Theory article

Subdirect Sums of GSDD1 matrices

  • Received: 29 January 2024 Revised: 20 May 2024 Accepted: 31 May 2024 Published: 19 June 2024
  • The class of generalized SDD1(GSDD1) matrices is a new subclass of H-matrices. In this paper, we focus on the subdirect sum of GSDD1 matrices, and some sufficient conditions to ensure that the subdirect sum of GSDD1 matrices with strictly diagonally dominant (SDD) matrices is in the class of GSDD1 matrices are given. Moreover, corresponding examples are given to illustrate our results.

    Citation: Jiaqi Qi, Yaqiang Wang. Subdirect Sums of GSDD1 matrices[J]. Electronic Research Archive, 2024, 32(6): 3989-4010. doi: 10.3934/era.2024179

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  • The class of generalized SDD1(GSDD1) matrices is a new subclass of H-matrices. In this paper, we focus on the subdirect sum of GSDD1 matrices, and some sufficient conditions to ensure that the subdirect sum of GSDD1 matrices with strictly diagonally dominant (SDD) matrices is in the class of GSDD1 matrices are given. Moreover, corresponding examples are given to illustrate our results.



    In 1999, the concept of k-subdirect sums of square matrices was proposed by Fallat and Johnson [1], which is a generalization of the usual sum of matrices [2]. The subdirect sum of matrices plays an important role in many areas, such as matrix completion problems, global stiffness matrices in finite elements and overlapping subdomains in domain decomposition methods [1,2,3,4,5].

    An important question for subdirect sums is whether the k-subdirect sum of two square matrices in one class of matrices lies in the same class. This question has attracted widespread attention in different classes of matrices and produced a variety of results. In 2005, Bru et al. gave sufficient conditions ensuring that the subdirect sum of two nonsingular M-matrices was also a nonsingular M-matrix [3]. Then the following year, they further came to the conclusion of the k-subdirect sum of S-SDD matrices is also an S-SDD matrix [2]. In [6], Chen and Wang succeeded in producing some sufficient conditions that the k-subdirect sum of SDD1 matrices is an SDD1 matrix. In [7], Li et al. gave some sufficient conditions such that the k-subdirect sum of doubly strictly diagonally dominant (DSDD) matrices is in the class of DSDD matrices. In addition, the k-subdirect sum of other classes of matrices were mentioned, such as Nekrasov matrices [8,9,10], quasi-Nekrasov (QN) matrices [11], SDD(p) matrices [12], weakly chained diagonally dominant matrices [13], Ostrowski-Brauer Sparse (OBS) matrices [14], {i0}-Nekrasov matrices [15], {p1,p2}-Nekrasov matrices [16], Dashnic-Zusmanovich (DZ) matrices [17], and B-matrices [18,19].

    GSDD1 matrices as a new subclass of H-matrices was proposed by Dai et al. in 2023 [20]. In this paper, we focus on the subdirect sum of GSDD1 matrices, and some sufficient conditions such that the k-subdirect sum of GSDD1 matrices with SDD matrices belong to GSDD1 matrices are given. Numerical examples are presented to illustrate the corresponding results.

    Now, some definitions are listed as follows.

    Definition 1.1. ([2]) Let A and B be two square matrices of order n1 and n2, respectively, and k be an integer such that 1kmin{n1,n2}, and let A and B be partitioned into 2×2 blocks as follows:

    A=(A11A12A21A22),B=(B11B12B21B22), (1.1)

    where A22 and B11 are square matrices of order k. Following [1], we call the square matrix of order n=n1+n2k given by

    C=(A11A120A21A22+B11B120B21B22)

    the k-subdirect sum of A and B, denoted by C=AkB. We can use the elements in A and B to represent any element in C. Before that, let us define the following set of indices:

    S1={1,2,...,n1k},S2={n1k+1,n1k+2,...,n1},S3={n1+1,...,n}. (1.2)

    Obviously, S1S2S3=N:={1,2,...,n}. Denoting C=AkB=[cij], A=[aij] and B=[bij], then

    cij={aij,iS1,jS1S2,0,iS1,jS3,aij,iS2,jS1,aij+bin1+k,jn1+k,iS2,jS2,bin1+k,jn1+k,iS2,jS3,0,iS3,jS1,bin1+k,jn1+k,iS3,jS2S3.

    Definition 1.2. ([20]) Given a matrix A=[aij]Cn×n, where Cn×n is the set of complex matrices. Let

    ri(A)=jN,ji|aij|,iN.
    NA={i||aii|ri(A)},
    ¯NA={i||aii|>ri(A)}.

    It is easy to obtain that ¯NA is the complement of NA in N, i.e., ¯NA=NNA.

    Definition 1.3. ([6]) A matrix A=[aij]Cn×n is called a strictly diagonally dominant (SDD) matrix if

    |aii|>ri(A),iN.

    Definition 1.4. ([20]) A matrix A=[aij]Cn×n is called a GSDD1 matrix if

    {ri(A)>pi¯NA(A),i¯NA,(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A),i¯NA,jNA,

    where

    piNA(A):=jNA{i}|aij|,pi¯NA(A):=j¯NA{i}rj(A)|ajj||aij|,iN.

    Remark 1.1. From Definitions 1.3 and 1.4, it is easy to obtain that if a matrix A is an SDD matrix with ri(A)>0, then it is a GSDD1 matrix.

    First of all, a counterexample is given to show that the subdirect sum of two GSDD1 matrices may not necessarily be a GSDD1 matrix.

    Example 2.1. Consider the following GSDD1 matrices A and B, where

    A=(432143013.5),B=(2.5201212.31.84).

    and the 1-subdirect sum C=A1B is

    C=(43200143000162000121002.31.84).

    However, C is not a GSDD1 matrix because

    (r3(C)p3¯NC(C))(|c11|p1NC(C))=(30)(43)=3=3×1=p3NC(C)p1¯NC(C).

    Example 2.1 shows that the subdirect sum of GSDD1 matrices is not a GSDD1 matrix. Then, a meaningful discussion is concerned with: under what conditions will the subdirect sum of GSDD1 matrices is in the class of GSDD1 matrices?

    In order to obtain the main results, several lemmas are introduced that will be used in the sequel.

    Lemma 2.1. If matrix A=[aij]Cn×n is a GSDD1 matrix, then |ajj|pjNA(A)>0 holds for all jNA.

    Proof. According to the definition of GSDD1 matrices, we get

    (ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A).

    Since ri(A)pi¯NA(A)>0,piNA(A), and pj¯NA(A) are all nonnegative, |ajj|pjNA(A)>0 is obtained.

    Lemma 2.2. ([20]) If A=[aij]Cn×n is a GSDD1 matrix, then there is at least one entry aij0, ij, i¯NA, jN.

    Lemma 2.3. ([20]) If A=[aij]Cn×n is a GSDD1 matrix with NA=, then A is an SDD matrix, and there is at least one entry aij0, ij, i¯NA, j¯NA.

    Now, we consider the 1-subdirect sum of GSDD1 matrices.

    Theorem 2.1. Let A=[aij] and B=[bij] be square matrices of order n1 and n2 partitioned as in (1.1), respectively. And let k=1, S1={1,2,,n11}, S2={n1}, and S3={n1+1,n1+2,,n1+n21}. We assume that A is a GSDD1 matrix, and B is an SDD matrix with ri(B)>0 for all i¯NB. If all diagonal entries of A22 and B11 are positive (or all negative), n1¯NA and

    rn1(A)|an1,n1|rn1(A)+r1(B)|an1,n1+b11|,

    then the 1-subdirect sum C=A1B is a GSDD1 matrix.

    Proof. According to the 1-subdirect sum C=A1B, we have

    rn1(C)=rn1(A)+r1(B).

    From n1¯NA, we know |an1,n1|>rn1(A). Because all diagonal entries of A22 and B11 are positive (or negative), we have

    |cn1,n1|=|an1,n1+b11|=|an1,n1|+|b11|>rn1(A)+r1(B)=rn1(C).

    Since A is a GSDD1 matrix, B is an SDD matrix with ri(B)>0 for all i¯NB, C=A1B, and according to Lemmas 2.2 and 2.3, we know that ri(C)0 for all i¯NC. Therefore, for any i¯NC,

    ri(C)=jN{i}|cij|>j¯NC{i}rj(C)|cjj||cij|=pi¯NC(C).

    For any jNC, we easily get jNCS1=NAS1S1. For the three different selection ranges of i, that is, i¯NCS1=¯NAS1S1, i¯NCS2={n1}, and i¯NCS3S3, therefore, we divide the proof into three cases.

    Case 1. For i¯NCS1=¯NAS1S1, jNC, we have

    ri(C)=ri(A),
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS2rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=j¯NA{i},jS1rj(A)|ajj||aij|+rn1(A)+r1(B)|an1,n1+b11||ai,n1|+0j¯NA{i},jS1rj(A)|ajj||aij|+rn1(A)|an1,n1||ai,n1|=pi¯NA(A),
    |cjj|=|ajj|, (2.1)
    pjNC(C)=jNC{j}|cjj|=jNA{j}|ajj|=pjNA(A), (2.2)
    piNC(C)=jNC{i}|cij|=jNA{i}|aij|=piNA(A),
    pj¯NC(C)=j¯NC{j},jS1rj(C)|cjj||cjj|+j¯NC{j},jS2rj(C)|cjj||cjj|+j¯NC{j},jS3rj(C)|cjj||cjj|=j¯NA{j},jS1rj(A)|ajj||ajj|+rn1(A)+r1(B)|an1,n1+b11||aj,n1|+0j¯NA{j},jS1rj(A)|ajj||ajj|+rn1(A)|an1,n1||aj,n1|=pj¯NA(A). (2.3)

    Therefore, we obtain that

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A)piNC(C)pj¯NC(C).

    Case 2. For i¯NCS2={n1}, jNC,

    rn1(C)=rn1(A)+r1(B),
    pn1NC(C)=jNC{n1}|cn1,j|=jNA{n1}|an1,j|=pn1NA(A).
    pn1¯NC(C)=j¯NC{n1},jS1rj(C)|cjj||cn1,j|+j¯NC{n1},jS3rj(C)|cjj||cn1,j|=j¯NA{n1}rj(A)|ajj||an1,j|+j¯NB{1}rj(B)|bjj||b1j|=pn1¯NA(A)+p1¯NB(B).

    We know that the results of the |cjj|, pjNC(C), and pj¯NC(C) are the same as (2.1), (2.2), and (2.3). Because B is an SDD matrix with ri(B)>0 for all i¯NB, we clearly get

    r1(B)p1¯NB(B)>0.

    Hence,

    (rn1(C)pn1¯NC(C))(|cjj|pjNC(C))=(rn1(A)+r1(B)pn1¯NA(A)p1¯NB(B))(|ajj|pjNA(A))>(rn1(A)pn1¯NA(A))(|ajj|pjNA(A))>pn1NA(A)pj¯NA(A)pn1NC(C)pj¯NC(C).

    Case 3. For i¯NCS3S3, jNC, in particular, we obtain that

    piNC(C)=jNC{i}|cij|=0.

    So we easily come up with

    (rn1(C)pn1¯NC(C))(|cjj|pjNC(C))=(rn1(C)pn1¯NC(C))(|ajj|pjNA(A))>0=piNC(C)pj¯NC(C).

    From Cases 1–3, we have that for any i¯NC and jNC, the C matrix satisfies the definition of the GSDD1 matrix. The conclusion is as follows.

    Theorem 2.2. Let A=[aij] and B=[bij] be square matrices of order n1 and n2 partitioned as in (1.1), respectively. And let k, S1, S2, and S3 be as in Theorem 2.1. Likewise, we assume A is a GSDD1 matrix, and B is an SDD matrix with ri(B)>0 for all i¯NB. If all diagonal entries of A22 and B11 are positive (or all negative), n1NA, rn1(A)+r1(B)|an1,n1|+|b11| and

    min2ln2(rl(B)pl¯NB(B))maxm¯NA(rm(A)pm¯NA(A)),
    minm¯NApmNA(A)max2ln2|bl1|,

    then C=A1B is a GSDD1 matrix.

    Proof. Since A is a GSDD1 matrix, B is an SDD matrix with ri(B)>0 for all i¯NB, n1NA, and rn1(A)+r1(B)|an1,n1|+|b11|, we get n1NC and then NC=NA.

    For any i¯NC, by Lemmas 2.2 and 2.3, we have ri(C)0 and then

    ri(C)=jN{i}|cij|>j¯NC{i}rj(C)|cjj||cij|=pi¯NC(C).

    Since n1NC, i.e., i¯NCS2=, we prove it according to the two different selection ranges of i, namely i¯NCS1=¯NAS1S1 and i¯NCS3S3. For any jNC, that is, jNCS1=NAS1S1 and jNCS2=NAS2={n1}. Therefore, we prove it from the following cases.

    Case 1. For i¯NCS1=¯NAS1S1, jNCS1=NAS1S1, we obtain that

    ri(C)=ri(A), (2.4)
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=j¯NA{i}rj(A)|ajj||aij|+0=pi¯NA(A), (2.5)
    |cjj|=|ajj|, (2.6)
    pjNC(C)=jNC{j}|cjj|=jNA{j}|ajj|=pjNA(A), (2.7)
    piNC(C)=jNC{i}|cij|=jNA{i}|aij|=piNA(A), (2.8)
    pj¯NC(C)=j¯NC{j},jS1rj(C)|cjj||cjj|+j¯NC{j},jS3rj(C)|cjj||cjj|=j¯NA{j}rj(A)|ajj||ajj|+0=pj¯NA(A). (2.9)

    Therefore,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))=(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A)=piNC(C)pj¯NC(C).

    Case 2. For i¯NCS1=¯NAS1S1, jNCS2=NAS2={n1}, we know that ri(C), piNC(C), and pi¯NC(C) have the same results as (2.4), (2.8), and (2.5). Moreover,

    |cn1,n1|=|an1,n1+b11|=|an1,n1|+|b11|, (2.10)
    pn1NC(C)=jNC{n1}|cn1,j|=jNA{n1}|an1,j|=pn1NA(A), (2.11)
    pn1¯NC(C)=j¯NC{n1},jS1rj(C)|cjj||cn1,j|+j¯NC{n1},jS3rj(C)|cjj||cn1,j|=j¯NA{n1}rj(A)|ajj||an1,j|+j¯NB{1}rj(B)|bjj||b1j|=pn1¯NA(A)+p1¯NB(B). (2.12)

    Hence, we obtain that

    (ri(C)pi¯NC(C))(|cn1,n1|pn1NC(C))=(ri(A)pi¯NA(A))(|an1,n1|+|b11|pn1NA(A))=(ri(A)pi¯NA(A))(|an1,n1|pn1NA(A))+(ri(A)pi¯NA(A))|b11|>piNA(A)pn1¯NA(A)+piNA(A)p1¯NB(B)=piNC(C)pn1¯NC(C).

    Case 3. For i¯NCS3S3, jNCS1=NAS1S1, we have

    ri(C)=rin1+1(B)=rl(B), (2.13)
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=0+j¯NB{l},j{2,,n2}rj(B)|bjj||blj|pl¯NB(B), (2.14)
    piNC(C)=jNC{i},jS1|cij|+jNC{i},j=n1|ci,n1|=0+|bl1|=|bl1|, (2.15)

    where l=in1+1. We have the same values of |cjj|, pjNC(C), and pj¯NC(C) as (2.6), (2.7), and (2.9). Therefore,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(rl(B)pl¯NB(B))(|ajj|pjNA(A))(rm(A)pm¯NA(A))(|ajj|pjNA(A))>pmNA(A)pj¯NA(A)|bl1|pj¯NA(A)=piNC(C)pj¯NC(C).

    Case 4. For i¯NCS3S3, jNCS2=NAS2={n1}, we get that the values of ri(C), piNC(C), and pi¯NC(C) are the same as (2.13), (2.15), and (2.14). Moreover, the results of |cjj|, pjNC(C), and pj¯NC(C) are the same as (2.10), (2.11), and (2.12). Hence, we obtain that

    (ri(C)pi¯NC(C))(|cn1,n1|pn1NC(C))(rl(B)pl¯NB(B))(|an1,n1|+|b11|pn1NA(A))=(rl(B)pl¯NB(B))|b11|+(rl(B)pl¯NB(B))(|an1,n1|pn1NA(A))(rm(A)pm¯NA(A))|b11|+(rm(A)pm¯NA(A))(|an1,n1|pn1NA(A))>pmNA(A)p1¯NB(B)+pmNA(A)pn1¯NA(A)|bl1|(p1¯NB(B)+pn1¯NA(A))=piNC(C)pn1¯NC(C).

    From Cases 1–4, we definitively get that C is a GSDD1 matrix.

    The following Example 2.2 shows that Theorem 2.1 may not necessarily hold when k2.

    Example 2.2. Consider the following matrices:

    A=(311.71141122410113),B=(311021103),

    where A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. It is easy to verify that A and B satisfy the conditions of Theorem 2.1 and A1B is a GSDD1 matrix. However, C=A2B is not a GSDD1 matrix. In fact,

    C=(311.71014110227210115100103).

    By computation,

    ¯NC={2,4,5},NC={1,3},
    (r5(C)p5¯NC(C))(|c11|p1NC(C))=(10)(31.7)=1.3<1.35=1×1.35=p5NC(C)p1¯NC(C).

    Therefore, C=A2B is not a GSDD1 matrix.

    The following Example 2.3 shows that Theorem 2.2 may not necessarily hold when k2.

    Example 2.3. Consider the following matrices:

    A=(5221040100321112),B=(3201154.30.95.117),

    where A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. It is easy to verify that A and B satisfy the conditions of Theorem 2.2 and A1B is a GSDD1 matrix. However, C=A2B is not a GSDD1. In fact,

    C=(522100401000600112174.3000.95.117).

    By computation, r3(C)p3¯NC(C)=0, therefore, C=A2B is not a GSDD1 matrix.

    Those are sufficient conditions to ensure that the 1-subdirect sum of GSDD1 matrices with SDD matrices is a GSDD1 matrix. In fact, as the value of k increases, the situation becomes more complicated, so that the adequate conditions we give will also be more complicated.

    Next, some sufficient conditions ensuring that the k-subdirect (k2) sum of GSDD1 matrices with SDD matrices is a GSDD1 matrix are given.

    Theorem 2.3. Let A=[aij] and B=[bij] be square matrices of order n1 and n2 partitioned as in (1.1), respectively. And let 2kmin{n1,n2}, S1, S2, and S3 be as in (1.2). We assume A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. If all diagonal entries of A22 and B11 are positive (or all negative), i¯NA for any iS2 and

    j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij|j¯NA{i},jS2rj(A)|ajj||aij|,(iS1S2)
    j¯NB{in1+k}j{1,,k}λj+n1k|aj+n1k,j+n1k+bjj||bin1+k,j|j¯NB{in1+k}j{1,,k}rj(B)|bjj||bin1+k,j|,(iS2)
    λiri(A)+pin1+k¯NB(B),(iS2)

    where λi=ri(A)+rin1+k(B)+n1j=n1k+1ji|aij+bin1+k,jn1+k|n1j=n1k+1ji(|aij|+|bin1+k,jn1+k|), then the k-subdirect sum C=AkB is a GSDD1 matrix.

    Proof. Since A is a GSDD1 matrix with i¯NA for any iS2, we get |aii|>ri(A). According to the k-subdirect sum C=AkB, we have ri(C)=λiri(A)+rin1+k(B). Because all diagonal entries of A22 and B11 are positive (or negative), we get |cii|=|aii|+|bin1+k,in1+k|. Therefore, we obtain that |cii|>ri(C), that is, for any iS2, i¯NC. Since A is a GSDD1 matrix, B is an SDD matrix with ri(B)>0 for all i¯NB, and C=AkB, by Lemmas 2.2 and 2.3 we know that ri(C)0 for i¯NCS1S3=¯NAS1S3. For iS2, by sufficient conditions, we have λiri(A)+pin1+k¯NB(B), which means that λi>0. Therefore, for any i¯NC, we obtain that

    ri(C)=jN{i}|cij|>j¯NC{i}rj(C)|cjj||cij|=pi¯NC(C).

    Moreover, for any jNC, we get jNCS1=NAS1S1. For any i¯NC, similarly, we prove it from the following three cases, which are i¯NCS1=¯NAS1S1, i¯NCS2=¯NAS2S2, and i¯NCS3S3.

    Case 1. For i¯NCS1=¯NAS1S1, jNC, we have

    ri(C)=ri(A),
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS2rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=j¯NA{i},jS1rj(A)|ajj||aij|+j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij|+0j¯NA{i},jS1rj(A)|ajj||aij|+j¯NA{i},jS2rj(A)|ajj||aij|=pi¯NA(A),
    |cjj|=|ajj|, (2.16)
    pjNC(C)=jNC{j}|cjj|=jNA{j}|ajj|=pjNA(A), (2.17)
    piNC(C)=jNC{i}|cij|=jNA{i}|aij|=piNA(A),
    pj¯NC(C)=j¯NC{j},jS1rj(C)|cjj||cjj|+j¯NC{j},jS2rj(C)|cjj||cjj|+j¯NC{j},jS3rj(C)|cjj||cjj|=j¯NA{j},jS1rj(A)|ajj||ajj|+j¯NA{j},jS2λj|ajj+bjn1+k,jn1+k||ajj|+0j¯NA{j},jS1rj(A)|ajj||ajj|+j¯NA{j},jS2rj(A)|ajj||ajj|=pj¯NA(A). (2.18)

    Therefore,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A)piNC(C)pj¯NC(C).

    Case 2. For i¯NCS2=¯NAS2S2, jNC, we obtain that

    ri(C)=λi,
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS2rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=j¯NA{i},jS1rj(A)|ajj||aij|+j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij+bin1+k,jn1+k|+j¯NB{in1+k}j{k+1,,n2}rj(B)|bjj||bin1+k,j|j¯NA{i},jS1rj(A)|ajj||aij|+j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij|+j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||bin1+k,jn1+k|+j¯NB{in1+k}j{k+1,,n2}rj(B)|bjj||bin1+k,j|j¯NA{i},jS1rj(A)|ajj||aij|+j¯NA{i},jS2rj(A)|ajj||aij|+j¯NB{in1+k}j{1,,k}λj+n1k|aj+n1k,j+n1k+bjj||bin1+k,j|+j¯NB{in1+k}j{k+1,,n2}rj(B)|bjj||bin1+k,j|pi¯NA(A)+j¯NB{in1+k}j{1,,k}rj(B)|bjj||bin1+k,j|+j¯NB{in1+k}j{k+1,,n2}rj(B)|bjj||bin1+k,j|=pi¯NA(A)+pin1+k¯NB(B),
    piNC(C)=jNC{i}|cij|=jNA{i}|aij|=piNA(A).

    We know that |cjj|, pjNC(C), and pj¯NC(C) are the same as (2.16), (2.17), and (2.18). Therefore,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(λipi¯NA(A)pin1+k¯NB(B))(|ajj|pjNA(A))ri(A)+pin1+k¯NB(B)pi¯NA(A)pin1+k¯NB(B)×(|ajj|pjNA(A))=(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A)piNC(C)pj¯NC(C).

    Case 3. For i¯NCS3S3, jNC, specifically, we obtain that

    pNCi(C)=jNC/{i}|cij|=0.

    Hence,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))=(ri(C)pi¯NC(C))(|ajj|pjNA(A))>0=piNC(C)pj¯NC(C).

    Therefore, we get that ri(C)pi¯NC(C)>0 and (ri(C)pi¯NC(C))(|cjj|pjNC(C))>piNC(C)pj¯NC(C) for any i¯NC, jNC.

    Corollary 2.1. Let A=[aij] and B=[bij] be square matrices of order n1 and n2 partitioned as in (1.1), respectively. And let 2kmin{n1,n2}, S1, S2, and S3 be as in (1.2). We assume A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. If all diagonal entries of A22 and B11 are positive (or all negative), i¯NA for any iS2 and

    λj|ajj+bjn1+k,jn1+k|min{rj(A)|ajj|,rjn1+k(B)|bjn1+k,jn1+k|},(jS2)
    λiri(A)+pin1+k¯NB(B),

    where λi is the same as λi of Theorem 2.3 and iS2, then the k-subdirect sum C=AkB is a GSDD1 matrix.

    Proof. For the inequality

    λi|aii+bin1+k,in1+k|ri(A)|aii|,

    multiplying both sides of this inequality by |aij| (iS1S2,ji) and summing for every j¯NA{i} (jS2), we have

    j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij|j¯NA{i},jS2rj(A)|ajj||aij|.

    Similarly, for iS2, we obtain that

    j¯NB{in1+k}j{1,,k}λj+n1k|aj+n1k,j+n1k+bjj||bin1+k,j|j¯NB{in1+k}j{1,,k}rj(B)|bjj||bin1+k,j|.

    By Theorem 2.3, we obtain that the k-subdirect sum C=AkB is a GSDD1 matrix.

    Theorem 2.4. Let A=[aij] and B=[bij] be square matrices of order n1 and n2 partitioned as in (1.1), respectively. And let 2kmin{n1,n2}, S1, S2, and S3 be as in (1.2). We assume A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. If all diagonal entries of A22 and B11 are positive (or all negative), iNA for any iS2, |aii|+|bin1+k,in1+k|λi and

    |bjn1+k,jn1+k|j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|pjn1+k¯NB(B),(jS2)
    mink+1ln2(rl(B)pl¯NB(B))maxm¯NA(rm(A)pm¯NA(A)),
    minm¯NApmNA(A)maxk+1ln2j{1,,k}|blj|,

    where λi is the same as λi of Theorem 2.3, then the k-subdirect sum C=AkB is a GSDD1 matrix.

    Proof. Since A is a GSDD1 matrix with iNA for any iS2 and |aii|+|bin1+k,in1+k|λi, we have |aii|ri(A) and |cii|=|aii|+|bin1+k,in1+k|λi=ri(C), that is, for any iS2, we have iNC. Moreover, we know that iNCS1=NAS1S1, which means that NC=NA. Combining Lemmas 2.2 and 2.3, we get that ri(C)0 for i¯NCS1S3=¯NAS1S3. Therefore, for any i¯NC, we obtain that

    ri(C)=jN{i}|cij|>j¯NC{i}rj(C)|cjj||cij|=pi¯NC(C).

    Since i¯NCS2=, we prove it from the following two aspects, which are i¯NCS1=¯NAS1S1 and i¯NCS3S3. For any jNC, that is, jNCS1=NAS1S1 and jNCS2=NAS2S2. Therefore, we prove it from the following cases.

    Case 1. For i¯NCS1=¯NAS1S1, jNCS1=NAS1S1, we get

    ri(C)=ri(A), (2.19)
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=j¯NA{i}rj(A)|ajj||aij|+0=pi¯NA(A), (2.20)
    |cjj|=|ajj|, (2.21)
    pjNC(C)=jNC{j}|cjj|=jNA{j}|ajj|=pjNA(A), (2.22)
    piNC(C)=jNC{i}|cij|=jNA{i}|aij|=piNA(A), (2.23)
    pj¯NC(C)=j¯NC{j},jS1rj(C)|cjj||cjj|+j¯NC{j},jS3rj(C)|cjj||cjj|=j¯NA{j}rj(A)|ajj||ajj|+0=pj¯NA(A). (2.24)

    Therefore, we obtain that

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))=(ri(A)pi¯NA(A))(|ajj|pjNA(A))>piNA(A)pj¯NA(A)=piNC(C)pj¯NC(C).

    Case 2. For i¯NCS1=¯NAS1S1, jNCS2=NAS2S2, we know that ri(C) and (2.19)} are equal, pi¯NC(C) and (2.20) are equal, and piNC(C) and (2.23) are equal. Moreover,

    |cjj|=|ajj+bjn1+k,jn1+k|=|ajj|+|bjn1+k,jn1+k|, (2.25)
    pjNC(C)=jNC{j},jS1|cjj|+jNC{j},jS2|cjj|=jNA{j},jS1|ajj|+j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|, (2.26)
    pj¯NC(C)=j¯NC{j},jS1rj(C)|cjj||cjj|+j¯NC{j},jS3rj(C)|cjj||cjj|=j¯NA{j}rj(A)|ajj||ajj|+j{k+1,,n2}rj(B)|bjj||bjn1+k,j|pj¯NA(A)+pjn1+k¯NB(B). (2.27)

    Hence,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))=(ri(A)pi¯NA(A))×(|ajj|+|bjn1+k,jn1+k|jNA{j},jS1|ajj|j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|)=(ri(A)pi¯NA(A))(|ajj|jNA{j},jS1|ajj|)+(ri(A)pi¯NA(A))(|bjn1+k,jn1+k|j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|)(ri(A)pi¯NA(A))(|ajj|pjNA(A))+piNA(A)pjn1+k¯NB(B)>piNA(A)pj¯NA(A)+piNA(A)pjn1+k¯NB(B)piNC(C)pj¯NC(C).

    Case 3. For i¯NCS3S3, jNCS1=NAS1S1, we obtain that

    ri(C)=rin1+k(B)=rl(B), (2.28)
    pi¯NC(C)=j¯NC{i},jS1rj(C)|cjj||cij|+j¯NC{i},jS3rj(C)|cjj||cij|=0+j¯NB{l}j{k+1,,n2}rj(B)|bjj||blj|pl¯NB(B), (2.29)
    piNC(C)=jNC{i},jS1|cij|+jNC{i},jS2|cij|=0+j{1,,k}|bin1+k,j|=j{1,,k}|blj|, (2.30)

    where l=in1+k. We know that |cjj|, pjNC(C), and pj¯NC(C) are the same as (2.21), (2.22), and (2.24). Therefore,

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(rl(B)pl¯NB(B))(|ajj|pjNA(A))(rm(A)pm¯NA(A))(|ajj|pjNA(A))>pmNA(A)pj¯NA(A)j{1,,k}|blj|pj¯NA(A)=piNC(C)pj¯NC(C).

    Case 4. For i¯NCS3S3, jNCS2=NAS2S2, we obtain that the values of ri(C), piNC(C), and pi¯NC(C) are equal to (2.28), (2.30), and (2.29). Moreover, the results of |cjj|, pjNC(C), and pj¯NC(C) are the same as (2.25), (2.26), and (2.27). Hence, we arrive at

    (ri(C)pi¯NC(C))(|cjj|pjNC(C))(rl(B)pl¯NB(B))×(|ajj|+|bjn1+k,jn1+k|(jNA{j},jS1|ajj|+j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|))=(rl(B)pl¯NB(B))(|ajj|jNA{j},jS1|ajj|)+(rl(B)pl¯NB(B))(|bjn1+k,jn1+k|j¯NB{jn1+k}j{1,,k}|aj,j+n1k+bjn1+k,j|)(rm(A)pm¯NA(A))(|ajj|pjNA(A))+(rm(A)pm¯NA(A))pjn1+k¯NB(B)>pmNA(A)pj¯NA(A)+pmNA(A)pjn1+k¯NB(B)=pmNA(A)(pj¯NA(A)+pjn1+k¯NB(B))j{1,,k}|blj|(pj¯NA(A)+pjn1+k¯NB(B))piNC(C)pj¯NC(C).

    In conclusion, for any i¯NC, jNC, we successfully derive that ri(C)pi¯NC(C)>0 and} (ri(C)pi¯NC(C))(|cjj|pjNC(C))>piNC(C)pj¯NC(C). Therefore, C=AkB is a GSDD1 matrix.

    Example 2.4. Consider the following matrices:

    A=(7.512212.5170.3120.21.11.3510.810.410.26.51.20.90.310.20.96.61.40.70.90.11.218),B=(651.5211.51.2662.31.60.91.42671.31.233.42660.60.42.111.877),

    where A is a GSDD1 matrix with i¯NA for all iS2, and B is an SDD matrix with ri(B)>0 for all i¯NB. By computation, we derive NA={1,3},¯NA={2,4,5,6}. Moreover,

    λ4|a44+b11|=5.571.50.077<0.5693.76.5=r4(A)|a44|,λ5|a55+b22|=5.272.60.072<0.5763.86.6=r5(A)|a55|,
    λ6|a66+b33|=5.475=0.072<0.4883.98=r6(A)|a66|,

    we get that j¯NA{i},jS2λj|ajj+bjn1+k,jn1+k||aij|j¯NA{i},jS2rj(A)|ajj||aij| is true for iS1S2.

    λ4|a44+b11|0.077<0.092665=r1(B)|b11|,λ5|a55+b22|0.072<0.091666=r2(B)|b22|,
    λ6|a66+b33|=0.072<0.0885.967=r3(B)|b33|,

    we have that the second sufficient condition in Theorem 2.3 is true.

    λ4=5.5>4.252=3.7+0.552r4(A)+p1¯NB(B),λ5=5.2>4.393=3.8+0.593r5(A)+p2¯NB(B),
    λ6=5.4>4.471=3.9+0.571r6(A)+p3¯NB(B),

    we get that the third sufficient condition in Theorem 2.3 is met. Therefore, by Theorem 2.3, C=A3B is a GSDD1 matrix. In fact,

    C=(7.512212.500170.3120.2001.11.3510.81000.410.271.50.31.111.50.310.20.372.60.91.60.90.70.90.10.21751.31.200033.42660.60000.42.111.877),

    where NC={1,3},¯NC={2,4,5,6,7,8}. By computation,

    r2(C)=4.5,p2¯NC(C)0.235,p2NC(C)=1.3;r4(C)=5.5,p4¯NC(C)0.983,p4NC(C)=0.6;
    r5(C)=5.2,p5¯NC(C)1.011,p5NC(C)=0.5;r6(C)=5.4,p6¯NC(C)0.925,p6NC(C)=0.8;
    r7(C)=9,p7¯NC(C)0.66,p7NC(C)=0;r8(C)=5.3,p8¯NC(C)0.499,p8NC(C)=0;
    |c11|=7.5,p1NC(C)=2,p1¯NC(C)1.048;|c33|=5,p3NC(C)=1.1,p3¯NC(C)1.042.

    It is not difficult to find that ri(C)pi¯NC(C)>piNC(C) and |cjj|pjNC(C)>pj¯NC(C) when i¯NC, jNC. So we deduce that ri(C)>pi¯NC(C) and (ri(C)pi¯NC(C))(|cjj|pjNC(C))>piNC(C)pj¯NC(C) are true when i¯NC, jNC. Thus, C=A3B is a GSDD1 matrix.

    Example 2.5. Consider the following matrices:

    A=(620.5110.81.20.180.70.311.30.80.50.87.71.11.20.30.12.11.50.981.80.61.70.30.71.418.42.52.81.62.5211.79.21.10.81.21.62.41.81.59),B=(40210.71520.81.24532.51919.22.52.1541.12126.91.82.40.9612530.80.511.30.265101.40.40.70.59.968),

    where A is a GSDD1 matrix and B is an SDD matrix with ri(B)>0 for all i¯NB. By computation, NA={1,4,5,6,7},S2={4,5,6,7},

    |a44|+|b11|=48<48.1=λ4,|a55|+|b22|=53.4<53.6=λ5,
    |a66|+|b33|=63.2<63.5=λ6,|a77|+|b44|=70<70.2=λ7.

    Moreover,

    |b11|j¯NB{1},j{1,...,4}|a4,j+3+b1j|=32.2>10.633p1¯NB(B),
    |b22|j¯NB{2},j{1,...,4}|a5,j+3+b2j|=32>14.101p2¯NB(B),
    |b33|j¯NB{3},j{1,...,4}|a6,j+3+b3j|=44.5>14.965p3¯NB(B),
    |b44|j¯NB{4},j{1,...,4}|a7,j+3+b4j|=50.2>15.908p4¯NB(B).
    mink+1ln2(rl(B)pl¯NB(B))=r6(B)p6¯NB(B)7.944>3.836r2(A)p2¯NA(A)=maxm¯NA(rm(A)pm¯NA(A)),
    minm¯NApmNA(A)=p3NA(A)=3.2>3=j{1,...,4}|b5j|=j{1,...,4}|b6j|=maxk+1ln2j{1,,k}|blj|.

    Hence, the conditions in Theorem 2.4 are met. By Theorem 2.4, C=A4B is a GSDD1 matrix. In fact,

    C=(620.5110.81.2000.180.70.311.30.8000.50.87.71.11.20.30.1002.11.50.9483.81.62.41520.80.30.71.42.253.45.55.31919.21.62.523.53.863.22.22126.90.81.21.64.24.22.4702530.80000.511.30.265100001.40.40.70.59.968).

    By computation, NC={1,4,5,6,7}, ¯NC={2,3,8,9}. Moreover,

    r2(C)=4.2,p2¯NC(C)0.364,p2NC(C)=3.5;r3(C)=4,p3¯NC(C)=0.42,p3NC(C)=3.2;
    r8(C)=13,p8¯NC(C)1.897,p8NC(C)=3;r9(C)=12.9,p9¯NC(C)=1.98,p9NC(C)=3.
    |c11|=6,p1NC(C)=4,p1¯NC(C)1.31;|c44|=48,p4NC(C)=9.9,p4¯NC(C)8.201;
    |c55|=53.4,p5NC(C)=13.3,p5¯NC(C)8.537;|c66|=63.2,p6NC(C)=11.1,p6¯NC(C)11.655;
    |c77|=70,p7NC(C)=11.6,p7¯NC(C)12.304.

    We see that ri(C)pi¯NC(C)>piNC(C) and |cjj|pjNC(C)>pj¯NC(C) when i¯NC, jNC. Therefore, we obtain that ri(C)>pi¯NC(C) and (ri(C)pi¯NC(C))(|cjj|pjNC(C))>piNC(C)pj¯NC(C) are true when i¯NC, jNC. Therefore, C=A4B is a GSDD1 matrix.

    Remark 2.1. Since the subdirect sum of matrices does not satisfy the commutative law, if we change "A is a GSDD1 matrix, and B is an SDD matrix" to "A is an SDD matrix, and B is a GSDD1 matrix", then we will obtain new sufficient conditions by using similar proofs in this paper.

    In this paper, some sufficient conditions are given to show that the subdirect sum of GSDD1 matrices with SDD matrices is in the class of GSDD1 matrices, and these conditions are only dependent on the elements of the given matrices. Furthermore, some numerical examples are also presented to illustrate the corresponding theoretical results.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was partly supported by the National Natural Science Foundation of China (31600299), Natural Science Basic Research Program of Shaanxi, China (2020JM-622), and the Postgraduate Innovative Research Project of Baoji University of Arts and Sciences (YJSCX23YB33).

    The authors declare there is no conflicts of interest.



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