# The Expression of the Drazin Inverse with Rank Constraints

Authors: Linlin Zhao
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

By using the matrix decomposition and the reverse order law, we provide some new expressions of the Drazin inverse for any2×2 block matrix with rank constraints.


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## Body

## 1. Introduction

LetA be a square complex matrix. The symbols r(A) and A† stand for the rank and the Moore-Penrose inverse of the matrix A, respectively. The Drazin inverse AD of A is the unique matrix satisfying
(1.1)Ak+1AD=Ak,ADAAD=AD,AAD=ADA,
where k=ind(A) is the index of A, the smallest nonnegative integer such that r(Ak+1)=r(Ak). We write Aπ=I-AAD.The Drazin inverse of a square matrix plays an important role in various fields like singular differential equations and singular difference equations, Markov chains, and iterative methods.The problem of finding explicit representations for the Drazin inverse of a complex block matrix,(1.2)M=(ABCD)
in terms of its blocks was posed by Campbell and Meyer [1, 2] in 1979. Many authors have considered this problem and have provided formulas for MD under some specific conditions [3–6].In this paper, under rank constraints, we will present some new representations ofMD which have not been discussed before.


## 2. Preliminary

Lemma 2.1 (see [4]).
LetP and Q be square matrices of the same order.
IfPQ=0, then
(2.1)(P+Q)D=Qπ[∑i=0k-1‍Qi(PD)i]PD+QD[∑i=0k-1‍(QD)iPi]Pπ,
where max{ind(P),ind(Q)}≤k≤ind(P)+ind(Q).
IfPQ=0 and QP=0, then (P+Q)D=PD+QD.Lemma 2.2 (see [7]).
Let(2.2)M1=(A0CB),M2=(BC0A),
where A, B are square matrices with ind(A)=r, ind(B)=s. Then
(2.3)M1D=(AD0XBD),M2D=(BDX0AD),
where
(2.4)X=(BD)2[∑i=0r-1‍(BD)iCAi]Aπ+Bπ(∑i=0s-1‍BiC(AD)i)(AD)2-BDCAD.Lemma 2.3 (see [8]).
LetA=A1A2⋯An, X=AnDAn-1D⋯A1D. Then X=AD if and only if A1,A2,…,An, A satisfy
(2.5)r((-1)nA2k+1AkE-E-TAkN-)=r(Ak)+r(A1k)+⋯+r(Ank),
where Ai∈Cm×m,i=1,2,…,n, E-=(0,…,0,Im)∈Cm×m(n+1) and
(2.6)N-=(A12k+1A1k⋮⋮An2k+1AnkAn-1kAnk),
with k=max{ind(Ai),ind(A)},1≤i≤n.Lemma 2.4 (see [9]).
LetA∈Cm×n,B∈Cm×k,C∈Cl×n. Then
(2.7)r(ABC0)=r(B)+r(C)+r[(Im-BB†)A(In-C†C)].


## 3. Main Results

In this section, with rank equality constraints, we consider the Drazin inverse of block matrices.LetM=(ABCD), where A∈Cm×m is invertible and D-CA-1B∈Cn×n is singular. It is easy to verify that M can be decomposed as
(3.1)M=(Im0CA-1In)(A00D-CA-1B)(ImA-1B0In)=M1M2M3.
Let
(3.2)E=(0,0,0,Im+n),N=(M12k+1M1kM22k+1M2kM1kM32k+1M3kM2kM3k),
where k=max{ind(M2),ind(M)}. According to Lemma 2.3, we have the following theorem.Theorem 3.1.
LetM=(ABCD), where A∈Cm×m is invertible and D-CA-1B∈Cn×n is singular. If
(3.3)r(Mk)+r(M2kM1-1FMk00EMkM3-1M2k)=r(M2k),
where k=max{ind(M2),ind(M)}, EMk=I-Mk(Mk)†, FMk=I-(Mk)†Mk, then MD has the following form:
(3.4)MD=(AD+A-1B(D-CA-1B)DCA-1-A-1B(D-CA-1B)D-(D-CA-1B)DCA-1(D-CA-1B)D).Proof.
From Lemma2.3 and (3.1), we know that if
(3.5)r(-M2k+1MkEETMkN)=r(Mk)+r(M1k)+r(M2k)+r(M3k),
where k=max{ind(M2),ind(M)}, then
(3.6)MD=M3-1M2DM1-1=(Im-A-1B0In)(AD00(D-CA-1B)D)(Im0-CA-1In)=(AD+A-1B(D-CA-1B)DCA-1-A-1B(D-CA-1B)D-(D-CA-1B)DCA-1(D-CA-1B)D).
Note that
(3.7)r(-M2k+1MkEETMkN)=r(-M2k+1000Mk000M12k+1M1k00M22k+1M2kM1k00M32k+1M3kM2k00MkM3k000)=r(0Mk+1M3k00Mk000M12k+1M1k00M22k+1M2kM1k00M32k+1M3kM2k00MkM3k000)=r(0000Mk0-M1kMM3k0M12k+1M1k00M22k+1M2kM1k00M32k+1M3kM2k00MkM3k000).
Let
(3.8)G=(0,0,0,Mk),J=(000Mk),H=(-M1kMM3k0M12k+1M1k0M22k+1M2kM1k0M32k+1M3kM2k00M3k000).
From Lemma 2.4, we have
(3.9)r(0GJH)=r(HJG0)=r(J)+r(G)+r[(I4m-JJ†)H(I4m-G†G)]=2r(Mk)+r[(I4m-JJ†)H(I4m-G†G)].
Note thatJ†=(0,0,0,(Mk)†), G†=(000(Mk)†). Then we get
(3.10)(I4m-JJ†)H(I4m-G†G)=(-M1kMM3k0M12k+1M1k[I4m-(Mk)†Mk]0M22k+1M2kM1k0M32k+1M3kM2k00[I4m-Mk(Mk)†]M3k000).
LetEMk=I4m-Mk(Mk)†, F=I4m-(Mk)†Mk. Then (I4m-JJ†)H(I4m-G†G) can be rewritten as the following three matrix products:
(3.11)(M1k0000Im0000M3k0000Im)(-M0M1FMk0M22k+1M2k0M3M2k00EMk000)(M3k0000Im0000M1k0000Im).
SinceM1,M3 is nonsingular, then
(3.12)r[(I4m-JJ†)H(I4m-G†G)]=r(-M0M1FMk0M22k+1M2k0M3M2k00EMk000)=r(-M0Im+nFMk0M22k+1M2kM1-10Im+nM3-1M2k00EMk000)=r(0M1M2k+1Im+nFMk0M22k+1M2kM1-10Im+nM3-1M2k000-EMkM3-1M2k00)=r(0M1M2k+1Im+nFMk000-M2kM1-1FMkIm+nM3-1M2k000-EMkM3-1M2k00)=r(00Im+n0000-M2kM1-1FMkIm+n0000-EMkM3-1M2k00)=2(m+n)+r(M2kM1-1FMk00EMkM3-1M2k).
Thus, we have
(3.13)r(-M2k+1MkEETMkN)=r(0GJH)=2r(Mk)+2(m+n)+r(M2kM1-1FMk00EMkM3-1M2k).
From the above equality and the condition (3.3), (3.5) is easily verified.LetM=(ABCD), where A∈Cm×m, D∈Cn×n. It is easy to verify that the matrix M can be decomposed as
(3.14)M=(Im0CADIn)(AAπBCAπS)(ImADB0In)=A1A2A3,
where S=D-CADB is the generalized Schur complement of A in M.Let(3.15)N1=(A12k+1A1kA22k+1A2kA1kA32k+1A3kA2kA3k),
where k=max{ind(A2),ind(M)}. Then we have the following theorem.Theorem 3.2.
IfAAπB=0,CAπB=0, and the matrices A1,A2,A3, M satisfy
(3.16)r(Mk)+r(A2kA1-1FMk00EMkA3-1A2k)=r(A2k),
then,
(3.17)MD=(AD+AπBXAD+(AπBSD-ADB)YAπB(SD)2-ADBSDYSD),
where S=D-CADB, Y=(SD)2[∑i=0ind(A)-1‍(SD)iCAπAi]Aπ-SDCAD.Proof.
From Lemma2.3 and (3.14), we get that if
(3.18)r(-M2k+1MkEETMkN1)=r(Mk)+r(A1k)+r(A2k)+r(A3k),
then
(3.19)MD=A3-1A2DA1-1=(Im-ADB0In)(AAπBCAπS)D(Im0-CADIn).
Similar to the proof of Theorem3.1, we derive that the rank condition (3.18) can be simplified as (3.16).
Next, we will give the representation forA2D. Let
(3.20)A2=(A0CAπS)+(0AπB00)=P+Q.
SinceAAπB=0, CAπB=0, and AπAπ=Aπ, then PQ=0. From Lemma 2.2, we get
(3.21)QD=0,PD=(AD0XSD),whereX=(SD)2[∑i=0ind(A)-1‍(SD)iCAπAi]Aπ.
From Lemma2.1 and the fact Qi=0,i≥2, it follows that
(3.22)A2D=(P+Q)D=(Im+n-QQD)(Im+n+QPD)PD=[Im+n+(0AπB00)(AD0XSD)](AD0XSD)=(Im+AπBXAπBSD0In)(AD0XSD)=(AD+AπB(XAD+SDX)AπB(SD)2XSD).
Substituting A2D in (3.19), the conclusion can be obtained.From Theorem3.2, we can easily obtain the following corollaries.Corollary 3.3.
IfAπB=0 and the rank equality (3.16) hold, then
(3.23)MD=(AD-ADB(X-SDCAD)-ADBSDX-SDCADSD),
where S and X are the same as in Theorem 3.2.Corollary 3.4.
IfAπB=0,CAπ=0, and the rank equality (3.16) hold, then
(3.24)MD=(AD+ADBSDCAD-ADBSD-SDCADSD),
where S=D-CADB.Next, we will consider another decomposition ofM involving the generalized Schur complement SB=C-DBDA.LetM=(ABCD), where A,D∈Cm×m. Then M can be decomposed as
(3.25)M=(Im0DBDIm)(BπABSBDBπ)(Im0BDAIm)=V1V2V3,
where SB=C-DBDA is the generalized Schur complement of B in M.Let(3.26)E1=(0,0,0,I2m),N2=(V12k+1V1kV22k+1V2kV1kV32k+1V3kV2kV3k),
where k=max{ind(V2),ind(M)}. Then we have the following theorem.Theorem 3.5.
LetM=(ABCD), where A,D∈Cm×m. If BπA=0, CB=DBDAB and the matrices V1,V2,V3, M satisfy the following rank equality:
(3.27)r(Mk)+r(V2kV1-1FMk00EMkV3-1V2k)=r(V2k),
then
(3.28)MD=(BX2(XSB-DBD)BX2(I-BDABX)X(XSB-DBD)(I-BDABX)X),
where X=(DBπ)D.Proof.
From Lemma2.3 and (3.25), we get that if the following rank condition
(3.29)r(-M2k+1MkEETMkN2)=r(Mk)+r(V1k)+r(V2k)+r(V3k)
holds, then M=V3-1V2DV1-1. From the same method used in Theorem 3.1, we can verify that the above condition (3.29) can be reduced to (3.27).
Next, we will give the representation forV2D. For BπA=0, we write
(3.30)V2=(00SBDBπ)+(0B00)=P1+Q1,
where SB=C-DBDA.
From the conditionCB=DBDAB, we get P1Q1=0, according to Lemma 2.2, we have
(3.31)Q1D=0,P1D=[00((DBπ)D)2SB(DBπ)D].
Let X=(DBπ)D. By Lemma 2.1 and the fact Q1i=0,i≥2, we get
(3.32)V2D=(I2m-Q1Q1D)(I2m+Q1P1D)P1D=[I2m+(0B00)(00X2SBX)](00X2SBX)=(Im+BX2SBBX0Im)(00X2SBX)=(BX3SBXBX2X2SBX).
Therefore, we get
(3.33)MD=(Im0-BDAIm)(BX3SBXBX2X2SBX)(Im0-DBDIm)=(BX2(XSB-DBD)BX2(I-BDABX)X(XSB-DBD)(I-BDABX)X).Remark 3.6.
In addition to the decompositions ofM in (3.14) and (3.25), the matrix M also can be decomposed as other matrix products involving the generalized Schur complements SC=B-ACDD or SD=A-BDDC. In these cases, new formulas for MD would be derived by the method used in this paper.


## 4. Conclusion

In this paper, we mainly discuss the Drazin inverse of block matrices under rank equality constraints. Comparing with the existing results, it is obvious that our results have more strong restrictions, but the methods used in this paper are different from those in previous relevant paper.



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