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The reverse order laws for {1,2,3M}- and {1,2,4N}- inverse of three matrix products

  • 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 ACm×n, and let MCm×m, NCn×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)=AM,N.

    For a given matrix ACm×n, the sets A{1,2,3M}- and A{1,2,4N} inverses of A are

    A{1,2,3M}={XCn×m|AXA=A,XAX=X,(MAX)=MAX},

    A{1,2,4N}={XCn×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 BA=(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 AiCmi×mi+1,i=1,2,3, MiCmi×mi,i=1,2,3, NiCmi×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=AR(A)L, (1.2)
    APL,M=AN(A)M. (1.3)

    Lemma 1.2. [1,3] Let ACm×n, XCn×m, and let M, N be two positive definite Hermitian matrices of order m and n, respectively, then

    XA{1,2,3M}AMAX=AMandr(X)=r(A), (1.4)
    XA{1,2,4N}XAN1A=N1Aandr(X)=r(A), (1.5)
    XA{1,2,4N}XA{1,2,3N1}. (1.6)

    Lemma 1.3. [21] Let ACm×n, BCm×k, CCl×n, DCl×k, and let MCm×m, NCn×n be two positive definite Hermitian matrices, then

    maxA(1,2,3M) r(DCA(1,2,3M)B)=min{ r(AMAAMBCD)r(A),r(AMBD)}, (1.7)
    minA(1,2,3M) r(DCA(1,2,3M)B)=r(AMAAMBCD)+r(AMBD)r(AOOAMBCD), (1.8)
    maxA(1,2,4N) r(DCA(1,2,4N)B)=min{ r(CN1A,D),r(AN1ABCN1AD)r(A)}, (1.9)
    minA(1,2,4N) r(DCA(1,2,4N)B)=r(CN1A,D)+r(AN1ABCN1AD)r(AOBOCN1AD). (1.10)

    Lemma 1.4. [22] Let ACm×n, BCm×k and CCp×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=ImAA, EB=ImBB, FA=InAA, FC=InCC.

    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 AiCmi×mi+1, A(1,2,3Mi)iAi{1,2,3Mi} and i{1,2,3}. Let MiCmi×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( A3OOA2A3A2A1M1A1A2M13A3A2A1M1A1M12)=r(A2)+r(A3)andr(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r( A2OOA3A1M12A1A2M13). (2.1)

    Proof. According to the formula (1.4) in Lemma 1.2, for any A(1,2,3Mi)iAi{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

    A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1=A3A2A1M1and
    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(A3A2A1M1A3A2A1M1A1A2A3A(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=A3A2A1M1 and C=A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2, we have

    maxA(1,2,3M1)1r(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A1M1A1A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A3A2A1M1)r(A1),r(A1M1A3A2A1M1)}=min{r(A3A2A1M1A1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2),r(A1)}=r(A3A2A1M1A1A3A2A1M1A1A2A3A(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=A3A2A1M1A1, and C=A3A2A1M1A1A2A3A(1,2,3M3)3, we have

    maxA(1,2,3M2)2,A(1,2,3M1)1r(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=maxA(1,2,3M2)2r(A3A2A1M1A1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2)=min{r(A2M2A2A2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1)r(A2),r(A2M2A3A2A1M1A1)}. (2.5)

    According to the formulas (1.12) and (1.13) of Lemma 1.4, we have

    min{r(A2M2A2A2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1)r(A2),r(A2M2A3A2A1M1A1)}=min{r(OA2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2A3A2A1M1A1)r(A2),r(A2M2A3A2A1M1A1)}

    and

    r(OA2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2A3A2A1M1A1)r(A2)r(A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2)+r(A2M2A3A2A1M1A1)r(A2)r(A3A2A1M1A1A2(A3A(1,2,3M3)3Im3))+r(A2M2A3A2A1M1A1)r(A2)r(A2)+r(A2M2A3A2A1M1A1)r(A2)=r(A2M2A3A2A1M1A1) (2.6)

    and

    r(OA2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2A3A2A1M1A1)r(A2)=r(OA2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2A3A2A1M1A1M12)r(A2)=r(A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2,A3A2A1M1A1M12FA2). (2.7)

    Combining (2.5), (2.6), and (2.7), we have

    maxA(1,2,3M2)2,A(1,2,3M1)1r(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=min{r(A2M2A2A2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1)r(A2),r(A2M2A3A2A1M1A1)}=r(OA2M2A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2A3A2A1M1A1)r(A2)=r(A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2,A3A2A1M1A1M12FA2). (2.8)

    From (2.8) and again by formula (1.7) in Lemma 1.3 with A=A3, B=(Im3,O), C=A3A2A1M1A1A2A3, and D=(A3A2A1M1A1A2,A3A2A1M1A1M12FA2), we have

    r(A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2,A3A2A1M1A1M12FA2)=r((A3A2A1M1A1A2,A3A2A1M1A1M12FA2)A3A2A1M1A1A2A3A(1,2,3M3)3(Im3,O))

    and

    maxA(1,2,3M3)3,A(1,2,3M2)2,A(1,2,3M1)1r(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=maxA(1,2,3M3)3r(A3A2A1M1A1A2A3A(1,2,3M3)3A3A2A1M1A1A2,A3A2A1M1A1M12FA2)=min{r(A3M3A3A3M3OA3A2A1M1A1A2A3A3A2A1M1A1A2A3A2A1M1A1M12FA2)r(A3),r(A3M3OA3A2A1M1A1A2A3A2A1M1A1M12FA2)}=min{r(OA3M3OOA3A2A1M1A1A2A3A2A1M1A1M12FA2)r(A3),r(A3M3OA3A2A1M1A1A2A3A2A1M1A1M12FA2)}=r(A3M3OA3A2A1M1A1A2A3A2A1M1A1M12FA2)r(A3)=r(A3A2A1M1A1A2M13FA3,A3A2A1M1A1M12FA2), (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(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=r(A3A2A1M1A1A2M13FA3,A3A2A1M1A1M12FA2)=r( A3OOA2A3A2A1M1A1A2M13A3A2A1M1A1M12)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(A3A2A1M1A3A2A1M1A1A2A3A(1,2,3M3)3A(1,2,3M2)2A(1,2,3M1)1)=0

    if and only if

    r( A3OOA2A3A2A1M1A1A2M13A3A2A1M1A1M12)=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(A1M1A1A1M1A(1,2,3M3)3A(1,2,3M2)2O)r(A1),r(A1M1O)}=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(A2M2A2A2M2A(1,2,3M3)3O)r(A2),r(A2M2O)}),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(A3M3A3A3M3A(1,2,3M2)2A(1,2,3M1)1Im4O)+r(A3M3A(1,2,3M2)2A(1,2,3M1)1O)r(A3OOA3M3A(1,2,3M2)2A(1,2,3M1)1Im4O)=r(A3M3A(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=A3M3, 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(A3M3A(1,2,3M2)2A(1,2,3M1)1)=r(A2M2A2A2M2A(1,2,3M1)1A3M3O)+r(A2M2A(1,2,3M1)1O)r(A2OOA2M2A(1,2,3M1)1A3M3O)=r(A2M2A2M13FA3,A2M2A(1,2,3M1)1)+r(A3)r(A2A3M3). (2.18)

    By formula (1.12) of Lemma 1.4, we have

    r(A2M2A2A2M2A(1,2,3M1)1A3M3O)=r(A3M3OA2M2A2A2M2A(1,2,3M1)1)=r(A3OA2M2A2M13A2M2A(1,2,3M1)1)=r(A2M2A2M13FA3,A2M2A(1,2,3M1)1)+r(A3) (2.19)

    and

    r(A2OOA2M2A(1,2,3M1)1A3M3O)=r(A2A3M3)+r(A2M2A(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=A2M2), and (D=(A2M2A2M13FA3,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(A2M2A2M13FA3,A2M2A(1,2,3M1)1))+r(A3)r(A2A3M3)=(minA(1,2,3M1)1r[(A2M2A2M13FA3,O)A2M2A(1,2,3M1)1(O,Im1)])+r(A3)r(A2A3M3)=r(A1M1A1OA1M1A2M2A2M2A2M13FA3O)+r(OA1M1A2M2A2M13FA3O)r(A1OOOOA1M1A2M2A2M2A2M13FA3O)+r(A3)r(A2A3M3)=r(A1M1A1A1M1A1A2M13FA3A1M1A2M2OO)+r(A1)+r(A2M2A2M13FA3)r(A1)r(A1OA2M2A2M2A2M13FA3)+r(A3)r(A2A3M3), (2.21)

    where

    r(A1M1A1A1M1A1A2M13FA3A1M1A2M2OO)=r(A1M1A1M12A1M1A1A2M13FA3A1M1A2OO)=r(A2)+r(A1M1,A1M1A1M12FA2,A1M1A1A2M13FA3) (2.22)

    and

    r(A2M2A2M13FA3)=r((M1/22A2)(M1/22A2)M13FA3)r((M1/22A2)M13FA3)=r((M1/22A2)(M1/22A2)(M1/22A2)M13FA3)=r((M1/22A2)((M1/22A2)M1/22A2)(M1/22A2)(M1/22A2)M13FA3)((M1/22A2)(M1/22A2)M13FA3)=r(A2M2A2M13FA3). (2.23)

    By the formula (2.23), we have

    r(A2M2A2M13FA3)=r((M1/22A2)M13FA3)=r(A2M13FA3). (2.24)

    By the formula (1.12) of Lemma 1.4, we have

    r(A2A3M3)=r(A2M13A3)=r(A3)+r(A2M13FA3). (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(A1M1,A1M1A1M12FA2,A1M1A1A2M13FA3)r(A1OA2M2A2M2A2M13FA3)=r(A1)+r(A2)r(A1OA2M2A2M2A2M13FA3)=r(A1)+r(A2)r(A1A1A2M13FA3A2M2O)=r(A1)r(A1M12FA2,A1A2M13FA3)=3i=1r(Ai)r(A2OOA3A1M12A1A2M13). (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 AiCmi×mi+1, A(1,2,3Mi)iAi{1,2,3Mi}, where i{1,2,3}. Let MiCmi×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(A3M3OOA2M2A3A2A1M1A1A2A3A2A1M1A1)=r(A2)+r(A3) (2.28)

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(A2M2OOA3M3A1A1A2); (2.29)

    (3)

    R(A2A1M1A1A2A3)R(M3A3), (2.30)
    R(A1M1A1A2A3)R(M2A2) (2.31)

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(A2M2OOA3M3A1A1A2);

    (4)

    A3(A3)M3,Im4M13A2A1M1A1A2A3=M13A2A1M1A1A2A3, (2.32)
    A2(A2)M2,Im3M12A1M1A1A2A3=M12A1M1A1A2A3 (2.33)

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(A2M2OOA3M3A1A1A2).

    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(M13A2A1M1A1A2A3)R(A3) and R(M12A1M1A1A2A3)R(A2). Using Lemma 1.1, we have

    R(A2A1M1A1A2A3)R(M3A3)andR(A1M1A1A2A3)R(M2A2)R(M13A2A1M1A1A2A3)R(A3)andR(M12A1M1A1A2A3)R(A2)r(A3OM13A2A1M1A1A2A3OA2M12A1M1A1A2A3)=r(A2)+r(A3)r(A3M3OOA2M2A3A2A1M1A1A2A3A2A1M1A1)=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(A3M3OOA2M2A3A2A1M1A1A2A3A2A1M1A1)=r(A2)+r(A3)r(A3OM13A2A1M1A1A2A3OA2M12A1M1A1A2A3)=r(A2)+r(A3)r(EA3M13A2A1M1A1A2A3EA2M12A1M1A1A2A3)+r(A2)+r(A3)=r(A2)+r(A3)r(EA3M13A2A1M1A1A2A3EA2M12A1M1A1A2A3)=0

    and

    A3A3M13A2A1M1A1A2A3=M13A2A1M1A1A2A3

    and

    A2A2M12A1M1A1A2A3=M12A1M1A1A2A3.

    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(A2A1M1A1A2A3)R(M3A3)R(M13A2A1M1A1A2A3)R(A3)

    and

    A3(A3)M3,Im4M13A2A1M1A1A2A3=PR(A3(A3)M3,Im4),N(A3(A3)M3,Im4)M13A2A1M1A1A2A3=PR(A3),N(A3(A3)M3,Im4)M13A2A1M1A1A2A3=M13A2A1M1A1A2A3

    and

    R(A1M1A1A2A3)R(M2A2)R(M12A1M1A1A2A3)R(A2)
    A2(A2)M2,Im3M12A1M1A1A2A3=PR(A2(A2)M2,Im3),N(A2(A2)M2,Im3)M12A1M1A1A2A3=PR(A2),N(A2(A2)M2,Im3)M12A1M1A1A2A3=M12A1M1A1A2A3.

    So we have that (1), (2), (3), and (4) are equivalent.

    From Lemma 1.2, we have XA{1,2,4N} if and only if XA{1,2,3N1}. 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 AiCmi×mi+1, A(1,2,4Ni+1)iAi{1,2,4Ni+1}, where i{1,2,3}. Let NiCmi×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( A1ON2A2A3N14A3A2A1OA2N3A3N14A3A2A1)=r(A1)+r(A2)andr(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r( A2ON3A3OA1N2A2A3).

    From Lemma 1.1, Lemma 1.4, and Theorem 2.2, we immediately obtain the corollary as follows:

    Corollary 2.2. Let AiCmi×mi+1, A(1,2,4Ni+1)iAi{1,2,4Ni+1}, where i{1,2,3}. Let NiCmi×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(N12A1OA2A3N14A3A2A1ON13A2A3N14A3A2A1)=r(A1)+r(A2)

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(N13A2OA3ON12A1A2A3).

    (3)

    R(A2A3N14A3A2A1)R(N12A1),
    R(A3N14A3A2A1)R(N13A2)

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(N13A2OA3ON12A1A2A3).

    (4)

    (A1)Im1,N2A1A2A3N14A3A2A1=A2A3N14A3A2A1,
    (A2)Im2,N3A3A3N14A3A2A1=A3N14A3A2A1

    and

    r(A1A2A3)=min{r(A1),r(A2),r(A3)}=3i=1r(Ai)r(N13A2OA3ON12A1A2A3).

    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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