Inﬁnity norm upper bounds for the inverse of S DD 1 matrices

: In this paper, a new proof that S DD 1 matrices is a subclass of H -matrices is presented, and some properties of S DD 1 matrices are obtained. Based on the new proof, some upper bounds of the inﬁnity norm of inverse of S DD 1 matrices and S DD matrices are given. Moreover, we show that these new bounds of S DD matrices are better than the well-known Varah bound for S DD matrices in some cases. In addition, some numerical examples are given to illustrate the corresponding results


Introduction
Let n be an integer number, N = {1, 2, . . . , n}, and C n×n be the set of all complex matrices of order n. A matrix A = [a i j ] ∈ C n×n (n ≥ 2) is called a strictly diagonally dominant (S DD) matrix if It was shown that S DD matrices is a subclass of H-matrices [1], where a matrix A = [a i j ] ∈ C n×n is an H-matrix if and only if there exists a positive diagonal matrix X such that AX is an S DD matrix [1].
In 2011, a new subclass of H-matrices was proposed by J. M. Peña, which is called S DD 1 matrices [2], and the definition of S DD 1 matrix is given as follows. Definition 1. In [2], J. M. Peña "proved" the following result: Theorem 1. ([2 Theorem 2.3]) If a matrix A = [a i j ] ∈ C n×n is an S DD 1 matrix by rows, then it is an H-matrix.
From the definition of H-matrix and Theorem 1, given an S DD 1 matrix A, there exists a correspondingly positive diagonal matrix D, such that AD is an S DD matrix. The great interest of the constitution of positive diagonal matrix D was commented in the introduction in [2], and divided it into two cases, that is, the given S DD 1 matrix has a unique row i strictly diagonally dominant and at least two rows i and j strictly diagonally dominant, to constitute positive diagonal matrix. However, Dai in [3] found that the proof of Theorem 1 is incorrect, and a counter example was given as follows.
From the proof of Theorem 1 in [2], it is easy to obtain that D = diag{ 3 4 , 1 3 , 1 4 , 1}, however, AD is not an S DD matrix by rows.
Dai found that the proof of the case that the given S DD 1 matrix has at least two rows i and j strictly diagonally dominant is incorrect, and a correct proof of Theorem 1 was presented in [3]. The correct proof of Theorem 1 divides the case that S DD 1 matrices have at least two rows i and j strictly diagonally dominant into S = ∅ and S ∅, where S is given as follows: However, when we use the correct proof to give the upper bound for the infinity norm of the inverse of S DD 1 matrices, the upper bound needs to be considered in different cases. Therefore, in order to avoid the difficult, we need to improve the proof of Theorem 1.
In addition, it was shown that upper bound of the infinity norm of inverse of a given nonsingular matrix has many potential applications in computational mathematics, such as for bounding the condition number and for proving the convergence of iteration methods. Moreover, upper bounds of the infinity norm of inverse for different classes of matrix have been widely studied, such as Nekrasov matrices [4][5][6], S -Nekrasov matrices [7,8], QN-Nekrasov matrices [8], {p 1 , p 2 }-Nekrasov matrices [9,10], DZT matrices [11,12], S -S DD matrices [13], S -S DDS matrices [14] and so on. However, the estimation of upper bounds of the infinity norm of inverse for S DD 1 matrices has never been reported.
In this paper, a new proof of Theorem 1 is given firstly. Secondly, some properties of S DD 1 matrices are presented. Finally, based on the new proof, some upper bounds of the infinity norm of inverse of S DD 1 matrices and S DD matrices are obtained. Moreover, it is shown that these new bounds of S DD matrices works better than the well-known Varah bound in some cases, and numerical examples are given to illustrate the corresponding results.

Main results
Firstly, some notations and a lemma are listed. D = diag{d 1 , d 2 , . . . , d n } denotes a diagonal matrix. (AD) i j denotes the entry (i, j) of matrix AD, and (AD) ii denotes the diagonal element of the ith row of matrix AD.
Proof. From Definition 1, we get that |a ii | > p i (A) for any i ∈ N 1 (A) and for any i ∈ N 2 (A), thus, we obtain that a matrix A is an S DD 1 matrix if and only if |a ii | > p i (A) for all i ∈ N.
Next, a new proof of Theorem 1 is given as follows.
Proof. It is sufficient to prove that each S DD 1 matrix A is an H-matrix. In order to do that, let us define the diagonal matrix as and if j∈N 2 (A)\{i} |a i j | = 0, then the corresponding fraction is defined to be ∝ .
Since matrix A is an S DD 1 matrix, D is a positive diagonal matrix. In the following, we prove that AD is an S DD matrix, and divided it into two cases. Case 1: for any i ∈ N 1 (A), it is easy to obtain that |(AD) ii | = |a ii |, and From Cases 1 and 2, we obtain that Since the definition of S DD 1 matrix was proposed, some properties of S DD 1 matrices were obtained, such as Schur complements of S DD 1 matrices [2], subdirect sums of S DD 1 matrices [15]. Next, some new properties of S DD 1 matrices are listed as follows.
Proof. Suppose on the contrary that for each i ∈ N 1 (A), a i j = 0, where j ∈ N 2 (A) and j i, then it is easy to obtain that p i (A) = r i (A) for any i ∈ N 1 (A) from Definition 1, thus we obtain that |a ii | > p i (A) = r i (A) ≤ |a ii |, which does not hold, hence for each i ∈ N 1 (A), there is at least one a i j 0, where j ∈ N 2 (A) and j i.
is an S DD 1 matrix by rows, and for each i ∈ N 2 (A), there is at least one a i j 0, where j ∈ N 2 (A) and j i, then |a ii | > p i (A) > 0 for any i ∈ N and Proof. From the Lemma 1, we get that |a ii | > p i (A) for any i ∈ N and |a ii | > r i (A) ≥ p i (A) for all i ∈ N 2 (A).
Since A is an S DD 1 matrix, and from the condition that for each i ∈ N 2 (A), A has at least one a i j 0, where j ∈ N 2 (A) and j i, it is easy to obtain that We next prove that |a ii | > p i (A) > 0 for any i ∈ N, and consider the following two cases separately. Case 1: if N 1 (A) = ∅, then A is an S DD matrix, and from the condition that for each i ∈ N 2 (A), A has at least one a i j 0, where j ∈ N 2 (A) and j i, thus it is easy to get |a ii | > p i (A) > 0 for any i ∈ N = N 2 (A). Case 2: if N 1 (A) ∅, then from Theorem 2 and the condition that for each i ∈ N 2 (A), A has at least one a i j 0, where j ∈ N 2 (A) and j i, we obtain that |a ii | > p i (A) > 0 for all i ∈ N.
From Cases 1 and 2, we obtain that |a ii | > p i (A) > 0 for any i ∈ N.
Theorem 4. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD 1 matrix by rows, and for each i ∈ N 2 (A), A has at least one a i j 0, where j ∈ N 2 (A) and j i, then there exists a diagonal matrix D = diag{d 1 , d 2 , . . . , d n }, where d i = p i (A) |a ii | , i = 1, 2, . . . , n, such that AD is an S DD matrix.
Proof. In order to prove that matrix AD is an S DD matrix, we need to prove that matrix AD satisfies the following inequalities: |(AD) ii | > r i (AD) f or any i ∈ N.
Since for each i ∈ N 2 (A), there is at least one a i j 0, where j ∈ N 2 (A) and j i, from Theorems 2 and 3, we obtain that |a ii | > p i (A) > 0 for any i ∈ N and |a ii | > r i (A) > p i (A) > 0 for all i ∈ N 2 (A).
Therefore, for any i ∈ N, it is easy to get |(AD) ii | = p i (A), and from 0 < |a j j | < 1 for any j ∈ N 2 (A), Theorems 2 and 3, we get that Obviously, for any i ∈ N, we get |(AD) ii | > r i (AD), that is, AD is an S DD matrix.
Finally, some upper bounds of the infinity norm of inverse of S DD 1 matrices and S DD matrices are established. Before that, a theorem which will be used later is listed. .
Proof. From the new proof of Theorem 1, we obtain that there exists a positive diagonal matrix D such that AD is an S DD matrix, where D is defined as Eq (2.2). Therefore, we have the following inequality: Since the matrix D is positive diagonal, it is easy to obtain that where ε satisfy inequality (2.3).
Since AD is an S DD matrix, by Theorem 5, we obtain Thus, we get Based the new proof, the upper bound of the infinity norm of inverse of S DD 1 matrix is presented, and since S DD matrices is a subclass of S DD 1 matrices, from Theorem 6, it is easy to obtain the following Corollary 1. Corollary 1. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD matrix, then where Example 2. Considering the following S DD 1 matrices Obviously, A 1 and A 2 are also S DD matrices. By calculation, we have p 1 (A 1 ) = 1, p 2 (A 1 ) = 1, p 3 (A 1 ) = 1.5 p 4 (A 1 ) = 1.5 and 0 < ε 1 < 1, and p 1 (A 2 ) = 1, p 2 (A 2 ) = 1, p 3 (A 2 ) = 1, p 4 (A 2 ) = 0 and 0 < ε 2 < 1.5.
The following Theorem 7 shows that the bound in Corollary 1 is better than in Theorem 5 in some conditions. Theorem 7. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD matrix, if where M i is given as in Eq (2.7) and ε satisfy inequality (2.8).
Proof. From the condition it is easy to obtain that thus, from combining similar terms at the left end of the above inequality, we obtain the following inequality Since A is an S DD matrix, we have |a ii | > r i (A) and M i > 0 f or any i ∈ N.
Therefore, from inequality (2.9), it is easy to have max i∈N p i (A) and thus from Corollary 1, we have The following Example 3 also illustrates the Theorem 7. Example 3. This is the previous Example 2. For the matrix A 1 , by a simple calculation, we obtain It is easy to verify that max i∈N p i (A 1 ) |a ii | min i∈N (|a ii | − r i (A 1 )) = 0.25 < 0.375 = min i∈N j∈N\{i} that is, the matrix A 1 satisfies the conditions of Theorem 7. Therefore, from Theorem 7, we obtain that for any 0 < ε 1 < 1, However, the upper bound (2.5) contains the parameter ε. Next, based on the Theorem 4, a upper bound of the infinity norm of inverse of S DD 1 matrices is presented as follows, and this upper bound only depends on the elements of given matrices. Theorem 8. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD 1 matrix, and for each i ∈ N 2 (A), there is at least one a i j 0, where j ∈ N 2 (A) and j i, then Proof. By Theorem 4, we obtain that there exists a positive diagonal matrix D such that AD is an S DD matrix, where D is defined as Theorem 4. Therefore, we get the following inequality: Since the matrix D is positive diagonal, it is easy to obtain that Since AD is an S DD matrix, by Theorem 5, we obtain Thus, we get that Since S DD matrices is a subclass of S DD 1 matrices, from Theorem 8, it is easy to obtain the following corollary. Corollary 2. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD matrix, if r i (A) 0 for all i ∈ N, then The following Theorems 9 and 10 show that the bound in Corollary 2 is better than in Theorem 5 in some conditions. Theorem 9. Let A = (a i j ) ∈ C n×n (n ≥ 2) be an S DD matrix, if r i (A) 0 for all i ∈ N and min i∈N j∈N\{i} .
Proof. Since A is an S DD matrix, it is easy to get that and from the condition min i∈N j∈N\{i} Therefore, from Corollary 2, we get .
The following Example 4 also illustrates the Theorem 9. Example 4. Considering the following S DD matrix 2.5 2 0.4 0 2 5.5 3 0 1 2 3.5 0 1 2 0 3.5 By the Varah bound (2.4) of Theorem 5, we obtain that ||A −1 3 || ∞ ≤ 10. By a simple calculation, we obtain Proof. Since A is an S DD matrix, we have |a ii | > r i (A) f or any i ∈ N.
From the condition r i (A) 0 for all i ∈ N and Theorem 4, it is easy to obtain that j∈N\{i} r j (A) − p j (A) |a j j | a i j > 0 f or any i ∈ N.
Therefore, from the condition .
The following Example 5 also illustrates the Theorem 10. Example 5. Considering the following the following S DD matrix