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1.   Introduction

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 $SD{D_1}$ matrices is an $SD{D_1}$ matrix. In ^[7], Li et al. gave some sufficient conditions such that the $k$-subdirect sum of doubly strictly diagonally dominant $\left( {DSDD} \right)$ 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], $\{ {i_0}\}$-Nekrasov matrices ^[15], $\{ {p_1},{p_2}\}$-Nekrasov matrices ^[16], Dashnic-Zusmanovich $(DZ)$ matrices ^[17], and $B$-matrices ^[18,19].

$GSD{D_1}$ 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 $GSD{D_1}$ matrices, and some sufficient conditions such that the $k$-subdirect sum of $GSD{D_1}$ matrices with $SDD$ matrices belong to $GSD{D_1}$ 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 ${n_1}$ and ${n_2}$, respectively, and $k$ be an integer such that $1 \le k \le \min \left\{ {{n_1}, {n_2}} \right\}$, and let $A$ and $B$ be partitioned into $2 \times 2$ blocks as follows:

where ${A_{22}}$ and ${B_{11}}$ are square matrices of order $k$. Following [1], we call the square matrix of order $n = {n_1} + {n_2} - k$ given by

the $k$-subdirect sum of $A$ and $B$, denoted by $C = A{ \oplus _k}B$. 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:

Obviously, ${S_1} \cup {S_2} \cup {S_3} = N: = \left\{ {1, 2, ..., n} \right\}.$ Denoting $C = A{ \oplus _k}B = \left[ {{c_{ij}}} \right]$, $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$, then

Definition 1.2. (^[20]) Given a matrix $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$, where ${C^{n \times n}}$ is the set of complex matrices. Let

It is easy to obtain that $\overline {{N_A}}$ is the complement of ${N_A}$ in $N$, i.e., $\overline {{N_A}} = N\backslash {N_A}.$

Definition 1.3. (^[6]) A matrix $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$ is called a strictly diagonally dominant ($SDD$) matrix if

Definition 1.4. (^[20]) A matrix $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$ is called a $GSD{D_1}$ matrix if

where

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 ${r_i}(A) > 0$, then it is a $GSD{D_1}$ matrix.

2.   Main results

First of all, a counterexample is given to show that the subdirect sum of two $GSD{D_1}$ matrices may not necessarily be a $GSD{D_1}$ matrix.

Example 2.1. Consider the following $GSD{D_1}$ matrices $A$ and $B$, where

and the 1-subdirect sum $C = A{ \oplus _1}B$ is

However, $C$ is not a $GSD{D_1}$ matrix because

Example 2.1 shows that the subdirect sum of $GSD{D_1}$ matrices is not a $GSD{D_1}$ matrix. Then, a meaningful discussion is concerned with: under what conditions will the subdirect sum of $GSD{D_1}$ matrices is in the class of $GSD{D_1}$ matrices?

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

Lemma 2.1. If matrix $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$ is a $GSD{D_1}$ matrix, then $\left| {{a_{jj}}} \right| - {p_j}^{{N_A}}\left(A \right) > 0$ holds for all $j \in {N_A}$.

Proof. According to the definition of $GSD{D_1}$ matrices, we get

Since ${r_i}\left(A \right) - {p_i}\overline {^{{N_A}}} \left(A \right) > 0, \enspace {p_i}^{{N_A}}\left(A \right)$, and ${p_j}\overline {^{{N_A}}} \left(A \right)$ are all nonnegative, $\left| {{a_{jj}}} \right| - {p_j}^{{N_A}}\left(A \right) > 0$ is obtained. □

Lemma 2.2. (^[20]) If $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$ is a $GSD{D_1}$ matrix, then there is at least one entry ${a_{ij}} \ne 0$, $i \ne j$, $i \in \overline {{N_A}}$, $j \in N$.

Lemma 2.3. (^[20]) If $A = \left[ {{a_{ij}}} \right] \in {C^{n \times n}}$ is a $GSD{D_1}$ matrix with ${N_A} = \emptyset$, then $A$ is an $SDD$ matrix, and there is at least one entry ${a_{ij}} \ne 0$, $i \ne j$, $i \in \overline {{N_A}}$, $j \in \overline {{N_A}}$.

Now, we consider the 1-subdirect sum of $GSD{D_1}$ matrices.

Theorem 2.1. Let $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ be square matrices of order ${n_1}$ and ${n_2}$ partitioned as in (1.1), respectively. And let $k = 1$, ${S_1} = \left\{ {1, 2, \ldots, {n_1} - 1} \right\}$, ${S_2} = \left\{ {{n_1}} \right\}$, and ${S_3} = \left\{ {{n_1} + 1, {n_1} + 2, \ldots, {n_1} + {n_2} - 1} \right\}$. We assume that $A$ is a $GSD{D_1}$ matrix, and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. If all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or all negative), ${n_1} \in \overline {{N_A}}$ and

then the 1-subdirect sum $C = A{ \oplus _1}B$ is a $GSD{D_1}$ matrix.

Proof. According to the 1-subdirect sum $C = A{ \oplus _1}B$, we have

From ${n_1} \in \overline {{N_A}}$, we know $\left| {{a_{{n_1}, {n_1}}}} \right| > {r_{{n_1}}}\left(A \right)$. Because all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or negative), we have

Since $A$ is a $GSD{D_1}$ matrix, $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$, $C = A{ \oplus _1}B$, and according to Lemmas 2.2 and 2.3, we know that ${r_i}\left(C \right) \ne 0$ for all $i \in \overline {{N_C}}$. Therefore, for any $i \in \overline {{N_C}}$,

For any $j \in {N_C}$, we easily get $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$. For the three different selection ranges of $i$, that is, $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $i \in \overline {{N_C}} \cap {S_2} = \left\{ {{n_1}} \right\}$, and $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, therefore, we divide the proof into three cases.

Case 1. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C}$, we have

Therefore, we obtain that

Case 2. For $i \in \overline {{N_C}} \cap {S_2} = \left\{ {{n_1}} \right\}$, $j \in {N_C}$,

We know that the results of the $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}^{\overline {{N_C}} }\left(C \right)$ are the same as (2.1), (2.2), and (2.3). Because $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$, we clearly get

Hence,

Case 3. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C}$, in particular, we obtain that

So we easily come up with

From Cases 1–3, we have that for any $i \in \overline {{N_C}}$ and $j \in {N_C}$, the $C$ matrix satisfies the definition of the $GSD{D_1}$ matrix. The conclusion is as follows. □

Theorem 2.2. Let $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ be square matrices of order ${n_1}$ and ${n_2}$ partitioned as in (1.1), respectively. And let $k$, ${S_1}$, ${S_2}$, and ${S_3}$ be as in Theorem 2.1. Likewise, we assume $A$ is a $GSD{D_1}$ matrix, and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. If all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or all negative), ${n_1} \in {N_A}$, ${r_{{n_1}}}\left(A \right) + {r_1}\left(B \right) \ge \left| {{a_{{n_1}, {n_1}}}} \right| + \left| {{b_{11}}} \right|$ and

then $C = A{ \oplus _1}B$ is a $GSD{D_1}$ matrix.

Proof. Since $A$ is a $GSD{D_1}$ matrix, $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$, ${n_1} \in {N_A}$, and ${r_{{n_1}}}\left(A \right) + {r_1}\left(B \right) \ge \left| {{a_{{n_1}, {n_1}}}} \right| + \left| {{b_{11}}} \right|$, we get ${n_1} \in {N_C}$ and then ${N_C} = {N_A}$.

For any $i \in \overline {{N_C}}$, by Lemmas 2.2 and 2.3, we have ${r_i}\left(C \right) \ne 0$ and then

Since ${n_1} \in {N_C}$, i.e., $i \in \overline {{N_C}} \cap {S_2} = \emptyset$, we prove it according to the two different selection ranges of $i$, namely $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$ and $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$. For any $j \in {N_C}$, that is, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$ and $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} = \left\{ {{n_1}} \right\}$. Therefore, we prove it from the following cases.

Case 1. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$, we obtain that

Therefore,

Case 2. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} = \left\{ {{n_1}} \right\}$, we know that ${r_i}\left(C \right)$, ${p_i}^{{N_C}}\left(C \right)$, and ${p_i}^{\overline {{N_C}} }\left(C \right)$ have the same results as (2.4), (2.8), and (2.5). Moreover,

Hence, we obtain that

Case 3. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$, we have

where $l = i - {n_1} + 1$. We have the same values of $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}\overline {^{{N_C}}} \left(C \right)$ as (2.6), (2.7), and (2.9). Therefore,

Case 4. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} = \left\{ {{n_1}} \right\}$, we get that the values of ${r_i}\left(C \right)$, ${p_i}^{{N_C}}\left(C \right)$, and ${p_i}^{\overline {{N_C}} }\left(C \right)$ are the same as (2.13), (2.15), and (2.14). Moreover, the results of $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}\overline {^{{N_C}}} \left(C \right)$ are the same as (2.10), (2.11), and (2.12). Hence, we obtain that

From Cases 1–4, we definitively get that $C$ is a $GSD{D_1}$ matrix. □

The following Example 2.2 shows that Theorem 2.1 may not necessarily hold when $k \ge 2$.

Example 2.2. Consider the following matrices:

where $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. It is easy to verify that $A$ and $B$ satisfy the conditions of Theorem 2.1 and $A{ \oplus _1}B$ is a $GSD{D_1}$ matrix. However, $C = A{ \oplus _2}B$ is not a $GSD{D_1}$ matrix. In fact,

By computation,

Therefore, $C = A{ \oplus _2}B$ is not a $GSD{D_1}$ matrix.

The following Example 2.3 shows that Theorem 2.2 may not necessarily hold when $k \ge 2$.

Example 2.3. Consider the following matrices:

where $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. It is easy to verify that $A$ and $B$ satisfy the conditions of Theorem 2.2 and $A{ \oplus _1}B$ is a $GSD{D_1}$ matrix. However, $C = A{ \oplus _2}B$ is not a $GSD{D_1}$. In fact,

By computation, ${r_3}\left(C \right) - {p_3}^{\overline {{N_C}} }\left(C \right) = 0$, therefore, $C = A{ \oplus _2}B$ is not a $GSD{D_1}$ matrix.

Those are sufficient conditions to ensure that the 1-subdirect sum of $GSD{D_1}$ matrices with $SDD$ matrices is a $GSD{D_1}$ 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 ($k \ge 2$) sum of $GSD{D_1}$ matrices with $SDD$ matrices is a $GSD{D_1}$ matrix are given.

Theorem 2.3. Let $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ be square matrices of order ${n_1}$ and ${n_2}$ partitioned as in (1.1), respectively. And let $2 \le k \le \min \left\{ {{n_1}, {n_2}} \right\}$, ${S_1}$, ${S_2}$, and ${S_3}$ be as in (1.2). We assume $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. If all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or all negative), $i \in \overline {{N_A}}$ for any $i \in {S_2}$ and

where ${\lambda _i} = {r_i}\left(A \right) + {r_{i - {n_1} + k}}\left(B \right) + \sum\limits_{\mathop {j = {n_1} - k + 1}\limits_{j \ne i} }^{{n_1}} {\left| {{a_{ij}} + {b_{i - {n_1} + k, j - {n_1} + k}}} \right|} - \sum\limits_{\mathop {j = {n_1} - k + 1}\limits_{j \ne i} }^{{n_1}} {\left({\left| {{a_{ij}}} \right| + \left| {{b_{i - {n_1} + k, j - {n_1} + k}}} \right|} \right)},$ then the $k$-subdirect sum $C = A{ \oplus _k}B$ is a $GSD{D_1}$ matrix.

Proof. Since $A$ is a $GSD{D_1}$ matrix with $i \in \overline {{N_A}}$ for any $i \in {S_2}$, we get $\left| {{a_{ii}}} \right| > {r_i}(A).$ According to the $k$-subdirect sum $C = A{ \oplus _k}B$, we have ${r_i}(C) = {\lambda _i} \le {r_i}(A) + {r_{i - {n_1} + k}}(B)$. Because all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or negative), we get $\left| {{c_{ii}}} \right| = \left| {{a_{ii}}} \right| + \left| {{b_{i - {n_1} + k, i - {n_1} + k}}} \right|.$ Therefore, we obtain that $\left| {{c_{ii}}} \right| > {r_i}(C)$, that is, for any $i \in {S_2}$, $i \in \overline {{N_C}}$. Since $A$ is a $GSD{D_1}$ matrix, $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$, and $C = A{ \oplus _k}B$, by Lemmas 2.2 and 2.3 we know that ${r_i}\left(C \right) \ne 0$ for $i \in \overline {{N_C}} \cap {S_1} \cup {S_3} = \overline {{N_A}} \cap {S_1} \cup {S_3}$. For $i \in {S_2}$, by sufficient conditions, we have ${\lambda _i} \ge {r_i}\left(A \right) + {p_{i - {n_1} + k}}^{\overline{{N_B}}}\left(B \right)$, which means that ${\lambda _i} > 0$. Therefore, for any $i \in \overline {{N_C}}$, we obtain that

Moreover, for any $j \in {N_C}$, we get $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$. For any $i \in \overline {{N_C}}$, similarly, we prove it from the following three cases, which are $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $i \in \overline {{N_C}} \cap {S_2} = \overline {{N_A}} \cap {S_2} \subset {S_2}$, and $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$.

Case 1. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C}$, we have

Therefore,

Case 2. For $i \in \overline {{N_C}} \cap {S_2} = \overline {{N_A}} \cap {S_2} \subset {S_2}$, $j \in {N_C}$, we obtain that

We know that $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}\overline {^{{N_C}}} \left(C \right)$ are the same as (2.16), (2.17), and (2.18). Therefore,

Case 3. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C}$, specifically, we obtain that

Hence,

Therefore, we get that ${r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right) > 0$ and $\left({{r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right)} \right)\left({\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}\left(C \right)} \right) > {p_i}^{{N_C}}\left(C \right){p_j}^{\overline {{N_C}} }\left(C \right)$ for any $i \in \overline {{N_C}}$, $j \in {N_C}$. □

Corollary 2.1. Let $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ be square matrices of order ${n_1}$ and ${n_2}$ partitioned as in (1.1), respectively. And let $2 \le k \le \min \left\{ {{n_1}, {n_2}} \right\}$, ${S_1}$, ${S_2}$, and ${S_3}$ be as in (1.2). We assume $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. If all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or all negative), $i \in \overline {{N_A}}$ for any $i \in {S_2}$ and

where ${\lambda _i}$ is the same as ${\lambda _i}$ of Theorem 2.3 and $i \in {S_2}$, then the $k$-subdirect sum $C = A{ \oplus _k}B$ is a $GSD{D_1}$ matrix.

Proof. For the inequality

multiplying both sides of this inequality by $\left| {{a_{ij}}} \right|$ $\left({i \in {S_1} \cup {S_2}, \enspace j \ne i} \right)$ and summing for every $j \in \overline {{N_A}} \backslash \left\{ i \right\}$ $\left({j \in {S_2}} \right)$, we have

Similarly, for $i \in {S_2}$, we obtain that

By Theorem 2.3, we obtain that the $k$-subdirect sum $C = A{ \oplus _k}B$ is a $GSD{D_1}$ matrix. □

Theorem 2.4. Let $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ be square matrices of order ${n_1}$ and ${n_2}$ partitioned as in (1.1), respectively. And let $2 \le k \le \min \left\{ {{n_1}, {n_2}} \right\}$, ${S_1}$, ${S_2}$, and ${S_3}$ be as in (1.2). We assume $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. If all diagonal entries of ${A_{22}}$ and ${B_{11}}$ are positive (or all negative), $i \in {N_A}$ for any $i \in {S_2}$, $\left| {{a_{ii}}} \right| + \left| {{b_{i - {n_1} + k, i - {n_1} + k}}} \right| \le {\lambda _i}$ and

where ${\lambda _i}$ is the same as ${\lambda _i}$ of Theorem 2.3, then the $k$-subdirect sum $C = A{ \oplus _k}B$ is a $GSD{D_1}$ matrix.

Proof. Since $A$ is a $GSD{D_1}$ matrix with $i \in {N_A}$ for any $i \in {S_2}$ and $\left| {{a_{ii}}} \right| + \left| {{b_{i - {n_1} + k, i - {n_1} + k}}} \right| \le {\lambda _i}$, we have $\left| {{a_{ii}}} \right| \le {r_i}\left(A \right)$ and $\left| {{c_{ii}}} \right| = \left| {{a_{ii}}} \right| + \left| {{b_{i - {n_1} + k, i - {n_1} + k}}} \right| \le {\lambda _i} = {r_i}\left(C \right)$, that is, for any $i \in {S_2}$, we have $i \in {N_C}$. Moreover, we know that $i \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$, which means that ${N_C} = {N_A}$. Combining Lemmas 2.2 and 2.3, we get that ${r_i}\left(C \right) \ne 0$ for $i \in \overline {{N_C}} \cap {S_1} \cup {S_3} = \overline {{N_A}} \cap {S_1} \cup {S_3}$. Therefore, for any $i \in \overline {{N_C}}$, we obtain that

Since $i \in \overline {{N_C}} \cap {S_2} = \emptyset$, we prove it from the following two aspects, which are $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$ and $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$. For any $j \in {N_C}$, that is, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$ and $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} \subset {S_2}$. Therefore, we prove it from the following cases.

Case 1. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$, we get

Therefore, we obtain that

Case 2. For $i \in \overline {{N_C}} \cap {S_1} = \overline {{N_A}} \cap {S_1} \subset {S_1}$, $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} \subset {S_2}$, we know that ${r_i}\left(C \right)$ and (2.19)} are equal, ${p_i}^{\overline {{N_C}} }\left(C \right)$ and (2.20) are equal, and ${p_i}^{{N_C}}\left(C \right)$ and (2.23) are equal. Moreover,

Hence,

Case 3. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C} \cap {S_1} = {N_A} \cap {S_1} \subset {S_1}$, we obtain that

where $l = i - {n_1} + k$. We know that $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}^{\overline {{N_C}} }\left(C \right)$ are the same as (2.21), (2.22), and (2.24). Therefore,

Case 4. For $i \in \overline {{N_C}} \cap {S_3} \subset {S_3}$, $j \in {N_C} \cap {S_2} = {N_A} \cap {S_2} \subset {S_2}$, we obtain that the values of ${r_i}\left(C \right)$, ${p_i}^{{N_C}}\left(C \right)$, and ${p_i}^{\overline {{N_C}} }\left(C \right)$ are equal to (2.28), (2.30), and (2.29). Moreover, the results of $\left| {{c_{jj}}} \right|$, ${p_j}^{{N_C}}\left(C \right)$, and ${p_j}\overline {^{{N_C}}} \left(C \right)$ are the same as (2.25), (2.26), and (2.27). Hence, we arrive at

In conclusion, for any $i \in \overline {{N_C}}$, $j \in {N_C}$, we successfully derive that ${r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right) > 0$ and} $\left({{r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right)} \right)\left({\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}\left(C \right)} \right) > {p_i}^{{N_C}}\left(C \right){p_j}^{\overline {{N_C}} }\left(C \right).$ Therefore, $C = A{ \oplus _k}B$ is a $GSD{D_1}$ matrix.

Example 2.4. Consider the following matrices:

where $A$ is a $GSD{D_1}$ matrix with $i \in \overline {{N_A}}$ for all $i \in {S_2}$, and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. By computation, we derive ${N_A} = \left\{ {1, 3} \right\}, \enspace \overline {{N_A}} = \left\{ {2, 4, 5, 6} \right\}$. Moreover,

we get that $\sum\limits_{j \in \overline {{N_A}} \backslash \left\{ i \right\}, j \in {S_2}} {\frac{{{\lambda _j}}}{{\left| {{a_{jj}} + {b_{j - {n_1} + k, j - {n_1} + k}}} \right|}}} \left| {{a_{ij}}} \right| \le \sum\limits_{j \in \overline {{N_A}} \backslash \left\{ i \right\}, j \in {S_2}} {\frac{{{r_j}\left(A \right)}}{{\left| {{a_{jj}}} \right|}}} \left| {{a_{ij}}} \right|$ is true for $i \in {S_1} \cup {S_2}$.

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

we get that the third sufficient condition in Theorem 2.3 is met. Therefore, by Theorem 2.3, $C = A{ \oplus _3}B$ is a $GSD{D_1}$ matrix. In fact,

where ${N_C} = \left\{ {1, 3} \right\}, \enspace \overline {{N_C}} = \left\{ {2, 4, 5, 6, 7, 8} \right\}$. By computation,

It is not difficult to find that ${r_i}(C) - {p_i}^{\overline {{N_C}} }\left(C \right) > {p_i}^{{N_C}}(C)$ and $\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}(C) > {p_j}^{\overline {{N_C}} }(C)$ when $i \in \overline {{N_C}}$, $j \in {N_C}$. So we deduce that ${r_i}\left(C \right) > {p_i}^{\overline {{N_C}} }\left(C \right)$ and $\left({{r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right)} \right)\left({\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}\left(C \right)} \right) > {p_i}^{{N_C}}\left(C \right){p_j}^{\overline {{N_C}} }\left(C \right)$ are true when $i \in \overline {{N_C}}$, $j \in {N_C}$. Thus, $C = A{ \oplus _3}B$ is a $GSD{D_1}$ matrix.

Example 2.5. Consider the following matrices:

where $A$ is a $GSD{D_1}$ matrix and $B$ is an $SDD$ matrix with ${r_i}\left(B \right) > 0$ for all $i \in \overline {{N_B}}$. By computation, ${N_A} = \left\{ {1, 4, 5, 6, 7} \right\}, {S_2} = \left\{ {4, 5, 6, 7} \right\}$,

Moreover,

Hence, the conditions in Theorem 2.4 are met. By Theorem 2.4, $C = A{ \oplus _4}B$ is a $GSD{D_1}$ matrix. In fact,

By computation, ${N_C} = \{ 1, 4, 5, 6, 7\}$, $\overline {{N_C}} = \{ 2, 3, 8, 9\}$. Moreover,

We see that ${r_i}(C) - {p_i}^{\overline {{N_C}} }\left(C \right) > {p_i}^{{N_C}}(C)$ and $\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}(C) > {p_j}^{\overline {{N_C}} }(C)$ when $i \in \overline {{N_C}}$, $j \in {N_C}$. Therefore, we obtain that ${r_i}\left(C \right) > {p_i}^{\overline {{N_C}} }\left(C \right)$ and $\left({{r_i}\left(C \right) - {p_i}^{\overline {{N_C}} }\left(C \right)} \right)\left({\left| {{c_{jj}}} \right| - {p_j}^{{N_C}}\left(C \right)} \right) > {p_i}^{{N_C}}\left(C \right){p_j}^{\overline {{N_C}} }\left(C \right)$ are true when $i \in \overline {{N_C}}$, $j \in {N_C}$. Therefore, $C = A{ \oplus _4}B$ is a $GSD{D_1}$ matrix.

Remark 2.1. Since the subdirect sum of matrices does not satisfy the commutative law, if we change "$A$ is a $GSD{D_1}$ matrix, and $B$ is an $SDD$ matrix" to "$A$ is an $SDD$ matrix, and $B$ is a $GSD{D_1}$ matrix", then we will obtain new sufficient conditions by using similar proofs in this paper.

3.   Conclusions

In this paper, some sufficient conditions are given to show that the subdirect sum of $GSD{D_1}$ matrices with $SDD$ matrices is in the class of $GSD{D_1}$ 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.

Use of AI tools declaration

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

Acknowledgments

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).

Conflict of interest

The authors declare there is no conflicts of interest.