BP) and R(BP)  R(B), we have R(B) = R(BP), then R(P2) = R(TBP) = TR(BP) = TR(B) = R(TB) = R(P). This implies that rank(P) = rank(P2). So P exists. Let G=G1G2G3G4, whereG1=PAπ+PπAX-1-PπAZBPAπ,G2=-PAAX-1+PπAX-2+PπAZBPAAX-1,G3=-BPAX-1+B(P)2Aπ+BPAZBPAπ,G4=-BPAX-2-B(P)2AA

Examples

Now, we provide some examples to show that our results are generalizations of other results in papers [15], [16], [18].

Example 4.1

Let the real quaternion skew field K = {a + bi + cj + dk}, where a, b, c, d are real numbers, andM=XA+YBAB0,A=ij00K2×2,B=1ii-1K2×2,X=ij20-i2K2×2,Y=1jikK2×2.It is easy to verify that XA = AX and X is invertible. By computation, we know that A, B exist, andA=-i-j00,Aπ=0k01,P=(AπY-X-1AA)B=j-k-j-ki+j-k-1,P=i-12-i-12k-12-i+j2.Since rank(B) = rank(BP) = 1, then by Theorem 1.1, we know

Acknowledgments

The authors thank two anonymous reviewers for their helpful comments.

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    Supported by Education Commission of Heilongjiang Province of China (No. 11551365), Natural Science Foundation of Heilongjiang Province of China (No. 201013) and Nation Science Foundation of China (No. 10671026).

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