# Some Results for the Drazin Inverses of the Sum of Two Matrices and Some Block Matrices

Authors: Abdul Shakoor; Hu Yang; Ilyas Ali
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We give a formula of(P+Q)D under the conditions P2Q+QPQ=0, P3Q=0, and PQPQ=0. Then, we apply it to give some expressions for the Drazin inverse of block matrix M=(ABCD) (A and D are square matrices) under some conditions,  generalizing some recent results in the literature. Finally, numerical examples are given to illustrate our results.


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## Body

## 1. Introduction

Letℂm×n denote the set of m×n complex matrices. The Drazin inverse of A∈ℂn×n is the unique matrix AD, satisfying the following equation:
(1)Ak+1AD=Ak,ADAAD=X,AAD=ADA,
where k=ind(A) is the index of A, the smallest nonnegative integer for which rank(Ak+1)=rank(Ak) (see [1]). In particular, when ind(A)=1, the Drazin inverse of A is called the group inverse of A. If A is nonsingular, it is clear that ind(A)=0 and AD=A-1. Throughout this paper, we denote by Aπ=I-AAD and define A0=I, where I is the identity matrix with proper sizes.The importance of the Drazin inverse and its applications to singular differential equations and difference equations, Morkov chains and iterative methods, cryptography, numerical analysis, to structured matrices, and perturbation bounds for the relative eigenvalue problems can be found in [2–5].In 1958, Drazin [6] gave a result of (P+Q)D, where P and Q are square matrices, and proved that
(2)(P+Q)D=PD+QDwhenPQ=QP=0.In 2001, Hartwig et al. [7] derived a result of (P+Q)D when PQ=0. In 2005, Castro-González [8] provided the representation of (P+Q)D when PDQ=0, PQD=0, and QπPQPπ=0. In 2008, Castro-González et al. [9] determined the result of (P+Q)D when P2Q=0 and PQ2=0. In 2009, Martínez-Serrano and Castro-González [10] derived a formula of (P+Q)D when P2Q=0 and Q2=0. In 2011, Yang and Liu [11] established some expressions of (P+Q)D when PQ2=0 and PQP=0, and in 2012, Bu et al. [12] got the explicit representation of (P+Q)D when P2Q=0, Q2P=0, P3Q=0, and QPQ=0, QP2Q=0, respectively. Other related results have been studied in [4, 13–19].On the other hand, a related topic is to discuss a representation of the Drazin inverse of block matrixM=(ABCD)∈ℂn×n, where A and D are square matrices. Campbell and Meyer Jr. [2] first proposed an open problem to find an explicit formula of the Drazin inverse of block matrix M=(ABCD)∈ℂn×n, (where A and D are square matrices), in terms of A, B, C, and D. To find the Drazin inverse of (P+Q) and M in terms of P, Q, PD, and QD and A, B, C, and D, respectively, without side conditions is quite complicated and it has not been solved till now. However, many papers have been studied some special cases of this open problem and gave the representations for the Drazin inverse of (P+Q) and M under some conditions. Here, we list some cases of Drazin inverse of block matrix M:(i)
BC=0, DC=0 (or BD=0) and D is nilpotent (see [5]);(ii)
BCA=0, BD=0, and DC=0 (or BC is nilpotent) (see [9]);(iii)
BCA=0, BCB=0, DCA=0, and DCB=0 (see [11]);(iv)
BC=0, BD=0, and DC=0 (see [20]);(v)
BC=0 and DC=0 (see [21]);(vi)
BCA=0, BCB=0, ABD=0, and CBD=0 (see [22]).The generalized Schur complement ofA in M, which is stated as S=D-CADB, is very important to find the Drazin inverse of M, when the generalized Schur complement is either zero or nonsingular. Martínez-Serrano and Castro-González [10] gave results under the conditions A2AπB=0, CAAπB=0, and BCAπB=0 and generalized that Schur complement is equal to zero. Also, they derived the expressions of MD under the assumptions A2AπB=0, CAAπB=0, and CAπB=0 and generalized Schur complement is nonsingular. The Drazin inverse of M has been studied in [10, 23], when generalized Schur complement is equal to zero and also has been studied in [5, 10, 24, 25], when the generalized Schur complement is nonsingular. Some representations for the Drazin inverse of M when the generalized Schur complement is nonsingular, including generalizations of the above mentioned results, will be derived in Section 4 under some conditions.This paper is organized as follows. In Section2, some helpful lemmas will be given. In Section 3, we give the explicit formula of (P+Q)D under the conditions P3Q=0, P2Q+QPQ=0, and PQPQ=0 and also give a numerical example to demonstrate our result. In Section 4, we use our result to find the Drazin inverse of block matrix M=(ABCD) and also to find the expression for MD when the generalized Schur complement is nonsingular, which can be regarded as the generalizations of some results given in [5, 20]. Finally, in Section 5, we give two numerical examples to illustrate our results of block matrices.


## 2. Some Lemmas

In order to prove the main results, first we need the following lemmas.Lemma 1 (see [1]).
LetA∈ℂm×n, and let B∈ℂn×m. Then, (AB)D=A((BA)2)DB.Lemma 2 (see [7]).
LetP,Q∈ℂn×n. If PQ=0, then
(3)(P+Q)D=Qπ∑i=0t-1‍Qi(PD)i+1+∑i=0t-1‍(QD)i+1PiPπ,
where t=max{ind(P),ind(Q)}.Lemma 3 (see [26]).
LetM1=(A0CB), and let M2=(BC0A), where A and B are square matrices with ind(A)=r and ind(B)=s. Then,
(4)M1D=(AD0XBD),M2D=(BDX0AD),
where X=∑i=0r-1‍(BD)i+2CAiAπ+∑i-0s-1‍BπBiC(AD)i+2-BDCAD.Lemma 4 (see [24]).
LetM=(ABCD)∈Cn×n (A and D are square matrices). If S=D-CADB is nonsingular, AπB=0, and CAπ=0, then
(5)MD=(AD+ADBS-1CAD-ADBS-1-S-1CADS-1).


## 3. The Drazin Inverse of the Sum of Two Matrices

In this section, we first give the formula for the Drazin inverse ofP+Q under some conditions.Theorem 5.
LetP,Q∈ℂn×n. If P3Q=0, P2Q+QPQ=0, and PQPQ=0, then
(6)(P+Q)D=PD+QD+QXP+P(QD)2+P2(QD)3+PQ(PD)3+QP(QD)3+P2QXPD+P2QDXP+PQ2XPD+PQQDXP+QPQXPD+QPQDXP,
where
(7)X=∑i=0t-1‍(QD)2i+4(P+Q)P2iPπ+∑i=0t-1‍QπQ2i(P+Q)(PD)2i+4-(QD)2(P+Q)(PD)2,t=max{ind(P2),ind(Q2)}.Proof.
Using Lemma1, (AB)D=A((BA)2)DB, we have
(8)(P+Q)D=((IQ)(PI))D=(IQ)((PPQIQ)2)D(PI)=(IQ)(P2+PQP2Q+PQ2P+QQ2+PQ)D(PI).
Let
(9)M=(P2+PQP2Q+PQ2P+QQ2+PQ)=E+F,
where
(10)E=(P20P+QQ2),F=(PQP2Q+PQ20PQ).
From P3Q=0, P2Q+QPQ=0 and PQPQ=0, we get EF=0, and F2=0. Then, applying Lemma 2, we obtain
(11)MD=ED+F(ED)2.
By applying Lemma 3, we have
(12)ED=((PD)20X(QD)2),
where
(13)X=∑i=0t-1‍(QD)2i+4(P+Q)P2iPπ+∑i=0t-1‍QπQ2i(P+Q)(PD)2i+4-(QD)2(P+Q)(PD)2,t=max{ind(P2),ind(Q2)}.
Substituting (12) into (11), we get
(14)MD=((PD)2+PQ(PD)4+P2QX(PD)200000+P2QDX+PQ2X(PD)2+PQQDXP2(QD)3+P(QD)2X+PQX(PD)2+PQDX(QD)2+P(QD)3).
Substituting (14) into (8), we obtain the result.Similarly, we give a symmetrical form of Theorem5.Theorem 6.
LetP,Q∈ℂn×n. If PQ3=0, PQ2+PQP=0, and PQPQ=0, then
(15)(P+Q)D=PD+QD+QXP+(PD)2Q+(PD)3Q2+(PD)3QP+(QD)3PQ+QDXPQ2+QDXP2Q+QXPDQ2+QXPDPQ+QDXPQP+QXPDQP,
where
(16)X=∑i=0t-1‍(QD)2i+4(P+Q)P2iPπ+∑i=0t-1‍QπQ2i(P+Q)(PD)2i+4-(QD)2(P+Q)(PD)2,t=max{ind(P2),ind(Q2)}.Next, we present a numerical example to illustrate Theorem5. This numerical example describes neither the matrices P and Q which do not satisfy the conditions of [10, Theorem 2.2] nor the conditions of [11, Theorem 2.1], but they satisfy the conditions of Theorem 5. Therefore, we can apply the formula given in Theorem 5 to obtain the Drazin inverse of P+Q.Numerical Example. Consider the matrices P,Q∈ℂ4×4, where
(17)P=(1000000100000000),Q=(0000001000101000).
Since P2Q=0, Q2≠0 and PQ2=0, PQP≠0, we know that the conditions of Theorem  2.2 in [10] and Theorem 2.1 in [11] do not hold, respectively. But it satisfies P3Q=0, P2Q+QPQ=0, and PQPQ=0. Also, we have
(18)ind(P2)=1,ind(Q2)=1,PD=(1000000000000000),QD=(0000001000100000),X=(1000001000101000).
So, applying Theorem 5, we get
(19)(P+Q)D=PD+QD+PQ(PD)3+QXP=(1000101000101000).Remark 7.
The above example shows that the conditions given in Theorem5 are satisfied, but the conditions given in [10, 11] are not satisfied.


## 4. Drazin Inverse of Some Block Matrices

In this section, we apply our formula to give the representations for the Drazin inverse of block matrixM=(ABCD) (A and D are square matrices). First, we give the expression of MD under the conditions ABC=0, BDC=0, D2C=0, and CBC=0, which generalizes the results in [5, 20].Theorem 8.
LetM=(ABCD) (A and D are square matrices), such that ind(A)=r and ind(D)=s. If ABC=0, BDC=0, D2C=0, and CBC=0, then
(20)MD=PD+(00C0)(PD)2+(BC0DC0)(PD)3,
where
(21)(PD)i=((AD)i∑j=0i-1‍(AD)jX(DD)i-j-10(DD)i),fori≥1,X=∑i=0s-1‍(AD)i+2BDiDπ+∑i=0r-1‍AπAiB(DD)i+2-ADBDD.Proof.
Let(22)M=(ABCD)=P+Q,
where
(23)P=(AB0D),Q=(00C0).
From ABC=0, BDC=0, D2C=0, and CBC=0, we have P3Q=0, P2Q+QPQ=0, and PQPQ=0. We can see that Q is 2-nilpotent, so we get QD=0 and Qπ=I. Applying Theorem 5, we have
(24)MD=PD+Q(PD)2+PQ(PD)3.
Using Lemma 3, we obtain the following result:
(25)(PD)i=((AD)i∑j=0i-1‍(AD)jX(DD)i-j-10(DD)i),fori≥1,
where X=∑i=0s-1‍(AD)i+2BDiDπ+∑i=0r-1‍AπAiB(DD)i+2-ADBDD.
Substituting (25) into (24), we get the result.Similarly, we consider another splitting of the block matrixM and state another theorem.Theorem 9.
LetM=(ABCD) (A and D are square matrices), such that ind(A)=r and ind(D)=s. If DCB=0, BCB=0, A2B=0, and CAB=0, then
(26)MD=PD+(0B00)(PD)2+(0AB0CB)(PD)3,
where
(27)(PD)i=((AD)i0∑j=0i-1‍(DD)jX(AD)i-j-1(DD)i),fori≥1,X=∑i=0r-1‍(DD)i+2CAiAπ+∑i=0s-1‍DπDiC(AD)i+2-DDCAD.Proof.
Let(28)M=(ABCD)=P+Q,
where
(29)P=(A0CD),Q=(0B00).
The remaining proof follows directly from Theorem 8.Now, we give the representation forMD when the generalized Schur complement is nonsingular, which generalizes the result in [5].Theorem 10.
LetM=(ABCD) (A and D are square matrices). If S=D-CADB is nonsingular, BCAπB=0, DCAπB=0, A2AπB=0, and CAAπB=0, then
(30)MD=PD+(0AπB00)(PD)2+(0AAπB0CAπB)(PD)3,
where
(31)(PD)j=(P1D)j+∑i=1k‍(P1D)i+j(AiAπ0CAi-1Aπ0),P1D=(AD+ADBS-1CAD-ADBS-1-S-1CADS-1),k=ind(A).Proof.
Let(32)M=(ABCD)=P+Q,
where
(33)P=(AAADBCD),Q=(0AπB00).
From BCAπB=0, DCAπB=0, A2AπB=0, and CAAπB=0, it is obvious that P3Q=0, P2Q+QPQ=0, and PQPQ=0. We can see that Q is 2-nilpotent, so we get QD=0 and Qπ=I. Applying Theorem 5, we have
(34)MD=PD+Q(PD)2+PQ(PD)3.
Let P=P1+P2, where
(35)P1=(A2ADAADBCAADD),P2=(AAπ0CAπ0).
Obviously, P2P1=0 and P2k+1=0, where k=ind(A). By Lemma 2, we have
(36)(PD)j=(P1D)j+∑i=1k‍(P1D)i+jP2i,P2i=(AiAπ0CAi-1Aπ0),00000000000000000000000000000000000000000i≥1.
For P1, we get that S=D-CAAD(A2AD)DAADB=D-CADB is nonsingular, (A2AD)πAADB=0, and CAAD(A2AD)π=0. Using Lemma 4, we obtain the following result:
(37)P1D=(AD+ADBS-1CAD-ADBS-1-S-1CADS-1).
From the above equation, we obtain the result in Theorem 10.In the same way, we consider another splitting of the block matrixM and present next theorem.Theorem 11.
LetM=(ABCD) (A and D are square matrices). If S=D-CADB is nonsingular, ABCAπ=0, BDCAπ=0, D2CAπ=0, and AπBCAπ=0, then
(38)MD=PD+(00CAπ0)(PD)2+(BCAπ0DCAπ0)(PD)3,
where
(39)(PD)j=(P1D)j+∑i=1k‍(AiAπAi-1AπB00)(P1D)i+j,P1D=(AD+ADBS-1CAD-ADBS-1-S-1CADS-1),k=ind(A).Proof.
Let(40)M=(ABCD)=P+Q,
where
(41)P=(ABCAADD),Q=(00CAπ0).
The remaining proof is similar to that of Theorem 10.


## 5. Numerical Examples

In this section, two numerical examples are given to illustrate Theorems8 and 10.Example 1.
Consider the block matrixM=(ABCD)∈ℂ4×4, where
(42)A=(1000),B=(0001),C=(0010),D=(1010).
Since BC≠0 and BCA≠0, the representation for MD fail to apply in [9, 11, 20–22], respectively. But it satisfies ABC=0, BDC=0, D2C=0, and CBC=0. Also, we have
(43)AD=(1000),Aπ=(0001),DD=(1010),Dπ=(00-11).
Then, applying Theorem 8, we obtain
(44)MD=(1000101000101010).Remark 12.
The above example shows that the conditions given in Theorem8 are satisfied, but the conditions given in [9, 11, 20–22] are not satisfied.Example 2.
Consider the block matrixM=(ABCD)∈ℂ5×5, where
(45)A=(1100),B=(121000),C=(121001),D=(221131001).
By computing, we get that S=D-CADB is nonsingular and BCAπB=0, DCAπB=0, A2AπB=0, and CAAπB=0. Also, we have
(46)ind(A)=1,AD=(1100),Aπ=(0-101),S-1=(100010001),P1D=(44-1-2-100000-1-1100-1-101000001),PD=(44-1-2-100000-10100-1-201001001).
Then, applying Theorem 10, we get
(47)MD=(44-1-2-100000-10100-1-201001001).



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