Subdirect Sums of Matrices

Abstract

In this paper, the question of when the subdirect sum of matrices is in the class of matrices is studied. Some sufficient conditions are given. Moreover, these sufficient conditions only depend on the elements of given matrices; thus, they provide some simple criteria for application. Numerical examples are also presented to illustrate the corresponding results.

1. Introduction

The concept of -subdirect sums of square matrices, which was introduced by Fallat and Johnson [1], is a generalization of the usual sum of matrices [2]. It is found that the subdirect sum of matrices is widely used in many subjects, such as matrix completion problems, overlapping subdomains in domain decomposition methods, analysis of matrix classes, and global stiffness matrices in finite elements [1–3]. Many properties of the subdirect sum were analysed [1–9].

For the subdirect sum, an important problem is that whether the -subdirect sums of matrices belongs to the same class or not, which has been widely concerned for different classes of matrices, such as strictly diagonally dominant matrices [5], nonsingular -matrices [3], -strictly diagonally dominant matrices [2], doubly diagonally dominant matrices [4], -strictly diagonally dominant matrices [8], QN-matrices [10], and Nekrasov matrices which are studied in [6, 7].

In this paper, we concentrate on the subdirect sum of matrices which are a subclass of -matrices [11], and some sufficient conditions ensuring that the subdirect sums of matrices is in the class of matrices are given. Numerical examples are presented to illustrate the corresponding results.

Now, some notations, definitions, and lemmas are listed.

Definition 1 (see [2]). Let and be square matrices of order and , respectively, and be an integer such that . Let and be partitioned into a block as follows:where and are square matrices of order . Following [1], we call the square matrix of order given bythe -subdirect sum of and , and denote it by . As shown in [2], if we let , , and , thenwhereObviously, ; therefore, matrix can be expressed as follows:

Definition 2 (see [2]). Given a matrix , let the th deleted absolute row sum be

Definition 3 (see [12]). Matrix is called a strictly diagonally dominant () matrix if for each ,

Definition 4 (see [11] and [13]). Matrix is called a matrix if for each ,wherewhich is the subset of indices of .

Lemma 1 (see [11]). If matrix is a matrix by rows, then must have at least one row which is strictly diagonally dominant.

Remark 1. From Definitions 3 and 4 and Lemma 1, it is easy to obtain that the class of strictly diagonally dominant matrices is a subclass of matrices, that is, .

Lemma 2 (see [12]). Matrix is an -matrix if and only if there exists a positive diagonal matrix such that is an SDD matrix.

Theorem 1 (see [11]). If matrix is a matrix by rows, then it is also an -matrix. If, in addition, has positive diagonal entries, then det .

Remark 2. From Remark 1, Lemma 2, and Theorem 1, we have the following relationship:

2. Subdirect Sums of Matrices

In the following, we give an example to show that the subdirect sum of two matrices may not be a matrix.

Example 1. We consider the following matrices and :where the 1-subdirect sum givesBy calculation,Note that . Therefore, is not a matrix.
Example 1 shows that the subdirect sum of matrices is not necessarily a matrix; then, a meaningful discussion is concerned: under what conditions, the subdirect sum of matrices is in the class of matrices?
Firstly, we study the 1-subdirect sum of matrices.

Theorem 2. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andthen the 1-subdirect sum is a matrix.

Proof. Since is a matrix with , we have for any , andSince is a matrix, we have ,According to the 1-subdirect sum , we haveIn addition, from the condition that all diagonal entries of and are positive (or negative), we haveFrom inequalities (15) and (16) and equalities (17) and (18), we havethat is, , which means that .
For any , it is easy to obtain that . Therefore, we haveFor the conditionlet us multiply by on both sides of this inequality, and then add on both sides, simultaneously; multiply both sides by , in addition, and then from (17) and (18), we haveHence,Therefore, we can draw a conclusion that, for any , that is, is a matrix.

Theorem 3. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), then the 1-subdirect sum is a matrix.

Proof. Since is a matrix with , we have for any , andSince is a matrix, we have inequality (16) .
From inequalities (16) and (24) and equalities (17) and (18), we havethat is, . Therefore, the proof for can be divided into two cases as follows: Case 1: if , then , which means that . For any , we have . Therefore, we obtain that Case 2: if , then , which means that . For any , we have . Therefore, we obtain that For , we have . Therefore, we obtain thatFrom Cases 1 and 2, we get that , which implies the 1-subdirect sum is a matrix.
The following Example 2 shows that Theorem 2 does not hold when .

Example 2. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 2 by computation. Hence, is a matrix. However, is not a matrix. In fact,By calculation,Note that . Therefore, is not a matrix.
The following Example 3 shows that Theorem 3 does not hold when .

Example 3. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 2 by computation. Therefore, is a matrix. However, is not a matrix. In fact,By calculation, is not a matrix because .
Examples 2 and 3 motivate the search for other conditions such that is a matrix, where is a matrix and is a matrix.
Next, we give some sufficient conditions ensuring that the 2-subdirect sum of matrices and matrices is a matrix.

Theorem 4. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andthen the 2-subdirect sum is a matrix.

Proof. Since is a matrix with , we getand .
Since is a matrix, we get inequality (16) .
According to the 2-subdirect sum , we getIn addition, from the condition that all diagonal entries of and are positive (or negative), we getFor any and , we haveFrom equalities (38) and (40) and inequalities (16), (36), and (42), we obtain thatSimilarly, it is easy to obtain thatThat is, , which means that .
For any , it is easy to obtain that . Therefore, we haveFrom the conditionwe haveThus,and substituting equalities (38), (39), (40), and (41) into this inequality, we obtain thatHence,Therefore, it is obvious that, for any , that is, the 2-subdirect sum is a matrix.

Corollary 1. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative),then the 2-subdirect sum is a matrix.

Proof. Since is a matrix with , we obtain that with inequalities (36) and (37). Since is a matrix, we obtain inequality (16) .
From equalities (38) and (40) and inequalities (16), (36), and (42), we obtain that . In a similar manner, we obtain that . That is, , which means that .
For any , it is easy to get that . Therefore, we obtain thatFor the conditionlet us multiply by on both sides of this inequality, and then add on both sides, simultaneously; multiply both sides by , in addition, and then from equations (37) and (39), we obtain thatIn a similar manner, it is easy to get thatHence,Therefore, we get that, for any , that is, the 2-subdirect sum is a matrix.

Theorem 5. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), then the 2-subdirect sum is a matrix.

Proof. Since is a matrix with , we get. Since is a matrix, we get inequality (16) .
From equalities (38) and (40) and inequalities (16), (42), and (57), we obtain that , that is, . In a similar manner, we obtain , that is, . Hence, the proof for any can be divided into four cases as follows: Case 1: if , then , which means that . For any , we have . Therefore, we have For , we have . Therefore, we obtain that For , we have . Similar to the proof of of Case 1 in Theorem 5, it is easy to prove that, for . Case 2: if , then , which means that . For any , we have . Therefore, we have For , we have . Therefore, we obtain that Case 3: if , then .Similar to the proof of Case 2 in Theorem 5, it is easy to prove that, for any . Case 4: if , then , which means that . For any , we have . Therefore, we haveFrom Cases 1, 2, 3, and 4, we get that, for any , which implies that is a matrix.

Theorem 6. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andthen the 2-subdirect sum is a matrix.

Proof. Since is a matrix with , we get that with inequalities (37) and (57). Since is a matrix, we get inequality (16) .
From equalities (38) and (40) and inequalities (16), (42), and (57), we get that , that is, . And from equalities (39) and (41) and inequalities (16), (37), and (42), we have , that is, . Hence, the proof for any can be divided into two cases as follows: Case 1: if , then , which means that . For any , it is easy to get that . Therefore, we have For the condition let us multiply by on both sides of this inequality, and then add on both sides, simultaneously; multiply both sides by , in addition, and then from equations (39) and (40), we obtain that Hence, For , it is easy to get that . Therefore, we obtain that Multiplying both sides of inequality (66) by and then adding on both sides, in addition, combining , we obtain Therefore, Case 2: if , then , which means that . For any , it is easy to get that . So, we obtain thatFrom Cases 1 and 2, we get that, for any , , so the 2-subdirect sum is a matrix.

Theorem 7. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andthen the 2-subdirect sum is a matrix.

Proof. Using the same technique as in the proof of Theorem 6, it is easy to prove that ; hence, the 2-subdirect sum is a matrix.
In Theorems 4–7 and Corollary 1, we obtain some sufficient conditions that ensure the 2-subdirect sum of matrices is a matrix. The following four examples are presented to illustrate the corresponding results, that is, under these conditions, the 2-subdirect sum is a matrix, where is a matrix and is a matrix.

Example 4. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 4 by computation. Therefore, is a matrix. However, is not a matrix. In fact,By calculation,Note that . Therefore, is not a matrix.

Example 5. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 5 by computation. Therefore, is a matrix. However, is not a matrix. In fact,By calculation,Note that . Therefore, is not a matrix.

Example 6. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 6 by computation. Therefore, is a matrix. However, is not a matrix. In fact,By calculation, is not a matrix because .

Example 7. We consider the following matrices:where is a matrix and is a matrix. It is easy to verify that matrices and satisfy the condition of Theorem 7 by calculation. Therefore, is a matrix. However, is not a matrix. In fact,By calculation,Note that , which implies that is not a matrix.
Examples 4–7 show that is a matrix; however, is not necessarily a matrix.
Finally, we search for other sufficient conditions such that (, is an integer) is a matrix, where is a matrix and is a matrix.

Theorem 8. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andwherethen the k-subdirect sum is a matrix.

Proof. Let be a positive integer, and .
Since is a matrix with , we getand for any .
Since is a matrix, we get inequality (16) .
According to the k-subdirect sum , we getIn addition, from the condition that all diagonal entries of A₂₂ and B₁₁ are positive (or negative), we haveFor any and , we obtain thatFrom equalities (88) and (89) and inequalities (16), (87), and (90), we get that , that is, for any , which means that .
For any , it is easy to get that . Therefore, we get thatFrom the conditionwe obtain thatThus,and substituting equalities (88) and (89) into this inequality, we getHence,Therefore, we can draw a conclusion that for any , which implies that the -subdirect sum is a matrix.

Corollary 2. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andwhere and is as in Theorem 8, then the -subdirect sum is a matrix.

Proof. For the conditionmultiplying both sides of this inequality by , we obtain thatThus, it is easy to get thatTherefore, by Theorem 8, we obtain that the -subdirect sum is a matrix. □

Theorem 9. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), then the -subdirect sum is a matrix.

Proof. Let be a positive integer, and .
Since is a matrix with , we get that for any , andSince is a matrix, we obtain inequality (16) .
From equalities (88) and (89) and inequalities (16), (90), and (101), we obtain that, for any , , that is, for any , .
SupposeObviously, ; then, . Hence, the proof for any can be divided into three cases as follows: Case 1: if , then , which means that . For any , we have . Therefore, we get that Case 2: if , then , which means that . For any , we have . Therefore, we get that For , we have . Therefore, we obtain that Case 3: if , then , which means that . For any , we have . Therefore, we obtain that For , we know . Therefore, we get thatFrom Cases 1, 2, and 3, we get that, for any , , which implies the -subdirect sum is a matrix.

Theorem 10. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andwhere and is as in Theorem 8, then the k-subdirect sum is a matrix.

Proof. Let be a positive integer, and .
Since is a matrix, we obtain inequality (87) or (101) for any and for any .
SupposeObviously,Since , we have .
Since is a matrix, we obtain inequality (16) .
From equalities (88) and (89) and inequalities (16), (87), and (90), we obtain that, for any , . And from equalities (88) and (89) and inequalities (16), (90), and (101), we obtain that, for any , .
Therefore, supposeObviously,Thus, we obtain or or . Hence, the proof for any can be divided into three cases as follows: Case 1: if , then ; thus, , which means that . For any , we have . Therefore, we obtain that From the condition we obtain that Thus, it is easy to obtain that and substituting equalities (88) and (89) into this inequality, we obtain that Thus, we obtain that Hence, Case 2: if , then ; thus, , which means that . For any , we know . Therefore, we obtain that For , we have . Therefore, we obtain that It follows from inequality (118) that Then, Therefore, Case 3: if , , which mean that . For any , we have . Therefore, we obtain that For , we have . Therefore, we haveFrom Cases 1, 2, and 3, we obtain that for any , which implies that the -subdirect sum is a matrix.

Corollary 3. Let and be square matrices of order and , respectively, which are partitioned as in (1). Let , , where . We assume that is a matrix with , and is a matrix. If all diagonal entries of and are positive (or negative), andwhere and is as in Theorem 8, then the -subdirect sum is a matrix.

Proof. For the conditionmultiplying both sides of this inequality by , we obtain that, for any ,and by summing for every , we obtainBy Theorem 10, we obtain that the -subdirect sum is a matrix.

3. Conclusions

In this paper, some sufficient conditions such that the subdirect sum of matrices is in the class of matrices are given. Moreover, these sufficient conditions only depend on the elements of given matrices; thus, they provide some simple criteria for application. Numerical examples are also presented to illustrate the corresponding results.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors jointly worked on the results, and they read and approved the final manuscript.

Acknowledgments

This work was partly supported by the National Natural Science Foundations of China (31600299), Natural Science Basic Research Program of Shaanxi, China (2017JQ3020 and 2020JM-622), the Scientific Research Program funded by the Shaanxi Provincial Education Department (18JK0044), the Science and Technology Project of Baoji (2017JH2-21 and 2017JH2-24), and the Key Project of Baoji University of Arts and Sciences (ZK2017021 and ZK16050).