# Subdirect Sums ofSDD1 Matrices

Authors: Xiaoyong Chen; Yaqiang Wang
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, the question of when the subdirect sum ofSDD1 matrices is in the class of SDD1 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.


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## Body

## 1. Introduction

The concept ofk-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 thek-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 M-matrices [3], S-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 ofSDD1 matrices which are a subclass of H-matrices [11], and some sufficient conditions ensuring that the subdirect sums of SDD1 matrices is in the class of SDD1 matrices are given. Numerical examples are presented to illustrate the corresponding results.Now, some notations, definitions, and lemmas are listed.Definition 1 (see [2]).
LetA and B be square matrices of order n1 and n2, respectively, and k be an integer such that 1≤k≤minn1,n2. Let A and B be partitioned into a 2×2 block as follows:(1)A=A11A12A21A22,B=B11B12B21B22,where A22 and B11 are square matrices of order k. Following [1], we call the square matrix of order n=n1+n2−k given by(2)C=A11A120A21A22+B11B120B21B22,the k-subdirect sum of A and B, and denote it by C=A⊕kB. As shown in [2], if we let C=A⊕kB=cij, A=aij, and B=bij, then(3)cij=aij,i∈S1,j∈S1∪S2,0,i∈S1,j∈S3,aij,i∈S2,j∈S1,aij+bi−n1+k,j−n1+k,i∈S2,j∈S2,bi−n1+k,j−n1+k,i∈S2,j∈S3,0,i∈S3,j∈S1,bi−n1+k,j−n1+k,i∈S3,j∈S2∪S3,where(4)S1=1,2,…,n1−k,S2=n1−k+1,n1−k+2,…,n1,S3=n1+1,n1+2,…,n.
Obviously,S1∪S2∪S3=1,2,…,n; therefore, matrix C can be expressed as follows:(5)C=a11⋯a1,n1−ka1,n1−k+1⋯a1n10⋯0⋮⋱⋮⋮⋱⋮⋮⋱⋮an1−k,1⋯an1−k,n1−kan1−k,n1−k+1⋯an1−k,n10⋯0an1−k+1,1⋯an1−k+1,n1−kan1−k+1,n1−k+1+b11⋯an1−k+1,n1+b1kb1,k+1⋯b1n2⋮⋱⋮⋮⋱⋮⋮⋱⋮an11⋯an1,n1−kan1,n1−k+1+bk1⋯an1n1+bkkbk,k+1⋯bkn20⋯0bk+1,1⋯bk+1,kbk+1,k+1⋯bk+1,n2⋮⋱⋮⋮⋱⋮⋮⋱⋮0⋯0bn21⋯bn2kbn2,k+1⋯bn2n2.Definition 2 (see [2]).
Given a matrixA=aij∈Rn×n, let the ith deleted absolute row sum be(6)riA=∑j=1,j≠ij=naij,i=1,2…,n.Definition 3 (see [12]).
MatrixA=aij∈Rn×n is called a strictly diagonally dominant (SDD) matrix if for each i∈1,2,…,n,(7)aii>riA.Definition 4 (see [11] and [13]).
MatrixA=aij∈Rn×n is called a SDD1 matrix if for each i∈N1A,(8)aii>piA,where(9)piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij,N1A=iaii≤riA,which is the subset of indices of 1,2,…,n.Lemma 1 (see [11]).
If matrixA=aij∈Rn×n is a SDD1 matrix by rows, then A must have at least one row which is strictly diagonally dominant.Remark 1.
From Definitions3 and 4 and Lemma 1, it is easy to obtain that the class of strictly diagonally dominant matrices is a subclass of SDD1 matrices, that is, SDD⊆SDD1.Lemma 2 (see [12]).
MatrixA=aij∈ℂn×nis an H-matrix if and only if there exists a positive diagonal matrix X such that AX is an SDD matrix.Theorem 1 (see [11]).
If matrixA=aij∈Rn×n is a SDD1 matrix by rows, then it is also an H-matrix. If, in addition, A has positive diagonal entries, then det A>0.Remark 2.
From Remark1, Lemma 2, and Theorem 1, we have the following relationship:(10)SDD⊆SDD1⊆H.


## 2. Subdirect Sums ofSDD1 Matrices

In the following, we give an example to show that the subdirect sum of twoSDD1 matrices may not be a SDD1 matrix.Example 1.
We consider the followingSDD1 matrices A and B:(11)A=9−5−1−25−10−53.5,B=311−1101−152,where the 1-subdirect sum C=A⊕1B gives(12)C=A⊕1B=9−5−100−25−1000−56.51100−110100−152.
By calculation,(13)N1C=3,5,p3C=4.2,p5C=2.
Note thatc55=2=p5C. Therefore, C=A⊕1B is not a SDD1 matrix.
Example1 shows that the subdirect sum of SDD1 matrices is not necessarily a SDD1 matrix; then, a meaningful discussion is concerned: under what conditions, the subdirect sum of SDD1 matrices is in the class of SDD1 matrices?
Firstly, we study the 1-subdirect sum ofSDD1 matrices.Theorem 2.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=1, S1=1,2,…,n1−1,S2=n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−1. We assume that A is a SDD1 matrix with n1∉N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(14)rn1Aan1n1≥r1Bb11,then the 1-subdirect sum C=A⊕1B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1∉N1A, we have aii>piA for any i∈N1A, and(15)an1n1>rn1A.
SinceB is a SDD matrix, we have for anyi∈1,2,…,n2,(16)bii>riB.
According to the 1-subdirect sumC=A⊕1B, we have(17)rn1C=rn1A+r1B.
In addition, from the condition that all diagonal entries ofA22 and B11 are positive (or negative), we have(18)cn1n1=an1n1+b11=an1n1+b11.
From inequalities (15) and (16) and equalities (17) and (18), we have(19)cn1n1=an1n1+b11>rn1C,that is, n1∉N1C, which means that N1C=N1A.
For anyi∈N1C, it is easy to obtain that i∈N1C∩S1=N1A∩S1⊂S1. Therefore, we have(20)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\iaij+∑j∉N1A∪i,n1rjAajjaij+rn1Aan1n1ain1.
For the condition(21)rn1Aan1n1≥r1Bb11,let us multiply by an1n1b11 on both sides of this inequality, and then add rn1Aan1n1 on both sides, simultaneously; multiply both sides by 1/an1n1an1n1+b11, in addition, and then from (17) and (18), we have(22)rn1Aan1n1≥rn1Ccn1n1.
Hence,(23)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪i,n1rjAajjaij+rn1Aan1n1ain1≥∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
Therefore, we can draw a conclusion that, for anyi∈N1C,cii>riC, that is, C=A⊕1B is a SDD1 matrix.Theorem 3.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=1, S1=1,2,…,n1−1,S2=n1,S3=n1+1,n1+2,…,n, where n=n1+n2−1. We assume that A is a SDD1 matrix with n1∈N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), then the 1-subdirect sum C=A⊕1B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1∈N1A, we have aii>piA for any i∈N1A, and(24)an1n1≤rn1A.
SinceB is a SDD matrix, we have inequality (16) for anyi∈1,2,…,n2.
From inequalities (16) and (24) and equalities (17) and (18), we have(25)cn1n1=an1n1+b11>rn1Corcn1n1=an1n1+b11≤rn1C,that is, n1∉N1Corn1∈N1C. Therefore, the proof for cii>riCfor anyi∈N1C can be divided into two cases as follows:Case 1: ifan1n1+b11>rn1C, then n1∉N1C, which means that N1C=N1A\n1.For anyi∈N1C, we have i∈N1A\n1⊂S1. Therefore, we obtain that(26)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\i,n1aij+ain1+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+cin1≥∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+rjCcjjcin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Case 2: ifan1n1+b11≤rn1C, then n1∈N1C, which means that N1C=N1A.For anyi∈N1C∩S1, we have i∈N1A∩S1⊂S1. Therefore, we obtain that(27)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1∈N1C∩S2, we have i=n1∈N1A∩S2⊂S2. Therefore, we obtain that(28)cn1n1=an1n1+b11>pn1A+r1B=∑j∈N1A\n1an1j+∑j∉N1A∪n1rjAajjan1j+∑j≠1n2b1j≥∑j∈N1A\n1an1j+∑j∉N1A∪n1rjAajjan1j+∑j≠1n2rjBbjjb1j=∑j∈N1C\n1cn1j+∑j∉N1C∪n1rjCcjjcn1j=pn1C.
From Cases 1 and 2, we get thatcii>piCfor anyi∈N1C, which implies the 1-subdirect sum C=A⊕1B is a SDD1 matrix.
The following Example2 shows that Theorem 2 does not hold when k=2.Example 2.
We consider the following matrices:(29)A=2.563−1121−116,B=811241−2−25,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 2 by computation. Hence, A⊕1B is a SDD1 matrix. However, C=A⊕2B is not a SDD1 matrix. In fact,(30)C=A⊕2B=2.5630−12021−131010−2−25.
By calculation,(31)N1C=1,p1C=2.7.
Note thatc11=2.5<2.7=p1C. Therefore, C=A⊕2B is not a SDD1 matrix.
The following Example3 shows that Theorem 3 does not hold when k=2.Example 3.
We consider the following matrices:(32)A=3−1−1−1101−1103,B=971252045,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 2 by computation. Therefore, A⊕1B is a SDD1 matrix. However, C=A⊕2B is not a SDD1 matrix. In fact,(33)C=A⊕2B=3−1−10−11981−112820045.
By calculation,(34)N1C=3,p3C≈8.58.
C=A⊕2B is not a SDD1 matrix because c33=8<p3C.
Examples2 and 3 motivate the search for other conditions such that C=A⊕2B is a SDD1 matrix, where A is a SDD1 matrix and B is a SDD matrix.
Next, we give some sufficient conditions ensuring that the 2-subdirect sum ofSDD1 matrices and SDD matrices is a SDD1 matrix.Theorem 4.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=2, S1=1,2,…,n1−2,S2=n1−1,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−2. We assume that A is a SDD1 matrix with n1−1∉N1Aandn1∉N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(35)rn1−1Ab11an1−1,n1−1an1−1,n1−1+b11ai,n1−1+rn1Ab22an1n1an1n1+b22ain1≥r1B+an1−1,n1+b12−an1−1,n1−b12an1−1,n1−1+b11ai,n1−1+r2B+an1,n1−1+b21−an1,n1−1−b21an1n1+b22ain1,then the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1−1∉N1Aandn1∉N1A, we get(36)an1−1,n1−1>rn1−1A,(37)an1n1>rn1A,and aii>piAfor anyi∈N1A.
SinceB is a SDD matrix, we get inequality (16) for anyi∈1,2,…,n2.
According to the 2-subdirect sumC=A⊕2B, we get(38)rn1−1C=rn1−1A+r1B+an1−1,n1+b12−an1−1,n1−b12,(39)rn1C=rn1A+r2B+an1,n1−1+b21−an1,n1−1−b21.
In addition, from the condition that all diagonal entries ofA22 and B11 are positive (or negative), we get(40)cn1−1,n1−1=an1−1,n1−1+b11=an1−1,n1−1+b11,(41)cn1n1=an1n1+b22=an1n1+b22.
For anyan1−1,n1 and b12, we have(42)an1−1,n1+b12≤an1−1,n1+b12.
From equalities (38) and (40) and inequalities (16), (36), and (42), we obtain that(43)cn1−1,n1−1=an1−1,n1−1+b11>rn1−1C.
Similarly, it is easy to obtain that(44)cn1,n1=an1,n1+b22>rn1C.
That is,n1−1∉N1C,n1∉N1C, which means that N1C=N1A.
For anyi∈N1C, it is easy to obtain that i∈N1C∩S1=N1A∩S1⊂S1. Therefore, we have(45)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\iaij+∑j∉N1A∪i,n1−1,n1rjAajjaij+rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1.
From the condition(46)rn1−1Ab11an1−1,n1−1an1−1,n1−1+b11ai,n1−1+rn1Ab22an1n1an1n1+b22ain1≥r1B+an1−1,n1+b12−an1−1,n1−b12an1−1,n1−1+b11ai,n1−1+r2B+an1,n1−1+b21−an1,n1−1−b21an1n1+b22ain1,we have(47)rn1−1Aan1−1,n1−1+b11−rn1−1Aan1−1,n1−1an1−1,n1−1an1−1,n1−1+b11ai,n1−1+rn1Aan1n1+b22−rn1Aan1n1an1n1an1n1+b22ain1≥r1B+an1−1,n1+b12−an1−1,n1−b12an1−1,n1−1+b11ai,n1−1+r2B+an1,n1−1+b21−an1,n1−1−b21an1n1+b22ain1.
Thus,(48)rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1≥rn1−1A+r1B+an1−1,n1+b12−an1−1,n1−b12an1−1,n1−1+b11ai,n1−1+rn1A+r2B+an1,n1−1+b21−an1,n1−1−b21an1n1+b22ain1,and substituting equalities (38), (39), (40), and (41) into this inequality, we obtain that(49)rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1≥rn1−1Ccn1−1,n1−1ai,n1−1+rn1Ccn1n1ain1.
Hence,(50)|cii|=|aii|>piA=∑j∈N1A\iaij+∑j∉N1A∪i,n1−1,n1rjAajjaij+rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1≥∑j∈N1C\icij+∑j∉N1C∪i,n1−1,n1rjCcjjcij+rn1−1Ccn1−1,n1−1ci,n1−1+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
Therefore, it is obvious that, for anyi∈N1C,cii>riC, that is, the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Corollary 1.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=2, S1=1,2,…,n1−2,S2=n1−1,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−2. We assume that A is a SDD1 matrix with n1−1∉N1Aandn1∉N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative),(51)rn1−1Aan1−1,n1−1≥r1B+an1−1,n1+b12−an1−1,n1−b12b11,rn1Aan1n1≥r2B+an1,n1−1+b21−an1,n1−1−b21b22,then the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1−1∉N1Aandn1∉N1A, we obtain that aii>piAfor anyi∈N1A with inequalities (36) and (37). Since B is a SDD matrix, we obtain inequality (16) for anyi∈1,2,…,n2.
From equalities (38) and (40) and inequalities (16), (36), and (42), we obtain that cn1−1,n1−1=an1−1,n1−1+b11>rn1−1C. In a similar manner, we obtain that cn1n1=an1n1+b22>rn1C. That is, n1−1∉N1C,n1∉N1C, which means that N1C=N1A.
For anyi∈N1C, it is easy to get that i∈N1C∩S1=N1A∩S1⊂S1. Therefore, we obtain that(52)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\iaij+∑j∉N1A∪i,n1−1,n1rjAajjaij+rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1.
For the condition(53)rn1−1Aan1−1,n1−1≥r1B+an1−1,n1+b12−an1−1,n1−b12b11,let us multiply by an1−1,n1−1b11 on both sides of this inequality, and then add rn1−1Aan1−1,n1−1 on both sides, simultaneously; multiply both sides by 1/an1−1,n1−1an1−1,n1−1+b11, in addition, and then from equations (37) and (39), we obtain that(54)rn1−1Aan1−1,n1−1≥rn1−1Ccn1−1,n1−1.
In a similar manner, it is easy to get that(55)rn1Aan1n1≥rn1Ccn1n1.
Hence,(56)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪i,n1−1,n1rjAajjaij+rn1−1Aan1−1,n1−1ai,n1−1+rn1Aan1n1ain1≥∑j∈N1C\icij+∑j∉N1C∪i,n1−1,n1rjCcjjcij+rn1−1Ccn1−1,n1−1ci,n1−1+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
Therefore, we get that, for anyi∈N1C,cii>riC, that is, the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Theorem 5.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=2, S1=1,2,…,n1−2,S2=n1−1,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−2. We assume that A is a SDD1 matrix with n1−1∈N1Aandn1∈N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), then the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1−1∈N1Aandn1∈N1A, we get(57)an1−1,n1−1≤rn1−1A,an1n1≤rn1A,andaii>piAfor anyi∈N1A. Since B is a SDD matrix, we get inequality (16) for anyi∈1,2,…,n2.
From equalities (38) and (40) and inequalities (16), (42), and (57), we obtain that cn1−1,n1−1=an1−1,n1−1+b11>rn1−1Corcn1−1,n1−1=an1−1,n1−1+b11≤rn1−1C, that is, n1−1∉N1Corn1−1∈N1C. In a similar manner, we obtain cn1n1=an1n1+b22>rn1Corcn1n1=an1n1+b22≤rn1C, that is, n1∉N1Corn1∈N1C. Hence, the proof for any i∈N1C,cii>piC can be divided into four cases as follows:Case 1: ifan1−1,n1−1+b11≤rn1−1Candan1n1+b22≤rn1C, then n1−1∈N1C,n1∈N1C, which means that N1C=N1A.For anyi∈N1C∩S1, we have i∈N1A∩S1⊂S1. Therefore, we have(58)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−1∈N1C∩S2, we have i=n1−1∈N1A∩S2⊂S2. Therefore, we obtain that(59)cn1−1,n1−1=an1−1,n1−1+b11>pn1−1A+r1B=∑j∈N1A\n1−1an1−1,j+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j≠1n2b1j=∑j∈N1A\n1−1,n1an1−1,j+an1−1,n1+b12+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j=3n2b1j≥∑j∈N1A\n1−1,n1an1−1,j+an1−1,n1+b12+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j=3n2rjBbjjb1j=∑j∈N1C\n1−1,n1cn1−1,j+cn1−1,n1+∑j∉N1C∪n1−1rjCcjjcn1−1,j=∑j∈N1C\n1−1cn1−1,j+∑j∉N1C∪n1−1rjCcjjcn1−1,j=pn1−1C.Fori=n1∈N1C∩S2, we have i=n1∈N1A∩S2⊂S2.Similar to the proof ofi=n1−1∈N1C∩S2 of Case 1 in Theorem 5, it is easy to prove that, for n1∈N1C,cn1n1>pn1C.Case 2: ifan1−1,n1−1+b11≤rn1−1Candan1n1+b22>rn1C, then n1−1∈N1C,n1∉N1C, which means that N1C=N1A\n1.For anyi∈N1C∩S1, we have i∈N1A\n1∩S1⊂S1. Therefore, we have(60)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\i,n1aij+ain1+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+cin1≥∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−1∈N1C∩S2, we have i∈N1A\n1∩S2⊂S2. Therefore, we obtain that(61)cn1−1,n1−1=an1−1,n1−1+b11>pn1−1A+r1B=∑j∈N1A\n1−1an1−1,j+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j≠1n2b1j=∑j∈N1A\n1−1,n1an1−1,j+an1−1,n1+b12+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j=3n2b1j≥∑j∈N1A\n1−1,n1an1−1,j+an1−1,n1+b12+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j=3n2rjBbjjb1j=∑j∈N1C\n1−1cn1−1,j+cn1−1,n1+∑j∉N1C∪n1−1,n1rjCcjjcn1−1,j≥∑j∈N1C\n1−1cn1−1,j+rn1Ccn1n1cn1−1,n1+∑j∉N1C∪n1−1,n1rjCcjjcn1−1,j=∑j∈N1C\n1−1cn1−1,j+∑j∉N1C∪n1−1rjCcjjcn1−1,j=pn1−1C.Case 3: ifan1−1,n1−1+b11>rn1−1Candan1n1+b22≤rn1C, then n1−1∉N1C,n1∈N1C.
Similar to the proof of Case 2 in Theorem5, it is easy to prove that, for any i∈N1C,|cii|>piC.Case 4: ifan1−1,n1−1+b11>rn1−1Candan1n1+b22>rn1C, then n1−1∉N1C,n1∉N1C, which means that N1C=N1A\n1−1,n1.For anyi∈N1C, we have i∈N1A\n1−1,n1⊂S1. Therefore, we have(62)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\i,n1−1,n1aij+ai,n1−1+ain1+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∉N1C∪i,n1−1,n1rjCcjjcij+ci,n1−1+cin1≥∑j∈N1C\icij+∑j∉N1C∪i,n1−1,n1rjCcjjcij+rn1−1Ccn1−1,n1−1ci,n1−1+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
From Cases 1, 2, 3, and 4, we get that, for anyi∈N1C,cii>piC, which implies that C=A⊕2B is a SDD1 matrix.Theorem 6.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=2, S1=1,2,…,n1−2,S2=n1−1,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−2. We assume that A is a SDD1 matrix with n1−1∈N1Aandn1∉N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(63)rn1Aan1n1≥r2B+an1,n1−1+b21−an1,n1−1−b21b22,then the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Proof.
SinceA is a SDD1 matrix with n1−1∈N1Aandn1∉N1A, we get that aii>piAfor anyi∈N1A with inequalities (37) and (57). Since B is a SDD matrix, we get inequality (16) for anyi∈1,2,…,n2.
From equalities (38) and (40) and inequalities (16), (42), and (57), we get that cn1−1,n1−1=an1−1,n1−1+b11>rn1−1Corcn1−1,n1−1=an1−1,n1−1+b11≤rn1−1C, that is, n1∉N1Corn1∈N1C. And from equalities (39) and (41) and inequalities (16), (37), and (42), we have cn1n1=an1n1+b22>rn1C, that is, n1∉N1C. Hence, the proof for any i∈N1C,cii>piC can be divided into two cases as follows:Case 1: ifan1−1,n1−1+b11≤rn1−1C, then n1−1∈N1C, which means that N1C=N1A.For anyi∈N1C∩S1, it is easy to get that i∈N1A∩S1⊂S1. Therefore, we have(64)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\iaij+∑j∉N1A∪i,n1rjAajjaij+rn1Aan1n1ain1.For the condition(65)rn1Aan1n1≥r2B+an1,n1−1+b21−an1,n1−1−b21b22,let us multiply byan1n1b22 on both sides of this inequality, and then add rn1Aan1n1 on both sides, simultaneously; multiply both sides by 1/an1n1an1n1+b22, in addition, and then from equations (39) and (40), we obtain that(66)rn1Aan1n1≥rn1Ccn1n1.Hence,(67)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪i,n1rjAajjaij+rn1Aan1n1ain1≥∑j∈N1C\icij+∑j∉N1C∪i,n1rjCcjjcij+rn1Ccn1n1cin1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−1∈N1C∩S2, it is easy to get that i∈N1A∩S2⊂S2. Therefore, we obtain that(68)cn1−1,n1−1=an1−1,n1−1+b11>pn1−1A+r1B=∑j∈N1A\n1−1an1−1,j+∑j∉N1A∪n1−1rjAajjan1−1,j+∑j≠1n2b1j≥∑j∈N1A\n1−1an1−1,j+∑j∉N1A∪n1−1,n1rjAajjan1−1,j+rn1Aan1n1an1−1,n1+b12+∑j=3n2rjBbjjb1j.Multiplying both sides of inequality (66) by an1−1,n1 and then adding b12 on both sides, in addition, combining cn1−1,n1=an1−1,n1+b12≤an1−1,n1+b12, we obtain(69)rn1Aan1n1an1−1,n1+b12≥rn1Ccn1n1an1−1,n1+b12≥rn1Ccn1n1an1−1,n1+b12≥rn1Ccn1n1cn1−1,n1.Therefore,(70)cn1−1,n1−1=an1−1,n1−1+b11>pn1−1A+r1B≥∑j∈N1A\n1−1an1−1,j+∑j∉N1A∪n1−1,n1rjAajjan1−1,j+rn1Aan1n1an1−1,n1+b12+∑j=3n2rjBbjjb1j≥∑j∈N1C\n1−1cn1−1,j+∑j∉N1C∪n1−1,n1rjCcjjcn1−1,j+rn1Ccn1n1cn1−1,n1=∑j∈N1C\n1−1cn1−1,j+∑j∉N1C∪n1−1rjCcjjcn1−1,j=pn1−1C.Case 2: ifan1−1,n1−1+b11>rn1−1C, then n1−1∉N1C, which means that N1C=N1A\n1−1.For anyi∈N1C, it is easy to get that i∈N1A\n1−1⊂S1. So, we obtain that(71)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\i,n1−1aij+ai,n1−1+∑j∉N1A∪i,n1rjAajjaij+rn1Aan1n1ain1≥∑j∈N1C\icij+ci,n1−1+∑j∉N1C∪i,n1−1,n1rjCcjjcij+rn1Ccn1n1cin1≥∑j∈N1C\icij+∑j∉N1C∪i,n1−1rjCcjjcij+rn1−1Ccn1−1,n1−1ci,n1−1=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
From Cases 1 and 2, we get that, for anyi∈N1C, cii>piC, so the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Theorem 7.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k=2, S1=1,2,…,n1−2,S2=n1−1,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−2. We assume that A is a SDD1 matrix with n1−1∉N1Aandn1∈N1A, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(72)rn1−1Aan1−1,n1−1≥r1B+an1−1,n1+b12−an1−1,n1−b12b11,then the 2-subdirect sum C=A⊕2B is a SDD1 matrix.Proof.
Using the same technique as in the proof of Theorem6, it is easy to prove that cii|>piCfor anyi∈N1C; hence, the 2-subdirect sum C=A⊕2B is a SDD1 matrix.
In Theorems4–7 and Corollary 1, we obtain some sufficient conditions that ensure the 2-subdirect sum of SDD1 matrices is a SDD1 matrix. The following four examples are presented to illustrate the corresponding results, that is, under these conditions, the 2-subdirect sum A⊕2B is a SDD1 matrix, where A is a SDD1 matrix and B is a SDD matrix.Example 4.
We consider the following matrices:(73)A=9.555−6−12012−2150−21712,B=1611−21411211511−3218,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 4 by computation. Therefore, A⊕2B is a SDD1 matrix. However, C=A⊕3B is not a SDD1 matrix. In fact,(74)C=A⊕3B=9.555−60−13623−2−22911−23827110−3218.
By calculation,(75)N1C=1,p1C≈9.78.
Note thatc11=9.5<p1C. Therefore, C=A⊕3B is not a SDD1 matrix.Example 5.
We consider the following matrices:(76)A=82111810881584426,B=1154101081251010214,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 5 by computation. Therefore, A⊕2B is a SDD1 matrix. However, C=A⊕3B is not a SDD1 matrix. In fact,(77)C=A⊕3B=82110119641882516146716100214.
By calculation,(78)N1C=3,4,p3C≈25.80,p4C≈13.54.
Note thatc33=25<p3C. Therefore, C=A⊕3B is not a SDD1 matrix.Example 6.
We consider the following matrices:(79)A=16−8480831−882081158,B=8150691−14410−10−1810,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 6 by computation. Therefore, A⊕2B is a SDD1 matrix. However, C=A⊕3B is not a SDD1 matrix. In fact,(80)C=A⊕3B=16−8480016460−814299−115918−100−1810.
By calculation,(81)N1C=1,3,p1C≈16.11,p3C=25.65.
C=A⊕3B is not a SDD1 matrix because c11=16<p1C.Example 7.
We consider the following matrices:(82)A=16−88408311581−88820,B=816018504410−10−1−810,where A is a SDD1 matrix and B is a SDD matrix. It is easy to verify that matrices A and B satisfy the condition of Theorem 7 by calculation. Therefore, A⊕2B is a SDD1 matrix. However, C=A⊕3B is not a SDD1 matrix. In fact,(83)C=A⊕3B=16−8840016470161660−8121230−100−1−810.
By calculation,(84)N1C=1,4,p1C=16,p4C=26.9.
Note thatc11=16=p1C, which implies that C=A⊕3B is not a SDD1 matrix.
Examples4–7 show that A⊕2B is a SDD1 matrix; however, C=A⊕3B is not necessarily a SDD1 matrix.
Finally, we search for other sufficient conditions such thatC=A⊕kB (3≤k≤minn1,n2, k is an integer) is a SDD1 matrix, where A is a SDD1 matrix and B is a SDD matrix.Theorem 8.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k≥3, S1=1,2,…,n1−k,S2=n1−k+1,…,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−k. We assume that A is a SDD1 matrix with i∉N1Afor anyi∈S2, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(85)∑j=n1−k+1n1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j=n1−k+1n1εj−rjAajj+bj−n1+k,j−n1+kaij,where(86)εj=rjA+rj−n1+kB+∑i=n1−k+1i≠jn1aji+bj−n1+k,i−n1+k−∑i=n1−k+1i≠jn1aji+bj−n1+k,i−n1+k,then the k-subdirect sum C=A⊕kB is a SDD1 matrix.Proof.
Letm be a positive integer, and 1≤m≤k.
SinceA is a SDD1 matrix with i∉N1Afor anyi=n1−k+m∈S2, we get(87)an1−k+m,n1−k+m>rn1−k+mA,and for any i∈N1A,aii>piA.
SinceB is a SDD matrix, we get inequality (16) for anyi∈1,2,…,n2.
According to thek-subdirect sum C=A⊕kB, we get(88)rn1−k+mC=εn1−k+m.
In addition, from the condition that all diagonal entries of A22 and B11 are positive (or negative), we have(89)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm=an1−k+m,n1−k+m+bmm.
For anyan1−k+m,j and bm,j−n1+k, we obtain that(90)an1−k+m,j+bm,j−n1+k<an1−k+m,j+bm,j−n1+k.
From equalities (88) and (89) and inequalities (16), (87), and (90), we get that cn1−k+m,n1−k+m>rn1−k+mC, that is, for any i=n1−k+m∈S2,i∉N1C, which means that N1C=N1A.
For anyi∈N1C, it is easy to get that i∈N1C∩S1=N1A∩S1⊂S1. Therefore, we get that(91)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A\iaij+∑j∉N1A∪i∪S2rjAajjaij+∑j∈S2\irjAajjaij.
From the condition(92)∑j=n1−k+1n1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j=n1−k+1n1εj−rjAajj+bj−n1+k,j−n1+kaij,we obtain that(93)∑j=n1−k+1n1rjAajj+bj−n1+k,j−n1+k−rjAajjajjajj+bj−n1+k,j−n1+kaij≥∑j=n1−k+1n1εj−rjAajj+bj−n1+k,j−n1+kaij.
Thus,(94)∑j=n1−k+1n1rjAajjaij≥∑j=n1−k+1n1εjajj+bj−n1+k,j−n1+kaij,and substituting equalities (88) and (89) into this inequality, we get(95)∑j∈S2rjAajjaij≥∑j∈S2rjCcjjaij.
Hence,(96)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪i∪S2rjAajjaij+∑j∈S2\irjAajjaij≥∑j∈N1C\icij+∑j∉N1C∪i∪S2rjCcjjcij+∑j∈S2\irjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.
Therefore, we can draw a conclusion thatcii>piC for any i∈N1C, which implies that the k-subdirect sum C=A⊕kB is a SDD1 matrix.Corollary 2.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k≥3, S1=1,2,…,n1−k,S2=n1−k+1,…,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−k. We assume that A is a SDD1 matrix with i∉N1Afor anyi∈S2, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(97)rjAajj≥εj−rjAbj−n1+k,j−n1+k,where j∈S2 and εj is as in Theorem 8, then the k-subdirect sum C=A⊕kB is a SDD1 matrix.Proof.
For the condition(98)rjAajj≥εj−rjAbj−n1+k,j−n1+k,multiplying both sides of this inequality by bj−n1+k,j−n1+k/ajj+bj−n1+k,j−n1+k, we obtain that(99)rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+k≥εj−rjAajj+bj−n1+k,j−n1+k.
Thus, it is easy to get that(100)∑j=n1−k+1n1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j=n1−k+1n1εj−rjAajj+bj−n1+k,j−n1+kaij.
Therefore, by Theorem8, we obtain that the k-subdirect sum C=A⊕kB is a SDD1 matrix. □Theorem 9.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k≥3, S1=1,2,…,n1−k,S2=n1−k+1,…,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−k. We assume that A is a SDD1 matrix with i∈N1Afor anyi∈S2, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), then the k-subdirect sum C=A⊕kB is a SDD1 matrix.Proof.
Letm be a positive integer, and 1≤m≤k.
SinceA is a SDD1 matrix with i∈N1Afor anyi=n1−k+m∈S2, we get that aii>piA for any i∈N1A, and(101)an1−k+m,n1−k+m≤rn1−k+mA.
SinceB is a SDD matrix, we obtain inequality (16) for anyi∈1,2,…,n2.
From equalities (88) and (89) and inequalities (16), (90), and (101), we obtain that, for any m, cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>rn1−k+mCorcn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm≤rn1−k+mC, that is, for any m, n1−k+m∈N1Corn1−k+m∉N1C.
Suppose(102)N2C1=n1−k+man1−k+m,n1−k+m+bmm≤rn1−k+mC,N3C1=n1−k+man1−k+m,n1−k+m+bmm>rn1−k+mC.
Obviously,N2C1∪N3C1=S2=n1−k+1,…,n1; then, N2C1=∅orN2C1=S2orN2C1≠∅andN2C1≠S2. Hence, the proof for any i∈N1C,cii>piC can be divided into three cases as follows:Case 1: ifN2C1=∅, then N3C1=S2, which means that N1C=N1A\N3C1.For anyi∈N1C, we have i∈N1A\N3C1⊂S1. Therefore, we get that(103)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N1A∩S2\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1C\icij+∑j∈N3C1\icij+∑j∉N1C∪i∪N3C1rjCcjjcij≥∑j∈N1C\icij+∑j∉N1C∪i∪N3C1rjCcjjcij+∑j∈N3C1\irjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Case 2: ifN2C1=S2, then N3C1=∅, which means that N1C=N1A.For anyi∈N1C∩S1, we have i∈N1A∩S1⊂S1. Therefore, we get that(104)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N1A∩S2\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1C∩S1\icij+∑j∈N2C1\icij+∑j∉N1C∪irjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−k+m∈N1C∩S2, we have i=n1−k+m∈N1A∩S2⊂S2. Therefore, we obtain that(105)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>pn1−k+mA+rmB=∑j∈N1A\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=1j≠mn2bmj≥∑j∈N1A∩S1\n1−k+man1−k+m,j+∑j∈N1A∩S2\n1−k+man1−k+m,j+∑j=1j≠mkbmj+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=k+1n2rjBbjjbmj≥∑j∈N1C∩S1\n1−k+mcn1−k+m,j+∑j∈N2C1\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+mrjCcjjcn1−k+m,j=∑j∈N1C\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+mrjCcjjcn1−k+m,j=pn1−k+mC.Case 3: ifN2C1≠∅andN2C1≠S2, then N3C1≠∅andN3C1≠S2, which means that N1C=N1A\N3C1.For anyi∈N1C∩S1, we have i∈N1A\N3C1∩S1⊂S1. Therefore, we obtain that(106)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N1A∩S2\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1C∩S1\icij+∑j∈N2C\icij+∑j∈N3C\icij+∑j∉N1C∪i∪N3CrjCcjjcij≥∑j∈N1C\icij+∑j∈N3C1\irjCcjjcij+∑j∉N1C∪i∪N3CrjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−k+m∈N1C∩S2, we know i=n1−k+m∈N1A\N3C1∩S2⊂S2. Therefore, we get that(107)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>pn1−k+mA+rmB=∑j∈N1A\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=1j≠mn2bmj≥∑j∈N1A∩S1\n1−k+man1−k+m,j+∑j∈N1A∩S2\n1−k+man1−k+m,j+∑j=1j≠mkbmj+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=k+1n2rjBbjjbmj≥∑j∈N1C∩S1\n1−k+mcn1−k+m,j+∑j∈N2C1\n1−k+mcn1−k+m,j+∑j∈N3C1\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+m∪N3C1rjCcjjcn1−k+m,j≥∑j∈N1C\n1−k+mcn1−k+m,j+∑j∈N3C1\n1−k+mrjCcjjcn1−k+m,j+∑j∉N1C∪n1−k+m∪N3C1rjCcjjcn1−k+m,j=∑j∈N1C\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+mrjCcjjcn1−k+m,j=pn1−k+mC.
From Cases 1, 2, and 3, we get that, for anyi∈N1C, cii>piC, which implies the k-subdirect sum C=A⊕kB is a SDD1 matrix.Theorem 10.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k≥3, S1=1,2,…,n1−k,S2=n1−k+1,…,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−k. We assume that A is a SDD1 matrix with N1A∩S2≠∅andN1A∩S2≠S2, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(108)∑j∉N1A∪i∪S1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j∉N1A∪i∪S1εj−rjAajj+bj−n1+k,j−n1+kaij,where i∈S1∪S2 and εj is as in Theorem 8, then the k-subdirect sum C=A⊕kB is a SDD1 matrix.Proof.
Letm be a positive integer, and 1≤m≤k.
SinceA is a SDD1 matrix, we obtain inequality (87) or (101) for any m and aii>piA for any i∈N1A.
Suppose(109)N2A1=iaii>riA,i∈S2,N3A1=iaii≤riA,i∈S2.
Obviously,(110)N2A1∪N3A1=S2=n1−k+1,…,n1.
SinceN1A∩S2≠∅andN1A∩S2≠S2, we have N3A1≠∅andN3A1≠S2.
SinceB is a SDD matrix, we obtain inequality (16) for anyi∈1,2,…,n2.
From equalities (88) and (89) and inequalities (16), (87), and (90), we obtain that, for any n1−k+m∈N2A1, cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>rn1−k+mC. And from equalities (88) and (89) and inequalities (16), (90), and (101), we obtain that, for any n1−k+m∈N3A1, cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>rn1−k+mCorcn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm≤rn1−k+mC.
Therefore, suppose(111)N2C1=iaii+bi−n1+k,i−n1+k>riA,i∈N2A1,N3C1=iaii+bi−n1+k,i−n1+k>riA,i∈N3A1,N4C1=iaii+bi−n1+k,i−n1+k≤riA,i∈N3A1.
Obviously,(112)N2C1∪N3C1∪N4C1=S2,N2C1=N2A1,N3C1∪N4C1=N3A1.
Thus, we obtainN3C1=N3A1 or N4C1=N3A1 or N3C1∪N4C1=N3A1. Hence, the proof for any i∈N1C,cii>piC can be divided into three cases as follows:Case 1: ifN3C1=N3A1, then N4C1=∅; thus, N2C1∪N3C1=S2, which means that N1C=N1A\N3C1.For anyi∈N1C, we have i∈N1A\N3C1⊂S1. Therefore, we obtain that(113)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N3A1\iaij+∑j∉N1A∪i∪N2A1rjAajjaij+∑j∈N2A1\irjAajjaij.From the condition(114)∑j∉N1A∪i∪S1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j∉N1A∪i∪S1εj−rjAajj+bj−n1+k,j−n1+kaij,we obtain that(115)∑j∉N1A∪i∪S1rjA|ajj|+|bj−n1+k,j−n1+k|−rjA|ajjajj||ajj|+|bj−n1+k,j−n1+k||aij|≥∑j∉N1A∪i∪S1εj−rjAajj|+|bj−n1+k,j−n1+k|aij|.Thus, it is easy to obtain that(116)∑j∉N1A∪i∪S1rjAajjaij≥∑j∉N1A∪i∪S1εjajj+bj−n1+k,j−n1+kaij,and substituting equalities (88) and (89) into this inequality, we obtain that(117)∑j∉N1A∪i∪S1rjAajjaij≥∑j∉N1A∪i∪S1rjCcjjaij.Thus, we obtain that(118)∑j∈N2A1\irjAajjaij≥∑j∈N2A1\irjCcjjaij.Hence,(119)cii=aii>piA=∑j∈N1A∩S1\iaij+∑j∈N3A1\iaij+∑j∉N1A∪i∪N2A1rjAajjaij+∑j∈N2A1\irjAajjaij≥∑j∈N1C\icij+∑j∈N3C1\icij+∑j∉N1C∪i∪N2C1∪N3C1rjCcjjcij+∑j∈N2C1\irjCcjjcij≥∑j∈N1C\icij+∑j∈N3C1\irjCcjjcij+∑j∉N1C∪i∪N3C1rjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Case 2: ifN4C1=N3A1, then N3C1=∅; thus, N2C1∪N4C1=S2, which means that N1C=N1A.For anyi∈N1C∩S1, we know i∈N1A∩S1⊂S1. Therefore, we obtain that(120)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N3A1\iaij+∑j∉N1A∪i∪N2A1rjAajjaij+∑j∈N2A1\irjAajjaij≥∑j∈N1C∩S1\icij+∑j∈N4C1\icij+∑j∉N1C∪i∪N2C1rjCcjjcij+∑j∈N2C1\irjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−k+m∈N1C∩S2, we have i=n1−k+m∈N1A∩S2⊂S2. Therefore, we obtain that(121)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>pn1−k+mA+rmB=∑j∈N1A\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=1j≠mn2bmj≥∑j∈N1A∩S1\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+m∪N2A1rjAajjan1−k+m,j+∑j∈N3A1\n1−k+man1−k+m,j+∑j=1j≠mkbmj+∑j∈N2A1\n1−k+mrjAajjan1−k+m,j+∑j=k+1n2rjBbjjbmj.It follows from inequality (118) that(122)∑j∈N2A1\n1−k+mrjAajjan1−k+m,j≥∑j∈N2A1\n1−k+mrjCcjjan1−k+m,j.Then,(123)∑j∈N2A1\n1−k+mrjAajjan1−k+m,j+bm,j−n1+k≥∑j∈N2A1\n1−k+mrjCcjjan1−k+m,j+bm,j−n1+k≥∑j∈N2A1\n1−k+mrjCcjjan1−k+m,j+bm,j−n1+k≥∑j∈N2A1\n1−k+mrjCcjjcn1−k+m,j.Therefore,(124)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>pn1−k+mA+rmB≥∑j∈N1A∩S1\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+m∪N2A1rjAajjan1−k+m,j+∑j∈N3A1\n1−k+man1−k+m,j+∑j=1j≠mkbmj+∑j∈N2A1\n1−k+mrjAajjan1−k+m,j+∑j=k+1n2rjBbjjbmj≥∑j∈N1C∩S1\n1−k+mcn1−k+m,j+∑j∈N4C1\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+m∪N2C1rjCcjjcn1−k+m,j+∑j∈N2C1\n1−k+mrjCcjjcn1−k+m,j=∑j∈N1C\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+mrjCcjjcn1−k+m,j=pn1−k+mC.Case 3: ifN3C1∪N4C1=N3A1, N3C1≠∅andN4C1≠∅, which mean that N1C=N1A\N3C1.For anyi∈N1C∩S1, we have i∈N1A\N3C1∩S1⊂S1. Therefore, we obtain that(125)cii=aii>piA=∑j∈N1A\iaij+∑j∉N1A∪irjAajjaij=∑j∈N1A∩S1\iaij+∑j∈N3A1\iaij+∑j∉N1A∪i∪N2A1rjAajjaij+∑j∈N2A1\irjAajjaij≥∑j∈N1C∩S1\icij+∑j∈N4C1\icij+∑j∈N3C1\icij+∑j∉N1C∪i∪N2C1∪N3C1rjCcjjcij+∑j∈N2C1\irjCcjjcij≥∑j∈N1C\icij+∑j∈N3C1\irjCcjjcij+∑j∉N1C∪i∪N3C1rjCcjjcij=∑j∈N1C\icij+∑j∉N1C∪irjCcjjcij=piC.Fori=n1−k+m∈N1C∩S2=N4C1, we have i=n1−k+m∈N1A\N3C1∩S2⊂S2. Therefore, we have(126)cn1−k+m,n1−k+m=an1−k+m,n1−k+m+bmm>pn1−k+mA+rmB=∑j∈N1A\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+mrjAajjan1−k+m,j+∑j=1j≠mn2bmj≥∑j∈N1A∩S1\n1−k+man1−k+m,j+∑j∉N1A∪n1−k+m∪N2A1rjAajjan1−k+m,j+∑j∈N3A1\n1−k+man1−k+m,j+∑j=1j≠mkbmj+∑j∈N2A1\n1−k+mrjAajjan1−k+m,j+∑j=k+1n2rjBbjjbmj≥∑j∈N1C∩S1\n1−k+mcn1−k+m,j+∑j∈N4C1\n1−k+mcn1−k+m,j+∑j∈N3C1\n1−k+mcn1−k+m,j+∑j∈N2C1\n1−k+mrjCcjjcn1−k+m,j+∑j∉N1C∪n1−k+m∪N2C1∪N3C1rjCcjjcn1−k+m,j≥∑j∈N1C\n1−k+mcn1−k+m,j+∑j∈N3C\n1−k+mrjCcjjcn1−k+m,j+∑j∉N1C∪n1−k+m∪N3C1rjCcjjcn1−k+m,j=∑j∈N1C\n1−k+mcn1−k+m,j+∑j∉N1C∪n1−k+mrjCcjjcn1−k+m,j=pn1−k+mC.
From Cases 1, 2, and 3, we obtain thatcii>piC for any i∈N1C, which implies that the k-subdirect sum C=A⊕kB is a SDD1 matrix.Corollary 3.
LetA=aij and B=bij be square matrices of order n1 and n2, respectively, which are partitioned as in (1). Let k≥3, S1=1,2,…,n1−k,S2=n1−k+1,…,n1,andS3=n1+1,n1+2,…,n, where n=n1+n2−k. We assume that A is a SDD1 matrix with N1A∩S2≠∅andN1A∩S2≠S2, and B is a SDD matrix. If all diagonal entries of A22 and B11 are positive (or negative), and(127)rjAajj≥εj−rjAbj−n1+k,j−n1+k,where j∉N1A∪S1 and εj is as in Theorem 8, then the k-subdirect sum C=A⊕kB is a SDD1 matrix.Proof.
For the condition(128)rjAajj≥εj−rjAbj−n1+k,j−n1+k,multiplying both sides of this inequality by aijbj−n1+k,j−n1+k/ajj+bj−n1+k,j−n1+kj≠i, we obtain that, for any j∉N1A∪S1,(129)rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥εj−rjAajj+bj−n1+k,j−n1+kaij,and by summing for every j∉N1A∪i∪S1, we obtain(130)∑j∉N1A∪i∪S1rjAbj−n1+k,j−n1+kajjajj+bj−n1+k,j−n1+kaij≥∑j∉N1A∪i∪S1εj−rjAajj+bj−n1+k,j−n1+kaij.
By Theorem10, we obtain that the k-subdirect sum C=A⊕kB is a SDD1 matrix.


## 3. Conclusions

In this paper, some sufficient conditions such that the subdirect sum ofSDD1 matrices is in the class of SDD1 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.



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