# On the Drazin Inverse of the Sum of Two Matrices

Authors: Xiaoji Liu; Shuxia Wu; Yaoming Yu
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We deduce the explicit expressions for(P+Q)D and (PQ)D of
two matrices P and Q under the conditions P2Q=PQP and Q2P=QPQ. Also,
we give the upper bound of |P+QD-PD|2.


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## Body

## 1. Introduction

The symbolℂm×n stands for the set of m×n complex matrices, and In (for short I) stands for the n×n identity matrix. For A∈ℂn×n, its Drazin inverse, denoted by AD, is defined as the unique matrix satisfying(1.1)Ak+1AD=Ak,ADAAD=AD,AAD=ADA,
where k=Ind(A) is the index of A. In particular, if k=0, A is invertible and AD=A-1 (see,  e.g., [1–3] for details). Recall that for A∈ℂn×n withInd(A)=k, there exists an n×n nonsingular matrix X such that(1.2)A=X(C00N)X-1,
where C is a nonsingular matrix and N is nilpotent of index k, and(1.3)AD=X(C-1000)X-1
(see [1, 3]). It is well known that if A is nilpotent, then AD=0. We always write Aπ∶=I-AAD.Drazin [2] proved that in associative ring (A+B)D=AD+BD when A,B are Drazin invertible and AB=BA=0. In [4], Hartwig et al. relaxed the condition to AB=0 and put forward the expression for (A+B)D where A,B∈ℂn×n. In recent years, the Drazin inverse of the sum of two matrices or operators has been extensively investigated under different conditions (see, [5–15]). For example, in [7], the conditions are PQ=λQP and PQ=PQP, in [9] they are P3Q=QP and Q3P=PQ, and in [15], they are PQP=0 and PQ2=0. These results motivate us to investigate how to explicitly express the Drazin inverse of the sum P+Q under the conditions P2Q=PQP and Q2P=QPQ, which are implied by the condition PQ=QP.The paper is organized as follows. In Section2, we will deduce some lemmas. In Section 3, we will present the explicit expressions for (P+Q)D and (PQ)D of two matrices P and Q under the conditions P2Q=PQP and Q2P=QPQ. We also give the upper bound of ∥(P+Q)D-PD∥2.


## 2. Some Lemmas

In this section, we will make preparations for discussing the Drazin inverse of the sum of two matrices in next section. To this end, we will introduce some lemmas.The first lemma is a trivial consequence of [16, Theorem  3.2].Lemma 2.1.
LetA∈ℂn×n,B∈ℂm×m, and C∈ℂm×n with BC=0, and define
(2.1)M=(A0CB).
Then,
(2.2)MD=(AD0C(AD)2BD).Lemma 2.2.
LetP,Q∈ℂn×n. If P2Q=PQP, then, for any positive integers i,j,(i)
Pi+1Q=PiQP=PQPi,P2iQ=PiQPi,(ii)
PiQi=(PQ)i.
Moreover, ifQ2P=QPQ, then
(2.3)PQjPi=Pi+1Qj.Proof.
(i)
By induction, we can easily get the results.(ii)
Fori=1, it is evident. Assume that, for i=k, the equation holds, that is, PkQk=(PQ)k. When i=k+1, by (i), we have(2.4)Pi+1Qi+1=PQPiQi=PQ(PQ)i=(PQ)i+1.
Hence, by induction, we have PiQi=(PQ)i for any i.
AssumeQ2P=QPQ. By induction on j for (2.3). Obviously, when j=1, it holds by statement (i). Assume that it holds for j=k, that is, PQkPi=Pi+1Qk. When j=k+1,
(2.5)PQk+1Pi=PQk-1Q2PPi-1=PQk(PQPi-1)=PQkPiQ=Pi+1QkQ=Pi+1Qk+1.
Hence (2.3) holds for any j.Lemma 2.3.
LetP,Q∈ℂn×n. Suppose that P2Q=PQP and Q2P=QPQ. Then, for any positive integer m,
(2.6)(P+Q)m=∑i=0m-1Cm-1i(Pm-iQi+Qm-iPi),
where the binomial coefficient Cji=j!/i!(j-i)!,j≥i.
Moreover, ifP,Q are nilpotent with Ps=0 and Qt=0, then P+Q is nilpotent and its index is less than s+t.Proof.
We will show by induction that (2.6) holds. Trivially, (2.6) holds for m=1. Assume that (2.6) holds for m=k, that is,
(2.7)(P+Q)k=∑i=0k-1Ck-1i(Pk-iQi+Qk-iPi).
Then, for m=k+1, we have, by Lemma 2.2,
(2.8)(P+Q)k+1=∑i=0k-1Ck-1i(Pk-iQi+Qk-iPi)(P+Q)=∑i=0k-1Ck-1i(Pk+1-iQi+Pk-iQi+1+Qk-iPi+1+Qk+1-iPi)=Pk+1+∑i=1k-1(Ck-1i+Ck-1i-1)Pk+1-iQi+PQk+Qk+1+∑i=1k-1(Ck-1i+Ck-1i-1)Qk+1-iPi+QPk=Pk+1+∑i=1k-1CkiPk+1-iQi+PQk+Qk+1+∑i=1k-1CkiQk+1-iPi+QPk=∑i=0kCkiPk+1-iQi+∑i=0kCkiQk+1-iPi.
Hence (2.6) holds for any m≥1.
IfP,Q are nilpotent with Ps=0 and Qt=0, then taking m=s+t-1 in (2.6) yields (P+Q)s+t-1=0, that is, P+Q is nilpotent of index less than s+t.Lemma 2.4 (see [1, Theorem  7.8.4]).
LetP,Q∈ℂn×n. If PQ=QP, then (PQ)D=QDPD=PDQD and PDQ=QPD.Lemma 2.5.
LetP,Q∈ℂn×n and P be invertible. If PQ=QP, then
(2.9)(P+Q)D=(I+P-1Q)DP-1=P-1(I+P-1Q)D.
Moreover, ifQ is nilpotent of index t, then P+Q is invertible and
(2.10)(P+Q)-1=∑i=0t-1(-Q)iP-i-1=∑i=0t-1P-i-1(-Q)i.Proof.
SinceP+Q=P(I+P-1Q)=(I+P-1Q)P, by Lemma 2.4,
(2.11)(P+Q)D=(I+P-1Q)DP-1=P-1(I+P-1Q)D.
Note that the nilpotency ofQ with commuting with P implies that P-1Q is nilpotent of index t. Thus, I+P-1Q is invertible and so is P+Q, and
(2.12)(P+Q)-1=(I+P-1Q)-1P-1=∑i=0t-1(-Q)iP-i-1=∑i=0t-1P-i-1(-Q)i.Lemma 2.6.
LetP,Q∈ℂn×n with Q=Q1⊕Q2, where Q1 is invertible and Q2 is nilpotent of index t, and let
(2.13)P=(P1P3P4P2)
be partitioned conformably with Q. Suppose that Q2P=QPQ and P2Q=PQP. Then, P3=0 and
(2.14)Q1P1=P1Q1,(2.15)Q2P4=P2P4=0,(2.16)Q22P2=Q2P2Q2,(2.17)Pi2Qi=PiQiPi,i=1,2.
Moreover, ifP is nilpotent of index s, then P4P1s-1=0.Proof.
SinceQ2P=QPQ,Q2tP=QtPQt by Lemma 2.2, that is,
(2.18)(Q12t000)(P1P3P4P2)=(Q1t000)(P1P3P4P2)(Q1t000),
namely,
(2.19)(Q12tP1Q12tP300)=(Q1tP1Q1t000).
Thus, P3=0 because the invertibility of Q1. So from Q2P=QPQ and P2Q=PQP, it follows, respectively, that
(2.20)P1Q1=Q1P1,Q22P4=Q2P4Q1,Q22P2=Q2P2Q2,
and that
(2.21)P2P4Q1=P2Q2P4,Pi2Qi=PiQiPi,i=1,2.
Since Q2t=0,
(2.22)Q2P4=Q22P4Q1-1=Q2tP4Q1-t+1=0,
and then P2P4=P2Q2P4Q1-1=0. From this, we can easily verify
(2.23)Ps=(P1s0P4P1s-1P2s).
Therefore, if Ps=0, then P4P1s-1=0.


## 3. Main Results

In this section, we will give the explicit expressions for(P+Q)D and (PQ)D, under the conditions P2Q=PQP and Q2P=QPQ. Now, we begin with the following theorem.Theorem 3.1.
LetQ∈ℂn×n with Ind(Q)=t and P∈ℂn×n be nilpotent with Ps=0. If P2Q=PQP and Q2P=QPQ, then
(3.1)(P+Q)D=∑i=0s-1(QD)i+1(-P)i+QπP∑i=0s-2(-1)i(i+1)(QD)i+2Pi=QQD∑i=0s-1(-P)i(QD)i+1+QπPQQD∑i=0s-2(-1)i(i+1)Pi(QD)i+2.Proof.
Ift=0, then Q is invertible, and therefore QP=PQ. So, by Lemma 2.5, (3.1) holds.
Now assume thatt>0 and, without loss of generality, Q can be written as Q=Q1⊕Q2, where Q1 is invertible and Q2 is nilpotent of index t. So QD=Q1-1⊕0. Since P2Q=PQP and Q2P=QPQ, we can write P, partitioned conformably with Q, by Lemma 2.6,   as follows:
(3.2)P=(P10P4P2),
where P1,P2 are nilpotent since P is nilpotent. We also write I=I1⊕I2, partitioned conformably with Q.
SinceP1 is nilpotent and Q1 is invertible, by Lemma 2.5,
(3.3)(P1+Q1)-1=∑i=0s-1Q1-i-1(-P1)i=∑i=0s-1(-P1)iQ1-i-1.
Also, the nilpotency of P2,Q2 implies (P2+Q2)D=0 by Lemma 2.3.
By (2.15), (P2+Q2)P4=0. Hence, by Lemma 2.1, the argument above, and (2.14), we have
(3.4)(P+Q)D=(P1+Q10P4P2+Q2)D=(Q1-1(I1+Q1-1P1)-10P4Q1-2(I1+Q1-1P1)-20).
By (3.3), it is easy to verify that
(3.5)(I1+Q1-1P1)-2=∑i=0s-1(-1)i(i+1)Q1-iP1i=∑i=0s-1(-1)i(i+1)P1iQ1-i.
Since Ps=0, we have P4P1s-1=0 by Lemma 2.6, and, therefore, by (3.5),
(3.6)QπP∑i=0s-2(-1)i(i+1)(QD)i+2Pi=∑i=0s-2(-1)i(i+1)(000I2)(P10P4P2)(Q1-(i+2)000)(P1i0*P2i)=∑i=0s-2(-1)i(i+1)(00P4Q1-(i+2)P1i0)=(00P4Q1-2(I+Q1-1P1)-20).
Analogous to the argument above, we can see, by Lemma 2.5,
(3.7)∑i=0s-1(QD)i+1(-P)i=(Q1-1(I1+Q1-1P1)-1000).
Thus, putting (3.6) and (3.7) into (3.4) yields the first equation of (3.1).
Similar to the discussion of (3.6), we have
(3.8)QQD∑i=0s-1(-P)i(QD)i+1=(Q1-1(I+Q1-1P1)-1000),QπPQQD∑i=0s-2(-1)i(i+1)Pi(QD)i+2=(00P4Q1-2(I+Q1-1P1)-20),
and then putting them into (3.4) yields the second equation of (3.1).The following theorem is our main result, and Theorem3.1 and Lemma 2.5 can be regarded as its special cases.Theorem 3.2.
LetP,Q∈ℂn×n with Ind(P)=s≥1 and Ind(Q)=t. If P2Q=PQP and Q2P=QPQ. Then,(i)
(3.9)(PQ)D=PDQD=PPDQDPD=PQD(PD)2,(3.10)Q2PD=QPDQ,(3.11)(PD)2Q=PDQPD.(ii)
(3.12)(P+Q)D=PD(I+PDQ)D+PπQ[PD(I+PDQ)D]2+∑i=0s-1(QD)i+1(-P)iPπ+QπP∑i=0s-2(-1)i(i+1)(QD)i+2PiPπ.Proof.
Ifs=0, then P is invertible and PQ=QP. So, by Lemmas 2.4 and 2.5, (3.9) and (3.12) hold, respectively. Therefore, assume that s>0, and, without loss of generality, let P=P1⊕P2, where P1 is invertible and P2 is nilpotent of index s. From hypotheses, by Lemma 2.6, we can write
(3.13)Q=(Q10Q4Q2),
partitioned conformably with P, and those equations in Lemma 2.6 hold. By Lemma 2.1, therefore, we have
(3.14)QD=(Q1D0Q4(Q1D)2Q2D),Q2=(Q120Q4Q1Q22).(i)
By (2.14) and (2.15),(3.15)Q2PD=(Q12P1-10Q4Q1P1-10)=(Q1P1-1Q10Q4P1-1Q10)=QPDQ,(PD)2Q=(P1-2Q1000)=(P1-1Q1P1-1000)=PDQPD,PQD(PD)2=(P1Q1DP1-2000)=(P1-1Q1D000)=PDQD=(Q1DP1-1000)=PPDQDPD,(3.16)PQ=(P1Q100P2Q2)=(Q1P100P2Q2).
By (2.17) and Lemma 2.2, (P2Q2)s=P2sQ2s=0. By Lemma 2.4 and (3.16), we have
(3.17)(PQ)D=((Q1P1)D00(P2Q2)D)=(P1-1Q1D000)=PDQD.
As a result, (3.9) holds.(ii)
By Lemma2.6, (P2+Q2)Q4=0 and then, by Lemma 2.1, we have
(3.18)(P+Q)D=(P1+Q10Q4P2+Q2)D=((P1+Q1)D0Q4[(P1+Q1)D]2(P2+Q2)D).By Lemma 2.5, we have
(3.19)PD(I+PDQ)D=(P1-1000)(I1+P1-1Q100I2)D=(P1-1(I1+P1-1Q1)D000)=((P1+Q1)D000)
and, therefore,
(3.20)(00Q4[(P1+Q1)D]20)=(000I2)(Q10Q4Q2)([(P1+Q1)D]2000)=PπQ[PD(I+PDQ)D]2.
By (3.1), we have
(3.21)(000(P2+Q2)D)=(000∑i=0s-1(Q2D)i+1(-P2)i+Q2πP2∑i=0s-2(-1)i(i+1)(Q2D)i+2P2i)=∑i=0s-1(QD)i+1(-P)iPπ+QπP∑i=0s-2(-1)i(i+1)(QD)i+2PiPπ.
Thus, substituting (3.19), (3.21), and (3.20) in (3.18) yields (3.12).Note thatPQ=QP implies P2Q=PQP and Q2P=QPQ.Corollary 3.3 (see [14, Theorem 2]).
IfP,Q∈ℂn×n with PQ=QP and Ind(P)=s, then
(3.22)(P+Q)D=(I+PDQ)DPD+Pπ∑i=0s-1(QD)i+1(-P)i.Proof.
From (3.19) and Lemma 2.5, we can obtain
(3.23)(I+PDQ)DPD=PD(I+PDQ)D.SincePkQ=0 for some k,PDQ=0. Thus, we have the following corollary.Corollary 3.4.
LetP,Q∈ℂn×n with Ind(P)=s≥1 and Ind(Q)=t. Suppose P2Q=PQP and Q2P=QPQ. If there exist two positive integers k and h such that PkQ=0 and QhP=0, then
(3.24)(P+Q)D=PD+QD+Q(PD)2.
IfQ is a perturbation of P, then, we have the following result in which ∥(P+Q)D-PD∥2 has an upper bound. Before the theorem, let us recall that if ∥A∥2<1, then I+A is invertible and
(3.25)‖(I+A)-1‖2≤11-‖A‖2,‖I-(I+A)-1‖2≤‖A‖21-‖A‖2.Theorem 3.5.
LetP,Q∈ℂn×n with Ind(P)=s≥1 and Ind(Q)=t. Suppose P2Q=PQP and Q2P=QPQ. If ∥PDQ∥2<1, then
(3.26)‖(P+Q)D-PD‖2≤‖PD‖2‖PDQ‖21-‖PDQ‖2+‖Pπ‖2‖Q‖2‖PD‖22(1-‖PDQ‖2)2+‖QD‖2‖Pπ‖2(1-‖QD‖2s‖P‖2s)1-‖QD‖2‖P‖2+‖QD‖22‖Qπ‖2‖Pπ‖2‖P‖2(1-‖QD‖2‖P‖2)2×[1-s‖QD‖2s-1‖P‖2s-1+(s-1)‖QD‖2s‖P‖2s].Proof.
Since∥PDQ∥2<1, I+PDQ is invertible. Then by (3.12), we have
(3.27)(P+Q)D-PD=PD[(I+PDQ)-1-I]+PπQ[PD(I+PDQ)-1]2+∑i=0s-1(QD)i+1(-P)iPπ+QπP∑i=0s-2(-1)i(i+1)(QD)i+2PiPπ.
In order to verity (3.26), we need to calculate the 2-norms of the right-hand side of the above equation. By (3.25),
(3.28)‖PD[(I+PDQ)-1-I]‖2≤‖PD‖2‖PDQ‖21-‖PDQ‖2,‖PπQ[PD(I+PDQ)-1]2‖2≤‖Pπ‖2‖Q‖2‖PD‖22(1-‖PDQ‖2)2,‖∑i=0s-1(QD)i+1(-P)iPπ‖2≤∑i=0s-1‖QD‖2i+1‖P‖2i‖Pπ‖2=∑i=1s‖QD‖2i‖P‖2i-1‖Pπ‖2=‖QD‖2‖Pπ‖2(1-‖QD‖2s‖P‖2s)1-‖QD‖2‖P‖2,‖QπP∑i=0s-2(-1)i(i+1)(QD)i+2PiPπ‖2≤∑i=0s-2(i+1)‖QD‖2i+2‖P‖2i+1‖Qπ‖2‖Pπ‖2=∑i=1s-1i‖QD‖2i+1‖P‖2i‖Qπ‖2‖Pπ‖2.
Let q∶=∥QD∥2∥P∥2 and S∶=∑i=1s-1iqi. Then,
(3.29)(1-q)S=∑i=1s-1qi-(s-1)qs=q(1-qs-1)1-q-(s-1)qs=q-sqs+(s-1)qs+11-q.
Thus
(3.30)∑i=1s-1i‖QD‖2i+1‖P‖2i‖Qπ‖2‖Pπ‖2=‖QD‖22‖Qπ‖2‖Pπ‖2‖P‖2[1-s‖QD‖2s-1‖P‖2s-1+(s-1)‖QD‖2s‖P‖2s](1-‖QD‖2‖P‖2)2.
By the above argument, we can get (3.26).Finally, we give an example to illustrate our results.Example 3.6.
Consider the matrices(3.31)P=(1000010000010000),Q=(130000000000000013).
We observe that P2Q=PQP and Q2P=QPQ, but PQ≠QP. It is obvious that s=Ind(P)=2, and
(3.32)PD=(1000010000000000),QD=(3000000000000003).
Since ∥PDQ∥2=(1/3)<1, I+PDQ is invertible and
(3.33)(I+PDQ)-1=(34000010000100001).
By (3.12),
(3.34)(P+Q)D-PD=(-14000000000090003).
We can compute ∥(P+Q)D-PD∥2=310. On the other hand, it is easy to get that ∥P∥2=∥PD∥2=∥Pπ∥2=∥Qπ∥2=1, ∥Q∥2=1/3, ∥QD∥2=3. By (3.26), we get the upper bound of ∥(P+Q)D-PD∥2 is 16(1/4), it is bigger than and close to the exact norm.



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