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
♯.
$$AXA=A, XAX=X\quad \mbox{and}\quad AX=XA. $$
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].
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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].
. 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.
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 ℜ
m×n
(respectively, R
m×n
) be the set of all m×n matrices over ℜ (respectively, R). The rank of a matrix A∈ℜ
m×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.
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2 Some lemmas
The following three lemmas will be used in the paper.
Lemma 1
[12]
Let
A∈ℜ
n×n
. Then the following are equivalent:
(i)
A
♯
exists;
(ii)
A
2
X=A
for some
X∈ℜ
n×n
. In this case, A
♯=AX
2;
(iii)
YA
2=A
for some
Y∈ℜ
n×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∈ℜ
n×n
,B∈ℜ
n×m
,C∈ℜ
m×n
. If
CA=C, then
, where
A∈ℜ
n×n
,B∈ℜ
n×m
,C∈ℜ
m×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
, 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
$$ \left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )= \left ( \begin{array}{c@{\quad}c}A^{2}+BC&AB\\ C&CB \end{array} \right )= M^{2} $$
(1)
By Lemma 1, M
♯ exists if and only if M=YM
2 for some Y∈ℜ(n+m)×(n+m).
Let where Y
1∈ℜ
n×n
, so it follows that Y
1
A
2=A, by Lemma 1, so A
♯ exists.
$$Y\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right ), $$
$$\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ), $$
According to Lemma 1, M
♯ exists if and only if there exists a matrix
such that M
2
X=M, where X
1∈ℜ
n×n
.
such that M
2
X=M, where X
1∈ℜ
n×n
.By (1), we have
$$\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )X=M. $$
Hence,
$$\left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )\left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right )=\left ( \begin{array}{c@{\quad}c}I&0\\ -C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&-B\\ 0&I \end{array} \right )M. $$
Namely,
$$ {A^{2}X_{1}+(AB-BCB)X_{3}=A-BC}, $$
(2)
$$ {A^{2}X_{2}+(AB-BCB)X_{4}=B}, $$
(3)
$$ {CBCBX_{3}=CBC}, $$
(4)
$$ {CBCBX_{4}=-CB}. $$
(5)
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.
$$-A^{\pi}B(CB)^{\sharp}CBCBX_{4}=A^{\pi}B. $$
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 □
$$\begin{aligned} M^{\sharp} =& \left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp }B(CB)^{\sharp}C& A^{\sharp^{2}}B(CB)^{\pi}+A^{\sharp}B(CB)^{\sharp}\\ (CB)^{\sharp }C&-(CB)^{\sharp} \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}AA^{\sharp}-A^{\sharp}B(CB)^{\pi}C+A^{\pi}B(CB)^{\sharp}C& A^{\sharp}B(CB)^{\pi}-A^{\pi}B(CB)^{\sharp}\\ (CB)^{\pi}C&CB(CB)^{\sharp} \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp }B(CB)^{\sharp}C& A^{\sharp^{2}}B(CB)^{\pi}+A^{\pi}B((CB)^{\sharp})^{2}\\ -A^{\sharp}B(CB)^{\pi}C-A^{\pi}B((CB)^{\sharp})^{2}C&+A^{\sharp }B(CB)^{\sharp}+A^{\pi}B(CB)^{\sharp}\\ (CB)^{\pi}C+(CB)^{\sharp}C&-(CB)^{\sharp} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}M_{1}&M_{2}\\ M_{3}&M_{4} \end{array} \right ). \end{aligned}$$
Example for Theorem 1: Let ℜ be the integer ring, and let
, where
, where $$A=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&0&0\\ 0&-1&0\\ 0&0&0 \end{array} \right ),\qquad B=\left ( \begin{array}{c@{\quad}c} 1&0\\ 1&3\\ 2&0 \end{array} \right ),\qquad C=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&0&0\\ 2&0&0 \end{array} \right ). $$
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.
$$\begin{aligned} & A^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c} 1&0&0\\ 0&-1&0\\ 0&0&0 \end{array} \right ),\qquad A^{\pi}=\left ( \begin{array}{c@{\quad}c@{\quad}c}0&0&0\\ 0&0&0\\ 0&0&1 \end{array} \right ),\qquad (CB)^{\sharp}=CB=\left ( \begin{array}{c@{\quad}c}1&0\\ 2&0 \end{array} \right ),\\ & (CB)^{\pi}=\left ( \begin{array}{c@{\quad}c}0&0\\ -2&1 \end{array} \right ), \end{aligned}$$
$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&0&0&1&0\\ 7&-1&0&-13&3\\ -2&0&0&4&0\\ 1&0&0&-1&0\\ 2&0&0&-2&0 \end{array} \right ). $$
Theorem 2
Let
, where
A∈ℜ
n×n
,B∈ℜ
n×m
,C∈ℜ
m×n
. If
AB=B, then
, where
A∈ℜ
n×n
,B∈ℜ
n×m
,C∈ℜ
m×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
, 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
, 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
$$\begin{aligned} \left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&0\\ -CBC&CBCB \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A^{2}+BC&B\\ C&CB \end{array} \right ) =M^{2}. \end{aligned}$$
Let
$$X=\left ( \begin{array}{c@{\quad}c}I&0\\ -C&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&-B\\ 0&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right ). $$
Then
$$\left ( \begin{array}{c@{\quad}c}A^{2}&0\\ -CBC&CBCB \end{array} \right )\left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A&0\\ 0&-CB \end{array} \right ). $$
From above, we get
$$ {A^{2}X_{1}=A}, $$
(6)
$$ {A^{2}X_{2}=0}, $$
(7)
$$ {-CBCX_{1}+CBCBX_{3}=0}, $$
(8)
$$ {-CBCX_{2}+CBCBX_{4}=-CB}. $$
(9)
It is easy to get
$$\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&B-BCB\\ 0&CBCB \end{array} \right )=M^{2}. $$
From above equations and M=YM
2, let
, we have
, we have $$ {Y_{1}A^{2}=A} , $$
(10)
$$ {Y_{1}(B-BCB)+Y_{2}CBCB=B}, $$
(11)
$$ {Y_{3}A^{2}=C}, $$
(12)
$$ {Y_{3}(B-BCB)+Y_{4}CBCB=0}. $$
(13)
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
$$\begin{aligned} &X_{1}=A^{\sharp}, X_{2}=0, X_{3}=(CB)^{\sharp}C, X_{4}=-(CB)^{\sharp}. \\ &Y_{1}=A^{\sharp}, Y_{2}=B(CB)^{\sharp}, Y_{3}=C, Y_{4}=CB(CB)^{\sharp}-(CB)^{\sharp}. \end{aligned}$$
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
be a matrix over Z/(6Z), where $$A=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&-2&0\\ 0&1&0\\ 0&0&0 \end{array} \right ),\qquad B=\left ( \begin{array}{c@{\quad}c}3&1\\ 3&3\\ 0&0 \end{array} \right ),\qquad C=\left ( \begin{array}{c@{\quad}c@{\quad}c} 3&1&0\\ 3&2&0 \end{array} \right ). $$
It is easy to verify that A
2≠A, AB=B, CA=C. Furthermore, A
♯ and (CB)♯ exist.
By computation,
$$A^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&2&0\\ 0&1&0\\ 0&0&0 \end{array} \right ),\qquad (CB)^{\sharp}=CB=\left ( \begin{array}{c@{\quad}c}0&0\\ 3&3 \end{array} \right ). $$
By Theorem 3, M
♯ exists and
$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&1&0&3&1\\ 0&-2&0&3&3\\ 0&0&0&0&0\\ 3&1&0&0&0\\ 3&2&0&3&3 \end{array} \right ). $$
Remark 1
The following results extend the corresponding works of Zhao et al. [20].
Theorem 4
Let
, where
A,B∈ℜ
n×n
. If
B
♯
exists and
BAB
π
=0. Then
, where
A,B∈ℜ
n×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
, 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
$$M=\left ( \begin{array}{c@{\quad}c}A&B\\ B&0 \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}B^{\pi}A&B\\ B&0 \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ), $$
$$M^{2}=\left ( \begin{array}{c@{\quad}c}A^{2}+B^{2}&AB\\ BA&B^{2} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}AB^{\pi}A+B^{2}&AB\\ 0&B^{2} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ). $$
Since M
♯ exists, from Lemma 1 we know YM
2=M has a solution.
Let
.
.We have
$$ {Y_{1}AB^{\pi}A+Y_{1}B^{2}=B^{\pi}A}, $$
(14)
$$ {Y_{1}AB+Y_{2}B^{2}=B}, $$
(15)
$$ {Y_{3}AB^{\pi}A+Y_{3}B^{2}=B}, $$
(16)
$$ {Y_{3}AB+Y_{4}B^{2}=0}. $$
(17)
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].
. By Lemma 3, the sufficiency of (i) and the expression of M
♯ are similar to the proof in [20]. $$MX=\left ( \begin{array}{c@{\quad}c}A(AB^{\pi})^{\sharp}+BB^{\sharp}&-A(AB^{\pi})^{\sharp }AB^{\sharp}+B^{\pi}AB^{\sharp} \\ 0&BB^{\sharp} \end{array} \right )=XM. $$
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∈ℜ
n×n
. If
B
♯
exists and
B
π
AB=0, then
, where
A,B∈ℜ
n×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
, 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
, 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
$$M^{2}=\left ( \begin{array}{c@{\quad}c}I&AB^{\sharp}\\ 0&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}AB^{\pi}A+B^{2}&0\\ 0&B^{2} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ). $$
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).
$$X=\left ( \begin{array}{c@{\quad}c}I&0\\ -B^{\sharp}A&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ),\qquad Y=\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&-AB^{\sharp}\\ 0&I \end{array} \right ), $$
$$ AB^{\pi}=\bigl(AB^{\pi}A+B^2 \bigr)X_1, $$
(18)
$$ B=\bigl(AB^{\pi}A+B^{2}\bigr)X_2, $$
(19)
$$ B=B^{2}X_{3}, $$
(20)
$$ 0=B^{2}X_{4}, $$
(21)
$$ Y_{1}AB^{\pi}A+Y_{1}B^{2}=B^{\pi}A, $$
(22)
$$ Y_{2}B^{2}=B, $$
(23)
$$ Y_{3}AB^{\pi}A+Y_{3}B^{2}=B, $$
(24)
$$ Y_{4}B^{2}=0. $$
(25)
Let
$$X=\left ( \begin{array}{c@{\quad}c}(AB^{\pi})^{\sharp}&B^{\sharp}\\ B^{\sharp}&-B^{\sharp }AB^{\sharp} \end{array} \right ). $$
By Lemma 3,
$$MX=\left ( \begin{array}{c@{\quad}c}A(AB^{\pi})^{\sharp}+BB^{\sharp}&0\\ 0&BB^{\sharp} \end{array} \right )=XM. $$
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.
Acknowledgements
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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