# The Explicit Expression of the Drazin Inverse and Its Application

Authors: Xiaoji Liu; Shaowu Huang; Liang Xu; Yaoming Yu
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

We give explicit expressions of(P±Q)d of two matrices P and
Q, in terms of P, Q, Pd, and Qd, (P±I)d, under the condition that PQ=P, and apply the result to finding an explicit representation for the Drazin inverse of some 2×2 block matrix.


---

## Body

## 1. Introduction

In recent years, the representations and perturbation analysis of the Drazin inverse for matrices or operators have been investigated (see [1–6]). In [7], the author presented the presentations of the Drazin inverse of sum and product of two operators over Banach spaces on the condition of the commutativity up to a factor. And, in [8], the same author discussed explicit representations of Drazin inverses of sums and differences of two idempotents over Hilbert spaces.These investigations motivate us to deal with an explicit expression of the Drazin inverse of differences and sums of two matrices. The paper is organized as follows. In this section, we will introduce some notions and lemmas. In Section2, we will present these explicit expressions of differences and sums of two matrices P and Q under the condition PQ=P. In Section 3, we will deduce an explicit representation for the Drazin inverse of the 2×2 block matrix [ABCD] with A=BC and B=BD in terms of its subblocks and their Drazin inverses and (A+I)d. In Section 4, we will present a numerical example to demonstrate the main result in Section 2.Throughout this paper the symbolℂm×n stands for the set of m×n complex matrices, and I∈ℂn×n stands for the unit matrix. Let A∈ℂn×n; the Drazin inverse, denoted by Ad, of matrix A is defined as the unique matrix satisfying
(1)Ak+1Ad=Ak,AdAAd=Ad,AAd=AdA,
where k=Ind(A) is the index of A. In particular, if Ind(A)=1, then Ad is called the group inverse, denoted by Ag, of A. Apparently, if A is nonsingular, then Ind(A)=0; otherwise, Ind(A)≥1, especially Ind(0)=1. If A is nilpotent, then Ad=0. If X is nonsingular and B=XAX-1, then Bd=XAdX-1 (see [6, 9, 10]). For convenience, we write Aπ=I-AAd and use the convention ∑i=nm‍=0, if m<n.Before we start the discussion, we need some preparations.Lemma 1 (see [11, Theorem 3.2]).
LetM=[AB0D] with t=Ind(A) and l=Ind(D). Then,
(2)Md=[AdS0Dd],
where
(3)S=Aπ∑n=0t-1‍AnBDdn+2+∑n=0l-1‍Adn+2BDnDπ-AdBDd.Lemma 2.
LetP,Q∈ℂm×m with Ind(Q)=s. If PQ=P, then
(4)PQd=P,(QdQP)k=QdQPk,(QPπ)k=QkPπ,(QdPπ)k=QdkPπ,k≥1.Proof.
SincePQ=P and Ind(Q)=s,
(5)P=PQs=PQs+1Qd=PQd,
and then PQQd=P. So from this, by induction, it follows that (QdQP)k=QdQPk for k≥1.
Now, we will show inductively that(QPπ)k=QkPπ. It holds for k=1. Assume it holds for k=n; that is, (QPπ)n=QnPπ. Then,
(6)(QPπ)n+1=(QPπ)(QPπ)n=Q(I-PdP)QnPπ=Qn+1Pπ.
So it holds for any k≥1.
FromPQd=P, we can similarly show that (QdPπ)k=QdkPπ.Lemma 3.
LetP be nilpotent of index t>1 and S=∑i=0t-1‍ai[1]Pi. If ai[1]=1, then
(7)Sn=∑i=0t-1‍ai[n]Pi,n≥2,
where ai[n]=∑u=0i‍au[n-1], i=0,…,t-1.Proof.
SincePt=0, we can easily write Sn as
(8)Sn=∑i=0t-1‍ai[n]Pi,n≥2.
We will prove the relationship
(9)ai[n]=∑h=0i‍ah[n-1],i=0,…,t-1
by induction on n. Obviously, (9) holds for n=2. Assume inductively that (9) holds for n=k.
Sinceai[1]=1 and Pt=0,
(10)Sk+1=SkS=∑i=0t-1‍ai[k]Pi∑j=0t-1‍Pj=∑h=0t-1‍∑i+j=h‍ai[k]Pi+j=∑h=0t-1‍∑i=0h‍ai[k]Ph.
On the other hand,
(11)Sk+1=∑i=0t-1‍ai[k+1]Pi.
So, (9) holds for n=k+1. Hence, (9) holds for n≥2.Lemma 4 (see [9, Lemma 7.7.2]).
Let(12)N=[AB00]orN=[00BA].
Then, Ind(A)≤Ind(N)≤Ind(A)+1.


## 2. The Drazin Inverse of Differences and Sums of Two Matrices

In this section, we will investigate how to expressP±Q as a function of P, Q, Pd, and Qd, (P±I)d, under the condition PQ=P. We begin with the following theorem, in which P is assumed to be nilpotent.Theorem 5.
LetP,Q∈ℂn×n with Ind(Q)=s. If PQ=P and P is nilpotent of index t, then
(13)(P-Q)d=∑n=0s-1‍∑i=0t-2‍(ai[n+2]I-ai[n+1]Qd)QnPi+1-QdQs∑i=0t-2‍ai[s+1]Pi+1-Qd,
where ai[n]=∑k=0i‍ak[n-1] and ai[1]=1, i=0,…,t-1.Proof.
Ift=1, then, from the convention in Section 1, (13) clearly holds. Now assume t>1. From s=Ind(Q), there exists a nonsingular matrix W1 such that
(14)Q=W1[Q100Q2]W1-1,Qd=W1[Q1-1000]W1-1,
where Q1 is nonsingular and Q2 is nilpotent of index s. Partitioning W1-1PW1 conformably with W1-1QW1, we have
(15)P=W1[P1P4P3P2]W1-1.
Since PQ=P, PQd=P by Lemma 2, and then P2=0, P4=0, and PiQ1=Pi, i=1,3. Let t be the index of nilpotent matrix P. Then,
(16)Pt=W1[P1t0P3P1t-10]W1-1=0,
and therefore P1 is also nilpotent, and (P1-I)-1=-∑i=0t-1‍P1i. Thus,
(17)P-Q=W1[P1-Q10P3-Q2]W1-1=W1[(P1-I)Q10P3-Q2]W1-1.
By Lemma 1, we have
(18)(P-Q)d=W1[Q1-1(P1-I)-10X0]W1-1=W1[-Q1-1∑i=0t-1‍P1i0X0]W1-1,
where
(19)X=∑n=0s-1‍(-1)nQ2nP3[(P1-I)Q1]-(n+2).
By Lemma 3,
(20)(-1)nP3[(P1-I)Q1]-(n+2)=P3(Q1-1∑i=0t-1‍P1i)n+2=P3(∑i=0t-1‍P1i)n+2=P3∑i=0t-1‍ai[n+2]P1i=∑i=0t-2‍ai[n+2]P3P1i.
Thus,
(21)W1[00X0]W1-1=∑n=0s-1‍∑i=0t-2‍W1[00ai[n+2]Q2nP3P1i0]W1-1=∑n=0s-1‍∑i=0t-2‍ai[n+2]QπQnPi+1.
So, by (18),
(22)(P-Q)d=-∑i=0t-1‍QdPi+Qπ∑n=0s-1‍∑i=0t-2‍ai[n+2]QnPi+1=-∑i=0t-2‍QdPi+1-Qd+∑n=0s-1‍∑i=0t-2‍ai[n+2]QnPi+1-Qd∑n=1s‍∑i=0t-2‍ai[n+1]QnPi+1=-Qd+∑n=0s-1‍∑i=0t-2‍ai[n+2]QnPi+1-Qd∑n=0s‍∑i=0t-2‍ai[n+1]QnPi+1=∑n=0s-1‍∑i=0t-2‍(ai[n+2]I-ai[n+1]Qd)QnPi+1-QdQs∑i=0t-2‍ai[s+1]Pi+1-Qd.IfPQ=P, then (-P)Q=-P. So, it immediately follows from the above theorem.Corollary 6.
LetP,Q∈ℂn×n with Ind(Q)=s. If PQ=P and P is nilpotent of index k, then
(23)(P+Q)d=∑n=0s-1‍∑i=0t-2‍(-1)i(ai[n+2]I-ai[n+1]Qd)QnPi+1-QdQs∑i=0t-2‍(-1)iai[s+1]Pi+1+Qd,
where ai[n]=∑k=0i‍ak[n-1], ai[1]=1, i=0,…,t-1.If the nilpotency ofP is taken out in Theorem 5, then we can obtain our main result, a more general result.Theorem 7.
LetP,Q∈ℂn×n with Ind(QPπ)=s, Ind(P)=t, Ind[(P-I)PPd]=l, and Ind[(P-Q)Pπ]=k. If PQ=P, then
(24)(P-Q)d=∑n=0s-1‍∑i=0t-2‍(ai[n+2]I-ai[n+1]Qd)QnPi+1Pπ-QdPπ-∑n=0k-1‍(-1)nQnQπ(Q-I)PPd(P-I)dn+2+(Qπ+Qd)PPd(P-I)d+∑n=0l-1‍(-1)n+1Qdn(Qd-Qd2)PPd(P-I)n(P-I)π-QdQs∑i=0t-2‍ai[s+1]PπPi+1,
where ai[n]=∑k=0i‍ak[n-1], ai[1]=1, i=0,…,t-1.Proof.
There exists a nonsingular matrixW1 such that
(25)P=W1[P100P2]W1-1,Q=W1[Q1Q3Q4Q2]W1-1,
where P1 is nonsingular and P2 is nilpotent of index t. From PQ=P, we get
(26)Q=W1[I0Q4Q2]W1-1,
where P2Q4=0 and P2Q2=P2. So
(27)P-Q=W1[P1-I0-Q4P2-Q2]W1-1.
By Lemma 1, we have
(28)(P-Q)d=W1[(P1-I)d0X(P2-Q2)d]W1-1,
where
(29)X=∑n=0l-1‍(P2-Q2)dn+2(-Q4)(P1-I)n(P1-I)π+(P2-Q2)π∑n=0k-1‍(P2-Q2)n(-Q4)(P1-I)dn+2-(P2-Q2)d(-Q4)(P1-I)d,
and k=Ind(P2-Q2)=Ind[(P-Q)Pπ] and l=Ind(P1-I)=Ind[(P-I)PPd].
SinceP2Q4=0 and P2Q2=P2, by Theorem 5 and Lemma 2,
(30)(P2-Q2)d(Q2)djQ4=-∑n=0t-1‍(Q2)dP2n(Q2)djQ4+∑n=0s-1‍Q2πQ2nP2(∑i=0t-2‍P2i)n+2(Q2)djQ4=-(Q2)dj+1Q4,j≥0,(P2-Q2)πQ4=Q4+(P2-Q2)(Q2)dQ4=Q4-Q2(Q2)dQ4=Q2πQ4,P2Q2π=0,
and then
(31)(P2-Q2)π∑n=0k-1‍(P2-Q2)n(-Q4)=-∑n=0k-1‍(P2-Q2)nQ2πQ4=-∑n=0k-1‍(-Q2)nQ2πQ4.
Thus,
(32)X=∑n=0l-1‍(-1)n+1(Q2)dn+2Q4(P1-I)n(P1-I)π-∑n=0k-1‍(-1)nQ2nQ2πQ4(P1-I)dn+2-(Q2)dQ4(P1-I)d.
Since(33)P-I=W1[P1-I00P2-I]W1-1,
we have that, for n≥0,
(34)PPd(P-I)dn=W1[(P1-I)dn000]W1-1,(P-I)nPPd(P-I)π=W1[(P1-I)n(P1-I)π000]W1-1.
Obviously,
(35)(QPπ)d=W1[000(Q2)d]W1-1=QdPπ,(Q-I)PPd=W1[00Q40]W1-1,YnPπ=W1[000Y2n]W1-1,n≥0,
where the symbol Y denotes Q or P. Also,
(36)Pπ(QPπ)π=Pπ-PπQPπQdPπ=Pπ-QQdPπ=QπPπ.
Note that Pπ(Q-I)=(Q-I). Then, by (32),
(37)W1[00X0]W1-1=∑n=0l-1‍(-1)n+1Qdn+2Pπ(Q-I)PPd(P-I)nPPd(P-I)π-∑n=0k-1‍(-1)nQnPπ(QPπ)π(Q-I)PPd(P-I)dn+2-QdPπ(Q-I)PPd(P-I)d=∑n=0l-1‍(-1)n+1Qdn+2(Q-I)PPd(P-I)n(P-I)π-∑n=0k-1‍(-1)nQnQπ(Q-I)PPd(P-I)dn+2-Qd(Q-I)PPd(P-I)d.
Note thatPπQnPπ=QnPπ. So, by Theorem 5,
(38)W1[000(P2-Q2)d]W1-1=∑n=0s-1‍∑i=0t-2‍(ai[n+2]I-ai[n+1]QdPπ)QnPπPi+1Pπ-QdPπ-QdPπQsPπ∑i=0t-2‍ai[s+1]Pi+1Pπ=∑n=0s-1‍∑i=0t-2‍(ai[n+2]I-ai[n+1]Qd)QnPi+1Pπ-QdPπ-QdQs∑i=0t-2‍ai[s+1]PπPi+1.
Hence, putting (34), (37), and (38) into (28) yields (24).Corollary 8.
LetP,Q∈ℂn×n with Ind(QPπ)=s, Ind(P)=t, Ind[(P+I)PPd]=l, and Ind[(P+Q)Pπ]=k. If PQ=P, then
(39)(P+Q)d=∑n=0s-1‍∑i=0t-2‍(-1)i(ai[n+2]I-ai[n+1]Qd)QnPi+1Pπ+QdPπ+∑n=0k-1‍QnQπ(Q-I)PPd(P+I)dn+2+(Qπ+Qd)PPd(P+I)d+∑n=0l-1‍Qdn(Qd-Qd2)PPd(P+I)n(P+I)π-QdQs∑i=0t-2‍(-1)iai[s+1]PπPi+1,
where ai[n]=∑k=0i‍ak[n-1], ai[1]=1, i=0,…,t-1.


## 3. The Drazin Inverse of Some 2 × 2 Block Matrix

In this section, we will apply the results in Section2 to studying the representation for the Drazin inverse of a 2×2 block matrix, in terms of its subblocks.Theorem 9.
LetM=[ABCD]∈ℂn×n with Ind[(CB)πD]=s-1, Ind(A)=t-1, Ind[(A+I)AAd]=l-1, and Ind[[Aπ00(CB)π]M]=k. If A=BC and B=BD, then(40)Md=[(I-(A+I)d)(A+I)dAd0(Dd+Dπ)C(A+I)d2Ad+(Dd2-Dd)C(A+I)dAdDd2(CB)π]M+∑n=1s-1‍∑i=0t-2‍(-1)i[00(ai[n+2]I-ai[n+1]Dd)Dn-1CAiAπ0]M+∑n=1k-1‍[00Dπ(Dn-Dn-1)C(A+I)dn+2Ad0]M+∑n=0l-1‍[00(Ddn+2-Ddn+3)C(A+I)n(A+I)πAd0]M+∑i=0t-2‍(-1)i[ai[2]AiAπ0-(ai[1]Dd2+ai[s+1]DdDs-1)CAiAπ0]M,where ai[n]=∑k=0i‍ak[n-1], ai[1]=1, i=0,…,t-1.Proof.
Let(41)P=[AB00],Q=[00CD].
Then, M=P+Q, and, for n≥1,
(42)Qn=[00Dn-1CDn],Qdn=[00Ddn+1CDdn],Qπ=[I0-DdCDπ],Pn=[AnAn-1B00],Pd=[AdAd2B00],PPd=[AAdAdB00]=[Ad000]M,Pπ=[Aπ-AdB0I],(P+I)αn=[(A+I)αn*0I],
where the subscript α stands for d or its absence.
SinceWWd(W+I)=(W+I)WWd, WWd(W+I)dn=(W+I)dnWWd where W denotes P or A. Thus,
(43)PPd(P+I)αn=(P+I)αnPPd=[(A+I)αnAd000]M,(44)PPd(P+I)π=(P+I)πPPd=[(A+I)π*0I]PPd=[(A+I)πAd000]M,ai[n+2]I-ai[n+1]Qd=[ai[n+2]I00ai[n+2]I]-[00ai[n+1]Dd2Cai[n+1]Dd]=[ai[n+2]I0-ai[n+1]Dd2Cai[n+2]I-ai[n+1]Dd],Pi+1Pπ=[Ai+1AiB00][Aπ-AdB0I]=[Ai+1AπAiAπB00]=[AiAπ000]M,QnPi+1Pπ=[00Dn-1CDn]Pi+1Pπ=[00Dn-1CAiAπ0]M.
So, for n≥1,
(45)(ai[n+2]I-ai[n+1]Qd)QnPi+1Pπ=[00(ai[n+2]I-ai[n+1]Dd)Dn-1CAiAπ0]M,(ai[2]I-ai[1]Qd)Pi+1Pπ=[ai[2]AiAπ0-ai[1]Dd2CAiAπ0]M,ai[s+1]QdQsPπPi+1=ai[s+1][00Dd2CDd][00Ds-1CAiAπ0]M=[00ai[s+1]DdDs-1CAπAi0]M.
Also, forn≥1,
(46)Qπ(Qn+1-Qn)PPd(P+I)dn+2=[I0-DdCDπ][00(Dn-Dn-1)CDn+1-Dn]×[(A+I)dn+2Ad000]M=[00Dπ(Dn-Dn-1)C(A+I)dn+2Ad0]M,Qπ(Q-I)PPd(P+I)d2=[I0-DdCDπ][-I0CD-I][(A+I)d2Ad000]M=[-(A+I)d2Ad0(Dd+Dπ)C(A+I)d2Ad0]M,(Qπ+Qd)PPd(P+I)d=[I0Dd2C-DdCDπ+Dd][(A+I)dAd000]M=[(A+I)dAd0(Dd2-Dd)C(A+I)dAd0]M,
and, for n≥0,
(47)(Qdn+1-Qdn+2)PPd(P+I)n(P+I)π=[00(Ddn+2-Ddn+3)CDn+1-Dn+2]×[(A+I)nAd000]M[(A+I)πAd000]M=[00(Ddn+2-Ddn+3)C(A+I)n(A+I)πAd0]M.
Since(48)D-CAdB=D-CB(CB)d2CBD=(CB)πD,CAπ=C-CBCB(CB)d2C=(CB)πC,
we have
(49)QdPπ=[00Dd2CDd][Aπ-AdB0I]=[00Dd2(CB)πCDd2(CB)πD]=[000Dd2(CB)π]M,(50)QPπ=[00CAπD-CAdB]=[00(CB)πC(CB)πD],MPπ=[AAπB-AAdBCAπD-CAdB]=[AAπBAπ(CB)πC(CB)πD]=[Aπ00(CB)π]M,
and then, by Lemma 4, Ind(QPπ)≤Ind[(CB)πD]+1=s, and Ind[(P+Q)Pπ]≤Ind[[Aπ00(CB)π]M]=k.
By Lemma4, Ind(P)≤Ind(A)+1=t. Further, by (43), Ind[(P+I)PPd]≤Ind[(A+I)AAd]+1=l.
Hence, putting (45)~(49) into (39) yields (40).


## 4. Example

In this section, we present a numerical example to demonstrate Theorem7.Example 1.
TakingP,Q as follows:
(51)P=[0.13210.84590.2893-0.0597-0.28300.08020.09430.3066-0.33960.14621.11320.6179-0.62260.97640.20750.6745],Q=[0.54721.52830.1509-0.50940.00001.0000-0.000000.1132-0.38210.96230.1274-0.45281.52830.15090.4906],
we can get
(52)Pd=[0.28300.00310.2390-0.2233-0.28300.08020.09430.30660.84910.00940.7170-0.6698-0.84910.24060.28300.9198],Pπ=[0.9057-0.0566-0.30190.01890.28300.9198-0.0943-0.3066-0.2830-0.16980.09430.05660.8491-0.2406-0.28300.0802],Qd=[0.54721.52830.1509-0.50940.00001.0000-0.000000.1132-0.38210.96230.1274-0.45281.52830.15090.4906],Qπ=[0.4528-1.5283-0.15090.50940000-0.11320.38210.0377-0.12740.4528-1.5283-0.15090.5094],
and Ind(QPπ)=1, Ind(P)=2, Ind[(P-I)PPd]=2, and Ind[(P-Q)Pπ]=1. By Theorem 7, we have
(53)(P-Q)d=[-0.2264-0.73580.07550.2453-0.2830-0.91980.09430.30660.11320.3679-0.0377-0.1226-0.1698-0.55190.05660.1840].



---

*Source: 479260-2013-09-11.xml*