# Some Results on Characterizations of Matrix Partial Orderings

Authors: Hongxing Wang; Jin Xu
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

Some characterizations of the left-star, right-star, and star partial orderings between matrices of the same size are obtained. Based on those results, several characterizations of the star partial ordering between EP matrices are given. At last, one characterization of the sharp partial ordering between group matrices is obtained.


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## Body

## 1. Introduction

In this paper we use the following notation. LetCm×n be the set of complex m×n matrices. For any matrix A∈Cm×n, A*, R(A), and r(A) denote the conjugate transpose, the range, and the rank of A, respectively. The symbol In denotes the n×n identity matrix, and 0 denotes a zero matrix of appropriate size. The Moore-Penrose inverse of a matrix A∈Cm×n, denoted by A†, is defined to be the unique matrix X∈Cn×m satisfying the four matrix equations
(1)(1)AXA=A,(2)XAX=X,(3)(AX)*=AX,(4)(XA)*=XA,
and  A- denotes any solution to the matrix equation AXA=A with respect to X; A{1} denotes the set of A-; that is, A{1}={X∣AXA=A}. Moreover, A# denotes the group inverse of A with r(A2)=r(A), that is, the unique solution to
(2)(1)AXA=A,(2)XAX=X,(5)AX=XA.
It is well known that A# exists if and only if r(A2)=r(A), where case A is also called a group matrix. A matrix A is EP if and only if A is a group matrix with A#=A†. The symbols CGPn and CEP⁡n stand for the subset of Cn×n consisting of group matrices and EP matrices, respectively (see, e.g., [1, 2] for details).Five matrix partial orderings defined inCm×n are considered in this paper. The first of them is the minus partial ordering defined by Hartwig [3] and Nambooripad [4] independently in 1980:
(3)A≤B⟺A-A=A-B,AA==BA=,
where A-,A=∈A{1}. In [3] it was shown that
(4)A≤B⟺r(B-A)=r(B)-r(A).
The rank equality indicates why the minus partial ordering is also called the rank-subtractivity partial ordering. In the same paper [3] it was also shown that
(5)A≤*B⟺r[AB]=r[AB]=r(B),AB-A=A,
where B-∈B{1}.The second partial ordering of interest is the star partial ordering introduced by Drazin [5], which is determined by
(6)A≤*B⟺A†A=A†B,AA†=BA†.
It is well known that
(7)A≤*B⟺A*A=A*B,AA*=BA*.In 1991, Baksalary and Mitra [6] defined the left-star and right-star partial orderings characterized as
(8)A≤*B⟺A*A=A*B,R(A)⊆R(B),A≤*B⟺AA*=BA*,R(A*)⊆R(B*).The last partial ordering we will deal with in this paper is the sharp partial ordering, introduced by Mitra [7] in 1987, and is defined in the set CGPn by
(9)A≤#B⟺A#A=A#B,AA#=BA#.
A detailed discussion of partial orderings and their applications can be found in [1, 8–10].It is well known that rank of matrix is an important tool in matrix theory and its applications, and many problems are closely related with the ranks of some matrix expressions under some restrictions (see [11–15] for details). Our aim in this paper is to characterize the left-star, right-star, star, and sharp partial orderings by applying rank equalities. In the following, when A is considered below B with respect to one partial ordering, then the partial ordering should entail the assumption r(A)>r(B)≥1.


## 2. The Star Partial Ordering

LetA and B be m×n complex matrices with ranks a and b, respectively. Let A≤*B. Then there exist unitary matrices U∈Cm×m and V∈Cn×n such that
(10)U*AV=(Da000),U*BV=(Da000D0000),
where both the a×a matrix Da and the (b-a)×(b-a) matrix D are real, diagonal, and positive definite (see [16, Theorem 2]). In [1, Theorem 5.2.8], it was also shown that
(11)A≤*B⟺A†A=B†A,AA†=AB†.
In [17], Wang obtained the following characterizations of the left-star and right-star partial orderings for matrices:
(12)A≤*B⟺r[B*BA*ABA]=r(B),(13)A≤*B⟺r[BB*AA*B*A*]=r(B),(14)A≤*B⟺r[B*BA*ABA]=r(B),r[BB*AA*B*A*]=r(B).Theorem 1.
LetA,B∈Cm×n. Then (i)
(15)A≤*B⟺r[BB†AA†B*A*]=r(B);(ii)
(16)A≤*B⟺r[B†BA†ABA]=r(B);(iii)
(17)A≤*B⟺r[B†BA†AB*BA*ABA]=r(B);(iv)
(18)A≤*B⟺r[BB†AA†BB*AA*B*A*]=r(B).Proof.
From(19)r[BB†AA†B*A*]≥r([BB†AA†B*A*][B00A])=r[BAB*BA*A]≥r([BAB*BA*A][B†00A†])=r[BB†AA†B*A*],
we have
(20)r[B*BA*ABA]=r[BB†AA†B*A*].
Applying (12) gives (i).
In the same way, applying(21)r[B†BA†ABA]=r[BB*AA*B*A*]
and (13) gives (ii).
If(22)r[B†BA†AB*BA*ABA]=r(B),
then
(23)r[BB*AA*B*A*]=r(B),r[BB†AA†B*A*]=r(B).
Applying (i), (ii), and (14), we obtain A≤*B. Conversely, if A≤*B, by using (11) and (14), we have A†A-B†A=0, and
(24)r[B*BA*ABA]=r[0A†A-B†AB*BA*ABA]=r([In0B†0In000Im][0A†A-B†AB*BA*ABA])=r[B†BA†AB*BA*ABA],r(B)=r[B†BA†AB*BA*ABA].
Hence, we have (iii).
Similarly, applyingA≤*B, (11), and (14), we obtain AA†-AB†=0,   AB†=(AB†)*=(B*)†A*, and
(25)r[BB*AA*B*A*]=r[0AA†-AB†BB*AA*B*A*]=r([Im0(B*)†0In000In][0AA†-AB†BB*AA*B*A*])=r(B).
Then, we obtain (iv).In [9, Theorem 2.1], Benítez et al. deduce the characterizations of the left-star, right-star, and star partial orderings for matrices, when at least one of the two involved matrices is EP. When both A∈Cn×n and B∈Cn×n are EP matrices, [1, Theorems 5.4.15 and 5.4.2] give the following results:
(26)A≤*B⟺A≤B,AB*andB*AareHermitian.A≤*B⟺(AB)†=B†A†=A†B†=A†2.
In addition, it was also shown that A≤*B if and only if A and B have the form
(27)A=U[T00000000]U*,B=U[T000K0000]U*,
where T∈Cr(A)×r(A) is nonsingular, K∈C(r(B)-r(A))×(r(B)-r(A)) is nonsingular, and U∈Cn×n is unitary (see [1, Theorem 5.4.1]).Based on these results, we consider the characterizations of the star partial ordering for matrices in the set ofCEP⁡n.Theorem 2.
LetA,B∈CEP⁡n, r(B)≥r(A). Then(v)
(28)A≤*B⟺r[BAB2A2]=r(B);(vi)
(29)A≤*B⟺r[BB2AA2]=r(B).Proof.
ByA,B∈CEP⁡n, it is obvious that AA†=A†A and BB†=B†B. Then
(30)r[BAB2A2]=r(B)⟺r[B†BA†ABA]=r(B).
Hence, we have (v).
The proof of (vi) is similar to that of (v).Theorem 3.
LetA,B∈CEP⁡n. Then (vii)
(31)A≤*B⟺r[BB†AA†BAB*A*]=r(B);(viii)
(32)A≤*B⟺r[B*BA*ABAB*A*]=r(B);(ix)
(33)A≤*B⟺r[BBABABAAB]=r(B);(x)
(34)A≤*B⟺r[BBA†BA†BAA†B]=r(B);(xi)
(35)A≤*B⟺r[BBA*BA*BAA*B]=r(B).Proof.
ByA,B∈CEP⁡n, it is obvious that AA†=A†A and BB†=B†B. Applying (i), (ii), and the rank equality in (vii) we obtain
(36)r[B†BA†ABA]=r(B),r[BB†AA†B*A*]=r(B);
that is, A≤*B. Conversely, suppose that A≤*B. Applying A-AA†B=0 and B*BB†=B*, we obtain
(37)r(B)=r[BB†AA†BA]=r[BB†AA†BA0A*-B*AA†]=r[BB†AA†BAB*A*].
Applying (11), we obtain B*BB†B=B*B and B*BA†A=A*A and also (B*B)†B*B=B†B and (B*B)†A*A=A†A. Then
(38)r[BB†AA†BAB*A*]=r[B†BA†ABAB*A*]≥r([B*B000In000In][B†BA†ABAB*A*])=r[B*BA*ABAB*A*]≥r([(B*B)†000In000In][B*BA*ABAB*A*])=r[B†BA†ABAB*A*];
that is,
(39)r[BB†AA†BAB*A*]=r[B*BA*ABAB*A*].
Hence, we have (viii).
Suppose thatA≤*B. Since A,B∈CEP⁡n, applying (27), it is easy to check the rank equality in (ix). Conversely, under the rank equality in (ix), we have
(40)r[BBABAB]=r[BBA0AB-BA]=r(B)⟹AB=BA,r[BBAAAB]=r[B0AAB-AA2]=r(B)⟹AB=A2.
Since A is EP, there exists a unitary matrix U1∈Cn×n and a nonsingular matrix T∈Cr(A)×r(A) such that
(41)A=U1[T000]U1*.
Correspondingly denote P-1BP by
(42)B=U1[B1B2B3B4]U1*,
where B1∈Cr(A)×r(A). It follows that
(43)[TB1TB200]=[B1T0B3T0],[TB1TB200]=[T2000].
Since T is a unitary matrix,
(44)B1=T,B2=0,B3=0.
Thus
(45)B=U[T00B4]U*.
Since B is EP, B4 is EP, and there exists a unitary matrix U2∈C(n-r(A))×(n-r(A)) and a nonsingular matrix K∈C(r(B)-r(A))×(r(B)-r(A)) such that
(46)B4=U2[K000]U2*.
Write
(47)U=U1[000U2].
Then A and B have the form
(48)A=U[T00000000]U*,B=U[T000K0000]U*.
Applying (27), we have A≤*B.
The proofs of (x) and (xi) are similar to that of (ix).


## 3. The Sharp Partial Ordering

LetA,B∈CGPn with ranks a and b, respectively. It is well known that
(49)A≤#B⟺A2=AB=BA.
In addition, A≤#B if and only if A and B can be written as
(50)A=P[E00000000]P-1,B=P[E000E′0000]P-1,
where E∈Ca×a is nonsingular, E′∈C(b-a)×(b-a) is nonsingular, and P∈Cn×n is nonsingular (see [18]).In Theorem4, we give one characterization of the sharp partial ordering by using one rank equality.Theorem 4.
LetA,B∈CGPn. Then
(51)A≤#B⟺r[ABAABABA]=r(ABA).Proof.
LetA have the core-nilpotent decomposition (see [19, Exercise 5.10.12])
(52)A=P[Σ000]P-1,
with nonsingular matrices Σ∈Cr(A)×r(A) and P∈Cn×n. Correspondingly denote P-1BP by
(53)P-1BP=[B1B2B3B4],
where B1∈Cr(A)×r(A). It follows that
(54)r(ABA)=r(ΣB1Σ),r[ABAABABA]=r[Σ0B1Σ00B3ΣΣB1ΣB2ΣB1Σ]=r[Σ0000B3Σ0ΣB2ΣB1Σ-ΣB1Σ-1B1Σ]=r(Σ)+r[0B3ΣΣB2ΣB1Σ-ΣB1Σ-1B1Σ].
Applying (54) to the rank equality in (51), we obtain
(55)r[0B3ΣΣB2ΣB1Σ-ΣB1Σ-1B1Σ]+r(Σ)=r(ΣB1Σ).
Hence r(ΣB1Σ)=r(Σ), ΣB2=0, B3Σ=0, and ΣB1Σ=ΣB1Σ-1B1Σ. Since Σ∈Cr(A)×r(A) is invertible and B1∈Cr(A)×r(A), it follows immediately that
(56)r(B1)=r(Σ),B3=0,B2=0,B1=Σ.
Therefore
(57)B=P[Σ00B4]P-1.
Applying
(58)A2=P[Σ2000]P-1=P[Σ000]P-1P[Σ00B4]P-1=AB=P[Σ00B4]P-1P[Σ000]P-1=BA,
and (49), we obtain that A≤#B.
Conversely, it is a simple matter.



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