Filomat 31:16 (2017), 5151–5158 https://doi.org/10.2298/FIL1716151X

Published by Faculty of Sciences and Mathematics, University of Nisˇ, Serbia

Available at: http://www.pmf.ni.ac.rs/filomat

The Drazin Inverse of the Sum of Two Matrices and its Applications

Lingling Xia^a, Bin Deng^a

^aSchool of Mathematics, Hefei University of Technology, Hefei, 230009,China

Abstract. In this paper, we give the results for the Drazin inverse of P + Q, then derive a representation

•  
• !

AC
BD

• for the Drazin inverse of a block matrix M =
• under some conditions. Moreover, some alternative

representations for the Drazin inverse of M^D where the generalized Schur complement S = D − CA^DB is nonsingular. Finally, the numerical example is given to illustrate our results.

1. Introduction and preliminaries

Let C^n×n denote the set of n × n complex matrix. By R(A), N(A) and rank(A) we denote the range, the null space and the rank of matrix A. The Drazin inverse of A is the unique matrix A^D satisfying

• A^DAA^D = A^D,
• AA^D = A^DA,

A^k+1A^D = A^k.

(1) where k = ind(A) is the index of A, the smallest nonnegative integer k which satisfies rank(A^k+1) = rank(A^k). If ind(A) = 0, then we call A^D is the group inverse of A and denote it by A^]. If ind(A) = 0, then A^D = A⁻¹. In addition, we denote A^π = I − AA^D, and define A⁰ = I, where I is the identity matrix with proper sizes[1].
For A ∈ C^n×n, k is the index of A, there exists unique matrices C and N, such that A = C + N, CN = NC = 0,
N is the nilpotent of index k, and ind(C) = 0 or 1.We shall always use C, N in this context and will refer to A = C + N as the core-nilpotent decomposition of A, Note that A^D = C^D.
The Drazin inverse of a square matrix is widely applied in many fields, such as singular differential or difference equations, Markov chains, iterative method, cryptography and numerical analysis,which can be found in[2, 3]. The Drazin inverse in perturbation bounds for the relative eigenvalue problem has an important application value [4]. Accordingly, the Drazin inverse of 2 × 2 block matrix and its applications can be found in [3].
Suppose P, Q ∈ C^n×n such that PQ = QP = 0, then (P + Q)^D=P^D+Q^D. This result was firstly proved by
Drazin [5] in 1958. In 2001, Hartwig et al. [6] gave a formula for (P+Q)^D under the one side condition PQ = 0. In 2005, Castro-Gonza´lez [7] derived a result under the conditions P^DQ = 0, PQ^D = 0 and Q^πPQP^π = 0. In 2009, Mart´ınez-Serrano and Castro-Gonza´lez [8] extended these results to the case P²Q = 0, Q² = 0 and gave the formula for (P + Q)^D. Hartwig and Patricio [9] under the condition P²Q + PQ² = 0. In 2010, Wei and Deng [10] studied the additive result for generalized Drazin inverse under the commutative condition of PQ = QP on a Banach space. Liu et al. [11] gave the representations of the Drazin inverse of (P ± Q)^D with P³Q = QP and Q³P = PQ satisfied. In 2011, Liu et al. [12] extended the results to the case P²Q = 0, QPQ = 0. In 2012, Bu et al. [13] gave the representations of the Drazin inverse of (P + Q)^D under the following conditions:

(i)P²Q = 0, Q²P = 0; (ii)QPQ = 0, QP²Q = 0, P³Q = 0.

2010 Mathematics Subject Classification. 15A09

Keywords. Drazin inverse,Block matrix,Generalized Schur complement,Index Received: 25 September 2014; Revised: 07 January 2015; Accepted: 21 January 2015 Communicated by Dragana Cvetkovic´ - Ilic´ This work was supported by the Fundamental Research Funds for the Central Universities(J2014HGXJ0068)

Email address: [email protected] (Bin Deng)

L. Xia, B. Deng / Filomat 31:16 (2017), 5151–5158

5152
The results about the representation of (P + Q)^D are useful in computing the representations of the Drazin inverse for block matrices, analyzing a class of perturbation and iteration theory. The general questions of how

•  
• !

D

AC
BD

• to express (P + Q)^D by P, Q, P^D, Q^D and
• by A, B, C, D without side condition are very difficult and

have not been solved.
In this paper, we first give the formulas of (P + Q)^D under the conditions P²Q = 0, PQ + QP = 0 and

T

P^DQ = 0, PQ − QP = 0, N(P) N(Q) = 0. And similar reasoning is presented. In the second, we use the

•  
• !

AC
BD

• formulas of (P + Q)^D to give some representations for the Drazin inverse of block matrix M =
• (A and

D are square) under some conditions. Then we give the representation of M^D in which the generalized Schur complement S = D − CA^DB is nonsingular under new conditions. Finally, we take some numerical examples to illustrate our results.
Before giving the main results, we first introduce several lemmas as follows.

Lemma 1.1. ([14]) Let

•  
• !
•  
• !

AC

0

B
B

0

CA

M₁ =

,

M₂ =

,

where A and B are square matrices with ind(A)=r and ind(B)=s. Then

•  
• !
•  
• !

A^D

X

0

B^D

0

X

M^D₁ =

,

M^D₂ =

,

• B^D
• A^D

• P
• P

• r−1
• s−1

where X = _i=0 (B^D)ⁱ⁺²CAⁱA^π + B^π _i=0 BⁱC(A^D)ⁱ⁺² − B^DCA^D. Lemma 1.2. ([6]) Let P, Q ∈ C^n×n be such that ind(P)=r, ind(Q)=s and PQ = 0. Then

• s−1
• r−1

• X
• X

(P + Q)^D = Q^π Qⁱ(P^D)ⁱ⁺¹

+

(Q^D)ⁱ⁺¹PⁱP^π.

• i=0
• i=0

Lemma 1.3. ([8]) Let A, B ∈ C^n×n

,
(i) If R is nonsingular and B = RAR⁻¹, then B^D = RA^DR⁻¹.

•  
• !

A₁

0
0

A₂

• (ii) If ind(A)=k ≥ 0, then exists a nonsingular matrix R such that A = R
• R⁻¹, where A₁ ∈ C^r×r is

nonsingular and A₂ ∈ C^{(n−r)×(n−r)} is k-nilpotent. Relative to the above form, A^D and A^π = I − AA^D, are given by

•  
• !
•  
• !

A⁻₁ ¹

0
00
00
0

I
A^D = R

R⁻¹,

A^π = R

R⁻¹.

Lemma 1.4. ([5]) Let P, Q ∈ C^n×n be such that PQ = QP = 0, then (P + Q)^D = P^D + Q^D.

• ꢀ
• ꢁ

Lemma 1.5. ([1]) Let A ∈ C^m×n, B ∈ C^n×m, then (AB)^D = A (BA)^{2 D}B. Lemma 1.6. ([1]) Let A, B ∈ C^n×n, if AB = BA, then

(i) (AB)^D = B^DA^D = A^DB^D. (ii) A^DB = BA^D and AB^D = B^DA.

Lemma 1.7. ([20]) Let A, B ∈ C^n×n, suppose that c is such that (cA + B) is invertible, then
(i) (cA + B)⁻¹A and (cA + B)⁻¹B commute;

• ꢀ
• ꢁ
• ꢀ
• ꢁ

(ii) N(A) ∩ N(B) = {0} and N(A) = N (cA + B)⁻¹A , N(B) = N (cA + B)⁻¹B .

L. Xia, B. Deng / Filomat 31:16 (2017), 5151–5158

5153

2. Additive Results

In [10], Wei and Deng studied the additive result for generalized Drazin inverse under the commutative condition of PQ = QP on a Banach space. In this section, we will give the Drazin inverse of P + Q under the conditions that P²Q = 0, PQ + QP = 0 and P^DQ = 0, PQ − QP = 0, N(P) ∩ N(Q) = {0}, which will be the main tool in our following development.

Theorem 2.1. Let P, Q ∈ C^n×n be such that P²Q = 0, PQ + QP = 0, then

(P + Q)^D = P^D + (P + Q)(Q^D)².

(2)
Proo f. From the conditions of theorem , we can know P²Q = −PQP = QP² = 0.

•  
• !
•  
• !

P₁

0

P₂
Q₁ Q₁₂ Q₂₁ Q₂

• Let P = R
• R⁻¹, where P₁ is nonsingular and P₂ is nilpotent. Write Q = R
• R⁻¹. Form

0

P²Q = 0 it follows

Q₁ = 0, Q₁₂ = 0, P²₂Q₂₁ = 0, P₂²Q₂ = 0

Form QP² = 0 it follows
(3)

Q₂₁ = 0, Q₂P²₂ = 0.

(4) (5)
Form PQ + QP = 0 it follows P₂Q₂ + Q₂P₂ = 0. Now, using Lemma 1.1 we obtain

•  
• !
•  
• !

• D
• D

P₁

0
0

P⁻₁ ¹

0
0

• (P + Q)^D = R
• R⁻¹ = R

R⁻¹.

• (P₂ + Q₂)^D
• P₂ + Q₂

Now, we need compute (P₂ + Q₂)^D.

(P₂ + Q₂)² = P₂² + Q₂² + P₂Q₂ + Q₂P₂ = P²₂ + Q²₂, (P²₂)^D = (P₂^D)² = 0.

Applying (3) and (4) and Lemma 1.4, we get

• ꢀ
• ꢁ

D

(P₂ + Q₂)² = (P²₂ + Q²₂)^D = (Q²₂)^D.

Further,

• ꢀ
• ꢁ

D

(P₂ + Q₂)^D = (P₂ + Q₂) (P₂ + Q₂)² = (P₂ + Q₂)(Q²₂)^D.

(6)
By substituting (6) in (5), we get

•  
• !
•  
• !

P₁

0
00
00
0

• (P + Q)^D = R
• R⁻¹ + R
• R⁻¹ = P^D + (P + Q)(Q^D)².

(P₂ + Q₂)^D

Using the similar method as in the proof of Theorem 2.1, We get the following two results.

Theorem 2.2. Let P, Q ∈ C^n×n be such that P²Q = 0, QP^π = 0, then

(P + Q)^D = P^D + Q(P^D)² + PQ(P^D)³.

(7) (8)

Theorem 2.3. Let P, Q ∈ C^n×n be such that P^DQ = 0, QP^π = 0, s = indP, then

s−1

X

(P + Q)^D = P^D +

PⁱQ(P^D)ⁱ⁺²

.

i=0

L. Xia, B. Deng / Filomat 31:16 (2017), 5151–5158

5154
(9)

Theorem 2.4. Let P, Q ∈ C^n×n be such that PQ = QP, P^DQ = 0, N(P) ∩ N(Q) = {0}, then

k−1

X

(P + Q)^D = P^D +

(Q^D)ⁿ⁺¹(−P)ⁿ.

n=0

where k = ind(P).

•  
• !

P₁

0
0

P₂

Proo f. Let P = R

R⁻¹, where P₁ and R is nonsingular, and P₂ is nilpotent, then exists a positive real

•  
• !

Q₁ Q₁₂ Q₂₁ Q₂

number k, satisfy P^k₂ = 0, and we can easy know k = ind(P). Write Q = R follows Q₁ = 0, Q₂ = 0.
R⁻¹. From P^DQ = 0 it
Now, from PQ = QP it follows P₂Q₃ = Q₃P₁, P₂Q₄ = Q₄P₂. Then P^k₂Q₃ = Q₃P₁^k = 0. Thus Q₃ = 0 since P₁^k is invertible. Next we will show that Q₄ is invertible. If P₂ = 0, the assumption N(P) ∩ N(Q) = {0} implies N(Q₄) = {0}, and we are done. If P₂ , 0, suppose there exists a v , 0 such that v ∈ N(Q₄). Then