Subdirect Sums of Doubly Strictly Diagonally Dominant Matrices

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.


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 the k-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 SDD 1 matrices [9].
In this paper, we focus on the subdirect sum of doubly strictly diagonally dominant (shortly as DSDD) 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][12][13].
Let n be an integer number. C n×n is the set of complex matrices.
Definition 1 (see [1]). Let A and B be square matrices of orders n 1 and n 2 , respectively, and k be an integer number such that 1 ≤ k ≤ min n 1 , n 2 . Let A and B be partitioned in a 2 × 2 block as follows: where A 22 and B 11 are the square matrices of order k. We call the following square matrix of order n � n 1 + n 2 − k, 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, we can define the following set of indices: en, C can be expressed as follows: 1 · · · a t,t a t,p · · · a t,n 1 0 · · · 0 a p,1 · · · a p,t a p,p + b 11 · · · a p, a n 1 1 · · · a n 1 ,t a n 1 ,p + b n 1 −t,1 · · · a n 1 , where t � n 1 − k and p � t + 1.
Definition 2 (see [12]). Given a matrix A � [a ij ] ∈ C n×n , A is called (row) diagonally dominant (DD) if where If the inequality in (5) holds strictly for all i, we say that A is strictly diagonally dominant (SDD).

Subdirect Sums of DSDD Matrices
In general, the subdirect sum of two DSDD matrices is not always a DSDD matrix. We show this in the following example.
Example 1 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. Journal of Mathematics be DSDD square matrices of orders n 1 and n 2 partitioned as in (1), respectively. And, let k be an integer number such that 1 ≤ k ≤ min n 1 , n 2 , S 1 , S 2 , S 3 be defined as in (4), and all diagonal entries of A 22 and B 11 are positive (or all negative), with C � A⊕ k B, then, Proof. For i ∈ S 1 , we can write For k � 1, i ∈ S 2 � n 1 , we can obtain For k > 2, i ∈ S 2 , we obtain For the rest case of i ∈ S 3 , the proof is similar to the proof of i ∈ S 1 .

Theorem 1.
Let A and B be DSDD matrices of orders n 1 and n 2 partitioned as in (1), respectively, and k � 1. en, C �

A⊕ 1 B is a DSDD matrix if all diagonal entries of A 22 and B 11
are positive (or all negative) and for i ∈ S 1 , a ii > a n 1 ,n 1 , max Proof. Since k � 1 and A and B are the DSDD matrices of orders n 1 and n 2 respectively, it is obvious that Case 1: for i, j ∈ S 1 , from (a) of Lemma 1, we have Since A is DSDD, we obtain that for i, j ∈ S 1 , Case 2: for i ∈ S 1 and j ∈ S 2 � n 1 , from (a) of Lemma 1, it is easy to obtain c ii � a ii , Since all diagonal entries of A 22 and B 11 are positive (or all negative), we obtain c jj � a n 1 , Since A is DSDD, and from inequalities (13)-(15), we have From (b) of Lemma 1, it is easy to obtain that for i ∈ S 1 , j ∈ S 2 � n 1 , Case 3: for i ∈ S 1 and j ∈ S 3 , from (a) and (d) of Lemma 1, we have Journal of Mathematics 3 en, from inequalities (13)-(15), we have that for i ∈ S 1 and j ∈ S 3 , Case 4: for i ∈ S 2 � n 1 and j ∈ S 3 , from (b) and (d) of Lemma 1, we obtain c ii � a n 1 ,n 1 + b 11 , Since B is DSDD, and from inequality (15), for i ∈ S 2 � n 1 and j ∈ S 3 , we have Case 5: for i, j ∈ S 3 , from (d) of Lemma 1, we obtain Since B is DSDD, we have (27) erefore, we can draw a conclusion that for any are two DSDD matrices, and from eorem 1, it is easy to verify that However, is not DSDD since |c 11 ‖c 33 | � 0 < 3 � R 1 (C)R 3 (C).
Examples 2 motivates the search for other conditions such that C � A⊕ k B (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 the k-subdirect sum of DSDD matrices is a DSDD matrix are given. A and B be matrices of orders n 1 and n 2 partitioned as in (1), respectively, and k is an integer number such that 2 ≤ k ≤ min n 1 , n 2 . Let A and B be DSDD matrices, if all diagonal entries of A 22 and B 11 are positive (or all negative) and for any i ∈ S 1 ∪ S 2 ,

Theorem 2. Let
and then the k-subdirect sum C � A⊕ k B is DSDD.
Case 1: for i, j ∈ S 1 , from (a) of Lemma 1, we have c ii � a ii , Since A is DSDD, we obtain Case 2: for i ∈ S 1 and j ∈ S 2 , from (a) and (c) of Lemma 1, we obtain c ii � a ii , Since all diagonal entries of A 22 and B 11 are positive (or all negative), we obtain Since A is DSDD, and from inequality (31), we have From (c) of Lemma 1, it is easy to obtain that Case 3: for i ∈ S 1 and j ∈ S 3 , from (a) and (d) of Lemma 1, we conclude en, from inequality (31), we have Case 4: for i, j ∈ S 2 , from (c) of Lemma 1, we obtain Since A and B are DSDD, and from inequality (31), we conclude Journal of Mathematics (41) Case 5: for i ∈ S 2 and j ∈ S 3 , from (c) and (d) of Lemma 1, we obtain Since B is DSDD, and from inequality (31), we obtain Case 6: for i, j ∈ S 3 , from (d) of Lemma 1, we obtain Since B is DSDD, we obtain In conclusion, for any i, be two DSDD matrices. And from Definition 1, we obtain that and From Definitions 2 and 3, it is easy to show that SDD matrices are contained into DSDD matrices. erefore, from eorem 2, we obtain the following corollaries, which present sufficient conditions such that k-subdirect sum C � A⊕ k B is DSDD. A and B be square matrices of orders n 1 and n 2 partitioned as in (1), respectively, and k is an integer number such that 1 ≤ k ≤ min n 1 , n 2 . We assume that A is a DSDD matrix and B is a SDD matrix. If there exists an

Corollary 1. Let
And, all diagonal entries of A 22 and B 11 are positive (or all negative), then the k-subdirect sum C � A⊕ k B is a DSDD matrix.
Proof. Without loss of generality, we can assume i 0 � 1 such that max j∈S 2 Case 1: for i, j ∈ S 1 , from (a) of Lemma 1, we have c ii � a ii , c jj � a jj , Since A is DSDD, we obtain that for i, j ∈ S 1 , Case 2: for i ∈ S 1 and j ∈ S 2 , from (a) of Lemma 1, it is easy to obtain c ii � a ii , Since all diagonal entries of A 22 and B 11 are positive (or all negative), we obtain If i � 1, since A is DSDD, and from inequality (49), we can obtain If i � 2, 3, . . . , n 1 − k, since B is SDD, and from inequality (50), we can write Case 3: for i ∈ S 1 and j ∈ S 3 , from (a) and (d) of Lemma 1, we have If i � 1, from inequality (49), we have If i � 2, . . . , n 1 − k, since B is SDD, and from inequality (50), we can write Journal of Mathematics 7 (59) Case 4: for i, j ∈ S 2 , from (b) and (c) of Lemma 1, we obtain Since B is SDD, and from inequality (50), we can obtain Case 5: for i ∈ S 2 and j ∈ S 3 , from (c) and (d) of Lemma 1, we obtain Since B is SDD, and from inequality (50), we obtain Case 6: for i, j ∈ S 3 , from (d) of Lemma 1, we obtain Since B is SDD, we can obtain erefore, we can draw a conclusion that for any i, j ∈ 1, 2, . . . , n { }, |c ii ‖c jj | > R i (C)R j (C); that is, C � A⊕ k B is a DSDD matrix. □ Corollary 2. Let A and B be square matrices of orders n 1 and n 2 partitioned as in (1), respectively, and k is an integer number such that 1 ≤ k ≤ min n 1 , n 2 . We assume that A is a SDD matrix and B is a DSDD matrix. If there exists a j 0 ∈ S 2 ∪ S 3 such that And, all diagonal entries of A 22 and B 11 are positive (or all negative), then the k-subdirect sum C � A⊕ k B is a DSDD matrix.

(68)
Since A is SDD, we obtain that for i, j ∈ S 1 , c ii c jj � � � � � � a ii a jj � � � � � > R i (A)R j (A) � R i (C)R j (C). (69) Case 2: for i ∈ S 1 and j ∈ S 2 , from (a) of Lemma 1, it is easy to obtain c ii � a ii , Since all diagonal entries of A 22 and B 11 are positive (or all negative), we obtain c jj � a jj + b j−t,j−t .
Since A is SDD and from inequality (67), we have