The reverse order laws for weighted generalized inverses often appear in linear algebra problems of several applied fields, having attracted considerable attention. In this paper, by using the maximal and minimal ranks of the generalized Schur complement, we obtained some necessary and sufficient conditions for the reverse order laws
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1}
and
A3{1,2,4N4}A2{1,2,4N3}A1{1,2,4N2}⊆(A1A2A3){1,2,4N4}.
Citation: Baifeng Qiu, Yingying Qin, Zhiping Xiong. The reverse order laws for {1,2,3M}- and {1,2,4N}- inverse of three matrix products[J]. AIMS Mathematics, 2025, 10(1): 721-735. doi: 10.3934/math.2025033
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The reverse order laws for weighted generalized inverses often appear in linear algebra problems of several applied fields, having attracted considerable attention. In this paper, by using the maximal and minimal ranks of the generalized Schur complement, we obtained some necessary and sufficient conditions for the reverse order laws
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1}
and
A3{1,2,4N4}A2{1,2,4N3}A1{1,2,4N2}⊆(A1A2A3){1,2,4N4}.
At first, we will provide the following explanations of some of the notations for the convenience of readers.
Definition 1.1. Cm×n: the set of m×n complex matrices,
Ik: k-order identity matrix,
Om×n: zero matrix of order m×n,
R(A), N(A): the range space and the null space of A, respectively,
r(A), A∗: the rank and the conjugate transpose of A, respectively.
Definition 1.2. [1,2] Let A∈Cm×n, and let M∈Cm×m, N∈Cn×n be two positive definite Hermitian matrices. X is the weighted Moore-Penrose inverse of A when it satisfies
(1)AXA=A,(2)XAX=X,(3M)(MAX)∗=MAX,(4N)(NXA)∗=NXA, | (1.1) |
where X is denoted by X=A(1,2,3M,4N)=A†M,N.
For a given matrix A∈Cm×n, the sets A{1,2,3M}- and A{1,2,4N}− inverses of A are
A{1,2,3M}={X∈Cn×m|AXA=A,XAX=X,(MAX)∗=MAX},
A{1,2,4N}={X∈Cn×m|AXA=A,XAX=X,(NXA)∗=NXA};
more relevant theories can be found in [1,3].
The reverse order law for weighted generalized inverses is a key tool in the study of the weighted least squares problem, the weighted perturbation theory, optimization problems, and other related topics [4,5,6].
The reverse order laws for generalized inverses of matrix products are a class of interesting problems that are fundamental in the theory of generalized inverses [7,8,9,10]. In 1966, Greville [7] first gave an equivalent condition for the so-called reverse order law B†A†=(AB)†. Since then, many authors have studied this problem [11,12,13,14,15]. On studying the reverse order for any {i,j,⋯,k}-inverse of matrix products, one important relations problem is: under what conditions
A3{i,j,⋯,k}A2{i,j,⋯,k}A1{i,j,⋯,k}⊆(A1A2A3){i,j,⋯,k} |
holds, where {(i),(j),⋯,(k)}⊆{(1),(2),(3M),(4N)}.
In [16], some necessary and sufficient conditions were presented for the first times for several types of reverse order laws to hold for weighted generalized inverses. Since then, reverse order laws for weighted generalized inverses of matrix products have attracted considerable attention and some interesting results have been derived [17,18,19,20].
The purpose of this paper is to show some equivalent conditions for the following so-called reverse order laws
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1} |
and
A3{1,2,4N4}A2{1,2,4N3}A1{1,2,4N2}⊆(A1A2A3){1,2,4N4}, |
where Ai∈Cmi×mi+1,i=1,2,3, Mi∈Cmi×mi,i=1,2,3, Ni∈Cmi×mi,andi=2,3,4 are six positive definite Hermitian matrices.
The following lemmas are essential to the rest of this paper.
Lemma 1.1. [3] Let L, M be two complementary subspaces of Cm, and let PL,M be the projector on L along M, then
PL,MA=A⟺R(A)⊆L, | (1.2) |
APL,M=A⟺N(A)⊇M. | (1.3) |
Lemma 1.2. [1,3] Let A∈Cm×n, X∈Cn×m, and let M, N be two positive definite Hermitian matrices of order m and n, respectively, then
X∈A{1,2,3M}⟺A∗MAX=A∗Mandr(X)=r(A), | (1.4) |
X∈A{1,2,4N}⟺XAN−1A∗=N−1A∗andr(X)=r(A), | (1.5) |
X∈A{1,2,4N}⟺X∗∈A∗{1,2,3N−1}. | (1.6) |
Lemma 1.3. [21] Let A∈Cm×n, B∈Cm×k, C∈Cl×n, D∈Cl×k, and let M∈Cm×m, N∈Cn×n be two positive definite Hermitian matrices, then
maxA(1,2,3M) r(D−CA(1,2,3M)B)=min{ r(A∗MAA∗MBCD)−r(A),r(A∗MBD)}, | (1.7) |
minA(1,2,3M) r(D−CA(1,2,3M)B)=r(A∗MAA∗MBCD)+r(A∗MBD)−r(AOOA∗MBCD), | (1.8) |
maxA(1,2,4N) r(D−CA(1,2,4N)B)=min{ r(CN−1A∗,D),r(AN−1A∗BCN−1A∗D)−r(A)}, | (1.9) |
minA(1,2,4N) r(D−CA(1,2,4N)B)=r(CN−1A∗,D)+r(AN−1A∗BCN−1A∗D)−r(AOBOCN−1A∗D). | (1.10) |
Lemma 1.4. [22] Let A∈Cm×n, B∈Cm×k and C∈Cp×n, then
r(A,B)=r(A)+r(EAB)=r(B)+r(EBA), | (1.11) |
r(AC)=r(A)+r(CFA)=r(C)+r(AFC), | (1.12) |
r(AC)≤r(A)+r(C),r(A,B)≤r(A)+r(B), | (1.13) |
where the projectors EA=Im−AA†, EB=Im−BB†, FA=In−A†A, FC=In−C†C.
In this section, we will present some necessary and sufficient conditions for the reverse order laws for the weighted generalized inverses {1,2,3M}− and {1,2,4N}− of three matrix products. The following theorem is the main result in this section.
Theorem 2.1. Let Ai∈Cmi×mi+1, A(1,2,3Mi)i∈Ai{1,2,3Mi} and i∈{1,2,3}. Let Mi∈Cmi×mi, i∈{1,2,3} be three positive definite Hermitian matrices, then
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1}⟺r( A∗3OOA∗2A∗3A∗2A∗1M1A1A2M−13A∗3A∗2A∗1M1A1M−12)=r(A2)+r(A3)andr(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r( A∗2OOA∗3A1M−12A1A2M−13). | (2.1) |
Proof. According to the formula (1.4) in Lemma 1.2, for any A(1,2,3Mi)i∈Ai{1,2,3Mi}, i∈{1,2,3}, we can reach the conclusion that
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1} |
holds if and only if
A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1=A∗3A∗2A∗1M1and |
r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A1A2A3) |
holds, which are respectively equivalent to the following two identities
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=0 | (2.2) |
and
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=minA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A1A2A3). | (2.3) |
Using the formula (1.7) in Lemma 1.3 with A=A1, B=Im1, D=A∗3A∗2A∗1M1 and C=A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2, we have
maxA(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A∗1M1A1A∗1M1A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A∗3A∗2A∗1M1)−r(A1),r(A∗1M1A∗3A∗2A∗1M1)}=min{r(A∗3A∗2A∗1M1A1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2),r(A1)}=r(A∗3A∗2A∗1M1A1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2). | (2.4) |
From (2.4) and again by formula (1.7) in Lemma 1.3 with A=A2, B=Im2, D=A∗3A∗2A∗1M1A1, and C=A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3, we have
maxA(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=maxA(1,2,3M2)2r(A∗3A∗2A∗1M1A1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2)=min{r(A∗2M2A2A∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A∗3A∗2A∗1M1A1)−r(A2),r(A∗2M2A∗3A∗2A∗1M1A1)}. | (2.5) |
According to the formulas (1.12) and (1.13) of Lemma 1.4, we have
min{r(A∗2M2A2A∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A∗3A∗2A∗1M1A1)−r(A2),r(A∗2M2A∗3A∗2A∗1M1A1)}=min{r(OA∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)−r(A2),r(A∗2M2A∗3A∗2A∗1M1A1)} |
and
r(OA∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)−r(A2)≤r(A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2)+r(A∗2M2A∗3A∗2A∗1M1A1)−r(A2)≤r(A∗3A∗2A∗1M1A1A2(A3A(1,2,3M3)3−Im3))+r(A∗2M2A∗3A∗2A∗1M1A1)−r(A2)≤r(A2)+r(A∗2M2A∗3A∗2A∗1M1A1)−r(A2)=r(A∗2M2A∗3A∗2A∗1M1A1) | (2.6) |
and
r(OA∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)−r(A2)=r(OA∗2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12)−r(A2)=r(A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2). | (2.7) |
Combining (2.5), (2.6), and (2.7), we have
maxA(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A∗2M2A2A∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A∗3A∗2A∗1M1A1)−r(A2),r(A∗2M2A∗3A∗2A∗1M1A1)}=r(OA∗2M2A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)−r(A2)=r(A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2). | (2.8) |
From (2.8) and again by formula (1.7) in Lemma 1.3 with A=A3, B=(Im3,O), C=A∗3A∗2A∗1M1A1A2A3, and D=(A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2), we have
r(A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2)=r((A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2)−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3(Im3,O)) |
and
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=maxA(1,2,3M3)3r(A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3−A∗3A∗2A∗1M1A1A2,A∗3A∗2A∗1M1A1M−12FA∗2)=min{r(A∗3M3A3A∗3M3OA∗3A∗2A∗1M1A1A2A3A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12FA∗2)−r(A3),r(A∗3M3OA∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12FA∗2)}=min{r(OA∗3M3OOA∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12FA∗2)−r(A3),r(A∗3M3OA∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12FA∗2)}=r(A∗3M3OA∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1M−12FA∗2)−r(A3)=r(A∗3A∗2A∗1M1A1A2M−13FA∗3,A∗3A∗2A∗1M1A1M−12FA∗2), | (2.9) |
According to (2.8) and formula (1.12) of the Lemma 1.4, we have
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A∗3A∗2A∗1M1A1A2M−13FA∗3,A∗3A∗2A∗1M1A1M−12FA∗2)=r( A∗3OOA∗2A∗3A∗2A∗1M1A1A2M−13A∗3A∗2A∗1M1A1M−12)−r(A2)−r(A3). | (2.10) |
Combining (2.2) and (2.10), we have
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A∗3A∗2A∗1M1−A∗3A∗2A∗1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=0 |
if and only if
r( A∗3OOA∗2A∗3A∗2A∗1M1A1A2M−13A∗3A∗2A∗1M1A1M−12)=r(A2)+r(A3). | (2.11) |
In the rest of the section, we will find the equivalent conditions of (2.3). By Lemma 1.3(1.7) with A=A1, B=Im1, C=A(1,2,3M3)3A(1,2,3M2)2, and D=O, we have
maxA(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A∗1M1A1A∗1M1A(1,2,3M3)3A(1,2,3M2)2O)−r(A1),r(A∗1M1O)}=min{r(A(1,2,3M3)3A(1,2,3M2)2),r(A1)}. | (2.12) |
Form (2.12) and using Lemma 1.3(1.7) with A=A2, B=Im2, C=A(1,2,3M3)3, and D=O, we have
maxA(1,2,3M2)2,A(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{(maxA(1,2,3M2)2r(A(1,2,3M3)3A(1,2,3M2)2)),r(A1)}. | (2.13) |
From (2.13) and formula (1.7) of Lemma 1.3, we have
min{(maxA(1,2,3M2)2r(A(1,2,3M3)3A(1,2,3M2)2)),r(A1)}=min{(min{r(A∗2M2A2A∗2M2A(1,2,3M3)3O)−r(A2),r(A∗2M2O)}),r(A1)}=min{r(A(1,2,3M3)3),r(A2),r(A1)}. | (2.14) |
Combining (2.13) and (2.14), we have
maxA(1,2,3M2)2,A(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A(1,2,3M3)3),r(A2),r(A1)}. | (2.15) |
Since r(A(1,2,3M3)3)=r(A3), then
maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A(1,2,3M3)3),r(A2),r(A1)}=min{r(A3),r(A2),r(A1)}. | (2.16) |
On the other hand, according to (1.8) of Lemma 1.3 with A=A3, B=A(1,2,3M2)2A(1,2,3M1)1, C=Im4, and D=O, we have
minA(1,2,3M3)3r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A∗3M3A3A∗3M3A(1,2,3M2)2A(1,2,3M1)1Im4O)+r(A∗3M3A(1,2,3M2)2A(1,2,3M1)1O)−r(A3OOA∗3M3A(1,2,3M2)2A(1,2,3M1)1Im4O)=r(A∗3M3A(1,2,3M2)2A(1,2,3M1)1). | (2.17) |
From (2.17) and again by formula (1.8) in Lemma 1.3 with A=A2, B=A(1,2,3M1)1, C=A∗3M3, and D=O, we have
minA(1,2,3M2)2,A(1,2,3M3)3r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=minA(1,2,3M2)2r(A∗3M3A(1,2,3M2)2A(1,2,3M1)1)=r(A∗2M2A2A∗2M2A(1,2,3M1)1A∗3M3O)+r(A∗2M2A(1,2,3M1)1O)−r(A2OOA∗2M2A(1,2,3M1)1A∗3M3O)=r(A∗2M2A2M−13FA∗3,A∗2M2A(1,2,3M1)1)+r(A3)−r(A2A∗3M3). | (2.18) |
By formula (1.12) of Lemma 1.4, we have
r(A∗2M2A2A∗2M2A(1,2,3M1)1A∗3M3O)=r(A∗3M3OA∗2M2A2A∗2M2A(1,2,3M1)1)=r(A∗3OA∗2M2A2M−13A∗2M2A(1,2,3M1)1)=r(A∗2M2A2M−13FA∗3,A∗2M2A(1,2,3M1)1)+r(A3) | (2.19) |
and
r(A2OOA∗2M2A(1,2,3M1)1A∗3M3O)=r(A2A∗3M3)+r(A∗2M2A(1,2,3M1)1). | (2.20) |
Combining Eqs (2.19) and (2.20), we obtain Eq (2.18).
Using Eq (2.18) and formula (1.8) from Lemma 1.3 with (A=A1), (B=(O,−Im1)), (C=A∗2M2), and (D=(A∗2M2A2M−13FA∗3,O)), we derive
minA(1,2,3M1)1,A(1,2,3M2)2,A(1,2,3M3)3r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=(minA(1,2,3M1)1r(A∗2M2A2M−13FA∗3,A∗2M2A(1,2,3M1)1))+r(A3)−r(A2A∗3M3)=(minA(1,2,3M1)1r[(A∗2M2A2M−13FA∗3,O)−A∗2M2A(1,2,3M1)1(O,−Im1)])+r(A3)−r(A2A∗3M3)=r(A∗1M1A1O−A∗1M1A∗2M2A∗2M2A2M−13FA∗3O)+r(O−A∗1M1A∗2M2A2M−13FA∗3O)−r(A1OOOO−A∗1M1A∗2M2A∗2M2A2M−13FA∗3O)+r(A3)−r(A2A∗3M3)=r(A∗1M1A1A∗1M1A1A2M−13FA∗3A∗1M1A∗2M2OO)+r(A1)+r(A∗2M2A2M−13FA∗3)−r(A1)−r(A1OA∗2M2A∗2M2A2M−13FA∗3)+r(A3)−r(A2A∗3M3), | (2.21) |
where
r(A∗1M1A1A∗1M1A1A2M−13FA∗3A∗1M1A∗2M2OO)=r(A∗1M1A1M−12A∗1M1A1A2M−13FA∗3A∗1M1A∗2OO)=r(A2)+r(A∗1M1,A∗1M1A1M−12FA∗2,A∗1M1A1A2M−13FA∗3) | (2.22) |
and
r(A∗2M2A2M−13FA∗3)=r((M1/22A2)∗(M1/22A2)M−13FA∗3)≤r((M1/22A2)M−13FA∗3)=r((M1/22A2)(M1/22A2)†(M1/22A2)M−13FA∗3)=r((M1/22A2)((M1/22A2)∗M1/22A2)†(M1/22A2)∗(M1/22A2)M−13FA∗3)≤((M1/22A2)∗(M1/22A2)M−13FA∗3)=r(A∗2M2A2M−13FA∗3). | (2.23) |
By the formula (2.23), we have
r(A∗2M2A2M−13FA∗3)=r((M1/22A2)M−13FA∗3)=r(A2M−13FA∗3). | (2.24) |
By the formula (1.12) of Lemma 1.4, we have
r(A2A∗3M3)=r(A2M−13A∗3)=r(A3)+r(A2M−13FA∗3). | (2.25) |
Combining (2.21), (2.22), (2.24), and (2.25), we have
minA(1,2,3M1)1,A(1,2,3M2)2,A(1,2,3M3)3r(A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A2)+r(A∗1M1,A∗1M1A1M−12FA∗2,A∗1M1A1A2M−13FA∗3)−r(A1OA∗2M2A∗2M2A2M−13FA∗3)=r(A1)+r(A2)−r(A1OA∗2M2A∗2M2A2M−13FA∗3)=r(A1)+r(A2)−r(A1A1A2M−13FA∗3A∗2M2O)=r(A1)−r(A1M−12FA∗2,A1A2M−13FA∗3)=3∑i=1r(Ai)−r(A∗2OOA∗3A1M−12A1A2M−13). | (2.26) |
According to (2.2), (2.3), (2.11), (2.16), and (2.26), we have proved Theorem 2.1. ◻
From Lemma 1.1, Lemma 1.4, and Theorem 2.1, we immediately obtain the following corollary.
Corollary 2.1. Let Ai∈Cmi×mi+1, A(1,2,3Mi)i∈Ai{1,2,3Mi}, where i∈{1,2,3}. Let Mi∈Cmi×mi, i∈{1,2,3} be three positive definite Hermitian matrices. Then the following statements are equivalent:
(1)
A3{1,2,3M3}A2{1,2,3M2}A1{1,2,3M1}⊆(A1A2A3){1,2,3M1}; | (2.27) |
(2)
r(A∗3M3OOA∗2M2A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)=r(A2)+r(A3) | (2.28) |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(A∗2M2OOA∗3M3A1A1A2); | (2.29) |
(3)
R(A∗2A∗1M1A1A2A3)⊆R(M3A3), | (2.30) |
R(A∗1M1A1A2A3)⊆R(M2A2) | (2.31) |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(A∗2M2OOA∗3M3A1A1A2); |
(4)
A3(A3)†M3,Im4M−13A∗2A∗1M1A1A2A3=M−13A∗2A∗1M1A1A2A3, | (2.32) |
A2(A2)†M2,Im3M−12A∗1M1A1A2A3=M−12A∗1M1A1A2A3 | (2.33) |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(A∗2M2OOA∗3M3A1A1A2). |
Proof. According to Theorem 2.1, we have (1)⇔(2). Now, we will prove (3)⇒(2). By (2.30) and (2.31), we have R(M−13A∗2A∗1M1A1A2A3)⊆R(A3) and R(M−12A∗1M1A1A2A3)⊆R(A2). Using Lemma 1.1, we have
R(A∗2A∗1M1A1A2A3)⊆R(M3A3)andR(A∗1M1A1A2A3)⊆R(M2A2)⇒R(M−13A∗2A∗1M1A1A2A3)⊆R(A3)andR(M−12A∗1M1A1A2A3)⊆R(A2)⇒r(A3OM−13A∗2A∗1M1A1A2A3OA2M−12A∗1M1A1A2A3)=r(A2)+r(A3)⇒r(A∗3M3OOA∗2M2A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)=r(A2)+r(A3). |
Next, we will prove (2)⇒(3), that is, (2.28)⇒(2.30) and (2.31). According to (2.28) and the formula (1.11) of Lemma 1.4, we have
r(A∗3M3OOA∗2M2A∗3A∗2A∗1M1A1A2A∗3A∗2A∗1M1A1)=r(A2)+r(A3)⟺r(A3OM−13A∗2A∗1M1A1A2A3OA2M−12A∗1M1A1A2A3)=r(A2)+r(A3)⟺r(EA3M−13A∗2A∗1M1A1A2A3EA2M−12A∗1M1A1A2A3)+r(A2)+r(A3)=r(A2)+r(A3)⟺r(EA3M−13A∗2A∗1M1A1A2A3EA2M−12A∗1M1A1A2A3)=0 |
and
A3A†3M−13A∗2A∗1M1A1A2A3=M−13A∗2A∗1M1A1A2A3 |
and
A2A†2M−12A∗1M1A1A2A3=M−12A∗1M1A1A2A3. |
According to (1.2) of Lemma 1.1, we have (2)⇒ (3).
Finally, we will prove (3)⟺(4), that is (2.30)⇔(2.32) and (2.31)⇔(2.33). According to (1.2) of Lemma 1.1, we have
R(A∗2A∗1M1A1A2A3)⊆R(M3A3)⇔R(M−13A∗2A∗1M1A1A2A3)⊆R(A3) |
and
A3(A3)†M3,Im4M−13A∗2A∗1M1A1A2A3=PR(A3(A3)†M3,Im4),N(A3(A3)†M3,Im4)M−13A∗2A∗1M1A1A2A3=PR(A3),N(A3(A3)†M3,Im4)M−13A∗2A∗1M1A1A2A3=M−13A∗2A∗1M1A1A2A3 |
and
R(A∗1M1A1A2A3)⊆R(M2A2)⇔R(M−12A∗1M1A1A2A3)⊆R(A2) |
A2(A2)†M2,Im3M−12A∗1M1A1A2A3=PR(A2(A2)†M2,Im3),N(A2(A2)†M2,Im3)M−12A∗1M1A1A2A3=PR(A2),N(A2(A2)†M2,Im3)M−12A∗1M1A1A2A3=M−12A∗1M1A1A2A3. |
So we have that (1), (2), (3), and (4) are equivalent.
From Lemma 1.2, we have X∈A{1,2,4N} if and only if X∗∈A∗{1,2,3N−1}. Thus according to the results obtained in Theorem 2.1 and Corollary 2.1, we obtain the following theorem and corollary without any proof.
Theorem 2.2. Let Ai∈Cmi×mi+1, A(1,2,4Ni+1)i∈Ai{1,2,4Ni+1}, where i∈{1,2,3}. Let Ni∈Cmi×mi, i∈{1,2,3,4} be four positive definite Hermitian matrices. Then
A3{1,2,4N4}A2{1,2,4N3}A1{1,2,4N2}⊆(A1A2A3){1,2,4N4}⟺r( A∗1ON2A2A3N−14A∗3A∗2A∗1OA∗2N3A3N−14A∗3A∗2A∗1)=r(A1)+r(A2)andr(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r( A∗2ON3A3OA∗1N2A2A3). |
From Lemma 1.1, Lemma 1.4, and Theorem 2.2, we immediately obtain the corollary as follows:
Corollary 2.2. Let Ai∈Cmi×mi+1, A(1,2,4Ni+1)i∈Ai{1,2,4Ni+1}, where i∈{1,2,3}. Let Ni∈Cmi×mi, i∈{1,2,3,4} be four positive definite Hermitian matrices. Then the following statements are equivalent:
(1)
A3{1,2,4N4}A2{1,2,4N3}A1{1,2,4N2}⊆(A1A2A3){1,2,4N4}, |
(2)
r(N−12A∗1OA2A3N−14A∗3A∗2A∗1ON−13A∗2A3N−14A∗3A∗2A∗1)=r(A1)+r(A2) |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(N−13A∗2OA3ON−12A∗1A2A3). |
(3)
R(A2A3N−14A∗3A∗2A∗1)⊆R(N−12A∗1), |
R(A3N−14A∗3A∗2A∗1)⊆R(N−13A∗2) |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(N−13A∗2OA3ON−12A∗1A2A3). |
(4)
(A1)†Im1,N2A1A2A3N−14A∗3A∗2A∗1=A2A3N−14A∗3A∗2A∗1, |
(A2)†Im2,N3A3A3N−14A∗3A∗2A∗1=A3N−14A∗3A∗2A∗1 |
and
r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3∑i=1r(Ai)−r(N−13A∗2OA3ON−12A∗1A2A3). |
The reverse order law for the inverses {1,2,3M}- and {1,2,4N}- of matrix products has been studied in this article by using the ranks of the generalized Schur complement. The work performed in this paper is a useful tool for many algorithms for the computation of the weighted least squares technique of matrix equations.
Baifeng Qiu: Resources; Yingying Qin: Resources; Zhiping Xiong: Conceptualization, writing-review and editing. All authors have read and agree to the published version of the manuscript..
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the editors and the anonymous referees for their very detailed comments and constructive suggestions, which greatly improved the presentation of this paper. This work was supported by the project for characteristic innovation of 2018 Guangdong University (No: 2018KTSCX234) and the joint research and Development fund of Wuyi University, Hong Kong and Macao (No: 2019WGALH20) and the basic Theory and Scientific Research of Science and Technology Project of Jiangmen City, China (No: 2021030102610005049).
The authors declare that there are no conflicts of interest.
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