# Subdirect Sums of Doubly Strictly Diagonally Dominant Matrices

Authors: Yating Li; Xiaoyong Chen; Yi Liu; Lei Gao; Yaqiang Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, the question of when the subdirect sum of two doubly strictly diagonally dominant (DSDDs) matrices is addressed. Some sufficient conditions are given, and these sufficient conditions only depend on the elements of the given matrices. Moreover, examples are presented to illustrate the corresponding results.


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## Body

## 1. Introduction

In 1999, the concept of subdirect sums of square matrices was introduced by Fallat and Johnson, which is a generalization of the usual sum of matrices [1], and arises in several contexts, such as matrix completion problems, overlapping subdomains in domain decomposition problems, and global stiffness matrices in finite element methods [2].For a given class matrix, an important problem is that whether thek-subdirect sums of matrices belong to the same class or not, which has been widely concerned for differently classes of matrices, such as nonsingular M-matrices [3], S-strictly diagonally dominant matrices [4], ∑-strictly diagonally dominant matrices [5], doubly diagonally dominant matrices [6], Nekrasov matrices [7, 8], and SDD1 matrices [9].In this paper, we focus on the subdirect sum of doubly strictly diagonally dominant (shortly asDSDD) matrices, which is a subclass of H-matrices [10], and some sufficient conditions such that the k-subdirect sums of DSDD matrices belong to DSDD matrices are given, and these sufficient conditions only depend on the elements of the given matrices.Now, some notations and definitions are listed, which can also be found in [1, 11–13].Letn be an integer number. Cn×n is the set of complex matrices.Definition 1.
(see [1]). Let A and B be square matrices of orders n1 and n2, respectively, and k be an integer number such that 1≤k≤minn1,n2. Let A and B be partitioned in a 2×2 block as follows:(1)A=A11A12A21A22,B=B11B12B21B22,where A22 and B11 are the square matrices of order k. We call the following square matrix of order n=n1+n2−k,(2)C=A11A120A21A22+B11B120B21B22,the k-subdirect sum of A and B, and we denote it by C=A⊕kB.In order to more explicitly express each element ofC in terms of the ones of A and B, we can define the following set of indices:(3)S1=1,2,…,n1−k,S2=n1−k+1,n1−k+2,…,n1,S3=n1+1,n1+2,…,n.Then,C can be expressed as follows:(4)C=a11⋯a1,ta1,p⋯a1n10⋯0⋮⋱⋮⋮⋱⋮⋮⋱⋮at,1⋯at,tat,p⋯at,n10⋯0ap,1⋯ap,tap,p+b11⋯ap,n1+b1,n1−tb1,n1−t+1⋯b1,n−t⋮⋱⋮⋮⋱⋮⋮⋱⋮an11⋯an1,tan1,p+bn1−t,1⋯an1,n1+bn1−t,n1−tbn1−t,n1−t+1⋯bn1−t,n−t0⋯0bn1−t+1,1⋯bn1−t+1,n1−tbn1−t+1,n1−t+1⋯bn1−t+1,n−t⋮⋱⋮⋮⋱⋮⋮⋱⋮0⋯0bn−t,1⋯bn−t,n1−tbn−t,n1−t+1⋯bn−t,n−t,cij=aij,i∈S1,j∈S1∪S2,0,i∈S1,j∈S3,aij,i∈S2,j∈S1,aij+bi−t,j−t,i∈S2,j∈S2,bi−t,j−t,i∈S2,j∈S3,0,i∈S3,j∈S1,bi−t,j−t,i∈S3,j∈S2∪S3,where t=n1−kandp=t+1.Definition 2.
(see [12]). Given a matrix A=aij∈Cn×n, A is called (row) diagonally dominant (DD) if(5)aii≥RiA,i=1,2,…,n,where(6)RiA=∑j=1j≠inaij,i=1,2,…,n.
If the inequality in (5) holds strictly for all i, we say that A is strictly diagonally dominant (SDD).Definition 3.
(see [13]). The matrix A=aij∈Cn×n is a doubly strictly diagonally dominant (DSDD) matrix if(7)aiiajj>RiARjA,i,j=1,2,…,n,i≠j.


## 2. Subdirect Sums ofDSDD Matrices

In general, the subdirect sum of twoDSDD matrices is not always a DSDD matrix. We show this in the following example.Example 1.
Let(8)A=111−140−104,B=3−2−1−120114be two DSDD matrices, but(9)C=A⊕1B=1110−17−2−1−1−1600114is not a DSDD matrix, since c11c22=7<R1CR2C=8.Example1 shows that the subdirect sum of DSDD matrices is not a DSDD matrix; then, a meaningful discussion is concerned: under what conditions, the subdirect sum of DSDD matrices is in the class of DSDD matrices?In order to obtain the main results, we need the following lemma.Lemma 1.
LetA=aij and B=bij be DSDD square matrices of orders n1 and n2 partitioned as in (1), respectively. And, let k be an integer number such that 1≤k≤minn1,n2, S1,S2,S3 be defined as in (4), and all diagonal entries of A22 and B11 are positive (or all negative), with C=A⊕kB, then,(a)
Fori∈S1,we haveRiC=RiA(b)
Fork=1,i∈S2,we haveRiC=Rn1C=Rn1A+R1B(c)
Fork⩾2,i∈S2,we haveRiC⩽RiA+Ri−tB(d)
Fori∈S3,we haveRiC=Ri−tBProof.
Fori∈S1, we can write(10)RiC=∑j∈S1j≠iaij+∑j∈S2aij+0=RiA.
Fork=1,i∈S2=n1, we can obtain(11)RiC=Rn1C=∑j∈S1aij+∑j∈S3j−t≠1b1,j−t=Rn1A+R1B.
Fork>2,i∈S2, we obtain(12)RiC=∑j∈S1aij+∑j∈S2j≠iaij+bi−t,j−t+∑j∈S3bi−t,j−t≤∑j∈S1aij+∑j∈S2j≠iaij+∑j∈S2j≠ibi−t,j−t+∑j∈S3bi−t,j−t=RiA+Ri−tB.
For the rest case ofi∈S3, the proof is similar to the proof of i∈S1.
Firstly, we study the 1-subdirect sum ofDSDD matrices.Theorem 1.
LetA and B be DSDD matrices of orders n1 and n2 partitioned as in (1), respectively, and k=1. Then, C=A⊕1B is a DSDD matrix if all diagonal entries of A22 and B11 are positive (or all negative) and for i∈S1,(13)aii>an1,n1,(14)Rn1A≥RiA,(15)maxj∈S2∪S3Rj−tBbj−t,j−tRn1A≤an1,n1,t=n1−1.Proof.
Sincek=1 and A and B are the DSDD matrices of orders n1 and n2 respectively, it is obvious that S1=1,2,…,n1−1,S2=n1,S3=n1+1,n1+2,…,n.Case 1: fori,j∈S1, from (a) of Lemma 1, we have(16)cii=aii,cjj=ajj,RiC=RiA,RjC=RjA.SinceA is DSDD, we obtain that for i,j∈S1,(17)ciicjj=aiiajj>RiARjA=RiCRjC.Case 2: fori∈S1andj∈S2=n1, from (a) of Lemma 1, it is easy to obtain(18)cii=aii,RiC=RiA.Since all diagonal entries ofA22 and B11 are positive (or all negative), we obtain(19)cjj=an1,n1+b11.SinceA is DSDD, and from inequalities (13)–(15), we have(20)ciicjj=aiian1,n1+b11=aiian1,n1+aiib11>RiARn1A+an1,n1b11≥RiARn1A+b11maxj∈S2∪S3Rj−tBbj−t,j−tRn1A≥RiARn1A+Rn1AR1B≥RiARn1A+RiAR1B=RiARn1A+R1B.From (b) of Lemma1, it is easy to obtain that for i∈S1,j∈S2=n1,(21)ciicjj>RiCRjC.Case 3: fori∈S1andj∈S3, from (a) and (d) of Lemma 1, we have(22)cii=aii,RiC=RiA,cjj=bj−t,j−t,RjC=Rj−tB.Then, from inequalities (13)–(15), we have that for i∈S1andj∈S3,(23)ciicjj=aiibj−t,j−t>an1,n1bj−t,j−t≥bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRn1A≥Rn1ARj−tB≥RiARj−tB=RiCRjC.Case 4: fori∈S2=n1andj∈S3, from (b) and (d) of Lemma 1, we obtain(24)cii=an1,n1+b11,cjj=bj−t,j−t.SinceB is DSDD, and from inequality (15), for i∈S2=n1andj∈S3, we have(25)ciicjj=an1,n1+b11bj−t,j−t=an1,n1bj−t,j−t+b11bj−t,j−t>bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRn1A+R1BRj−tB≥Rn1ARj−tB+R1BRj−tB=Rn1A+R1BRj−tB=RiCRjC.Case 5: fori,j∈S3, from (d) of Lemma 1, we obtain(26)cii=bi−t,i−t,RiC=Ri−tB.SinceB is DSDD, we have(27)ciicjj=bi−t,i−tbj−t,j−t>Ri−tBRj−tB=RiCRjC.
Therefore, we can draw a conclusion that for anyi,j∈1,2,…,n,ciicjj>RiCRjC, that is, C=A⊕1B is a DSDD matrix.Example 2.
The matrices(28)A=30−1−140−202,B=31−1−1−20102are two DSDD matrices, and from Theorem 1, it is easy to verify that(29)C=A⊕1B=30−100−14000−2051−100−12000102is a DSDD matrix since
3=a11>a33=2,4=a22>a33=2,2=R3A>R1A=1,2=R3A>R2A=1, and maxj∈S2∪S3Rj−tB/bj−t,j−tRn1A=R3B/b33R3A=4/7×3<a33=2.
However,(30)C=A⊕2B=30−10−171−1−2−1000102is not DSDD since c11c33=0<3=R1CR3C.Examples2 motivates the search for other conditions such that C=A⊕kB (k⩾2) is also a DSDD matrix, where A is a DSDD matrix and B is a DSDD matrix.Next, some sufficient conditions ensuring that thek-subdirect sum of DSDD matrices is a DSDD matrix are given.Theorem 2.
LetA and B be matrices of orders n1 and n2 partitioned as in (1), respectively, and k is an integer number such that 2≤k≤minn1,n2. Let A and B be DSDD matrices, if all diagonal entries of A22 and B11 are positive (or all negative) and for any i∈S1∪S2,(31)maxj∈S2∪S3Rj−tBbj−t,j−tRiA<aii,t=n1−k,and then the k-subdirect sum C=A⊕kB is DSDD.Proof.
LetA and B be DSDD matrices of orders n1 and n2 respectively; thus, it is obvious that S1=1,2,…,n1−k,S2=n1−k+1,n1−k+2,…,n1,andS3=n1+1,n1+2,…,n.Case 1: fori,j∈S1, from (a) of Lemma 1, we have(32)cii=aii,RiC=RiA.SinceA is DSDD, we obtain(33)ciicjj=aiiajj>RiARjA=RiCRjC.Case 2: fori∈S1andj∈S2, from (a) and (c) of Lemma 1, we obtain(34)cii=aii,RiC=RiA,RjC≤RjA+Rj−tB.Since all diagonal entries ofA22 and B11 are positive (or all negative), we obtain(35)cjj=ajj+bj−t,j−t=ajj+bj−t,j−t.SinceA is DSDD, and from inequality (31), we have(36)ciicjj=aiiajj+bj−t,j−t=aiiajj+aiibj−t,j−t>RiARjA+bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRiA≥RiARjA+RiARj−tB=RiARjA+Rj−tB.From (c) of Lemma1, it is easy to obtain that(37)ciicjj>RiCRjC.Case 3: fori∈S1andj∈S3, from (a) and (d) of Lemma 1, we conclude(38)cii=aii,RiC=RiA,cjj=bj−t,j−t,RjC=Rj−tB.Then, from inequality (31), we have(39)ciicjj=aiibj−t,j−t>bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRiA≥RiARj−tB=RiCRjC.Case 4: fori,j∈S2, from (c) of Lemma 1, we obtain(40)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB.SinceA and B are DSDD, and from inequality (31), we conclude(41)ciicjj=aii+bi−t,i−tajj+bj−t,j−t=aiiajj+aiibj−t,j−t+ajjbi−t,i−t+bi−t,i−tbj−t,j−t>RiARjA+bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRiA+bi−t,i−tmaxi∈S2∪S3Ri−tBbi−t,i−tRjA+Ri−tBRj−tB≥RiARjA+RiARj−tB+RjARi−tB+Ri−tBRj−tB=RA+Ri−tBRjA+Rj−tB≥RiCRjC.Case 5: fori∈S2andj∈S3, from (c) and (d) of Lemma 1, we obtain(42)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB,cjj=bj−t,j−t,RjC=Rj−tB.SinceB is DSDD, and from inequality (31), we obtain(43)ciicjj=aii+bi−t,i−tbj−t,j−t=aiibj−t,j−t+bi−t,i−tbj−t,j−t>bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tRiA+Ri−tBRj−tB≥RiARj−tB+Ri−tBRj−tB=RiA+Ri−tBRj−tB≥RiCRjC.Case 6: fori,j∈S3, from (d) of Lemma 1, we obtain(44)cii=bi−t,i−t,RiC=Ri−tB.SinceB is DSDD, we obtain(45)ciicjj=bi−t,i−tbj−t,j−t>Ri−tBRj−tB=RiCRjC.
In conclusion,for anyi,j∈S1∪S2∪S3,ciicjj>RiCRjC. Therefore, C=A⊕kB is a DSDD matrix.Example 3.
Let(46)A=41415−20−12,B=201380−125be two DSDD matrices. And from Definition 1, we obtain that(47)C=A⊕2B=414017−21021000−125,and S1=1,S2=2,3,andS3=4. From Theorem 2, C=A⊕2B is a DSDD matrix since(48)maxj∈S2∪S3Rj−tBbj−t,j−tR1A=R3Bb33R1A=35×5<a11,maxj∈S2∪S3Rj−tBbj−t,j−tR2A=R3Bb33R2A=35×3<a22,maxj∈S2∪S3Rj−tBbj−t,j−tR3A=R3Bb33R3A=35×1<a33.From Definitions2 and 3, it is easy to show that SDD matrices are contained into DSDD matrices. Therefore, from Theorem 2, we obtain the following corollaries, which present sufficient conditions such that k-subdirect sum C=A⊕kB is DSDD.Corollary 1.
LetA and B be square matrices of orders n1 and n2 partitioned as in (1), respectively, and k is an integer number such that 1≤k≤minn1,n2. We assume that A is a DSDD matrix and B is a SDD matrix. If there exists an i0∈S1∪S2 such that(49)maxj∈S2∪S3Rj−tBbj−t,j−tRi0A<ai0,i0<Ri0A,(50)aii⩾RiA,i∈S1∪S2\i0.
And, all diagonal entries ofA22 and B11 are positive (or all negative), then the k-subdirect sum C=A⊕kB is a DSDD matrix.Proof.
Without loss of generality, we can assumei0=1 such that maxj∈S2∪S3Rj−tB/bj−t,j−tR1A<a11<R1A and aii≥RiA,i=2,3,…,n1.Case 1: fori,j∈S1, from (a) of Lemma 1, we have(51)cii=aii,cjj=ajj,RiC=RiA,RjC=RjA.SinceA is DSDD, we obtain that for i,j∈S1,(52)ciicjj=aiiajj>RiARjA=RiCRjC.Case 2: fori∈S1andj∈S2, from (a) of Lemma 1, it is easy to obtain(53)cii=aii,RiC=RiA.Since all diagonal entries ofA22 and B11 are positive (or all negative), we obtain(54)cjj=ajj+bj−t,j−t.Ifi=1, since A is DSDD, and from inequality (49), we can obtain(55)c11cjj=a11ajj+bj−t,j−t=a11ajj+a11bj−t,j−t>R1ARjA+bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tR1A≥R1ARjA+bj−t,j−tRj−tBbj−t,j−tR1A=R1ARjA+Rj−tBR1A=R1ARjA+Rj−tB≥R1CRjC.Ifi=2,3,…,n1−k, since B is SDD, and from inequality (50), we can write(56)ciicjj=aiiajj+bj−t,j−t>RiARjA+Rj−tB≥RiCRjC.Case 3: fori∈S1andj∈S3, from (a) and (d) of Lemma 1, we have(57)cii=aii,RiC=RiA,cjj=bj−t,j−t,RjC=Rj−tB.Ifi=1, from inequality (49), we have(58)c11cjj=a11bj−t,j−t>bj−t,j−tmaxj∈S2∪S3Rj−tBbj−t,j−tR1A≥bj−t,j−tRj−tBbj−t,j−tR1A=R1ARj−tB=R1CRjC.Ifi=2,…,n1−k, since B is SDD, and from inequality (50), we can write(59)ciicjj=aiibj−t,j−t>RiARj−tB=RiCRjC.Case 4: fori,j∈S2, from (b) and (c) of Lemma 1, we obtain(60)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB.SinceB is SDD, and from inequality (50), we can obtain(61)ciicjj=aii+bi−t,i−tajj+bj−t,j−t>RiA+Ri−tBRjA+Rj−tB≥RiCRjC.Case 5: fori∈S2andj∈S3, from (c) and (d) of Lemma 1, we obtain(62)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB,cjj=bj−t,j−t,RjC=Rj−tB.SinceB is SDD, and from inequality (50), we obtain(63)ciicjj=aii+bi−t,i−tbj−t,j−t>RiA+Ri−tBRj−tB≥RiCRjC.Case 6: fori,j∈S3, from (d) of Lemma 1, we obtain(64)cii=bi−t,i−t,RiC=Ri−tB.SinceB is SDD, we can obtain(65)ciicjj=bi−t,i−tbj−t,j−t>Ri−tBRj−tB=RiCRjC.
Therefore, we can draw a conclusion that for anyi,j∈1,2,…,n,ciicjj>RiCRjC; that is, C=A⊕kB is a DSDD matrix.Corollary 2.
LetA and B be square matrices of orders n1 and n2 partitioned as in (1), respectively, and k is an integer number such that 1≤k≤minn1,n2. We assume that A is a SDD matrix and B is a DSDD matrix. If there exists a j0∈S2∪S3 such that(66)maxi∈S1∪S2RiAaiiRj0−tB<bj0−t,j0−t<Rj0−tB,(67)aii≥Rj−tB,j∈S2∪S3\j0.
And, all diagonal entries ofA22 and B11 are positive (or all negative), then the k-subdirect sum C=A⊕kB is a DSDD matrix.Proof.
Without loss of generality, we can assumej0=n such that maxi∈S1∪S2RiA/aiiRn−tB<bn−t,n−t<Rn−tB and bj−t,j−t≥Ri−tB,j=n1−k+1,n1−k+2,…,n−1.Case 1: fori,j∈S1, from (a) of Lemma 1, we have(68)cii=aii,cjj=ajj,RiC=RiA,RjC=RjA.SinceA is SDD, we obtain that for i,j∈S1,(69)ciicjj=aiiajj>RiARjA=RiCRjC.Case 2: fori∈S1andj∈S2, from (a) of Lemma 1, it is easy to obtain(70)cii=aii,RiC=RiA.Since all diagonal entries ofA22 and B11 are positive (or all negative), we obtain(71)cjj=ajj+bj−t,j−t.SinceA is SDD and from inequality (67), we have(72)ciicjj=aiiajj+bj−t,j−t>RiARjA+Rj−tB≥RiCRjC.Case 3: fori∈S1andj∈S3, from (a) and (d) of Lemma 1, we have(73)cii=aii,RiC=RiA,cjj=bj−t,j−t,RjC=Rj−tB.Ifj=n, from inequality (66), we have(74)ciicnn=aiibn−t,n−t>aiimaxi∈S1∪S2RiAaiiRn−tB≥aiiRiAaiiRn−tB=RiARn−tB=RiCRnC.Ifj=n1−k+1,n1−k+2,…,n−1, since A is SDD, and from inequality (67), we can write(75)ciicjj=aiibj−t,j−t>RiARj−tB=RiCRjC.Case 4: fori,j∈S2, from (c) of Lemma 1, we obtain(76)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB.SinceA is SDD, and from inequality (67), we can obtain(77)ciicjj=aii+bi−t,i−tajj+bj−t,j−t>RiA+Ri−tBRjA+Rj−tB≥RiCRjC.Case 5: fori∈S2,j∈S3, from (c) and (d) of Lemma 1, we obtain(78)cii=aii+bi−t,i−t,RiC≤RiA+Ri−tB,cjj=bj−t,j−t,RjC=Rj−tB.Ifj=n, since B is DSDD, and from inequality (66), we have(79)ciicnn=aii+bi−t,i−tbn−t,n−t=aiibn−t,n−t+bi−t,i−tbn−t,n−t>aiimaxi∈S1∪S2RiAaiiRn−tB+Ri−tBRn−tB≥aiiRiAaiiRn−tB+Ri−tBRn−tB=RiARn−tB+Ri−tBRn−tB=RiA+Ri−tBRn−tB≥RiCRnC.Ifj=n1−k+1,n1−k+2,…,n−1, since A is SDD, and from inequality (67), we can write(80)ciicjj=aii+bi−t,i−tbj−t,j−t>RiA+Ri−tBRj−tB≥RiCRjC.Case 6: fori,j∈S3, from (d) of Lemma 1, we obtain(81)cii=bi−t,i−t,RiC=Ri−tB.SinceB is DSDD, we can obtain(82)ciicjj=bi−t,i−tbj−t,j−t>Ri−tBRj−tB=RiCRjC.
In conclusion, for anyi,j∈S1∪S2∪S3,ciicjj>RiCRjC. Therefore, C=A⊕kB is a DSDD matrix.


## 3. Conclusions

In this paper, some sufficient conditions such that the subdirect sum ofDSDD matrices is in the class of DSDD matrices are given. Moreover, numerical examples are also presented to illustrate the corresponding results.



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