Group inverse for the block matrices with an invertible subblock

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Abstract

Let M=ABCD (A is square) be a square block matrix with an invertible subblock over a skew field K. In this paper, we give the necessary and sufficient conditions for the existence as well as the expressions of the group inverse for M under some conditions.

Introduction

The group inverses of block matrices have numerous applications in many areas, such as singular differential and difference equations, Markov chains, iterative methods, cryptography and so on (see [1], [2], [3], [4], [5], [6]). In 1979, Campbell and Meyer proposed an open problem to find an explicit representation for the Drazin inverse of a 2×2 block matrix ABCD, where the blocks A and D are supposed to be square matrices but their sizes need not be the same (see [1]). Until now, this problem has not been solved completely, and there is even no known expression for the Drazin (group) inverse of ABC0 which was posed by Campbell in 1983 in [3]. However, there are many literatures about the existence and the representation of the Drazin (group) inverse for the block matrix M=ABCD under some conditions (see [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19]). For example, in [7], the existence of the group inverse of M over a field F was investigated under the condition: A and I+CA-2B are invertible. The representations of the Drazin inverse and the group inverse of M over the complex number field C were given in [12], when A=B=In and D=0. In [13], [14], [16], the authors studied the existence and representations of the group inverse of M over skew fields under different conditions: in [13], A is square, A=B,A2=A and D=0. In [14], A=In,D=0 and rank(CB)2=rank(B)=rank(C). And in [16], A is square and C=0.

Let Km×n be the set of all m×n matrices over a skew field K, In the n×n identity matrix over K and rank(A) the rank of AKm×n. For a square matrix A, the Drazin inverse of A is the matrix AD satisfyingAl+1AD=Al,ADAAD=AD,AAD=ADAfor all integerslk,where k=Ind(A) is the index of A, the smallest nonnegative integer such that rank(Ak)=rank(Ak+1). It is known that AD is existent and unique (see [20]). For AKn×n, if there exists a matrix XKn×n satisfying the matrix equationAXA=A,then we say that X is a {1}-inverse of A and it is denoted by A(1). Denote the set of all {1}-inverses of A by A{1} (see [20]). Furthermore, if the matrix X also satisfies the matrix equationsXAX=X,AX=XA,then we say that X is the group inverse of A and it is denoted by A. It is well known that if A exists, it is unique (see [20]). Notice that the group inverse of a matrix exists iff Ind(A)1.

In this paper, we study the necessary and sufficient conditions for the existence of the group inverse for the block matrix M=ABCDKn×n (A is square) when A,B,C and D satisfy one of the conditions listed below:

  • (1)

    A is invertible, S exists, where S=D-CA-1B;

  • (2)

    D is invertible, S exists, where S=A-BD-1C;

  • (3)

    B or C is invertible.

We also give the expressions of the group inverse for M under these conditions respectively.

Section snippets

Lemmas

Lemma 2.1

[20]

Let AKn×n, then A exists if and only if rank(A)=rank(A2).

Lemma 2.2

[21]

Let AKn×n, then there are invertible matrices P,QKn×n such thatA=PIr000Q,andA(1)=Q-1IrX1X2X3P-1,where rank(A)=r,X1Kr×(n-r), X2K(n-r)×r and X3K(n-r)×(n-r) are arbitrary.

Lemma 2.3

Let A=P000A1QKn×n, where P,QKn×n and A1Kr×r are invertible, rank(A)=r. ThenA(1)=Q-1X1X2X3A1-1P-1,where X1K(n-r)×(n-r), X2K(n-r)×r, and X3Kr×(n-r) are arbitrary.

Lemma 2.4

Let M=ABCDKm×m, where AKn×n is invertible, S=D-CA-1B. If S exists, Sπ=Im-n-SS and R=A2+BSπC is

Main results

Theorem 3.1

Let M=ABCDKm×m, where AKn×n is invertible, S=D-CA-1B. If S exists, then

  • (i)

    M exists if and only if R is invertible, where R=A2+BSπC and Sπ=Im-n-SS;

  • (ii)

    If M exists, then

M=XYZW,whereX=AR-1(A+BSC)R-1A,Y=AR-1(A+BSC)R-1BSπ-AR-1BS,Z=SπCR-1(A+BSC)R-1A-SCR-1A,W=SπCR-1(A+BSC)R-1BSπ-SCR-1BSπ-SπCR-1BS+S.

Proof

  • (i)

    Since A is invertible, then we haverank(M)=rankABCD=rankAB0D-CA-1B=rankA00S,where S=D-CA-1B, so rank(M)=rank(A)+rank(S).

    Moreover, as S exists, we getrank(M2)=rankA2+BCAB+BDCA+DCCB+D2=rankA2+BCAB+BD

Acknowledgement

The authors would like to thank two anonymous reviewers for their helpful comments.

References (21)

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Supported by Natural Science Foundation of the Heilongjiang Province, No. 159110120002.

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