vol21p63-75. 129KB Jun 04 2011 12:06:11 AM
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 21, pp. 63-75, October 2010
ELA
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GROUP INVERSE FOR THE BLOCK MATRIX WITH
TWO IDENTICAL SUBBLOCKS OVER SKEW FIELDS∗
JIEMEI ZHAO† AND CHANGJIANG BU§†‡
Abstract. Let K be a skew field and K n×n be the set of all n × n matrices over K. The
purpose of this paper is to give some necessary and sufficient
conditions for the existence and the
A B
under some conditions.
representations of the group inverse of the block matrix C
D
Key words. Skew fields, Block matrix, Group inverse.
AMS subject classifications. 15A09; 65F20.
1. Introduction. Let K be a skew field, C be the complex number field, K m×n
be the set of all m × n matrices over K, and In be the n × n identity matrix over K.
For a matrix A ∈ K n×n , the matrix X ∈ K n×n satisfying
Ak XA = Ak , XAX = X, AX = XA
is called the Drazin inverse of A and is denoted by X = AD , where k is the index
of A, i.e., the smallest non-negative integer such that rank(Ak ) = rank(Ak+1 ). We
denote such a k by Ind(A). It is well-known that AD exists and is unique (see [2]).
If Ind(A) = 1, AD is also called the group inverse of A and is denoted by A♯ . Then
A♯ exists if and only if rank(A) = rank(A2 ) (see [1, 3, 11-14, 24, 25, 29]). We denote
I − AA♯ by Aπ .
The group inverse of block matrices has numerous applications in matrix theory,
such as singular differential and difference equations, Markov chains, iterative methods and so on (see [12]-[14]). For instance, Y. Wei et al. studied the representation of
the group inverse of a real singular Toeplitz matrix which arises in scientific computing
and engineering (see [29]); in [25], S. Kirkland et al. investigated the representation
of the group inverse of the Laplacian matrix of an undirected weighted graph G on
n vertices; and in [24], G. Heinig studied the group inverse of the Sylvester transformation ϕ(X) = AX − XB, where A ∈ Cm×m and B ∈ Cn×n ; utilizing the theory of
Drazin inverse, the differential equation Ax′ + Bx = f is studied in linear systems,
∗ Received by the editors on June 2009. Accepted for publication on July 31, 2010 Handling
Editors: Roger A. Horn and Fuzhen Zhang.
† College of Automation, Harbin Engineering University, Harbin 150001, P. R. China. ([email protected]). Supported by the NSFH, No.159110120002.
‡ Dept. of Applied Math., College of Science, Harbin Engineering University, Harbin 150001, P.
R. China. § Corresponding author. Email: [email protected]
63
Electronic Journal of Linear Algebra ISSN 1081-3810
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where A, B are both n × n singular matrices and f is a vector-valued function (see
[13]); in [3], R. Bru et al. gave the general explicit solution which is obtained when
the symmetric singular linear control system satisfies the regularity condition.
In 1979, S. Campbell and C. Meyer proposed an open problem
to find an explicit
AB
representation for the Drazin inverse of a 2 × 2 block matrix C D , where the blocks
A and D are supposed to be square matrices but their sizes need not be the same (see
[12]). A simplified problem
tofind an explicit representation of the Drazin (group)
AB
inverse for block matrix C
(A is square, O is square null matrix) was proposed
O
by S. Campbell in 1983. This open problem was motivated in hoping to find general
expressions for the solutions of the second-order system of the differential equations
Ex′′ (t) + F x′ (t) + Gx(t) = 0 (t ≥ 0),
where E is a singular matrix. Detailed discussions of the importance of the problem
can be found in [11]. Until now, both problems have not been resolved. However,
there have been some studies about the representations of the Drazin(group) inverse
under certain conditions (see [4-10, 15-23, 26, 27]). For example, the formulae of
Drazin(group) inverse of an operator matrix under some conditions are investigated
in [18] and [19]. The representations for the Drazin(group) inverse of the sum of
several matrices are given in [15] and [17]. In particular, J. Chen et al. [15] presented
the representation of the group inverse of M = P + Q + S, where P Q = QP = O,
P S = SQ = O, P ♯ and Q♯ exist.
Some conditions for the representations of group
AB
inverse for block matrix C D are listed below:
(i) A♯ and (D − CA♯ B)♯ exist [18].
(ii) D = O, A = B = Id , Id , C ∈ Cd×d [10].
(iii) D = O, A, B, C ∈ {P, P ∗ , P P ∗ }, P ∈ Cn×n , P 2 = P , P ∗ is the conjugate
transpose of P [9].
(iv) C = O, A and D are square matrices over K [8].
(v) D = O, A = B = A2 , A, C ∈ K n×n [4].
(vi) D = O, A = B, rank(C) ≥ rank(A), A, C ∈ K n×n [5].
(vii) A is invertible and (D − CA−1 B)♯ exists; D is invertible and (A − BD−1 C)♯
exists; B or C is invertible, where A, B, C ∈ K n×n [6].
(viii) D = O, A = c1 B + c2 C, where non-zero elements c1 and c2 are in the center
of K; D = O, A = B k C l , where k and l are positive integers [7].
In order to further solve the problem proposed in [11], in this paper, we give the
sufficient conditions or the necessary and sufficient conditions
for the existence and
AB
the representations of the group inverse for block matrix C D (A, B ∈ K n×n ) when
A and B satisfy one of the following conditions:
(1) B ♯ and (B π A)♯ exist;
Electronic Journal of Linear Algebra ISSN 1081-3810
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(2) B ♯ and (AB π )♯ exist;
(3) B ♯ exists and BAB π = O;
(4) B ♯ exists and B π AB = O.
2. Some Lemmas.
Lemma 2.1. Let A ∈ K n×n . Then A has a group inverse if and
only if there exist
A1 O
P −1
nonsingular matrices P ∈ K n×n and A1 ∈ K r×r such that A = P
O O
A1 ♯ O
♯
P −1 , where rank(A) = r.
and A = P
O O
Proof. Theorem 2.2.2 of [28] holds over the complex number field and also over
skew fields.
Lemma 2.2. [7] Let A, G ∈ K n×n , Ind(A) = k. Then G = AD if and only if
Ak GA = Ak ,
Lemma 2.3. Let M =
A
B
AG = GA,
B
O
rank(G) ≤ rank(Ak ).
, S = B π AB π , A, B ∈ K n×n .
(i) If B ♯ and (B π A)♯ exist, then S ♯ and M ♯ exist.
(ii) If B ♯ exists and BAB π = O, then M ♯ exists if and only if (AB π )♯ exists.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
P ∈ K n×n and B1 ∈ K r×r
−1
such that
B1
O
B1 O
−1
♯
B=P
P −1 .
P
and B = P
O
O
O O
A1 A2
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
Let A = P
A3 A4
and A4 ∈ K (n−r)×(n−r) .
(i) Because (B π A)♯ exists, we have rank(B π A) = rank(B π A)2 , that is,
rank(A3 A4 ) = rank(A4 A3 A24 ).
Since rank(A3 A4 ) = rank(A4 (A3 A4 )) ≤ rank(A4 ) and rank(A3 A4 ) ≥ rank(A4 ),
we have rank(A3 A4 ) = rank(A4 ).
So there exists a matrix X ∈ K (n−r)×r such that A3 = A4 X.
Electronic Journal of Linear Algebra ISSN 1081-3810
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Hence, rank(A4 ) = rank(A24 ), i.e., A♯4 exists. Noting that
S
=
=
=
B π AB π
O
O
A1
P
O In−r
A3
O O
P
P −1 ,
O A4
we see that S ♯ exists. Since
A1
A3
rank(M ) = rank
B1
O
A2
A4
O
O
B1
O
O
O
A2
A4
O
O
O
O
= rank
O
O
O
In−r
O
O
B1
O
O
A4
O
O
P −1
B1
O
O
O
O
O
O
O
= 2r + rank(A4 )
and
2
rank(M )
= rank
= rank
2
A − ABB ♯ A + B 2
A2 + B 2 AB
= rank
2
O
BA
B
2
B1 + A2 A3 A2 A4 O O
O O
A4 A3
A24
,
O
O
B12 O
O
O
O O
O
B2
by A3 = A4 X, we get
B12
O
rank(M 2 ) = rank
O
O
O
A24
O
O
O
O
B12
O
O
O
= 2r + rank(A2 ),
4
O
O
and
rank(M ) = rank(M 2 ),
i.e., M ♯ exists.
(ii) If BAB π = O, then A2 = O, thus AB π = P
A1
A3
rank(M ) = rank
B1
O
O
A4
O
O
B1
O
O
O
O
O
O
O
= rank
O
O
O
A4
O
O
B1
O
P −1 ,
O
A4
O
O
B1
O
O
O
O
O
O
O
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= 2r + rank(A4 )
and
rank(M 2 ) =
=
=
2
A − ABB ♯ A + B 2 O
A2 + B 2 AB
=
rank
O
B
BA
B2
2
2
B1
O O O
B1 O O O
2
A4 A3 A24 O O
= rank O A4 O O
rank
2
2
O
O B1 O
O O B1 O
O
O O O
O O O O
2r + rank(A24 ).
rank
As (AB π )♯ exists, we get rank(A4 ) = rank(A24 ). Thus rank(M ) = rank(M 2 ); that
is, M ♯ exists, so the “if” part holds. Now we prove the “only if” part.
Since M ♯ exists if and only if rank(M ) = rank(M 2 ), rank(A4 ) = rank(A24 ), i.e.,
(AB π )♯ exists.
Lemma 2.4. Let A, B ∈ K n×n , S = B π AB π . Suppose B ♯ and (B π A)♯ exist.
Then S ♯ exists and the following conclusions hold:
(i) B π AS ♯ A = B π A;
(ii) B π AS ♯ = S ♯ AB π ;
(iii) BS ♯ = S ♯ B = B ♯ S ♯ = S ♯ B ♯ = O.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
B1 O
n×n
r×r
P −1 and
P ∈K
and B1 ∈ K
such that B = P
O O
♯
B =P
A1
A3
A2
Let A = P
A4
and A4 ∈ K (n−r)×(n−r) .
B1−1
O
O
O
P −1 .
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
From Lemma 2.3 (i) and the proof of Lemma 2.3 (i), we get S ♯ exists and
!
O
O
S♯ = P
P −1 .
O A♯4
π
♯
(i) B AS A = P
O
A3
O
A4
O
O
O
A♯4
!
A1
A3
A2
A4
P −1
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O
A4 A♯4 A3
=P
O
A4
!
P −1 .
By A3 = A4 X, X ∈ K (n−r)×r , we have
O
π
♯
B AS A = P
A3
(ii) B π AS ♯ = P
♯
(iii) BS = P
O
O
B1
O
O
A4 A♯4
!
O
O
O
O
O
A4
P −1 = B π A.
P −1 = S ♯ AB π .
O
A♯4
!
P −1 = O.
Similarly, we obtain S ♯ B = B ♯ S ♯ = S ♯ B ♯ = O.
Lemma 2.5. Let A, B ∈ K n×n . If BAB π = O, B ♯ and (AB π )♯ exist, then
(a) A(AB π )♯ AB π = AB π ;
(b) A(AB π )♯ = (AB π )♯ AB π ;
(c) B ♯ AB π = O, (AB π )♯ B = O, B(AB π )♯ = O.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
P ∈ K n×n and B1 ∈ K r×r
−1
such that
B1
O
B1 O
−1
♯
P −1 .
P
and B = P
B=P
O
O
O O
A1 A2
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
Let A = P
A3 A4
A1 O
and A4 ∈ K (n−r)×(n−r) . By BAB π = O, we get A = P
P −1 . Then
A3 A4
!
O O
A1 O
O O
π ♯
π
(a) A(AB ) AB = P
P −1
A3 A4
O A4
O A♯4
O O
=P
P −1 = AB π .
O A4
!
!
O O
O
O
A1 O
π ♯
−1
(b) A(AB ) = P
P =P
P −1
A3 A4
O A♯4
O A♯4 A4
= (AB π )♯ AB π .
−1
B1
O
O
(c) B ♯ AB π = P
O
O
O
O
A4
Similarly, we have (AB π )♯ B = O, B(AB π )♯ = O.
P −1 = O.
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3. Conclusions.
A
B
(B π A)♯ exist. Then M ♯ exists and
Theorem 3.1. Let M =
♯
M =
B
O
, where A, B ∈ K n×n . Suppose B ♯ and
U11
U21
U12
U22
,
where
U11
U12
U21
U22
S
=
=
=
=
=
S ♯ + (S ♯ A − I)BB ♯ AB π AS ♯ − (S ♯ A − I)BB ♯ AB π ;
B ♯ − S ♯ AB ♯ ;
B ♯ − B ♯ AS ♯ + B ♯ A(S ♯ A − I)BB ♯ AB π − B ♯ A(S ♯ A − I)BB ♯ AB π AS ♯ ;
B ♯ AS ♯ AB ♯ − B ♯ AB ♯ ;
B π AB π .
Proof. The existence of M ♯ and S ♯ have been given in Lemma 2.3 (i).
U11 U12
. Then
Let X =
U21 U22
AU11 + BU21 AU12 + BU22
U11 A + U12 B U11 B
MX =
.
and XM =
BU11
BU12
U21 A + U22 B U21 B
We prove X = M ♯ . Applying (i)–(iii) of Lemma 2.4, we get the following identities:
AU11 + BU21 = AS ♯ + A(S ♯ A − I)BB ♯ AB π AS ♯ − A(S ♯ A − I)BB ♯ AB π + BB ♯
− BB ♯ AS ♯ + BB ♯ A(S ♯ A − I)BB ♯ AB π − BB ♯ A(S ♯ A − I)BB ♯ AB π AS ♯
= BB ♯ + B π AS ♯ + B π A(S ♯ A − I)BB ♯ AB π AS ♯ − B π A(S ♯ A − I)BB ♯ AB π
(i)
= B π AS ♯ + BB ♯ ,
and
U11 A + U12 B = S ♯ A + (S ♯ A − I)BB ♯ AB π AS ♯ A − (S ♯ A − I)BB ♯ AB π A + BB ♯
(i)
− S ♯ ABB ♯
= S ♯ AB π + BB ♯
(ii)
= B π AS ♯ + BB ♯ .
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Hence,
AU11 + BU21 = U11 A + U12 B.
AU12 + BU22 = AB ♯ − B π AS ♯ AB ♯ − BB ♯ AB ♯
(i)
= AB ♯ − B π AB ♯ − BB ♯ AB ♯
= O,
U11 B = S ♯ B + (S ♯ A − I)BB ♯ AB π AS ♯ B − (S ♯ A − I)BB ♯ AB π B
= S ♯ B + (S ♯ A − I)BB ♯ AB π AS ♯ B
(iii)
= O.
Therefore, AU12 + BU22 = U11 B. Similarly, we can get
BU11 = U21 A + U22 B = BB ♯ AB π (I − AS ♯ ),
BU12 = U21 B = BB ♯ .
Consequently, M X = XM =
BB ♯ + B π AS ♯
O
BB ♯ AB π (I − AS ♯ ) BB ♯
It is easy to compute
BB ♯ + B π AS ♯
M XM =
BB ♯ AB π (I − AS ♯ )
O
BB ♯
A
B
B
O
.
=
A
B
B
O
= M.
Suppose rank(B) = r. By Lemma 2.1, there exist invertible matrices P ∈ K n×n
−1
B1
O
B1 O
P −1 .
P −1 and B ♯ = P
and B1 ∈ K r×r such that B = P
O O
O
O
A1 A2
Let A = P
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈
A3 A4
K (n−r)×r and A4 ∈ K (n−r)×(n−r) .
U11 U12
rank(X) = rank
U21 U22
♯
S + (S ♯ A − I)BB ♯ AB π AS ♯ − (S ♯ A − I)BB ♯ AB π (I − S ♯ A)B ♯
= rank
B♯
O
−1
O
O
O B1
♯
♯
♯
♯
♯
S
B − S AB
A4
O
O
O
= rank −1
= rank
♯
B
O
B1
O
O
O
O
O
O
O
= 2r + rank(A♯4 ) = 2r + rank(A4 ) = rank(M ).
From Lemma 2.2 , we get X = M ♯ .
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Now we present an example to show the representation of the group inverse for a
block matrix over the real quaternion skew field.
Example 3.2. Let the real quaternion skew field K = {a + bi + cj + dk}, where
0 0
A B
and
, where A =
a, b, c and d are real numbers. Let M =
0 k
B O
i j
.
B=
0 0
−i −j
♯
π
♯
♯
,
By computation, B and (B A) exist, and B =
0
0
0
1
.
S♯ =
0 −k
−i −j
0
1
, U21 =
, U12 =
By Theorem 3.1, M ♯ exists and U11 =
0
0
0 −k
0
1 −i −j
0 −k
0
0
0 0
−i 0
.
, so M ♯ =
, U22 =
−i
0
0
0
0 0
0 0
0
0
0
0
A B
, where A, B ∈ K n×n . Suppose B ♯ and
Theorem 3.3. Let M =
B O
(AB π )♯ exist. Then M ♯ exists and
♯
M =
U11
U21
U12
U22
,
where
U11 = S ♯ + S ♯ AB π ABB ♯ (AS ♯ − I) − B π ABB ♯ (AS ♯ − I);
U12 = B ♯ − S ♯ AB ♯ − S ♯ AB π ABB ♯ (AS ♯ − I)AB ♯ + B π ABB ♯ (I − AS ♯ )AB ♯ ;
U21 = B ♯ − B ♯ AS ♯ ;
U22 = B ♯ AS ♯ AB ♯ − B ♯ AB ♯ ;
S = B π AB π .
Proof. The proof is similar to that of Theorem 3.1.
A B
, where A, B ∈ K n×n . Suppose B ♯ exists
Theorem 3.4. Let M =
B O
and BAB π = O. Then
(i) M ♯ exists if and only if (AB π )♯ exists.
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(ii) If M ♯ exists, then
M♯ =
U
B♯
V
−B ♯ AB ♯
,
where
U = B π A(B ♯ )2 − (AB π )♯ AB π A(B ♯ )2 + (AB π )♯ ;
V = − B π A(B ♯ )2 AB ♯ + (AB π )♯ AB π A(B ♯ )2 AB ♯ − (AB π )♯ AB ♯ + B ♯ .
Proof. (i) The existence of M ♯ has been given in Lemma 2.3 (ii).
U
V
. Then
(ii) Let X =
B ♯ −B ♯ AB ♯
UA + V B
AU + BB ♯ AV − BB ♯ AB ♯
and XM =
MX =
B ♯ AB π
BU
BV
We prove M X = XM by Lemma 2.5.
UB
B♯B
AU + BB ♯ = AB π A(B ♯ )2 − A(AB π )♯ AB π A(B ♯ )2 + A(AB π )♯ + BB ♯
(a)
= AB π A(B ♯ )2 − AB π A(B ♯ )2 + A(AB π )♯ + BB ♯
= A(AB π )♯ + BB ♯ .
U A + V B = B π A(B ♯ )2 AB π − (AB π )♯ AB π A(B ♯ )2 AB π + (AB π )♯ AB π + B ♯ B
(c)
= (AB π )♯ AB π + B ♯ B
(b)
= A(AB π )♯ + BB ♯ .
Thus,
AU + BB ♯ = U A + V B.
AV − BB ♯ AB ♯ = − AB π A(B ♯ )2 AB ♯ + A(AB π )♯ AB π A(B ♯ )2 AB ♯ − A(AB π )♯ AB ♯
(a)
+ AB ♯ − BB ♯ AB ♯
♯
= − A (AB π ) AB ♯ + B π AB ♯
U B = B π AB ♯ − (AB π )♯ AB π AB ♯ + (AB π )♯ B
(b)
= B π AB ♯ − A(AB π )♯ AB ♯ + (AB π )♯ B
(c)
= − A(AB π )♯ AB ♯ + B π AB ♯ .
Hence, AV − BB ♯ AB ♯ = U B. Similarly, we can obtain
BU = B ♯ AB π = O, BV = B ♯ B.
.
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Consequently,
M X = XM =
A(AB π )♯ + BB ♯
O
−A(AB π )♯ AB ♯ + B π AB ♯
BB ♯
.
Applying Lemma 2.5, it is easy to compute
♯
A (AB π ) + BB ♯
O
M XM =
=
A
B
B
O
!
−A(AB π )♯ AB ♯ + B π AB ♯
BB ♯
A
B
B
O
= M.
Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices P ∈
K n×n and B1 ∈ K r×r such that
B=P
B1
O
O
O
P −1 and B ♯ = P
B1−1
O
O
O
P −1 .
A2
Let A = P
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r), A3 ∈ K (n−r)×r ,
A4
and A4 ∈ K (n−r)×(n−r). Then
U B♯
U
V
= rank
rank(X) = rank
B♯ O
B ♯ −B ♯ AB ♯
O
O B1−1 O
(AB π )♯ B ♯
A♯4
O
O
O
=
rank
= rank
−1
B♯
O
B1
O
O
O
O
O
O
O
A1
A3
= 2r + rank(A♯4 ).
Since rank(M ) = 2r + rank(A4 ), rank(X) = rank(M ), thus X = M ♯ by Lemma
2.2.
Example 3.5. Let M =
√
i i
, i = −1.
0 0
A
B
B
O
, where A =
1
i
1+i
0
and B =
−i −i
.
0
0
0
0
0 −i
♯
and −B ♯ AB ♯
,V =
From Theorem 3.4, M exists, U =
−i −i
0
i
By computation, B ♯ exists, BAB π = O and B ♯ =
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0
0
1+i 1+i
, then M ♯ =
=
−i
0
0
0
A B
Theorem 3.6. Let M =
B O
exists and B π AB = O. Then
−i
0
0
i
−i
−i
.
−i 1 + i 1 + i
0
0
0
, where A, B ∈ K n×n . Suppose that B ♯
(i) M ♯ exists if and only if (B π A)♯ exists.
(ii) If M ♯ exists, then
M♯ =
U
V
B♯
−B ♯ AB ♯
,
where
U = (B ♯ )2 AB π − (B ♯ )2 AB π A(B π A)♯ + (B π A)♯ ;
V = − B ♯ A(B ♯ )2 AB π + B ♯ A(B ♯ )2 AB π A(B π A)♯ − B ♯ A(B π A)♯ + B ♯ .
Proof. The proof is similar to that of Theorem 3.4.
Acknowledgments: The authors would like to thank the referee for his/her
helpful comments.
REFERENCES
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[9] C. Cao, X. Tang. Representations of the group inverse of some 2 × 2 block matrices.
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[12] S.L. Campbell, C.D. Meyer. Generalized inverses of linear transformations. Dover, New
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[22] R. Hartwig, X. Li, Y. Wei. Representations for the Drazin inverse of 2 × 2 block matrix.
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[23] R.E. Hartwig, J.M. Shoaf. Group inverse and Drazin inverse of bidiagonal and triangular
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A publication of the International Linear Algebra Society
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GROUP INVERSE FOR THE BLOCK MATRIX WITH
TWO IDENTICAL SUBBLOCKS OVER SKEW FIELDS∗
JIEMEI ZHAO† AND CHANGJIANG BU§†‡
Abstract. Let K be a skew field and K n×n be the set of all n × n matrices over K. The
purpose of this paper is to give some necessary and sufficient
conditions for the existence and the
A B
under some conditions.
representations of the group inverse of the block matrix C
D
Key words. Skew fields, Block matrix, Group inverse.
AMS subject classifications. 15A09; 65F20.
1. Introduction. Let K be a skew field, C be the complex number field, K m×n
be the set of all m × n matrices over K, and In be the n × n identity matrix over K.
For a matrix A ∈ K n×n , the matrix X ∈ K n×n satisfying
Ak XA = Ak , XAX = X, AX = XA
is called the Drazin inverse of A and is denoted by X = AD , where k is the index
of A, i.e., the smallest non-negative integer such that rank(Ak ) = rank(Ak+1 ). We
denote such a k by Ind(A). It is well-known that AD exists and is unique (see [2]).
If Ind(A) = 1, AD is also called the group inverse of A and is denoted by A♯ . Then
A♯ exists if and only if rank(A) = rank(A2 ) (see [1, 3, 11-14, 24, 25, 29]). We denote
I − AA♯ by Aπ .
The group inverse of block matrices has numerous applications in matrix theory,
such as singular differential and difference equations, Markov chains, iterative methods and so on (see [12]-[14]). For instance, Y. Wei et al. studied the representation of
the group inverse of a real singular Toeplitz matrix which arises in scientific computing
and engineering (see [29]); in [25], S. Kirkland et al. investigated the representation
of the group inverse of the Laplacian matrix of an undirected weighted graph G on
n vertices; and in [24], G. Heinig studied the group inverse of the Sylvester transformation ϕ(X) = AX − XB, where A ∈ Cm×m and B ∈ Cn×n ; utilizing the theory of
Drazin inverse, the differential equation Ax′ + Bx = f is studied in linear systems,
∗ Received by the editors on June 2009. Accepted for publication on July 31, 2010 Handling
Editors: Roger A. Horn and Fuzhen Zhang.
† College of Automation, Harbin Engineering University, Harbin 150001, P. R. China. ([email protected]). Supported by the NSFH, No.159110120002.
‡ Dept. of Applied Math., College of Science, Harbin Engineering University, Harbin 150001, P.
R. China. § Corresponding author. Email: [email protected]
63
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where A, B are both n × n singular matrices and f is a vector-valued function (see
[13]); in [3], R. Bru et al. gave the general explicit solution which is obtained when
the symmetric singular linear control system satisfies the regularity condition.
In 1979, S. Campbell and C. Meyer proposed an open problem
to find an explicit
AB
representation for the Drazin inverse of a 2 × 2 block matrix C D , where the blocks
A and D are supposed to be square matrices but their sizes need not be the same (see
[12]). A simplified problem
tofind an explicit representation of the Drazin (group)
AB
inverse for block matrix C
(A is square, O is square null matrix) was proposed
O
by S. Campbell in 1983. This open problem was motivated in hoping to find general
expressions for the solutions of the second-order system of the differential equations
Ex′′ (t) + F x′ (t) + Gx(t) = 0 (t ≥ 0),
where E is a singular matrix. Detailed discussions of the importance of the problem
can be found in [11]. Until now, both problems have not been resolved. However,
there have been some studies about the representations of the Drazin(group) inverse
under certain conditions (see [4-10, 15-23, 26, 27]). For example, the formulae of
Drazin(group) inverse of an operator matrix under some conditions are investigated
in [18] and [19]. The representations for the Drazin(group) inverse of the sum of
several matrices are given in [15] and [17]. In particular, J. Chen et al. [15] presented
the representation of the group inverse of M = P + Q + S, where P Q = QP = O,
P S = SQ = O, P ♯ and Q♯ exist.
Some conditions for the representations of group
AB
inverse for block matrix C D are listed below:
(i) A♯ and (D − CA♯ B)♯ exist [18].
(ii) D = O, A = B = Id , Id , C ∈ Cd×d [10].
(iii) D = O, A, B, C ∈ {P, P ∗ , P P ∗ }, P ∈ Cn×n , P 2 = P , P ∗ is the conjugate
transpose of P [9].
(iv) C = O, A and D are square matrices over K [8].
(v) D = O, A = B = A2 , A, C ∈ K n×n [4].
(vi) D = O, A = B, rank(C) ≥ rank(A), A, C ∈ K n×n [5].
(vii) A is invertible and (D − CA−1 B)♯ exists; D is invertible and (A − BD−1 C)♯
exists; B or C is invertible, where A, B, C ∈ K n×n [6].
(viii) D = O, A = c1 B + c2 C, where non-zero elements c1 and c2 are in the center
of K; D = O, A = B k C l , where k and l are positive integers [7].
In order to further solve the problem proposed in [11], in this paper, we give the
sufficient conditions or the necessary and sufficient conditions
for the existence and
AB
the representations of the group inverse for block matrix C D (A, B ∈ K n×n ) when
A and B satisfy one of the following conditions:
(1) B ♯ and (B π A)♯ exist;
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(2) B ♯ and (AB π )♯ exist;
(3) B ♯ exists and BAB π = O;
(4) B ♯ exists and B π AB = O.
2. Some Lemmas.
Lemma 2.1. Let A ∈ K n×n . Then A has a group inverse if and
only if there exist
A1 O
P −1
nonsingular matrices P ∈ K n×n and A1 ∈ K r×r such that A = P
O O
A1 ♯ O
♯
P −1 , where rank(A) = r.
and A = P
O O
Proof. Theorem 2.2.2 of [28] holds over the complex number field and also over
skew fields.
Lemma 2.2. [7] Let A, G ∈ K n×n , Ind(A) = k. Then G = AD if and only if
Ak GA = Ak ,
Lemma 2.3. Let M =
A
B
AG = GA,
B
O
rank(G) ≤ rank(Ak ).
, S = B π AB π , A, B ∈ K n×n .
(i) If B ♯ and (B π A)♯ exist, then S ♯ and M ♯ exist.
(ii) If B ♯ exists and BAB π = O, then M ♯ exists if and only if (AB π )♯ exists.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
P ∈ K n×n and B1 ∈ K r×r
−1
such that
B1
O
B1 O
−1
♯
B=P
P −1 .
P
and B = P
O
O
O O
A1 A2
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
Let A = P
A3 A4
and A4 ∈ K (n−r)×(n−r) .
(i) Because (B π A)♯ exists, we have rank(B π A) = rank(B π A)2 , that is,
rank(A3 A4 ) = rank(A4 A3 A24 ).
Since rank(A3 A4 ) = rank(A4 (A3 A4 )) ≤ rank(A4 ) and rank(A3 A4 ) ≥ rank(A4 ),
we have rank(A3 A4 ) = rank(A4 ).
So there exists a matrix X ∈ K (n−r)×r such that A3 = A4 X.
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Hence, rank(A4 ) = rank(A24 ), i.e., A♯4 exists. Noting that
S
=
=
=
B π AB π
O
O
A1
P
O In−r
A3
O O
P
P −1 ,
O A4
we see that S ♯ exists. Since
A1
A3
rank(M ) = rank
B1
O
A2
A4
O
O
B1
O
O
O
A2
A4
O
O
O
O
= rank
O
O
O
In−r
O
O
B1
O
O
A4
O
O
P −1
B1
O
O
O
O
O
O
O
= 2r + rank(A4 )
and
2
rank(M )
= rank
= rank
2
A − ABB ♯ A + B 2
A2 + B 2 AB
= rank
2
O
BA
B
2
B1 + A2 A3 A2 A4 O O
O O
A4 A3
A24
,
O
O
B12 O
O
O
O O
O
B2
by A3 = A4 X, we get
B12
O
rank(M 2 ) = rank
O
O
O
A24
O
O
O
O
B12
O
O
O
= 2r + rank(A2 ),
4
O
O
and
rank(M ) = rank(M 2 ),
i.e., M ♯ exists.
(ii) If BAB π = O, then A2 = O, thus AB π = P
A1
A3
rank(M ) = rank
B1
O
O
A4
O
O
B1
O
O
O
O
O
O
O
= rank
O
O
O
A4
O
O
B1
O
P −1 ,
O
A4
O
O
B1
O
O
O
O
O
O
O
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= 2r + rank(A4 )
and
rank(M 2 ) =
=
=
2
A − ABB ♯ A + B 2 O
A2 + B 2 AB
=
rank
O
B
BA
B2
2
2
B1
O O O
B1 O O O
2
A4 A3 A24 O O
= rank O A4 O O
rank
2
2
O
O B1 O
O O B1 O
O
O O O
O O O O
2r + rank(A24 ).
rank
As (AB π )♯ exists, we get rank(A4 ) = rank(A24 ). Thus rank(M ) = rank(M 2 ); that
is, M ♯ exists, so the “if” part holds. Now we prove the “only if” part.
Since M ♯ exists if and only if rank(M ) = rank(M 2 ), rank(A4 ) = rank(A24 ), i.e.,
(AB π )♯ exists.
Lemma 2.4. Let A, B ∈ K n×n , S = B π AB π . Suppose B ♯ and (B π A)♯ exist.
Then S ♯ exists and the following conclusions hold:
(i) B π AS ♯ A = B π A;
(ii) B π AS ♯ = S ♯ AB π ;
(iii) BS ♯ = S ♯ B = B ♯ S ♯ = S ♯ B ♯ = O.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
B1 O
n×n
r×r
P −1 and
P ∈K
and B1 ∈ K
such that B = P
O O
♯
B =P
A1
A3
A2
Let A = P
A4
and A4 ∈ K (n−r)×(n−r) .
B1−1
O
O
O
P −1 .
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
From Lemma 2.3 (i) and the proof of Lemma 2.3 (i), we get S ♯ exists and
!
O
O
S♯ = P
P −1 .
O A♯4
π
♯
(i) B AS A = P
O
A3
O
A4
O
O
O
A♯4
!
A1
A3
A2
A4
P −1
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O
A4 A♯4 A3
=P
O
A4
!
P −1 .
By A3 = A4 X, X ∈ K (n−r)×r , we have
O
π
♯
B AS A = P
A3
(ii) B π AS ♯ = P
♯
(iii) BS = P
O
O
B1
O
O
A4 A♯4
!
O
O
O
O
O
A4
P −1 = B π A.
P −1 = S ♯ AB π .
O
A♯4
!
P −1 = O.
Similarly, we obtain S ♯ B = B ♯ S ♯ = S ♯ B ♯ = O.
Lemma 2.5. Let A, B ∈ K n×n . If BAB π = O, B ♯ and (AB π )♯ exist, then
(a) A(AB π )♯ AB π = AB π ;
(b) A(AB π )♯ = (AB π )♯ AB π ;
(c) B ♯ AB π = O, (AB π )♯ B = O, B(AB π )♯ = O.
Proof. Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices
P ∈ K n×n and B1 ∈ K r×r
−1
such that
B1
O
B1 O
−1
♯
P −1 .
P
and B = P
B=P
O
O
O O
A1 A2
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈ K (n−r)×r
Let A = P
A3 A4
A1 O
and A4 ∈ K (n−r)×(n−r) . By BAB π = O, we get A = P
P −1 . Then
A3 A4
!
O O
A1 O
O O
π ♯
π
(a) A(AB ) AB = P
P −1
A3 A4
O A4
O A♯4
O O
=P
P −1 = AB π .
O A4
!
!
O O
O
O
A1 O
π ♯
−1
(b) A(AB ) = P
P =P
P −1
A3 A4
O A♯4
O A♯4 A4
= (AB π )♯ AB π .
−1
B1
O
O
(c) B ♯ AB π = P
O
O
O
O
A4
Similarly, we have (AB π )♯ B = O, B(AB π )♯ = O.
P −1 = O.
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3. Conclusions.
A
B
(B π A)♯ exist. Then M ♯ exists and
Theorem 3.1. Let M =
♯
M =
B
O
, where A, B ∈ K n×n . Suppose B ♯ and
U11
U21
U12
U22
,
where
U11
U12
U21
U22
S
=
=
=
=
=
S ♯ + (S ♯ A − I)BB ♯ AB π AS ♯ − (S ♯ A − I)BB ♯ AB π ;
B ♯ − S ♯ AB ♯ ;
B ♯ − B ♯ AS ♯ + B ♯ A(S ♯ A − I)BB ♯ AB π − B ♯ A(S ♯ A − I)BB ♯ AB π AS ♯ ;
B ♯ AS ♯ AB ♯ − B ♯ AB ♯ ;
B π AB π .
Proof. The existence of M ♯ and S ♯ have been given in Lemma 2.3 (i).
U11 U12
. Then
Let X =
U21 U22
AU11 + BU21 AU12 + BU22
U11 A + U12 B U11 B
MX =
.
and XM =
BU11
BU12
U21 A + U22 B U21 B
We prove X = M ♯ . Applying (i)–(iii) of Lemma 2.4, we get the following identities:
AU11 + BU21 = AS ♯ + A(S ♯ A − I)BB ♯ AB π AS ♯ − A(S ♯ A − I)BB ♯ AB π + BB ♯
− BB ♯ AS ♯ + BB ♯ A(S ♯ A − I)BB ♯ AB π − BB ♯ A(S ♯ A − I)BB ♯ AB π AS ♯
= BB ♯ + B π AS ♯ + B π A(S ♯ A − I)BB ♯ AB π AS ♯ − B π A(S ♯ A − I)BB ♯ AB π
(i)
= B π AS ♯ + BB ♯ ,
and
U11 A + U12 B = S ♯ A + (S ♯ A − I)BB ♯ AB π AS ♯ A − (S ♯ A − I)BB ♯ AB π A + BB ♯
(i)
− S ♯ ABB ♯
= S ♯ AB π + BB ♯
(ii)
= B π AS ♯ + BB ♯ .
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Hence,
AU11 + BU21 = U11 A + U12 B.
AU12 + BU22 = AB ♯ − B π AS ♯ AB ♯ − BB ♯ AB ♯
(i)
= AB ♯ − B π AB ♯ − BB ♯ AB ♯
= O,
U11 B = S ♯ B + (S ♯ A − I)BB ♯ AB π AS ♯ B − (S ♯ A − I)BB ♯ AB π B
= S ♯ B + (S ♯ A − I)BB ♯ AB π AS ♯ B
(iii)
= O.
Therefore, AU12 + BU22 = U11 B. Similarly, we can get
BU11 = U21 A + U22 B = BB ♯ AB π (I − AS ♯ ),
BU12 = U21 B = BB ♯ .
Consequently, M X = XM =
BB ♯ + B π AS ♯
O
BB ♯ AB π (I − AS ♯ ) BB ♯
It is easy to compute
BB ♯ + B π AS ♯
M XM =
BB ♯ AB π (I − AS ♯ )
O
BB ♯
A
B
B
O
.
=
A
B
B
O
= M.
Suppose rank(B) = r. By Lemma 2.1, there exist invertible matrices P ∈ K n×n
−1
B1
O
B1 O
P −1 .
P −1 and B ♯ = P
and B1 ∈ K r×r such that B = P
O O
O
O
A1 A2
Let A = P
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r) , A3 ∈
A3 A4
K (n−r)×r and A4 ∈ K (n−r)×(n−r) .
U11 U12
rank(X) = rank
U21 U22
♯
S + (S ♯ A − I)BB ♯ AB π AS ♯ − (S ♯ A − I)BB ♯ AB π (I − S ♯ A)B ♯
= rank
B♯
O
−1
O
O
O B1
♯
♯
♯
♯
♯
S
B − S AB
A4
O
O
O
= rank −1
= rank
♯
B
O
B1
O
O
O
O
O
O
O
= 2r + rank(A♯4 ) = 2r + rank(A4 ) = rank(M ).
From Lemma 2.2 , we get X = M ♯ .
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Now we present an example to show the representation of the group inverse for a
block matrix over the real quaternion skew field.
Example 3.2. Let the real quaternion skew field K = {a + bi + cj + dk}, where
0 0
A B
and
, where A =
a, b, c and d are real numbers. Let M =
0 k
B O
i j
.
B=
0 0
−i −j
♯
π
♯
♯
,
By computation, B and (B A) exist, and B =
0
0
0
1
.
S♯ =
0 −k
−i −j
0
1
, U21 =
, U12 =
By Theorem 3.1, M ♯ exists and U11 =
0
0
0 −k
0
1 −i −j
0 −k
0
0
0 0
−i 0
.
, so M ♯ =
, U22 =
−i
0
0
0
0 0
0 0
0
0
0
0
A B
, where A, B ∈ K n×n . Suppose B ♯ and
Theorem 3.3. Let M =
B O
(AB π )♯ exist. Then M ♯ exists and
♯
M =
U11
U21
U12
U22
,
where
U11 = S ♯ + S ♯ AB π ABB ♯ (AS ♯ − I) − B π ABB ♯ (AS ♯ − I);
U12 = B ♯ − S ♯ AB ♯ − S ♯ AB π ABB ♯ (AS ♯ − I)AB ♯ + B π ABB ♯ (I − AS ♯ )AB ♯ ;
U21 = B ♯ − B ♯ AS ♯ ;
U22 = B ♯ AS ♯ AB ♯ − B ♯ AB ♯ ;
S = B π AB π .
Proof. The proof is similar to that of Theorem 3.1.
A B
, where A, B ∈ K n×n . Suppose B ♯ exists
Theorem 3.4. Let M =
B O
and BAB π = O. Then
(i) M ♯ exists if and only if (AB π )♯ exists.
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(ii) If M ♯ exists, then
M♯ =
U
B♯
V
−B ♯ AB ♯
,
where
U = B π A(B ♯ )2 − (AB π )♯ AB π A(B ♯ )2 + (AB π )♯ ;
V = − B π A(B ♯ )2 AB ♯ + (AB π )♯ AB π A(B ♯ )2 AB ♯ − (AB π )♯ AB ♯ + B ♯ .
Proof. (i) The existence of M ♯ has been given in Lemma 2.3 (ii).
U
V
. Then
(ii) Let X =
B ♯ −B ♯ AB ♯
UA + V B
AU + BB ♯ AV − BB ♯ AB ♯
and XM =
MX =
B ♯ AB π
BU
BV
We prove M X = XM by Lemma 2.5.
UB
B♯B
AU + BB ♯ = AB π A(B ♯ )2 − A(AB π )♯ AB π A(B ♯ )2 + A(AB π )♯ + BB ♯
(a)
= AB π A(B ♯ )2 − AB π A(B ♯ )2 + A(AB π )♯ + BB ♯
= A(AB π )♯ + BB ♯ .
U A + V B = B π A(B ♯ )2 AB π − (AB π )♯ AB π A(B ♯ )2 AB π + (AB π )♯ AB π + B ♯ B
(c)
= (AB π )♯ AB π + B ♯ B
(b)
= A(AB π )♯ + BB ♯ .
Thus,
AU + BB ♯ = U A + V B.
AV − BB ♯ AB ♯ = − AB π A(B ♯ )2 AB ♯ + A(AB π )♯ AB π A(B ♯ )2 AB ♯ − A(AB π )♯ AB ♯
(a)
+ AB ♯ − BB ♯ AB ♯
♯
= − A (AB π ) AB ♯ + B π AB ♯
U B = B π AB ♯ − (AB π )♯ AB π AB ♯ + (AB π )♯ B
(b)
= B π AB ♯ − A(AB π )♯ AB ♯ + (AB π )♯ B
(c)
= − A(AB π )♯ AB ♯ + B π AB ♯ .
Hence, AV − BB ♯ AB ♯ = U B. Similarly, we can obtain
BU = B ♯ AB π = O, BV = B ♯ B.
.
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Consequently,
M X = XM =
A(AB π )♯ + BB ♯
O
−A(AB π )♯ AB ♯ + B π AB ♯
BB ♯
.
Applying Lemma 2.5, it is easy to compute
♯
A (AB π ) + BB ♯
O
M XM =
=
A
B
B
O
!
−A(AB π )♯ AB ♯ + B π AB ♯
BB ♯
A
B
B
O
= M.
Suppose rank(B) = r. Applying Lemma 2.1, there exist invertible matrices P ∈
K n×n and B1 ∈ K r×r such that
B=P
B1
O
O
O
P −1 and B ♯ = P
B1−1
O
O
O
P −1 .
A2
Let A = P
P −1 , where A1 ∈ K r×r , A2 ∈ K r×(n−r), A3 ∈ K (n−r)×r ,
A4
and A4 ∈ K (n−r)×(n−r). Then
U B♯
U
V
= rank
rank(X) = rank
B♯ O
B ♯ −B ♯ AB ♯
O
O B1−1 O
(AB π )♯ B ♯
A♯4
O
O
O
=
rank
= rank
−1
B♯
O
B1
O
O
O
O
O
O
O
A1
A3
= 2r + rank(A♯4 ).
Since rank(M ) = 2r + rank(A4 ), rank(X) = rank(M ), thus X = M ♯ by Lemma
2.2.
Example 3.5. Let M =
√
i i
, i = −1.
0 0
A
B
B
O
, where A =
1
i
1+i
0
and B =
−i −i
.
0
0
0
0
0 −i
♯
and −B ♯ AB ♯
,V =
From Theorem 3.4, M exists, U =
−i −i
0
i
By computation, B ♯ exists, BAB π = O and B ♯ =
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0
0
1+i 1+i
, then M ♯ =
=
−i
0
0
0
A B
Theorem 3.6. Let M =
B O
exists and B π AB = O. Then
−i
0
0
i
−i
−i
.
−i 1 + i 1 + i
0
0
0
, where A, B ∈ K n×n . Suppose that B ♯
(i) M ♯ exists if and only if (B π A)♯ exists.
(ii) If M ♯ exists, then
M♯ =
U
V
B♯
−B ♯ AB ♯
,
where
U = (B ♯ )2 AB π − (B ♯ )2 AB π A(B π A)♯ + (B π A)♯ ;
V = − B ♯ A(B ♯ )2 AB π + B ♯ A(B ♯ )2 AB π A(B π A)♯ − B ♯ A(B π A)♯ + B ♯ .
Proof. The proof is similar to that of Theorem 3.4.
Acknowledgments: The authors would like to thank the referee for his/her
helpful comments.
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