A note on the formulas for the Drazin inverse of the sum of two matrices

Xin Liu; Xiaoying Yang; Yaqiang Wang

doi:10.1515/math-2019-0015

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A note on the formulas for the Drazin inverse of the sum of two matrices

Xin Liu , Xiaoying Yang and Yaqiang Wang

Published/Copyright: March 26, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

In this paper we derive the formula of (P + Q)^D under the conditions Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0. Then, a corollary is given which satisfies the conditions (P + Q)P(P + Q) = 0 and QPQ² = 0. Meanwhile, we show that the additive formula provided by Bu et al. (J. Appl. Math. Comput. 38 (2012) 631-640) is not valid for all matrices which satisfies the conditions (P + Q)P(P + Q) = 0 and QPQ² = 0. Also, the representation can be simplified from Višnjić (Filomat 30 (2016) 125-130) which satisfies given conditions. Furthermore, we apply our result to establish a new representation for the Drazin inverse of a complex block matrix having generalized Schur complement equal to zero under some conditions. Finally, a numerical example is given to illustrate our result.

Keywords: Drazin inverse; Block matrix; Matrix index; Generalized Schur complement

MSC 2010: 15A09

1 Introduction

Let C^n×n denote the set of all n × n complex matrices. For A ∈ C^n×n, we call the smallest nonnegative integer k which satisfies rank(A^k+1) = rank(A^k) the index of A, and denote k by ind(A). Let A ∈ C^n×n with ind(A) = k, we call the matrix X ∈ C^n×n which satisfies

Ak+1X=Ak,XAX=X,XA=AX

the Drazin inverse of A and denote X by A^D (see [1]). The Drazin inverse of a square complex matrix always exists and is unique (see [1]). In this paper, we denote A^π = I − AA^D. A matrix A ∈ C^n×n is nilpotent if A^k = 0 for some integer k ≥ 0. The smallest such k is called the index of nilpotency of A.

The Drazin inverse of square complex matrices has applications in several areas, such as singular differential or difference equations, Markov chains and iterative method and so on (see [1, 2]). For applications of the Drazin inverse of a 2 × 2 block matrix, we refer the readers to [2, 3, 4].

Suppose P, Q ∈ C^n×n. In 1958, Drazin offered the formula (P + Q)^D = P^D + Q^D, when PQ = QP = 0. In the recent years many authors have considered this problem and provided the representations of (P + Q)^D with some specific conditions. In [5], the authors gave the representation of (P + Q)^D when PQ = 0. In [6], the authors derived the formula for (P + Q)^D under the conditions P²Q = 0, Q² = 0. The case when P²Q = 0, Q²P = 0 was studied in [7]. In [8], the authors gave the formula for (P + Q)^D when P²QP = 0, PQ²P = 0, P²Q² = 0, PQ³ = 0. In [9], the formula of (P + Q)^D under the conditions (P + Q)P(P + Q) = 0, QPQ² = 0 was given. In [10], the authors derived a result under the conditions P(P + Q)Q = 0.

In this short paper, we derive the formula of (P + Q)^D under the conditions Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0 in Section 3. We also get that a formula for (P + Q)^D from [9] is not valid for all matrices which satisfy conditions (P + Q)P(P + Q) = 0 and QPQ² = 0. Meanwhile, a corollary is given which is valid for all matrices under the mentioned conditions. Furthermore, we offer an example which shows that the formula from [9] is not valid for all matrices which satisfy conditions (P + Q)P(P + Q) = 0 and QPQ² = 0.

Another aim of this paper is to derive a representation of the Drazin inverse of 2 × 2 complex block matrix

M=ABCD (1.1)

where A and D are square matrices. This problem was firstly posed in 1979 by Compbell and Meyer [11]. No formula for M^D has yet been offered without any restrictions upon the blocks. Special cases of this problem have been studied. In some papers the expression of M^D is given under conditions which concern the generalized Schur complement of matrix M defined by S = D − CA^DB. Here we list some of them:

CA^π = 0, A^πB = 0, and S = 0 [12],
CA^πB = 0, AA^πB = 0, and S = 0 [3],
CA^πB = 0, CAA^π = 0, and S = 0 [3],
ABCA^π = 0, BCA^π is nilpotent, and S = 0 [6],
A^πBCA = 0, A^πBC is nilpotent, and S = 0 [6],
ABCA^π = 0, A^πABC = 0, and S = 0 [7],
ABCA^π = 0, CBCA^π = 0, and S = 0 [7],
ABCA^πA = 0, ABCA^πB = 0, and S = 0 [8],
AA^πBCA = 0, CA^πBCA = 0, and S = 0 [8].

In Section 4, we derive a new representation for M^D under the conditions ABCA^πA = 0, A^πBCA = 0, A^πBCB = 0 and S = 0.

2 Some lemmas

In order to give the main results, we first give some lemmas as follows.

Lemma 2.1

[11] Let A ∈ C^m×n, B ∈ C^n×m, then

(AB)D=A(BA)D2B.

Lemma 2.2

[4] Let M=A0CB, where A and B are square matrices with ind(A) = r and ind(B) = s. Then max{r, s} ≤ ind(M) ≤ r + s, and

MD=AD0XBD,

where X=∑k=0s−1BπBkC(AD)k+2+∑k=0r−1(BD)k+2CAkAπ−BDCAD.

Lemma 2.3

[5] Let P, Q ∈ C^n×n be such that ind(P) = r and ind(Q) = s. If PQ = 0, then

(P+Q)D=∑i=0s−1QπQi(PD)i+1+∑i=0r−1(QD)i+1PiPπ.

Lemma 2.4

[13] Let P, Q ∈ C^n×n, such that ind(P) = r, ind(Q) = s. If PQP = 0 and PQ² = 0, then

(P+Q)D=Y1+Y2+(Y1(PD)2+(QD)2Y2−QD(PD)2−(QD)2PD)PQ,

where Y1=∑i=0s−1QπQi(PD)i+1,Y2=∑i=0r−1(QD)i+1PiPπ.

Lemma 2.5

[6] Let P, Q ∈ C^n×n.

If PQ = QP = 0, then (P + Q)^D = P^D + Q^D.
If PQ = 0 and P is r-nilpotent, then ((P + Q)^D)^j = ∑i=0r−1 (Q^D)^i+jPⁱ, ∀j ≥ 1.
If PQ = 0 and Q is s-nilpotent, then ((P + Q)^D)^j = ∑i=0s−1 Qⁱ(P^D)^i+j, ∀j ≥ 1.

Lemma 2.6

[12] Let M be a matrix of the form (1.1), such that S = 0. If A^πB = 0, CA^π = 0, then

MD=ICAD((AW)D)2AIADB,

where W = AA^D + A^DBCA^D.

3 Additive results

In [9], Bu et al. gave the representation of (P + Q)^D when (P + Q)P(P + Q) = 0, QPQ² = 0 which is not correct for all matrices. In this paper we give the representation of (P + Q)^D when Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0 are satisfied, from which we will get the correct formula under the conditions (P + Q)P(P + Q) = 0 and QPQ² = 0.

Theorem 3.1

Let P, Q ∈ C^n×n, if Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0, then

(P+Q)D=(P+Q)3((Q2+QP)D)2=(P+Q)2((Q2+QP)D)2(P+Q),

where

((Q2+QP)D)2=∑i=0t−1QπQ2i((QP)D)i+2+∑i=0t−1(QD)2(i+2)(QP)i(QP)π−(QD)2(QP)D,

for t = max{ind(QP), ind(Q²)}.

Proof

Using definition of the Drazin inverse, we have that

(P+Q)D=(P+Q)((P+Q)2)D=(P+Q)IPQ+P2Q2+QPID. (3.1)

Denote by M=IPQ+P2Q2+QPI.

By Lemma 2.1 and Q(P + Q)P(P + Q) = 0, we have

MD=IPQ+P2Q2+QP0IPQ+P2D2⋅Q2+QPI. (3.2)

Denote by

N=Q2+QP0IPQ+P2.

From P(P + Q)P(P + Q) = 0, we get (PQ + P²)^D = 0. After applying Lemma 2.2 we get

ND=(Q2+QP)D0X0, (3.3)

where X = ((Q² + QP)^D)² + (PQ + P²) ((Q² + QP)^D)³.

Since QPQ² = 0 and Lemma 2.3, we obtain

(Q2+QP)D=∑i=0t−1QπQ2i((QP)D)i+1+∑i=0t−1(QD)2(i+1)(QP)i(QP)π,

where t = max{ind(QP), ind(Q²)}.

Substituting (3.3) into (3.2), we get

MD=(Q2+QP)D+(PQ+P2)(Q2+QP)D(Q2+QP).

By P(P + Q)P(P + Q) = 0, we have

MD=(Q2+QP)D+(PQ+P2)((Q2+QP)D)2. (3.4)

Substituting (3.4) into (3.1), we get

(P+Q)D=(P+Q)(Q2+QP)D+(P+Q)(PQ+P2)((Q2+QP)D)2=(P+Q)(Q2+QP)((Q2+QP)D)2+(P+Q)(PQ+P2)((Q2+QP)D)2=(P+Q)(P+Q)2((Q2+QP)D)2=(P+Q)3((Q2+QP)D)2.

Similarly, using definition of Drazin inverse, (P + Q)^D = ((P + Q)²)^D(P + Q).

So,

(P+Q)D=(P+Q)2((Q2+QP)D)2(P+Q).

The proof is completed.□

The next theorem is a symmetrical formulation of Theorem 3.1.

Theorem 3.2

Let P, Q ∈ C^n×n. If (P + Q)Q(P + Q)P = 0, (P + Q)P(P + Q)P = 0, Q²PQ = 0, then

(P+Q)D=(P+Q)3((Q2+PQ)D)2=(P+Q)2((Q2+PQ)D)2(P+Q),

where

((Q2+PQ)D)2=∑i=0t−1(PQ)π(PQ)i((Q)D)2(i+2)+∑i=0t−1((PQ)D)i+2Q2iQπ−(PQ)D(QD)2,

for t = max{ind(PQ), ind(Q²)}.

Notice that one special case of Theorem 3.1 is when matrices P, Q satisfy the conditions (P + Q)P(P + Q) = 0 and QPQ² = 0, we get following additive formula.

Corollary 3.3

Let P, Q ∈ C^n×n, such that max{ind(QP), ind(Q²)} = t, if (P + Q)P(P + Q) = 0, QPQ² = 0, then

(P+Q)D=(P+Q)2((Q2+QP)D)2(P+Q)=(P+Q)(Q2+QP)D,

where

(Q2+QP)D=∑i=0t−1QπQ2i((QP)D)i+1+∑i=0t−1(QD)2(i+1)(QP)i(QP)π,((Q2+QP)D)2=∑i=0t−1QπQ2i((QP)D)i+2+∑i=0t−1(QD)2(i+2)(QP)i(QP)π−(QD)2(QP)D.

Remark 3.4

In [9, Theorem 3.1], the authors studied conditions from Corollary 3.3 and obtained the formula

(P+Q)D=(P+Q)2(∑i=0t−1(QD)2i+4(QP)i(QP)π−(QD)2(QP)D)(P+Q). (3.5)

Notice that in Corollary 3.3 we have additional element

(P+Q)2(∑i=0t−1QπQ2i((QP)D)i+2)(P+Q).

In fact, this element doesn’t have to be equal to zero. So, the formula from [9] is not correct for all matrices which satisfy conditions (P + Q)P(P + Q) = 0 and QPQ² = 0.

Remark 3.5

In [10, Corollary 3.2], the author obtained the formula satisfy the same conditions from Corollary 3.3

(P+Q)D=(P+Q)2((Q2+QP)D)2(P+Q). (3.6)

Since (P + Q)P(P + Q) = 0, by Theorem 3.1, (3.6) can be checked easily that

(P+Q)D=(P+Q)3((Q2+QP)D)2=(P+Q)Q(P+Q)((Q2+QP)D)2+(P+Q)P(P+Q)((Q2+QP)D)2=(P+Q)(Q2+QP)D.

Next we consider matrices which satisfy conditions (P + Q)P(P + Q) = 0 and QPQ² = 0 and for which is (P+Q)2(∑i=0t−1QπQ2i((QP)D)i+2)(P+Q)≠0. Furthermore, we demonstrate the application of Corollary 3.3.

Example 3.6

Let P and Q be matrices: P=0−284,Q=0200.

It can be checked easily that (P + Q)P(P + Q) = 0, QPQ² = 0. Hence the conditions of Corollary 3.3 are satisfied. Also, we have that (P+Q)2(∑i=0t−1QπQ2i((QP)D)i+2)(P+Q)=001214≠0. Moreover, by the formula from Corollary 3.3, (P+Q)D=001214. But, if we apply the formula (3.5) from [9], we get that (P + Q)^D = 0.

Example 3.7

[9] Let P and Q be matrices: P=0100000000010000,Q=000011−10010111−10.

It is obvious that (P + Q)P(P + Q) = 0, QPQ² = 0. Hence the conditions of Corollary 3.3 are satisfied. After computation, we get

(QP)D=0,QD=0000−12−141214−12−3412−14−12−141214,(Q2+QP)D=0000−14−1414−141414−14−34−14−1414−14.

Moreover, by the formula from Corollary 3.3,

(P+Q)D=−14−1414−14−12−121212−34−3434−34−12−121212.

But, in [9], the authors got that

(P+Q)D=−140140−12−121212−340340−12−121212.

This is illustrated easily by the definition of Drazin inverse, the result of (P + Q)^D is not correct from [9].

4 A formula of the Drazin inverse for block matrix

In this section, we use the additive formula in section 3 to give representation for the Drazin inverse of a complex block matrix.

Let M=ABCD, where A and D are square matrices with generalized Schur complement S = D − CA^DB of matrix M equal to zero. In [6], Martinez-Serrano and Castro-Gonzalez gave the representation for the Drazin inverse of M under the condition ABC = 0. In [7], Bu et al. gave the formula for M^D under conditions ABCA^π = 0 and A^πABC = 0. In the following Theorem 4.1, we give the representation for the Drazin inverse of M under the conditions ABCA^πA = 0, A^πBCA = 0, A^πBCB = 0 and S = 0.

Theorem 4.1

Let M be a matrix of the form (1.1) such that S = 0. If ABCA^πA = 0, A^πBCA = 0 and A^πBCB = 0, then

MD=LD+(LD)20AπB00+(LD)3AπBC000,

where

LD=AAADBCCADB(QD)2+AAADBCCADB(QD)4AADBCAπ0CADBCAπ0,(QD)n=ICAD((AW)D)n+1AIADB+((AAπ)D)n000,W=AAD+ADBCAD,forn≥1.

Proof

Consider the splitting of matrix M,

M=ABCCADB=AAADBCCADB+0AπB00.

If we denote by L=AAADBCCADB and H=0AπB00, we get H² = 0, H^D = 0. From A^πBCA = 0, A^πBCB = 0, it is obvious that HLH = 0, HL² = 0. Hence, the conditions of Lemma 2.4 are satisfied, and

MD=(H+L)D=LD+(LD)2H+(LD)3HL. (4.1)

Let

Q=AAADBCAADCADB,P=00CAπ0,

then L = P + Q, P² = 0. From ABCA^πA = 0, we have Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0. Hence, the conditions of Theorem 3.1 are satisfied and

LD=(P+Q)D=(P+Q)(Q2+QP)D+(PQ+P2)((Q2+QP)D)2, (4.2)

where (Q² + QP)^D is defined as in Theorem 3.1.

Notice (QP)² = 0, so (QP)^D = 0. Thus ((Q² + QP)^D)ⁿ = (Q^D)²ⁿ + (Q^D)²ⁿ⁺²QP, n ≥ 1.

If we split matrix Q as

Q=A2ADAADBCAADCADB+AAπ000,

and denote by

Q1=A2ADAADBCAADCADB,Q2=AAπ000,

we get Q₂Q₁ = Q₁Q₂ = 0. After applying Lemma 2.5,

QD=(Q1)D+(Q2)D.

Let S₁ be the generalized Schur complement of Q₁, then we have

S1=CADB−CAAD(A2AD)DAADB=0,

and

(A2AD)πAADB=0,CAAD(A2AD)π=0,

so from Lemma 2.6, we get

((Q1)D)n=ICAD((AW)D)n+1AIADB,

where W = AA^D + A^DBCA^D, for n ≥ 1.

After computation we get

(QD)n=ICAD((AW)D)n+1AIADB+((AAπ)D)n000.

After substituting this expressions and (4.2) into (4.1), we complete the proof.□

5 Numerical example

The following numerical example illustrates the application of Theorem 4.1.

Example 5.1

Consider the matrix M=ABCD, where

A=1000000000000010,B=1000000000001000,C=0110100000000100,D=0000100000000000.

By computing we know the generalized Schur complement S = D − CA^DB is zero and M satisfies the condition ABCA^πA = 0, A^πBCA = 0, A^πBCB = 0, in Theorem 4.1 in the paper. We have ind(A) = 2, and

AD=1000000000000000,Aπ=0000010000100001,

then according to the formula in Theorem 4.1, we have

MD=1110100000000000000000000000000000000000111010000000000000000000.

6 Conclusions

In this paper we derive the formula of (P + Q)^D under the conditions Q(P + Q)P(P + Q) = 0, P(P + Q)P(P + Q) = 0 and QPQ² = 0. Then, a corollary is given which satisfies the conditions (P + Q)P(P + Q) = 0 and QPQ² = 0. Furthermore, we apply the new result of (P + Q)^D to establish a new representation for the Drazin inverse of complex block matrix having generalized Schur complement equal to zero under the conditions ABCA^πA = 0, A^πBCA = 0 and A^πBCB = 0.

Acknowledgement

This work is supported by the Natural Science Foundation of Education Department of Sichuan Province(18ZB0521).

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Received: 2018-08-23

Accepted: 2019-01-03

Published Online: 2019-03-26

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https://doi.org/10.1515/math-2019-0015

Keywords for this article

Drazin inverse; Block matrix; Matrix index; Generalized Schur complement

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