Abstract
Suppose ℜ is a right Ore domain with unity 1. In this paper, we investigate the existence of the group inverse of some anti-triangular block matrices over ℜ and obtain the sufficient and necessary conditions for such existence. Further, the representations of the group inverse for the following two classes are given. (i) , where CA=C; (ii) , where B ♯ exists and BA=BAB ♯ B.
The results extend the earlier works of Liu et al. (in Appl. Math. Comput. 218:8978–8986, 2012) and Zhao et al. (in E. J. Linear Algebra 21:63–75, 2010). Some results in special cases are also generalized to any ring.
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1 Introduction
For a square matrix A, the matrix X is called the group inverse of A if X satisfies the matrix equations
It is well known that if the group inverse X exists, it is unique, and is denoted by A ♯. Let A π be I−AA ♯.
The group inverse of block matrix is a very useful tool in many fields, such as iterative methods, Markov chains, singular differential and difference equations, see [1–8].
In 1983, in the context of differential equations, Campbell et al. [9] proposed the problem of finding the representation for the Drazin inverse (group inverse) of the anti-triangular block matrix . This problem remains open. However, there are many results in some special cases, see [10, 13–18, 20–22]. It is important to study them in a larger ring, see for example [11, 12, 19, 26].
Liu et al. [10] studied this problem under the conditions A 2=A,CA=C over complex fields. In this paper, we not only delete the condition A 2=A but also solve it for the matrices over right Ore domains by using matrix equations. This also generalizes the results of Ge et al. [27].
On the other hand, in [20], Zhao et al. characterized the existence and the representation of group inverse for block matrix over skew fields under the condition that B ♯ exists and BAB π=0. In this paper, we extend these results to right Ore domains. Some results in special cases are studied over rings with unity 1.
In this paper, let R be a ring with unity 1. A ring is called a right Ore domain (denoted by ℜ) if it possesses no zero divisors and every two elements of the ring have a right common multiple. Integral rings, polynomial rings in an indeterminate over field, noncommutative principal ideal domains and so on are right Ore domains. A left Ore domain is defined similarly. Every right (left) Ore domain ℜ can be embedded in the skew field (denoted by K ℜ) of quotients of itself. More details are found in [23–25]. Let ℜm×n (respectively, R m×n) be the set of all m×n matrices over ℜ (respectively, R). The rank of a matrix A∈ℜm×n (denoted by r(A)) is defined as the rank of A over K ℜ, i.e., the maximum order of all invertible subblocks of A over K ℜ. For convenience, we suppose the right Ore domain ℜ has identity 1.
2 Some lemmas
The following three lemmas will be used in the paper.
Lemma 1
[12]
Let A∈ℜn×n. Then the following are equivalent:
-
(i)
A ♯ exists;
-
(ii)
A 2 X=A for some X∈ℜn×n. In this case, A ♯=AX 2;
-
(iii)
YA 2=A for some Y∈ℜn×n. In this case, A ♯=Y 2 A.
Lemma 2
Let A∈R n×n. Then the following are equivalent:
-
(i)
A ♯ exists;
-
(ii)
A 2 X=A,YA 2=A for some X,Y over R. In this case, A ♯=Y 2 A=AX 2=YAX.
Lemma 3
Let A,B∈R n×n. If BAB π=0, B ♯ and (AB π)♯ exist, then
-
(i)
B ♯ AB π=0, (AB π)♯ B=0, B(AB π)♯=0;
-
(ii)
A(AB π)♯=(AB π)♯ AB π;
-
(iii)
A(AB π)♯ AB π=AB π.
Proof
(i) B ♯ AB π=(B ♯)2 BAB π=0, (AB π)♯ B=((AB π)♯)2 AB π B=0, similarly, B(AB π)♯=0.
(ii) A(AB π)♯=AB π(AB π)♯=(AB π)♯ AB π.
(iii) From (ii), we can easily show that (iii) holds. □
3 Main results
The following is the main result in this note.
Theorem 1
Let , where A∈ℜn×n,B∈ℜn×m,C∈ℜm×n. If CA=C, then
-
(i)
M ♯ exists if and only if (CB)♯ and A ♯ exist, A π B(CB)π=0;
-
(ii)
If M ♯ exists, then , where
$$\begin{aligned} M_{1} =&A^{\sharp}\,{-}\,(A^{\sharp})^{2}B(CB)^{\pi}C\,{-}\,A^{\sharp}B(CB)^{\sharp }C-A^{\sharp}B(CB)^{\pi}C\,{-}\,A^{\pi}B[(CB)^{\sharp}]^{2}C, \\ M_{2} =&(A^{\sharp})^{2}B(CB)^{\pi}+A^{\pi}B[(CB)^{\sharp}]^{2}+A^{\sharp }B(CB)^{\sharp}+A^{\pi}B(CB)^{\sharp}, \\ M_{3} =&(CB)^{\pi}C+(CB)^{\sharp}C, \\ M_{4} =& -(CB)^{\sharp}. \end{aligned}$$
Proof
(i) The “only if” part.
Since CA 2=CA=C, we have
By Lemma 1, M ♯ exists if and only if M=YM 2 for some Y∈ℜ(n+m)×(n+m).
Let
where Y 1∈ℜn×n, so
it follows that Y 1 A 2=A, by Lemma 1, so A ♯ exists.
According to Lemma 1, M ♯ exists if and only if there exists a matrix such that M 2 X=M, where X 1∈ℜn×n.
By (1), we have
Hence,
Namely,
From (5) and Lemma 1, we know (CB)♯ exists. By (3), we have −A π BCBX 4=A π B. i.e.,
Substitute (5) into the above equation, we have A π B(CB)♯ CB=A π B. Therefore, A π B(CB)π=0.
The “if” part.
Let \(X_{1}=A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp}B(CB)^{\sharp}C\), \(X_{2}=A^{\sharp^{2}}B(CB)^{\pi}+A^{\sharp}B(CB)^{\sharp}\), X 3=(CB)♯ C,X 4=−(CB)♯.
Note that A π B(CB)π=0. It is easy to verify that (2)–(5) hold. This implies M=M 2 X has a solution, so M ♯ exists.
(ii) By Lemma 1, A π B(CB)π=0 and CA ♯=CAA ♯=CA 2 A ♯=CA=C, the expression of M ♯ can be obtained from M ♯=MX 2. Next we can compute that
□
Example for Theorem 1: Let ℜ be the integer ring, and let , where
It is easy to verify A 2≠A,CA=C. Furthermore, A ♯ and (CB)♯ exist.
By computation,
so A π B(CB)π=0. By Theorem 1, M ♯ exists and
Similarly, we can prove the counterpart of Theorem 1.
Theorem 2
Let , where A∈ℜn×n,B∈ℜn×m,C∈ℜm×n. If AB=B, then
-
(i)
M ♯ exists if and only if (CB)♯ and A ♯ exist and (CB)π CA π=0;
-
(ii)
If M ♯ exists, then , where
$$\begin{aligned} M_{1} =&A^{\sharp}\,{-}\,B(CB)^{\pi}CA^{\sharp}\,{-}\,B(CB)^{\pi}C(A^{\sharp })^{2}\,{-}\,B(CB)^{\sharp}CA^{\sharp}\,{-}\,B[(CB)^{\sharp}]^{2}CA^{\pi}, \\ M_{2} =&B(CB)^{\sharp}+B(CB)^{\pi}, \\ M_{3} =&(CB)^{\sharp}CA^{\sharp}+(CB)^{\sharp}CA^{\pi}+[(CB)^{\sharp }]^{2}CA^{\pi}+(CB)^{\pi}C(A^{\sharp})^{2}, \\ M_{4} =&-(CB)^{\sharp}. \end{aligned}$$
Next we consider a special case of Theorems 1 and 2, and investigate it over any ring.
Theorem 3
Let , where A∈R n×n,B∈R n×m,C∈R m×n. If AB=B,CA=C, then
-
(i)
M ♯ exists if and only if (CB)♯ and A ♯ exist;
-
(ii)
If M ♯ exists, then
$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c} A^{\sharp}-2B(CB)^{\pi}C-B(CB)^{\sharp}C&B(CB)^{\sharp }+B(CB)^{\pi}\\ (CB)^{\sharp}C+(CB)^{\pi}C&-(CB)^{\sharp} \end{array} \right ). $$
Proof
(i) The “only if” part.
If M ♯ exists, then by Lemma 2 there exist matrices X and Y over R such that M=M 2 X and M=YM 2.
By AB=B and CA=C, we have
Let
Then
From above, we get
It is easy to get
From above equations and M=YM 2, let , we have
By Lemma 2, (6) and (10) imply A ♯ exists.
By (7) and (12), we have CX 2=0 and Y 3 B=CB. Substitute these identities into (9) and (13) respectively, we get CBCBX 4=−CB and CB=(I−Y 4)CBCB.
By Lemma 1, (CB)♯ exists.
The ‘if’ part. Let
It is easy to verify (6)–(13) hold. That implies M=M 2 X and M=YM 2 have solutions.
From Lemma 2, we know M ♯ exists.
(ii) By Lemma 2, the expression of M ♯ can get from M ♯=YMX. □
Example for Theorem 3: Let Z be the integer ring, and let be a matrix over Z/(6Z), where
It is easy to verify that A 2≠A, AB=B, CA=C. Furthermore, A ♯ and (CB)♯ exist.
By computation,
By Theorem 3, M ♯ exists and
Remark 1
From Theorem 1 and 2, we can obtain Theorem 3.1 and 3.2 of [27] and Theorem 2.1 and 2.2 of [10]. From Theorem 3, we can also obtain the Corollary 2.2 of [10]. In above two examples, we especially point that A 2≠A. This shows that the generalizations are true.
The following results extend the corresponding works of Zhao et al. [20].
Theorem 4
Let , where A,B∈ℜn×n. If B ♯ exists and BAB π=0. Then
-
(i)
M ♯ exists if and only if (AB π)♯ exists.
-
(ii)
If M ♯ exists, then , where
$$\begin{aligned} M_{1} =&B^{\pi}A(B^{\sharp})^{2}-(AB^{\pi})^{\sharp}AB^{\pi}A(B^{\sharp })^{2}+(AB^{\pi})^{\sharp}; \\ M_{2} =&-B^{\pi}A(B^{\sharp})^{2}AB^{\sharp}+(AB^{\pi})^{\sharp}AB^{\pi }A(B^{\sharp})^{2}AB^{\sharp} -(AB^{\pi})^{\sharp}AB^{\sharp}+B^{\sharp}; \\ M_{3} =&B^{\sharp}; \\ M_{4} =&-B^{\sharp}AB^{\sharp}. \end{aligned}$$
Proof
The “only if” part of (i).
It is easy to get
Since M ♯ exists, from Lemma 1 we know YM 2=M has a solution.
Let .
We have
By (14), we have Y 1(AB π)2=B π AB π=AB π. From Lemma 1, we know (AB π)♯ exists. Next, we prove the sufficiency of (i) and the expression of (ii):
Let . By Lemma 3, the sufficiency of (i) and the expression of M ♯ are similar to the proof in [20].
It is easy to verify that XMX=M,MXM=M.
So X=M ♯. □
Similarly, we state the symmetrical result of Theorem 4.
Theorem 5
Let , where A,B∈ℜn×n. If B ♯ exists and B π AB=0, then
-
(i)
M ♯ exists if only if (B π A)♯ exists.
-
(ii)
If M ♯ exists, then , where
$$\begin{aligned} M_{1} =&(B^{\sharp})^{2}AB^{\pi}-(B^{\sharp})^{2}AB^{\pi}A(B^{\pi }A)^{\sharp}+(B^{\pi}A)^{\sharp}; \\ M_{2} =&B^{\sharp}; \\ M_{3} =&-B^{\sharp}A(B^{\sharp})^{2}AB^{\pi}+B^{\sharp}A(B^{\sharp })^{2}AB^{\pi}A(B^{\pi}A)^{\sharp} -B^{\sharp}A(B^{\pi}A)^{\sharp}+B^{\sharp}; \\ M_{4} =&-B^{\sharp}AB^{\sharp}. \end{aligned}$$
Proof
The proof is similar to Theorem 4, so we omit it. □
Next we consider a special case of Theorem 4 and 5, and investigate it over any ring.
Theorem 6
Let , where A,B∈R n×n. If B ♯ exists and BAB π=0,B π AB=0, then
-
(i)
M ♯ exists if and only if (AB π)♯ exists.
-
(ii)
If M ♯ exists, then
$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c}(AB^{\pi})^{\sharp}&B^{\sharp}\\ B^{\sharp}&-B^{\sharp }AB^{\sharp} \end{array} \right ). $$
Proof
The “only if” part of (i).
Let
The decomposition of M is the same as in Theorem 4. By Lemma 2, M ♯ exists if and only if there exist X,Y over R such that MX 2=M and YM 2=M.
Let
then we have
By (18) and (22), we have AB π AB π X 1=B π AB π AB π X 1=B π AB π AX 1=B π AB π=AB π and Y 1 AB π AB π=AB π, respectively. From Lemma 2, we know (AB π)♯ exists. Next, we prove the sufficiency of (i) and the expression of (ii).
Let
By Lemma 3,
It is easy to verify that XMX=M, MXM=M.
So X=M ♯. □
Remark 2
We have expressed Theorems 1–2 and 4–5 over right Ore domains, but how to solve them in rings? This is still an open question.
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The authors would like to thank the editors and the referees for their very detailed comments and valuable suggestions which greatly improved our paper.
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Supported by National Natural Science Foundation of China (NO. 11371109).
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Cao, C., Zhang, H. & Ge, Y. Further results on the group inverse of some anti-triangular block matrices. J. Appl. Math. Comput. 46, 169–179 (2014). https://doi.org/10.1007/s12190-013-0744-3
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DOI: https://doi.org/10.1007/s12190-013-0744-3