Group inverses of matrices over right Ore domains
Introduction
The Drazin (group) inverse of matrices (2 × 2 block matrices) has applications in many areas, especially in singular differential and difference equations and finite Markov chains (see [1], [2], [3], [4], [5]). On the representation of the group inverse (or Drazin inverse) of 2 × 2 block matrices, there exist some results over skew fields (or the complex field). So what is the largest ring that these results remain true over is an interesting and challenging problem. In this paper, our main objective is to attempt to extend a classical result given in [11] to the largest ring that it remains true over, and give a representation of the group inverse of a new class of 2 × 2 block matrix.
In 1977, Meyer and Rose (see [6]), Hartwig and Shoaf (see [7]) gave the representation of the Drazin inverse of 2 × 2 block matrix over the field of complex numbers in terms of A, B and C, where A and C are square. Later, in 1979, Campbell and Meyer proposed an open problem of finding an explicit representation of the Drazin (group) inverse of block matrix in terms of A, B, C and D, where A and D are square (see [1]), which plays an important role in solving singular linear systems (see [3]). Although the problem has not been solved completely yet, there are many results over the field of complex numbers under some special conditions, see [8], [9], [10] and other relative papers. There are also some results over commutative rings, see [16] and references therein. However, relative results over non-commutative rings are rare, see [11], [12], [13], [14], [15].
On the representation of the group inverse of block matrix , Cao [11] gave the result over skew fields, and then Cao and Li [15] gave the result over Bézout domains by using the module theory over Bézout domains. In this paper, we firstly give some new equivalent existence conditions for the group inverse of a square matrix over a right (or left) Ore domain, and then using these existence conditions, extend the foregoing result in [15] to a right Ore domain, and infer that the technique given in [15] cannot be applied to matrices over right Ore domains. On the representation of block matrix , Bu et al. [12] gave the result under the condition A2 = A over skew fields, Bu et al. [13] gave the result under the condition rank(B) ⩾ rank(A) over skew fields, Cao and Li [15] extended the results given in [12] to Bézout domains. In this paper, we solve this problem completely over Bézout domains.
Section snippets
Preliminaries
In this section, some preliminaries are introduced.
A ring is called a right Ore domain (denoted by ) if it possesses no zero divisors and every two elements of the ring have a right common multiple. A left Ore domain is defined similarly. A Bézout domain (denoted by R) is a ring without zero divisors every two elements of which have a greatest common divisor that is a linear combination of them. Integral rings, non-commutative principal ideal domains, division rings, and so on are Bézout
Lemmas
In this section, we state and prove some lemmas which will be used to prove our main results, where Lemma 3.2 plays a central role. Lemma 3.1 Let A ∈ Rm×n, B ∈ Rp×n, C ∈ Rm×q. Then If and A is regular, then there exists a matrix Y ∈ Rp×msuch that B = YA; If and A is regular, then there exists a matrix X ∈ Rn×q such that C = AX.
Proof
- (1)
A is regular, then we havewhere P ∈ GLm(R) and Q ∈ GLn(R). Further,where . Thus r(A) = r(A) + r(B2), since . Hence B2 = 0. Let
Existence of the group inverse of a square matrix
Theorem 4.1 Let . Then the followings are equivalent: A has a group inverse; r(A) = r(A2) and A2 is regular; r(A) = r(A3) and A3 is regular; A has a {1}-inverse of the form ABA for some ; A2X = A for some . In this case, A♯ = AX2; YA2 = A for some . In this case, A♯ = Y2A. In particular if n = 1, (i) is also equivalent to A2 is regular.
Proof
(i) ⇒ (ii): Let G be a group inverse of A, then A = AGA = A2G, it follows that r(A) ⩽ r(A2). In any case, r(A2) ⩽ r(A). Thus r(A) = r(A2). Also we have A2G2A2 = AAGAGA = AAGA = A2, so A2 is
Acknowledgment
The authors thank the anonymous reviewer for his/her helpful comments.
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