# A Note on the Drazin Indices of Square Matrices

Authors: Lijun Yu; Tianyi Bu; Jiang Zhou
Journal: The Scientific World Journal
 (2014)

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## Abstract

For a square matrixA, the smallest nonnegative integer k such that rank (Ak) =
rank (Ak+1) is called the Drazin index of A. In this paper, we give some results on the Drazin indices of sum and product of square matrices.


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## Body

## 1. Introduction

ForA∈ℂn×n, the smallest nonnegative integer k such that rank⁡(Ak)=rank⁡(Ak+1) is called the Drazin index of A, denoted by ind⁡(A). The Drazin inverse of A is the unique matrix AD satisfying Ak+1AD=Ak,ADAAD=AD, and AAD=ADA, where k=ind(A) (see [1–3]). When ind⁡(A)⩽1, AD is called the group inverse of A, denoted by A#. If ind⁡(A)=0, then AD=A#=A-1. The theory of generalized inverses is an active research field in computational mathematics, and the Drazin index plays an important role in the study of the Drazin (group) inverse. Some results on the Drazin indices of matrices (operators) can be found in [4–9].For anyA∈ℂn×n, it is known [10] that there exist nonsingular matrices P,Δ and nilpotent matrix N such that A=P(Δ00N)P-1. In this case, AD=P(Δ-1000)P-1, and ind⁡(A) is the smallest nonnegative integer k such that Nk=0; that is, ind⁡(A)=ind⁡(N). Hence, ind⁡(A) is the smallest nonnegative integer k such that the group inverse of Ak exists.In this paper, we give some results on the Drazin indices of sum and product of square matrices.


## 2. Some Lemmas

In order to prove our main results, we give some lemmas as follows.Lemma 1 (see [7]).
For anyA∈ℂn×n and nonnegative integer p, the limit  lim⁡ε→0+εp(A+εI)-1 exists if and only if p⩾ind(A).Lemma 2 (see [8, 11]).
LetM=(AB0C) be a square complex matrix, where A is square. Then,
(1)max⁡{ind(A),ind(C)}⩽ind(M)⩽ind(A)+ind(C).Lemma 3 (see [8, 11]).
LetM=(A0BC) be a square complex matrix, where A is square. Then,
(2)max⁡{ind(A),ind(C)}⩽ind(M)⩽ind(A)+ind(C).Lemma 4 (see [12]).
LetM=(AB00) be a square complex matrix, where A is square. Then, M# exists if and only if A# exists and rank⁡(M)=rank⁡(A).Lemma 5 (see [12]).
LetM=(A0BC) be a square complex matrix, where A is square. Then, M# exists if and only if A#,C# exist and rank⁡(M)=rank⁡(A)+rank⁡(C).Lemma 6 (see [9]).
For anyA∈ℂm×n,B∈ℂn×m, |ind(AB)-ind(BA)|⩽1.


## 3. Main Results

For a square matrixA, let Aπ=I-AAD. We first give an upper bound for the Drazin index of the sum of two square matrices.Theorem 7.
LetP,Q be square complex matrices such that PDQ=0 and PQPπ=0. Then,
(3)ind(P+Q)⩽ind(P)+ind(Q)-1.Proof.
There exist nonsingular matricesU,Δ and nilpotent matrix N such that P=U(Δ00N)U-1. Then, PD=U(Δ-1000)U-1 and ind(P)=ind(N). Suppose that Q=U(Q1Q2Q3Q4)U-1, where Q1 is a square matrix with the same order as Δ. By PDQ=0 we get Q1=0, Q2=0. Then, Q=U(00Q3Q4)U-1. By PQPπ=0 we get NQ4=0.
Note thatind(P+Q) is the smallest nonnegative integer k such that the group inverse of (P+Q)k exists. Since P+Q=U(Δ0Q3N+Q4)U-1, by Lemma 5, we have ind(P+Q)=ind(N+Q4). Since NQ4=0, we have ε(N+Q4+εI)=(N+εI)(Q4+εI). Let m=ind(Q4)+ind(N)-1. By Lemma 1, the limit
(4)lim⁡ε→0+εm(N+Q4+εI)-1=lim⁡ε→0+εm+1(Q4+εI)-1(N+εI)-1
exists. So ind(P+Q)=ind(N+Q4)⩽m=ind(Q4)+ind(N)-1. By Lemma 3, we get ind(Q4)⩽ind(Q). Since ind(P)=ind(N), we have ind(P+Q)⩽ind(P)+ind(Q)-1.Now we give an example to show that the upper bound in Theorem7 can be attained.Example 8.
LetP=(1-110000200010000), Q=(1-12-21-12-21-11-11-11-1). Then, ind(P)=ind(Q)=2 and PD=(1-11-1000000000000). By computation, we have PDQ=0 and PQPπ=0. The Drazin index of P+Q=(2-23-21-1201-1101-11-1) is 3. So ind(P+Q)=ind(P)+ind(Q)-1.
We can obtain the following result from Theorem7.Corollary 9.
LetP,Q be square complex matrices such that PQ=0. Then,
(5)ind(P+Q)⩽ind(P)+ind(Q)-1.It is known thatind(ABCD)⩽ind(A)+ind(D)+2 under the conditions CA=0, CB=0 or BC=0, DC=0 (see [13]). A better upper bound is given as follows.Theorem 10.
LetM=(ABCD) be a square complex matrix, where A is square. Then, ind(M)⩽ind(A)+ind(D)+1 if one of the following holds:(1)
BC=0, DC=0;(2)
BC=0, BD=0;(3)
CA=0, CB=0;(4)
AB=0, CB=0.Proof.
We only prove part (1). Parts (2)–(4) can be obtained in the same way.
LetP=(AB0D), Q=(00C0); then, M=P+Q. By BC=0, DC=0, we get PQ=0. By Corollary 9, we have
(6)ind(M)⩽ind(P)+ind(Q)-1⩽ind(P)+1.
By Lemma 2, we have (7)ind(M)⩽ind(A)+ind(D)+1.Theorem 11.
ForA,B∈ℂn×n, if s⩾max⁡{ind(AB),ind(BA)}, then (AB)s and (BA)s are similar.Proof.
Clearly(AB)0 and (BA)0 are similar. So we only consider the case s⩾1. There exist nonsingular matrices P,Q such that A=P(Ir000)Q, where Ir is the identity matrix of order r=rank⁡(A). Suppose that B=Q-1(B1B2B3B4)P-1, where B1∈ℂr×r. Then,
(8)AB=P(B1B200)P-1,(AB)s=P(B1sB1s-1B200)P-1,BA=Q-1(B10B30)Q,(BA)s=Q-1(B1s0B3B1s-10)Q.
Since s⩾max⁡{ind(AB),ind(BA)}, the group inverses of (AB)s and (BA)s both exist. Lemmas 4 and 5 imply that there exist matrices X,Y such that B1sX=B1s-1B2 and YB1s=B3B1s-1. Then,
(9)(AB)s=P(B1sB1sX00)P-1=P(I-X0I)(B1s000)(IX0I)P-1,(BA)s=Q-1(B1s0YB1s0)Q=Q-1(I0YI)(B1s000)(I0-YI)Q.
Hence, (AB)s and (BA)s are similar.Corollary 12.
ForA,B∈ℂn×n, if ind(AB)=s, then (AB)l and (BA)l are similar for any l⩾s+1.Proof.
By Lemma6 and Theorem 11, (AB)l and (BA)l are similar for any l⩾s+1.The following corollary is a special case of Theorem11.Corollary 13.
ForA,B∈ℂn×n, if (AB)# and (BA)# both exist, then AB and BA are similar.



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*Source: 361349-2014-02-10.xml*