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A Note on the Formulas for the Drazin Inverse of the Sum of
Two Matrices and Its Applications
Xiaolan Qin
*, Ricai Luo
School of Mathematics and Statistics, Hechi University, Yizhou, China
Email address:
*Corresponding author
To cite this article:
Xiaolan Qin, Ricai Luo. A Note on the Formulas for the Drazin Inverse of the Sum of Two Matrices and Its Applications. Pure and Applied Mathematics Journal. Vol. 8, No. 3, 2019, pp. 54-71. doi: 10.11648/j.pamj.20190803.12
Received: July 6, 2019; Accepted: August 10, 2019; Published: August 30, 2019
Abstract:
The Drazin inverse has applications in a number of areas such as control theory, Markov chains, singular differential and difference equations, and iterative methods in numerical linear algebra. The study on representations for the Drazin inverse of block matrices stems essentially from finding the general expressions for the solutions to singular systems of differential equations, and then stimulated by a problem formulated by Campbell. In 1983, Campbell (Campbell et al. (1976)) established an explicit representation for the Drazin inverse of a 2 × 2 block matrix M in terms of the blocks of the partition, where the blocks A and D are assumed to be square matrices. Special cases of the problems have been studied. In 2009, Chunyuan Deng and Yimin Wei found an explicit representation for the Drazin inverse of an anti-triangular matrix M, where A and BC are generalized Drazin invertible, if AπAB=0 and BC (I–Aπ)=0. Afterwards, several authors have investigated this problem under some limited conditions on the blocks of M. In particular, a representation of the Drazin inverse of M, denoted by Md. In this paper, we consider the Drazin inverse of a sum of two matrices and we derive additive formulas under the conditions of ABAπ=0 and BAπ=0 respectively. Precisely, for a block matrix M, we give a new representation of Md under some conditions that AB=0 and DCAπ=0. Moreover, some particular cases of this result related to the Drazin inverse of block matrices are also considered.Keywords:
Drazin Inverse, Block Matrices, Drazin Index1. Introduction
Let A be a square complex matrix. We denote by R (A), N (A) and rank (A), the range, the null space and the rank of matrix A, respectively. In addition, the smallest integer k is called the Drazin index of A such that rank(Ak+1) =rank
(Ak). Let Cn n× be the set of all the n n× matrices over the
complex field, for every matrix A∈Cn n× , such that
( )
index A =k, there exists an unique matrixAd∈Cn n× , which
satisfies the relations:
1
, , .
d d d d d k d k
A AA =A A A= AA A + A = A
The matrixAd is called the Drazin inverse of A [1, 2]. The case index A( )=0 is valid if and only if A is nonsingular, so d
A reduces toA−1. By Aπ = −I AAd , we
denote the projection onN A( k)along R A( k). If the lower limit of a sum is greater than its upper limit, we always define
the sum to be 0. For example, the sum 2
1 *
k
l
−
=
=
∑
0 fork≤2. Weagree thatA0=I, for any matrix A.
The study on representations for the Drazin inverse of block matrices essentially originated from finding the general expressions for the solutions to singular systems of differential equations [3-5], and then stimulated by a problem formulated by Campbell [3]: establish an explicit representation for the Drazin inverse of 2×2 block matrices
, A B M
C D
=
(1)
of applications of the Drazin inverse of a 2×2 block matrix, we refer the reader to [5, 6]. Meyer and Rose [7], and independently Hartwig and Shoaf [8], first gave the formulas for block triangular matrices. Several authors have investigated this problem and they were able to find some partial answers (imposing some conditions on the blocks of M). Here, we list some cases of Drazin inverse of block matrix M:
(1) AB=0 and DC =0. See [9]. (2)BC=0 and BD =0. See [10]. (3)BC=0 and DC =0. See [11].
(4) AA Bπ =0, BCAAd =0and DC =0. See [12]. (5)BCA=0,BCB=0,DCA=0and DC B=0. See [13]. (6)BC A=0,C B D=0 ,
A BC
( )
π=
0
andD CB
( )
π=
0.
See [14].
The motivation for this article is [15]. In the paper, the authors considered some conditions on a, b∈Λ（Let Λ be a complex Banach algebra with the unit 1）that allowed them to express (a + b) d in terms of a, ad, b, bd. In this paper, we consider the Drazin inverse of a sum of two matrices and we derive additive formulas under the conditions of ABAπ =0 and BAπ =0 respectively. As an application we give some new representations for the Drazin inverse of a block matrix.
2. A New Additive Result for the Drazin
Inverse of Matrices
First, we will state some auxiliary lemmas.
Lemma 2.1 [16] Let A B, ∈Cn n× . If AB=0 an A is nilpotent, then
1 1
0
( ) ( ) ,
r
d d i i
i
A B B A
− + =
+ =
∑
ind A( )
=rLemma 2.2 [7] LetM1 and M2be complex matrices of the
form
1
0 , A M
C B
=
2 0 ,
B C M
A
=
where A and B are complex square matrices. Let
( )
r=ind A and s= ind B
( )
. Then max{ }
r s, ≤ ( i)ind M ≤ +r s, for i=1,2, and
1
0 , d d
d
A M
S B
=
2
, 0
d d
d
B S
M
A
=
where
1 1
2 2
0 0
=( ) ( ) ( ) ( ) .
r s
d d i i i d i d d d
i i
S B B CA Aπ Bπ B C A A B CA
− −
= =
+ −
∑
∑
The following theorem, we obtain the same expression for the Drazin inverse(A+B)d as in [15, Theorem2.3] for the Generalized Drazin inverse in a Banach Algebra.
Theorem 2.1. LetA B, ∈ℂn n× . If ABAπ =0, we have
1 1
1 1
0 0
1 1
2
0 0
1
2 0
1 1
1 1 2
0 0
( ) ( ) ( )
( ( ) )
( ) ( )
( ( ) ) ( ) ( ) ,
d r r
d d i i d i i d
i i
k r
d i j i j
j i
k
j d j
j
k r
d i i j d j
j i
A B W B A A B A A BW
B A A BW W
B A B A B W
B A A B A B W
π π
π π
π π
π
− −
+ +
= =
− −
+ + = =
−
+ =
− −
+ + +
= =
+ = + − +
+ +
− +
∑
∑
∑ ∑
∑
∑ ∑
where W= AAd
(
A+B)
and ind A( )
=r, ind B( )
=t ,{
}
max ind A ind B( ), ( ) ≤ ≤k ind A( )+ind B( ) , ind W( )≤
( ) ( )
ind A +ind B . Proof. If we represent
A as 1 1
2
0
= ,
0 A
A P P
A
−
wherePandA1 are nonsingular andA2 is nilpotent, then
1
1 1 0 0 0
d A
A P P
−
−
=
and ind A( )=ind A( 1). Let us write
1 2 1
3 4
,
B B
B P P
B B
−
=
where B1∈Cr r× being
r
the size of A1. FromABAπ =0, we have1 2
1 1 1 2 1
3 4
2 2 4
0 0 0 0
0 = .
0 0 0
B B
A A B
ABA P P P P
B B
A I A B
π − −
= =
1 1 1 1 1
3 4 3 2 4
0 0
, .
B A B
B P P A B P P
B B B A B
− + −
= + = +
(2)
Observe that
1
1 1
1 1 1 1 1 1
3 2
2
0
0 0 0
( ) ,
0 0 0 0 0
d A A A B A B
W AA A B P P P P
B A
A
−
− −
+ +
= + = =
which leads to
1 1 1
( ) 0
.
0 0
d
d A B
W =P + P−
Since A2is nilpotent and A B2 4 =0, Lemma 2.1 yields
1 1
2 4 4 2
0
( ) ( ) .
r
d d i i
i
A B B A
− + =
+ =
∑
(3) Thus, by lemma 2.2, we obtain1 1 1
2 4
( ) 0
( ) ,
( )
d d
d
A B
A B P P
S A B
−
+
+ =
+
and
1 1
2 2
2 4 3 1 1 1 1 2 4 2 4 3 1 1 2 4 3 1 1
0 0
(( ) ) ( ) ( ) ( ) ( ) (( ) ) ( ) ( ) ,
p q
d j j j d j d d
j j
S A B B A B A B π A B π A B B A B A B B A B
− −
+ +
= =
=
∑
+ + + +∑
+ + + − + + (4)wherep=ind A( 1+B1)and q=ind A( 2+B2).Observe that for any X∈Cm m× ,one has ind(X)≤m and if k≥ ind X( ), then =0
k
X Xπ (both affirmations can be proved by means of the Jordan canonical form of X). Thus, in (4) the upper limits of the summations can be replaced by simply n.
We have (we will write with an asterisk any entry whose exactly expression is not necessary)
1
1 1
1 1 1
1 1
4 2
4 2
0 0
0 0
( ) 0 0
( ) .
0 0 ( )
* ( ) 0
d i i
d i i
d i i
d i i
n r
B A
B A A P P P P
I B A
B A
π +
+ − −
+
+ −
= =
Hence, we get
1 1
1 1 1 1 1
1
4 2 2 4
0 0
0
0 0
0 0
( ) ( ) .
0 ( ) 0 ( )
r r
d i i d i i r
d i i d
i i
i
B A A B A A P P P P
B A A B
π π
− −
+ + − − −
+
= =
=
= = =
+
∑
∑
∑
Therefore,
1
1
1 1 1 1 1
3 4 2 4 3 1 1
2 4 0
0 0
0 0 0 ( ) 0
( ) .
( ) ( ) 0
0 ( ) 0 0
r d
d i i d
d d
d i
B A B
B A A BW P P P P
B B A B B A B
A B
π −
+ − −
=
+
= =
+ +
+
∑
In a similar way, we get for anyj∈N 1
1 2 1
2
2 4 3 1 1 1 1
0
0 0
( ( ) ) .
( ) ( ) ( ) 0
r
d i i j j
j
d j
i
B A A BW W P P
A B B A B A B
π π
π
−
+ + −
+ =
=
+ + +
∑
2 2 4 2 4( 4) 0.
d d
A B =A B B =
2 4 2 4 2 4
2 3 2 1
2 4 4 4 2 4 2 4 2
2 3 2 1
4 4 4 4 2 4 4 2 4 4 2
2 2 1 1
4 4 2 4 2 4 2
( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ,
d n r
d d d d r r
n r
d d d d r r
n r
d d d r r
A B I A B A B
I A B B B A B A B A
I B B B B A B B A B B A
B B A B A B A
π
π
−
− −
− −
− −
+ = − + +
= − + + + + +
= − + + + +
= − + + +
⋯
⋯
⋯
and so,
1
1 1 2 2 1 1
0
1
1 1
4 2 4 2
1
4 2 4
( ) ( ) ( )
0 0 0 0
0 0 ( )
0 0
.
0 ( )
r
d i i d d d r r
i
d d r r
B A A B AA B A A B A A
P P
B A B A
P P
B A B
π π π π
π π
−
+ + − −
=
− − −
−
= + + +
= + +
=
− +
∑
⋯⋯
In addition,
1 1 1
3 4 4
1 4
0 0 0
0 *
0 , *
d
d r
d n r
r
B I BB
I B B
P P
I B B B
I
P P
B
π
π
− −
− = −
= −
=
which implies
1
4 0 0
. 0
B A P P
B
π π
π −
=
Thus,
1
1 1 1 1
2 4 4 0
1
1 1
0 1
1 1
0
0 0 0 0
( )
0 ( ) 0
( )
( ( ) ) .
r
d i i
i r
d i i
i r
d i i
i
P P P P B A A
A B B
B A B A A
B B A A
π
π π
π π π
π π
−
− − + +
= −
+ + =
−
+ + =
= −
+
= − = −
∑
∑
∑
Hence for j∈N,
1 1
2 4 2 4 2 4 2 4
1
1 1
0
0 0 0 0 0 0
0 ( ) ( ) 0 ( ) 0 ( )
( ( ) ) ( ) .
j j
r
d i i j
i
P P P P
A B A B A B A B
B B A A A B A
π π
π π π
− −
−
+ + =
=
+ + + +
= − +
∑
But, as it is easy from (2), one has
( )j ( )j .
Aπ A B+ Aπ = A B+ Aπ
2 1
1
1 1 2 1 1 1
3 4
2 4 2 4
0
1 2
2 4 2 4 3 1 1
0 0 0 ( ) 0
( ( ) )( ) ( )
0 ( ) ( ) 0 0
0 0
.
( ) ( ) ( ) 0
j
r d
d i i j d j
j i
j
j d
B A B
B B A A B A B W P P
B B
A B A B
P P
A B A B B A B
π π
π
π
+ −
+ + + −
=
− +
+
− + =
+ +
=
+ + +
∑
Finally, let us observe that the expression
2 1
1 0
( )
j r
d i i
i
B A Aπ
+ −
+ =
∑
,
in effect, since
1
2 2
2 4 4 2
0
( ) ( )
r j
d d i j i
i
A B B A
−
+ + +
=
+ =
∑
,we have that
2
1 1
1 2
0 0
( ) ( )
j
r r
d i i d i j i
i i
B A Aπ B A Aπ
+
− −
+ + +
= =
=
∑
∑
The proof is finished.
The next result, we obtain the same expression for the Drazin inverse (A+B)d as in [15, Theorem2.6] for the Generalized
Drazin inverse in a Banach Algebra. Theorem 2.2. Let A B, ∈Cn n× and ind A
( )
=r. If BAπ =0, we have 12
0
( ) ( ) ,
r
d d i d i
i
A B W Aπ A B W
−
+ =
+ = +
∑
where W=AAd
(
A B+)
.Proof. We can represent A, d
A andB as in Theorem 2.1. From BAπ =0, we have
1 2 1 2 1
3 4 4
0 0 0
0 = .
0 0
B B B
BA P P P P
B B I B
π − −
= =
Therefore B2=B4 =0. Hence,
1 1
3
0 , 0 B
B P P
B
−
=
1 1 1
3 2
0 . A B
A B P P
B A
−
+
+ =
Observe that
1
1 1
1 1 1 1 1 1
3 2
2
0
0 0 0
( ) ,
0 0 0 0 0
d A A A B A B
W AA A B P P P P
B A
A
−
− −
+ +
= + = =
which leads to
1 1 1
( ) 0
.
0 0
d
d A B
W =P + P−
Thus, by Lemma 2.2, we obtain
1 1 1
( ) 0
( ) ,
0 d
d A B
A B P P
S
−
+
+ =
where
1
2 2 3 1 1 0
( ) ,
p
i
i d
i
S A B A B
− +
=
=
∑
+ and p=ind A
( )
2 .Also we have
2 1
1
2 1 1 1
2
3 2 3 1 1
2
0 0
0 0 0 0 (( ) ) 0
( ) .
0 0 0 0 0 ( ) 0
i d i
i d i
i
i d
i n r
B
A A B
A A B W P P P
I A B A B A B
π + + −
+ −
+
= =
+
Hence the Theorem follows.
3. Some Results on the Drazin Inverse of
2×2 Block Matrices
In this section we shall apply Theorem 2.1 and Theorem 2.2 to obtain some formulas for Mdunder some conditions when M is a 2×2 block matrix written as in (1). We assume that
( )
,ind A =r ind D
( )
=s , ind BC( )
=t ind CB,( )
=l, and( )
( )
( )
( )
{
, ,}
max ind A ind D ind BC ,ind CB ,
( )
( )
( )
( )
.k ind A ind D ind BC ind CB
≤ ≤ + + +
Next we will state some auxiliary lemmas.
Lemma 3.1 Let A∈Cm n× ,B∈Cn m× .
(
( )d) (
i ( )d)
iA BA = AB A for every integer i≥1 , and
( ) ( )
B ABπ = BAπB . Moreover, ind BA
( )
− ≤1 ind AB( )
( )
1ind BA
≤ + .
Proof. As in the proof of [5, Theorem7.8.4], we can obtain
2
(AB)d =A BA(( ) )d B. The results follow.
Lemma 3.2 Let X∈Cn n× . Then (XXπ)d =0, 2
(X Xd d) =Xd, (X X2 d)π =Xπ,and ind XX( π)= ind X( ),
2 ( d) 1
ind X X = .
Proof. The Jordan canonical form of X permits write
1
( )
X =S C⊕N S− , where S and C are nonsingular, and
Nis nilpotent. Evidently, Xd =S C( −1⊕0)S−1. Now, it is evident X X2 d =S C( ⊕0)S−1 and XXπ =S(0⊕N S) −1 , which leads to the affirmations of this lemma.
Lemma 3.3 [10]If M is matrix of a form (1) such that 0
BC= andBD=0, then
2
0 1
( ) ,
d d
d
d
A A B
M
D B
=
Σ + Σ
where
1 1
2 2 1 1
0 0 0
( ) ( ) ( ) ( ) , 0.
r s n
d i n i i d i n d i d n i
n
i i i
D CA Aπ Dπ D C A D C A n
− −
+ + + + + − +
= = =
Σ =
∑
+∑
−∑
≥ (5)Lemma 3.4[17]Let M be a matrix of a the form (1) withA=0andD=0, then
0 ( )
.
( ) 0
d d
d
B CB M
CB C
=
Furthermore, If ind BC( )=l, then ind M( )≤ +2l 1.
Lemma 3.5[18]If M is matrix of a form (1) such that AB=0andD=0, then ( )
, 0
d
d XA BC B
M
CX
=
where
1 2
1 1
2 2 2
0 0
( ) ( ) ( ) ( ) .
r
t i
i d i d i
i i
X BC π BC A BC A Aπ
−
− +
+
= =
=
∑
+∑
Theorem 3.1 Let M be given by (1). If AB=0 and DCAπ =0,then
2 0
2 1
1 2 0
2 ( )
( ) ( ) ( ) ( )
( ) ,
( ) ( ) 0
d d d
d
d d d
j d j
k
j d j j
A XA RD CA RD BC BD
M
CB Z CX SD CA C BC A XA CB D SD
I BCX RD A B BZ B D
C D
CXA CB SD D C A
π π π
π π π π π
π π
+ −
+ + =
+ − +
=
+ − + + +
− −
+
− −
∑
where
1 2
1
2 1 1
0 0
( ) , ( ) ( ) ,
r
n i
d i d i d n i
n
n i
X BC A A Zπ D C A
−
+
+ − +
= =
=
∑
= −∑
(6)1 1
2 2
1 1
2 2 1
0 0
( ) , ( ) .
s s
i i
d i d i
i i
R BC BD S CB D
− −
+ +
+
= =
=
∑
=∑
(7)Proof. We can split matrix MasM = +P Q, where
2 0
, ,
0 0
d
A A AA B
P Q
D C
π
= =
0 0
, .
0 0
d d
d
A A
P P
D D
π π
π
= =
From AB=0and DCAπ =0,we have PQPπ =0.Applying Theorem 2.1, we get
1 1 1 1 1
1 1 2 2
0 0 0 0 0
1 1
1 1 2
0 0
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
p p h p h
d d d i i d i i d d i j i j j d j
i i j i j
p h
d i i j d j
j i
M W Q P P Q P P QW Q P P QW W Q P Q P Q W
Q P P Q P Q W
π π π π π π
π
− − − − −
+ + + + +
= = = = =
− −
+ + +
= =
= + − + + +
− +
∑
∑
∑ ∑
∑
∑∑
(8)
where
( )
ind P = p,ind Q( )=q, ind Q( )=q,
(max
{
ind P ind Q( )
,( )
}
≤ ≤h ind P( )
+ind Q( )
,ind W( )
≤ind P( )
+ind Q( )
. Now we consider the matrices mentioned in the above equation. Clearly, for every integer i≥1,2 1 2 1
2 2 1
2 2 1
0 0
, .
0 0
i d i d
i i
i i
A A A A
P P
D D
+ +
+
+
= =
Since AB=0, we have AA Bπ =0and matrix Q satisfies the conditions of Lemma 3.5, so we get 0
( )
, ,
0 ( )
d
d XA BC B I BCX
Q Q
CX CXA CB
π
π
−
= =
−
where X is defined in (6).
( )d
XB= BC B,BC BC( )dX =X and X BC( )d =
(
(BC)d)
2. Then, by Lemma3.1, for every integer i≥1,1
2 ( ) 0
( ) ,
( ) ( )
i d
d i
i i
d d
BC X
Q
C BC XA CB
−
=
1
2 1 ( ) ( )
( ) .
( ) 0
i i
d d
d i
i d
BC XA BC B
Q
C BC X
+ +
=
FromAB =0, we have
2 2
2 2
0
( )= ,
d d d
d
d d d d
A A AA B A A
W PP P Q
DD C D D DD C D D
= + =
by Lemma 2.2 and Lemma 3.2, we obtain
0 1
0 ( ) 0 0
, ( ) ,
( ) 0
d d n
d d n
d d n
n
A A A
W W W
Z D Z D D
π π
π
−
= = =
where Znis defined in (6).
Since XAAπ =XA, we have, for i=0,
1
1 2
0
( )
( ) ( ) ,
( )
p d
d i i d d
d i
XA BC BD
Q P P Q P Q PP
CX CB DD
π
π π π
π
− + =
= + =
∑
since XAAd =0, we have, for every integer i≥1,
1 1
1 1 2 2 2 2
2 2 1
2 2 1 2
2 1 0
0 0
( ) ( )
0 ( )
( ) ,
( ) 0
0 0
s
p p i i
d i d d i
i
d i
d i i
i
d i d i
i i
BC XA A BC BD
BC BD
Q P
C BC XA A
+ −
− − +
+
+
+ =
= =
= =
∑
∑
∑
and
1 1 2 2
2 2
1 2 2 2 1
2
1
1 2 3 1 2 1 2 1
0 0
0
0 0
( ) 0
( ) = ,
0 ( )
( ) ( )
p p i
d i d
s
d i i
i
i i d i
d i d d i
n i
i
BC XA A
Q P
CB D
C BC XA A CB D
− − +
−
+ +
+
+ + + + +
= =
=
=
∑
∑
∑
And, similarly
1
1 2
2 1
2 1 2 1 2 1
0 0
0 ( )
( )
0 0
s p
i
d i
d i i
i i
BC BD
Q P
−
−
+
+ + +
= =
=
∑
∑
(9)1 2
1 2 2 2 2
2
1 2 2 0
0
0 0
( )
0 ( )
p
s
d i i
i
d i
i
i
Q P
CB BD
−
− + +
+ +
=
=
=
∑
∑
(10)Hence the first sum in (8) is,
1 2
1 2
1 1
1 2 2
0
1 2 1 2 2 2 2 1
1
0 0 0
2
1 2 1 0
( )
( ) ( ) ( ) ,
( ) s
i
p p d i
p
i
d i i d i i d i i
s
i i i
i
d i
i
XA BC BD D
XA RD
Q P P Q P Q P P
CX SD
CX CB D D
π
π
π π
π
π
−
+
− −
−
=
+ + + +
−
= = =
+ +
=
= + = =
∑
∑
∑
∑
∑
where RandSare defined in (7).
Next consider the second sum in (8). Note that
0 0
0
.
0 0
d d
d
d d
A BZ BD
AA B
QW
C Z D CA
π
= =
Since XB=(BC)dBand AB=0, using Lemma3.1, we get
1
0 1
0 0
0
( ) .
( ) ( )
0
p d d
d i i d
d d d d
d i
RD CA
XA RD BZ BD
Q P P QW
CB CB Z SD CA CB CB D
CX SD CA
π π
π
π π
− + =
== =
+
∑
Since DCAπ =0, we have
0 0 0 0 0 0 0
= ,
0 0
0 0 0 0
A
AA B
PP QW
DD C D DD CA
π π
π π
π π π π
= =
and
2 2
2 2
0 0 0 0 0
. 0 0 0
d d
d d d d
A A A A A A
WW
DD C D D D DD CA D D D
π π
π
π π π
= = =
Since XB=(BC)dB,AB =0 and DCAπ =0, we can prove
( )
1 1
2 2
0 0
2 2
0 ( )
( ) ( ) 0
( )
.
( ) ( ) ( ) ( ) ( ) 0
p h
d i j i j d
d d
j i
d
d d d d d
X AA A BD
Q P P QW W Q P QW
C BC XA CB D CA
XAA XBD XA BC BD
C BC XA A CB D CA C BC XAA BD C BC XA CB CA
π π π
π π π π
π π
π π π
π π π π π π
− −
+ + = =
= =
= =
+ +
∑ ∑
Observe that (9) and (10) yield
1 1
1 2 2
1 1 2 1 2 1 2 2 2 2
0 0 0
0
( ) ( ) ( ) ,
0
p p
p
d i i d i i d i i
i i i
RD
Q P Q P Q P
SD
− −
−
+ + + + + +
= = =
= + =
∑
∑
∑
1
1 1
2 1 1 2
0 0 0
2 1
1 2 0
( ) ( ) ( ) ( ) ( ) ,
( ) .
( ) ( ) 0
p
h h
j d j d i i j d j
j j i
j d j
k
j d j j
Q P Q P Q W Q P P Q P Q W
I BCX RD A B BZ B D
C D
CXA CB SD D C A
π π π
π π
−
− −
+ + + +
= = =
+ −
+ + =
+ − +
− −
=
− −
∑
∑∑
∑
The proof is finished
If we assume that ABDπ =0and DD Cd =0instead of AB =0, we will get another expression for Md. Theorem 3.2 Let M be given by (1). If ABDπ =0, DCAπ =0and DD Cd =0,then
(
)
(
)
2
1 1
1
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
d d
d
d d
d d
d d
A XA BC L BY CA DS
M
CX YDCA C BC A CB S YD S
Z XA BD AN B CB D BC N BY I CZ DR
D CX BD AN CB R YD I CZ DR
π
π π
π π
π
+ + − +
=
− + + −
− + + + + − −
− + + + − −
where
1 1
2 2
1 2 1 2
0 0
( ) , ( ) ,
r s
n n
d n d n
n n
X BC A A Yπ CB D Dπ
− −
+ +
= =
=
∑
=∑
1 1
0
( ) ( ) ,
n
d i d n i
n i
Z A B D
−
+ −
=
= −
∑
1 1
2 2 1 2 2 2 2 2 3
1 0 0
( ) ( ) ( ) ( ) ,
k n n
i n i d n i n i d n
n i i
L B CB D C A CB D C A
− −
− − + − +
= = =
= +
∑ ∑
∑
(
)
1 1
2 2 1 2 2 2 2 2 2
2 2 2 3
1 0 0 0
( ) ( ) ( ) ( ) ,
k n n n
i n i i n i i n i d d n
n n
n i i i
N B CB D CZ CB D CZ BC A Aπ B ABD D
− −
− − − − +
+ +
= = = =
= + +
∑ ∑
∑
∑
(11)(
)
1
2 2 2 2
1 0
( ) ( ) ,
k n
i n i d d n
n i
S CB D C DCA A
−
− +
= =
= +
∑ ∑
(
)
1 1
2 2 2 2 1 2 2 2 2 2 3
2 2 2 3
1 0 0 0
( ) ( ) ( ) ( ) ( )
k n n n
i n i i n i d n i n i d n
n n
n i i i
R CB D CZ DCZ CB CA A B Dπ CB CA A B Dπ
− −
− − − + − +
+ +
= = = =
= + + +
∑ ∑
∑
∑
Proof. We can split matrix M as M = +P Q, where 0
, 0 A P
D
=
0 , 0 B Q
C
=
0 , 0
d d
d
A P
D
=
0 . 0
A P
D
π π
π
=
(12)
From lemma 3.4, we have
0 ( )
,
( ) 0
d d
d
B CB Q
CB C
=
( ) 0
.
0 ( )
BC Q
CB
π π
π
=
(13)
From ABDπ =0and After the course, there were significant differences in the scores of heart sounds, lung sounds, mixed sound and total score between the two groups (P < 0.05), but there was no significant difference in the scores of bowl sound (P > 0.05).
1 1
1 1
0 0
1 1
1 1 1
2 2 1 1 2
0 0 0 0 0
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ,
p p
d d d i i d i i d
i i
p p
h h h
d i j i j j d j d i i j d j
j i j j i
M W Q P P Q P P QW
Q P P QW W Q P Q P Q W Q P P Q P Q W
π π
π π π π π
− −
+ +
= =
− −
− − −
+ + + + + +
= = = = =
= + −
+ + + − +
∑
∑
∑ ∑
∑
∑∑
where
( )
ind P =p,ind Q( )=q, ind Q( )=q, max
{
ind P ind Q( )
,( )
}
≤ ≤h ind P( )
+ind Q( )
and ind W( )
≤ind P( )
+ind Q( )
., From DD Cd =0, we obtain2
2
( ) .
d d
d
d d
A A AA B
W PP P Q
DD C D D
= + =
(14)
By Lemma 2.2 and Lemma 3.2, we have
1 , 0
d d
d
A Z
W
D
=
( )
( ) ,
0 ( )
d n n d n
d n
A Z
W
D
=
0 , 0
A W
D
π π
π
=
(15)
where Zn is defined in (11). Clearly, for every integer n≥1,
2 (( ) ) 0
( ) ,
0 (( ) )
d n
d n
d n
BC Q
CB
=
1 2 1
1
0 (( ) )
( ) .
(( ) ) 0
d n
d n
d n
B CB Q
CB C
+ +
+
=
Hence
1 1
1 2 2
1 2 1 2 2 2 2 1
0 0 0
1 1
2 2
1 2 1 1 2
0 0
1
2 2
1 2 1 2 1
0 0
( ) ( ) ( )
(( ) ) (( ) )
(( ) ) (( ) )
p p
p
d i i d i i d i i
i i i
r s
d i i d i i
n n
r s
d i i d i i
n n
Q P P Q P Q P P
BC A A B CB D D
C BC A A CB D D
π π
π π
π π
− −
−
+ + + +
= = =
− −
+ + +
= =
−
+ + +
= =
= +
=
∑
∑
∑
∑
∑
∑
1
,
XA BY
CX YD
−
=
∑
(16)
whereXand Yare defined in (11).
1 0
. d d
d
BD QW
CA CZ
=
(17)
By (16) and (17), we get
1
1 1
0 1 1
0
( ) .
p d d d
d i i d
d d d
i
BD BYCA XABD BYCZ
XA BY
Q P P QW
CX YD CA CZ YDCA CXBD YDCZ
π
− + =
+
= =
+
∑
2 2
2 2
0 0 0
. 0 0
0 0 0
d d d d
d d
A A AA B A A A A AA BD
WW
D D D D D D
π π π
π
π π
= = =
(18)
From ABDπand DCAπ, we get
0 0 0
. 0 0 0
AA BD PP QW
DD CA
π π
π π
π π
= =
Since W Wj π =0,PP QWπ π =0, ABDπ =0andDCAπ =0, we have
( )
( )
( )
1 1
2 2
0 0
( )
0 0 0 0
0
0 0
0
0 ( )
( ) 0
0 ( )
.
( ) 0
p h
d i j i j d
j i
d
d
d
d
d
d
Q P P QW W Q P QW
BC A B A
C
D D
CB
BC A BD
CB D CA
BC BD
CB CA
π π π π
π π
π π
π π
π π
π
π
− −
+ + = =
=
=
=
=
∑ ∑
The conditions ABDπ =0,DCAπ =0and DD Cd =0implies thatABD Cn =0andDCBD Cn =0,forn≥ 0, observe that (16) and (18) yield
1 1
2 2 1 2 2 2 2 1 2 2 2 2
2 2
0 0 0
2 2 2
1 1
2 2 2 2 2 2 2 2 1 2 2
2 2
0 0 0
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
n n n
i n i d n i n i i n i d n
n
i i i
i d i
n n n
i n i d n i n i i n i d n
n
i i i
B CB D C A B CB D CZ BC A A B D
P Q P Q W
CB D C A CB D CZ CB CA A B D
π π
π
− −
− − + − − − +
+
= = =
+
− −
− + − − − +
+
= = =
+
+ =
+
∑
∑
∑
∑
∑
∑
1
2 2 2 3 2 2 2 1 2 2 3
2 3
0 0 0
2 1 2 3
2 1 2 2 3 2 1 2 2 2 2 3
2 3
0 0 0
( ) ( ) ( ) ( ) ( )
( ) ( ) .
( ) ( ) ( ) ( ) ( )
n n n
i n i d n i n i i n i d n
n
i i i
n d n
n n n
i n i d n i n i i n i d n
n
i i i
B CB D C A B CB D CZ BC A A B D
P Q P Q W
CB D C A CB D CZ CB CA A B D
π π
π −
− + − + − +
+
= = =
+ +
+ − + + − − +
+
= = =
+
+ =
+
∑
∑
∑
∑
∑
∑
It follows that
1 1
2 2
1
2 2 2 2 2 1 2 3
0 0 0
( ) ( ) ( ) ( ) ( ) ( ) ,
h h
h
n d n n d n n d n
n n n
L N
P Q P Q W P Q P Q W P Q P Q W
S R
π π π
− −
−
+ + + +
= = =
+ = + + + =
∑
∑
∑
where L N S, , and R are defined in (11). Using the same way as (16), we have
1 1
1 2 2 2
1 1 2 1 2 1 2 2 2 2
2
0 0 0
( ) ( ) ( ) .
p p
p
d n n d n n d n n
n n n
XA BYD
Q P Q P Q P
CXA YD
− −
−
+ + + + + +
= = =
= + =
∑
∑
∑
1 2
1 1
2 1 1 2
2 2
0 0 0
( ) ( )
( ) ( ) ( ) ( ) ( ) .
( ) ( )
p
h h
j d j d i i j d j
j j i
BC L BYDS BC N XA N BYDR
Q P Q P Q W Q P P Q P Q W
CB S YD S CB R CXAN YD R
π π
π π π
π π
−
− −
+ + + +
= = =
− − −
+ − + =
− − −
∑
∑∑
The proof is finished.
The next result is a generalization of [9, Theorem 5].
Theorem 3.3 Let Mbe given by (1), if DD Cd =0 and BDπ =0. Then
2 1
2 0
1 1 1
2 1 2
0 1 1 0
( )
( ) ( ) n
r
d d d n d n
n d
s s n s
n d n d i n i d n n
n n i n
A A BD A A B D
M
D C A D D CA A B D D CX
π
π
+
−
+ =
− − −
+ − − +
= = = =
− +
=
+ +
∑
∑
∑∑
∑
where
1 1
0
( ) ( ) , 1.
n
d i d n i
n i
X A B D n
−
+ −
=
= −
∑
≥ (19) Proof. We can split matrixM as M = +P Q, where2
2 0
, 0
d
d
A A P
D D
=
= AA B ,
Q
C DD
π
π
0 , 0
d d
d
A P
D
=
0 . 0
A P
D
π π
π
=
From DD Cd =0 and BDπ =0, we have PQPπ =0. Applying Theorem 2.1, we get
1 1 1 1 1
1 1 2 2
0 0 0 0 0
1 1
1 1 2
0 0
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ,
p p h p h
d d d i i d i i d d i j i j j d j
i i j i j
p h
d i i j d j
j i
M W Q P P Q P P QW Q P P QW W Q P Q P Q W
Q P P Q P Q W
π π π π π π
π
− − − − −
+ + + + +
= = = = =
− −
+ + +
= =
= + − + + +
− +
∑
∑
∑ ∑
∑
∑∑
where ind P( )= p,ind Q( )=q, ind Q( )=q, max
{
ind P ind Q( )
,( )
}
≤ ≤h ind P( )
+ind Q( )
,ind W( )
≤ind P( )
+ind Q( )
. Since DD Cd =0 and BDπ =0, implies, for n≥0,BD D Cn π =BD Cn −BD DD Cn d =BD Cn =0.FromBDπ =0 and BC=0,the matrix Q satisfies Lemma 3.3, by Lemma3.2, we get
(
)
20 1
0 1
0 0
( ) ( )
,
( )
d d
d
d
AA AA B
Q
B
DD B
π π
π
= =∑ ∑
∑ +∑
where ∑n is defined in (5), since DD Cd =0, by Lemma3.2, we have ∑n =0. ThusQd =0 and Qπ =I. In this situation, we obtain
1
2 0
( ) ( ) .
h
d d j d j
j
M W P Q P Q Wπ
−
+ =
= +
∑
+Since DD Cd =0,we have
2 2
( ) .
0
d d
d
d
A A AA B
W PP P Q
D D
= + =
1 , 0
d d
d
A X
W
D
=
( )
( ) .
0 ( )
d n n d n
d n
A X
W
D
=
where Xn is defined in (19). Clearly, for every integern≥0,
2 2 2
2
0 ( )
( ) .
( ) j
d j d j
d j
A B D P Q W
C A CX
π π
+
+ +
+
=
Since DD Cd =0 and BD Ci =0, we can prove, for every integern≥0,
1
2 1
0 2
1 1 1
0 2 1 2
2
0 1 1 0
0 ( )
( ) ( )
( ) ( )
r
n d n
h
n
j d j
s r n s
j n d n i n i d n n
n
n n i n
A A B D
P Q P Q W
D C A D CA A B D D CX
π π
π −
+ −
= +
− − −
= + − − +
+
= = = =
+ =
+
∑
∑
∑
∑∑
∑
The proof is finished.
In the rest of the paper we will exploit Theorem 2.2 to obtain some representations of d
M under some weaker conditions. Firstly we will present the following result.
Theorem 3.4 Let M be given by (1), if BDπ =0, DD Cd =0 and CAAd =0.Then
1
2 1
0 1 1
1 2
1 0
( )
,
0 ( )
r
d n d n
n d
r n
d n i i d n
n i
A X A A B D
M
D D CA B D
π −
+ =
− −
− − +
= =
+
=
+
∑
∑∑
where
1 1
0
( ) ( ) , 1.
n
d i d n i
n i
X A B D n
−
+ −
=
= −
∑
≥ (20) Proof. We can split matrix M as M = +P Q, where0 0
, .
0 0
A B
P Q
C D
= =
From DD Cd =0and CAAd =0, by Lemma 2.2, we get
0 , 0
d d
d
A P
D
=
0 . 0
A P
D
π π
π
=
SinceDD Cd =0, we have, for every integer n≥1,
1 1 0 0
. 0
n
n n
n i i
i
A A B P P Q
D CA B
π
π −
− − =
=
∑
Since BDπ =0, we obtain QPπ =0. Applying Theorem 2.2, we get
1
2
0
( ) .
p
d d n d n
n
M W Pπ P Q W
−
+ =
From DD Cd =0and CAAd =0, we have
2 2
2 2
( ) .
0
d d d d
d
d d d
A A AA B A A AA B
W PP P Q
DD C D D D D
= + = =
Using the same way as Theorem 3.2, we obtain
1 , 0
d d
