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Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela SUBDIRECT SUMS OF DOUBLY DIAGONALLY DOMINANT MATRICES YAN ZHU AND TING-ZHU HUANG Abstract The problem of when the k-subdirect sum of a doubly diagonally dominant matrix (DDD matrix) is also a DDD matrix is studied Some sufficient conditions are given The same situation is analyzed for diagonally dominant matrices and strictly diagonally dominant matrices Additionally, some conditions are also derived when card(s)>card(s 1 ) which was not studied by Bru, Pedroche and Szyld [Electron J Linear Algebra, 15:201-209, 2006] Examples are given to illustrate the conditions presented Key words blocks Subdirect sum, Doubly diagonally dominant matrices, H-matrices, Overlapping AMS subject classifications 15A48 1 Introduction The concept of k-subdirect sums of square matrices was introduced by Fallat and Johnson [4], where many of their properties were analyzed They showed that the subdirect sum of positive definite matrices, or of symmetric M- matrices, is a positive definite matrix or symmetric M-matrix, respectively Subdirect sums of matrices are generalizations of the usual sum of matrices and arise naturally in several contexts For example, subdirect sums arise in matrix completion problems, overlapping subdomains in domain decomposition methods, and global stiffness matrices in finite elements (see [1, 2, 4]) In [1], Bru, Pedroche and Szyld have given sufficient conditions such that the subdirect sum of two nonsingular M-matrices is a nonsingular M-matrix Also in [3], they also showed that k-subdirect sum of S-strictly diagonally dominant matrices, which is a subclass of H-matrices, is an S-strictly diagonally dominant matrix (S- SDD matrix) Sufficient conditions are also given In this paper, we will give sufficient conditions such that the k-subdirect sum of matrices belongs to the same class We show this for certain strictly diagonally dominant (SDD) matrices and for doubly diagonally dominant (DDD) matrices which were introduced in [5]; see also [6] for further properties and analysis Furthermore, we also discuss the conditions such that the subdirect sum of S-SDD matrices is in the class of S-SDD matrices (for a fixed set S) when card(s)>card(s 1 )whichwas not studied in [3] Notice, the sets S and S 1 appear in section 2 2 Subdirect sums Let A and B be two square matrices of order m 1 and m 2, respectively, and let k be an integer such that 1 k min{m 1,m 2 }LetA and B be Received by the editors 28 December 2006 Accepted for publication 20 June 2007 Handling Editor: Daniel Szyld School of Applied Mathematics, University of Electronic Science and Technology of China, Chengdu, 610054, PRChina (tingzhuhuang@126com or tzhuang@uestceducn) Supported by NCET of Chinese Ministry of Education (NCET-04-0893), Sichuan Province Foundation for Leader of Sci and Tech and project of Academic Leader and group of UESTC 171

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 172 YanZhuandTing-ZhuHuang partitioned into 2 2 blocks as follows: (21) [ ] A11 A 12, B = A 21 A 22 [ ] B11 B 12, B 21 B 22 where A 22 and B 11 are square matrices of order k Following [4], we call the following square matrix of order N = m 1 + m 2 k, C = A 11 A 12 0 A 21 A 22 + B 11 B 12 0 B 21 B 22 the k-subdirect sum of A and B and we denote it by C = A k B In order to more explicitly express each element of C in terms of the ones of A and B, wecanwritec as follows, S 1 {}}{ a 11 a p1 C = a m1,1 0 S 2 {}}{ a 1p a 1,m1 a pp + b 11 a p,m1 + b 1,m1 t a m1,p + b m1 t,1 a m1,m 1 + b m1 t,m 1 t b N t,1 b N t,m1 t S 3 {}}{ 0 b p,n t b m1 t,n t b N t,n t (22) (23) S 1 = {1, 2,,m 1 k}, S 2 = {m 1 k +1,m 1 k +2,,m 1 }, S 3 = {m 1 +1,m 1 +2,,N} a ij or 0, i S 1,j S 1 S 2 S 3, c ij = a ij or a ij + b i t,j t or b i t,j t, i S 2,j S 1 S 2 S 3, b i t,j t or 0, i S 3,j S 1 S 2 S 3, N = m 1 + m 2 k, t = m 1 k, p = t +1 In general, the subdirect sum of two DDD matrices is not always a DDD matrix We show this in the following example Example 21 The matrices 1 1 1 2 4 0 1 0 4 3 2 1 and B = 1 2 0, 1 1 4

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela are two DDD matrices, but Subdirect Sums of Doubly Diagonally Dominant Matrices 173 C = A 2 B = 1 1 1 0 2 7 2 1 1 1 6 0 0 1 1 4 is not a DDD matrix 3 Subdirect sums of doubly diagonally dominant matrices We now formally introduce some notations and definitions which can be found in [5], [6] and [9] Let (a ij ) C n,n,n 2 For i =1, 2,,n,wedefine R i (A) = n j i,j=1 a ij Recall that A is called (row) diagonally dominant if for all i =1, 2,,n, (31) a ii R i (A) If the inequalities in (31) hold strictly for all i, wesaythata is strictly diagonally dominant Given any nonempty set of indices S N = {1, 2,,n}, wedenoteitscomplement in N by S = N\S Wehave R i (A) =Ri S (A)+R S i (A), Ri S (A) = a ij, i =1, 2,,n j i,j S Definition 31 The matrix (a ij ) C n,n is doubly diagonally dominant if (32) a ii a jj R i (A)R j (A), i,j =1, 2,,n, i j Furthermore, if the inequality in (32) is strict for all distinct i, j =1, 2,,n,we say A is strictly doubly diagonally dominant Definition 32 Let the matrix (a ij ) C n,n,n 2, and given any nonempty proper subset S of N ThenA is an S-strictly diagonally dominant matrix if the following two conditions hold: { 1) aii >Ri S (A) for at least one i S, 2) ( a ii Ri S(A))( a jj R S j (A)) >R S i (A)Rj S (A) for all i S, j S It was shown in [5] that SDD matrices are contained into DDD matrices In addition, if A is doubly diagonally dominant, there exists at most one index i 0 such that a i0i 0 <R i0 (A) It is easy to show that a doubly diagonally dominant matrix is not necessary an S-strictly diagonally dominant matrix We show this in the following example Example 33 The following matrix:

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 174 YanZhuandTing-ZhuHuang 1 1 0 0 1 1 0 05 1 is a DDD matrix, but is not an S-SDD matrix for any subset S of {1, 2, 3} We now give the following theorem, which presents sufficient condition such that C = A k B is a DDD matrix Theorem 34 Let A and B be square matrices of order m 1 and m 2 partitioned as in (21), respectively, and k be an integer such that 1 k min(m 1,m 2 ) which defines the sets S 1,S 2,S 3 as in (22) We assume that A is doubly diagonally dominant with R l t (B) max l S 2 S 3 b l t,l t R i 0 (A) a i0i 0 <R i0 (A) for some i 0,andB is strictly diagonally dominant If all diagonal entries of A 22 and B 11 are positive (or all negative), then the k-subdirect sum C = A k B is doubly diagonally dominant Proof Without loss of generality, we can assume a 11 <R 1 (A) anda ii R i (A), i =2, 3,,m 1 Case 1: i, l S 1,j S 1 S 2 S 3, using equation (23) we obtain c ij = a ij,r i (C) =R i (A), since A is doubly diagonally dominant, we obtain c ii c ll = a ii a ll R i (A)R l (A) =R i (C)R l (C) Case 2: i S 1, l S 2, j S 1 S 2 S 3, from equation (23), we have the following relations: R l (C) = a lj + j S 1 a lj + j S 1 = R l (A)+R l t (B) c ii = a ii,r i (C) =R i (A), j S 2,j l j S 2,j l a lj + b l t,j t + a lj + j S 2,j l b l t,j t j S 3 b l t,j t + j S 3 b l t,j t Since all diagonal entries of A 22 and B 11 are positive ( or all negative), we obtain If i =1, we have c ll = a ll + b l t,l t = a ll + b l t,l t c 11 c ll = a 11 ( a ll + b l t,l t ) = a 11 a ll + a 11 b l t,l t R 1 (A)R l (A)+ b l t,l t max l S 2 S 3 R 1 (A)R l (A)+ b l t,l t R l t(b) b l t,l t R 1(A) = R 1 (A)(R l (A)+R l t (B)) R 1 (C)R l (C) R l t(b) b l t,l t R 1(A)

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela Subdirect Sums of Doubly Diagonally Dominant Matrices 175 If i =2,,m 1 k, then we can write c ii c ll = a ii ( a ll + b l t,l t ) R i (A)(R l (A)+R l t (B)) R i (C)R l (C) Case 3: i S 1,l S 3,j S 1 S 2 S 3,weget c ij = a ij,r i (C) =R i (A), c lj = b l t,j t,r l (C) =R l t (B) Like the proof of Case 2, let i =1, we conclude c 11 c ll = a 11 b l t,l t max l S 2 S 3 R l t(b) b l t,l t R 1(A) b l t,l t R 1 (A) R l t(b) b l t,l t b l t,l t = R 1 (C)R l (C) If i =2,,m 1 k, obviously, the result still holds Case 4: i S 2,l S 2,j S 1 S 2 S 3, using equation (23), it follows that c ii = a ii + b i t,i t = a ii + b i t,i t, R i (C) = i j c ij R i (A)+R i t (B) Note that ( a ii R i (A)) since A is a DDD matrix and ( b i t,i t >R i t (B)) since B is an SDD matrix, thus we can write c ii c ll = a ii + b i t,i t a ll + b l t,l t > (R i (A)+R i t (B))(R l (A)+R l t (B)) R i (C)R l (C) For the rest of cases i S 2, l S 3, j S 1 S 2 S 3 and i S 3, l S 3, j S 1 S 2 S 3, the proofs are similar to the proofs of the Case 1 and Case 2 In the case i 0 S 2 the conclusion still holds and the proofs are similar to the cases above Therefore the details of the case are omitted Corollary 35 Let A 1 be a DDD matrix with existing only i 0 such that R max l1 t(a i) l a 1 S 2 S l1 t,l 3 1 t R i 0 (A 1 ) a i0i 0 <R i0 (A 1 ), i =2, 3,,andletA 2, A 3,, be SDD matrices If the diagonals of A i in the overlapping have the same sign pattern, then C =(A 1 k1 A 2 ) k2 A 3 is a DDD matrix Remark 36 Since A 2, A 3,are SDD we have that their diagonals are nonzero Since the overlapping blocks of A 2 and A 1 have the same sign pattern, this implies that overlapping block of A 1 has its diagonal nonzero

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 176 YanZhuandTing-ZhuHuang Theorem 37 Let A and B be matrices of order m 1 and m 2 partitioned as in (21), respectively, and k be an integer such that 1 k min(m 1,m 2 ), which defines the sets S 1,S 2,S 3 as in (22) LetA be diagonally dominant and B be strictly diagonally dominant If all diagonal entries of A 22 and B 11 are positive (or all negative), then the k-subdirect sum C = A k B is doubly diagonally dominant Proof Weonlyprovethefollowingcases Case 1: i, l S 1, j S 1 S 2 S 3 from equation (23), we obtain c ij = a ij,r i (C) =R i (A), so c ii c ll = a ii a ll R i (A)R l (A) =R i (C)R l (C) Case 2: i S 2,l S 2,j S 1 S 2 S 3, using equation (23), it follows that c il = a il + b i t,l t, R i (C) = c ij = a ij + a ij + b i t,j t + b i t,j t i j j S 1 i j,j S 2 j S 3 R i (A)+R i t (B) Then we can write c ii c ll = a ii + b i t,i t a ll + b l t,l t > (R i (A)+R i t (B))(R l (A)+R l t (B)) R i (C)R l (C) Finally, i S 1,l S 3,j S 1 S 2 S 3 ; i S 2,l S 3,j S 1 S 2 S 3 ;and i S 3,l S 3,j S 1 S 2 S 3 ; the proofs are similar to the cases above Corollary 38 Successive k-subdirect sums of the form (A 1 k1 A 2 ) k2 A 3 is doubly diagonally dominant if the diagonals of A i in the overlapping have the same sign pattern, where A 1 is diagonally dominant and A 2, A 3,, are strictly diagonally dominant Example 39 In this example, A is doubly diagonally dominant with R j t (B) max j S 2 S 3 b j t,j t R i 0 (A) > a i0i 0 for all i 0 =1, 2, 3, 4, and B is strictly diagonally dominant, but the subdirect sum C is not doubly diagonally dominant Let 10 03 04 05 20 04 05 06 08 16 02 03 06 23 08 08,B= 02 04 12 04 05 01 24 09 01 09 05 20 05 08 02 29

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela Subdirect Sums of Doubly Diagonally Dominant Matrices 177 However, C = A 2 B = 10 03 04 05 0 0 08 16 02 03 0 0 02 04 32 08 05 06 01 09 11 43 08 08 0 0 05 01 24 09 0 0 05 08 02 29 is not a DDD matrix, since a 11 =1< R2(B) b 22 R 1 (A) = 22 23 6 5 115 Example 310 Let A be doubly diagonally dominant with R j t (B) max j S 2 S 3 b j t,j t R i 0 (A) a i0i 0 for some i 0, and B be strictly diagonally dominant, 04 01 04 02,B= 05 02 05 1 2 10 04 04 02 12 04 04 06 14 Then C = A 2 B = 04 01 04 0 02 15 06 04 05 12 32 04 0 04 06 14 is a DDD matrix, since a 11 =04 = R1(B) b 11 R 1 (A) = 4 5 1 2 =04 As Bru, Pedroche and Szyld studied the result in section 4 of [3], we continue to consider A and B to be principal submatrices of a given doubly diagonally dominant matrix such that they have a common block with all positive (or negative) diagonal entries This situation, as well as a more general case outlined in Theorem 311 later, appears in many variants of additive Schwarz preconditioning (see [2, 7, 8]) Let (33) H = H 11 H 12 H 13 H 21 H 22 H 23 H 31 H 32 H 33 be doubly diagonally dominant with H 22 a square matrix of order k 1andlet (34) [ ] [ ] H11 H 12 H22 H,B= 23 H 21 H 22 H 32 H 33

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 178 YanZhuandTing-ZhuHuang be of order m 1 m 1 and m 2 m 2, respectively The k-subdirect sum of A and B is thus given by (35) C = A k B = H 11 H 12 O H 21 2H 22 H 23 O H 32 H 33 We now have the following theorem which shows that C is a DDD matrix Theorem 311 Let H be a DDD matrix partitioned as in (33), andleta and B be two overlapping principal submatrices given by (34) Then the k-subdirect sum C = A k B is a DDD matrix Proof For any i S 1,l S 2,j S 1 S 2 S 3, from equation (35), we have c ii = a ii = h ii, R i (C) =R i (A) R i (H) (ifh 13 = 0, equality holds), c ll = a ll + b l t,l t =2 a ll =2 h ll,and R l (C) = a lj + a lj + b l t,j t + b l t,j t j S 1 j S 2,j l j S 3 = h lj +2 h lj + h lj j S 1 j S 2,j l j S 3 < 2R l (H) Using now that H is doubly diagonally dominant, we can get that c ii c ll = a ii 2a ll =2 h ii h ll 2R i (H)R l (H) >R i (C)R l (C) The proofs of other cases are analogous Example 312 Let the following DDD matrix H be partitioned as 10 06 05 01 H = 07 29 03 05, 05 01 24 09 05 07 07 23

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela Subdirect Sums of Doubly Diagonally Dominant Matrices 179 where principal submatrices A, B are partitioned as 10 06 05 07 29 03 05 01 24 The 2-subdirect sum C = A 2 B, C = 29 03 05 and B = 01 24 09 07 07 23 10 06 05 0 07 58 06 05 05 02 48 09 0 07 07 23 is a DDD matrix, according to Theorem 311 Here, we consider consecutive principal submatrices defined by consecutive indices of the form {i, i +1,i+2,} Corollary 313 Let H be a DDD matrix with all positive (or negative) diagonal entries Let A i,i=1,,pbe consecutive principal submatrices of H of order n i n i Then each of the k i -subdirect sums C i is a DDD matrix In particular,, C i = C i 1 ki A i+1,i=1,,p 1 C p 1 = A 1 k1 A 2 k2 kp A p is a DDD matrix, where C 0 = A 1 and k i < min(n i,n i+1 ) Here, we discuss the conditions of the subdirect sum of S-strictly diagonally dominant matrices when card(s) >card(s 1 ) Theorem 314 Let A and B be square matrices of order m 1 and m 2 partitioned as in (21), respectively, and k be an integer such that 1 k min(m 1,m 2 ), which defines the sets S 1,S 2,S 3 as in (22) LetA be S-strictly diagonally dominant with card(s)>card(s 1 ), S be a set of indices of the form S = {1, 2,} and B be strictly diagonally dominant If all diagonal entries of A 22 and B 11 are positive (or all negative) and (36) a ii R i (A) >R i t (B) b i t,i t, for all i S 2,t= m 1 k, then the k-subdirect sum C = A k B is S-strictly diagonally dominant Proof We take S 2 = S S, S S = Let the set S = S 1 S (condition card(s)>card(s 1 )), S = S S 3,andleti S We first prove the condition 1) of Definition 32 for C to be an S-SDD matrix Observe that from (23) and the fact that A 22 and B 11 have positive diagonals (or both negative diagonals), we have c ii = a ii + b i t,i t = a ii + b i t,i t,

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 180 YanZhuandTing-ZhuHuang R S1 S i (C) = c il = a il + a il + b i t,l t l i,l S 1 S l S 1 l i,l S a il + b i t,l t = R S1 S i (A)+Ri t S (B) l i,l S 1 S l i,l S Therefore we can write c ii R S1 S i (C) a ii + b i t,i t R S1 S i (A) Ri t S (B) =( a ii R S1 S i (A)) + ( b i t,i t Ri t S (B)) wherewehaveusedthat( a ii R S1 S i (A)) is positive since A is S-strictly diagonally dominant Note that ( b i t,i t Ri t S (B)) is also positive since B is strictly diagonally dominant, and thus we can obtain c ii R S1 S i (C) > 0,for alli S Since S = S 1 S we have that condition 1) of Definition 32 holds In order to see that condition 2) holds we first study the case i S, j S Using equation (23) we obtain the following relations: R S S 3 j (C) = c jl = a jl + b j t,l t + b j t,l t l j,l S S 3 l j,l S l S 3 a jl + b j t,l t l j,l S l S S 3 = Rj S (A)+R S S 3 (B), j t R S1 S j (C) = c jl = a jl + a jl + b j t,l t l S 1 S l S 1 l S R S1 S j (A)+Rj t S (B), R S S 3 i (C) = c il = a il + b i t,l t + b i t,l t l S S 3 l S l S 3 Ri S (A)+R S S 3 (B), Therefore we can write i t c jj = a jj + b j t,j t = a jj + b j t,j t ( c ii R S1 S i (C))( c jj R S S 3 j (C)) =( a ii + b i t,i t R S1 S i (C))( a jj + b j t,j t R S S 3 j (C)) Now observe that: (37) a ii + b i t,i t R S1 S i (C) a ii R S1 S i (A)+ b i t,i t Ri t(b), S (38) a jj + b j t,j t R S S 3 j (C) a jj Rj S (A)+ b j t,j t R S S 3 j t (B),

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela Subdirect Sums of Doubly Diagonally Dominant Matrices 181 and from condition (36) we can write the inequality a ii R i (A) = a ii (R S1 S i (A)+Ri S (A)) >R i t (B) b i t,i t Therefore a ii R S1 S i (A) >Ri S (A)+R i t (B) b i t,i t, a jj Rj S (A) >R S1 S j (A)+R j t (B) b j t,j t, R i t (B) =Ri t S S 3 (B)+RS (B), R j t (B) =Rj t S S 3 (B)+RS j t (B), which jointly with (37) and (38) lead to the inequality ( c ii R S1 S i (C))( c jj R S S 3 j (C)) > (Ri S (A)+R S S 3 i t (B))(R S1 S j (A)+Rj t S (B)) R S S 3 i (C)R S1 S j (C) Therefore condition 2) of Definition 32 is fulfilled and the proof for the case i S,j S is completed For the case i S,j S 3, we have from equation (23) that R S S 3 j (C) = i t c ii = a ii + b i t,i t, c jj = b j t,j t, j l,l S S 3 c jl = j l,l S S 3 b j t,l t = R S S 3 j t R S1 S j (C) = c jl = b j t,l t = R S j t(b) l S 1 S l S 1 S Note that B is an SDD matrix, then we have (B), ( c ii R S1 S i (C))( c jj R S S 3 j (C)) > ( a ii R S1 S i (A)+ b i t,i t Ri t S (B))RS j t (B) >R S S 3 i (C)R S1 S j (C) Therefore condition 2) is fulfilled Condition 1) also holds, since as before, we have c ii R S1 S i (C) a ii + b i t,i t R S1 S i (A) Ri t S (B) > 0 The proof for the rest of cases i S 1,j S and i S 1,j S 3 are already included in the paper of Bru, Pedroche, and Szyld [3] Example 315 InthisexampleweshowamatrixA that is an S-SDD matrix with card(s)>card(s 1 ), a matrix B that is an SDD matrix, and such that the 3- subdirect sum C is an S-SDD matrix Let

Volume 16, pp 171-182, July 2007 http://mathtechnionacil/iic/ela 182 YanZhuandTing-ZhuHuang 10 03 04 05 20 02 03 01 08 29 02 05 09 16 04 07, B = 05 01 24 09 01 04 13 04 01 09 01 20 06 01 01 23 Here we have S 1 ={1}, S 2 ={2, 3, 4}, S 3 ={5} It is easy to show that A is an S-SDD matrix, with S={1, 2}, and therefore card(s)>card(s 1 ) Since B is an SDD matrix and a ii R i (A) >R i t (B) b i t,i t,for alli S 2, with t = m 1 k =4 3=1and S = {3, 4, 5}, wehavethatthe3-subdirectsum 10 03 04 05 0 09 36 02 10 01 C = A 3 B = 01 04 42 06 05 01 14 02 44 09 0 06 01 01 23 is a {1, 2}-SDD matrix, according to Theorem 314 Acknowledgment The authors would like to thank the referee and the editor, Professor DB Szyld, very much for their detailed and helpful suggestions for revising this manuscript REFERENCES [1] R Bru, F Pedroche, and DB Szyld Subdirect sums of nonsingular M-matrices and of their inverse Electron J Linear Algebra, 13:162 174, 2005 [2] R Bru, F Pedroche, and DB Szyld Additive Schwarz iterations for markov chains SIAM J Matrix Anal Appl, 27:445 458, 2005 [3] R Bru, F Pedroche, and DB Szyld Subdirect sums of S-strictly diagonally dominant matrices Electron J Linear Algebra, 15:201 209, 2006 [4] SM Fallat and CR Johnson Sub-direct sums and positivity classes of matrices Linear Algebra Appl, 288:149 173, 1999 [5] B Li and MJ Tsatsomeros Doubly diagonally dominant matrices Linear Algebra Appl, 261:221 235, 1997 [6] RA Horn and CR Johnson Matrix Analysis Cambridge University Press, New York, 1985 [7] A Frommer and DB Szyld Weighted max norms, splittings, and overlapping additive Schwarz iterations Numer Math, 83:259-278, 1999 [8] BF Smith, PE Bjorstad, and WD Gropp Domain Decomposition: Parallel Multilevel Methods for Elliptic Partial Differential Equations Cambridge University Press, Cambridge, 1996 [9] L Cvetković H-matrix theory vs eigenvalue localization Numer Algorithms, 42:229-245, 2006