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Academic Editor:Shi-Liang Wu
School of Mathematics and Statistics, Yunnan University, Kunming, Yunnan 650091, China
Received 1 May 2014; Accepted 24 May 2014; 12 June 2014
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
The class of Nekrasov matrices is a subclass of H -matrices. Estimating the infinity norm of the inverse of Nekrasov matrices can be used to prove the convergence of matrix splitting and matrix multisplitting iteration methods for solving large sparse systems of linear equations; see [1-4]. Here, we call a matrix A = ( a i j ) ∈ C n , n an H -matrix if its comparison matrix Y9; A YA; = [ m i j ] defined by [figure omitted; refer to PDF] is an M -matrix; that is, Y9; A YA; - 1 ...5; 0 [1, 5, 6], and a matrix A = [ a i j ] ∈ C n , n is called a Nekrasov matrix if for each i ∈ N , [figure omitted; refer to PDF] where h 1 ( A ) = ∑ j ...0; 1 | a 1 j | and h i ( A ) = ∑ j = 1 i - 1 ( | a i j | / | a j j | ) h j ( A ) + ∑ j = i + 1 n | a i j | , i = 2,3 , ... , n [2, 6].
In 1975, Varah [7] provided the following upper bound for strictly diagonally dominant (SDD) matrices as one most important subclass of Nekrasov matrices, consequently, H -matrices [2, 6, 8]. Here a matrix A = [ a i j ] ∈ C n , n is called SDD if for each i ∈ N = { 1,2 , ... , n } , [figure omitted; refer to PDF] where r i ( A ) = ∑ j ...0; i | a i j | .
Theorem 1 (see [7]).
Let A = [ a i j ] ∈ C n , n be SDD. Then [figure omitted; refer to PDF]
We call the bound in Theorem 1 the Varah's bound. As Cvetkovic et al. [2] said, Varah's bound works only for SDD matrices and even then it is not always good enough. To obtain new upper bounds for the infinity norm of the inverse of a wider class of matrices which sometimes works better in the SDD case, Cvetkovic et al. [2] give the following bound of Nekrasov matrices.
Theorem 2 (see [2, Theorem 2]).
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix. Then [figure omitted; refer to PDF] where z 1 ( A ) = 1 and z i ( A ) = ∑ j = 1 i - 1 ( | a i j | / | a j j | ) z j ( A ) + 1 , i = 2,3 ... , n .
In [9, Theorems 2.2 and 2.3], Kolotilina gave an improvement of these upper bounds in Theorem 2 (see Theorems 3 and 4).
Theorem 3 (see [9, Theorem 2.2]).
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix. Then [figure omitted; refer to PDF]
Theorem 4 (see [9, Theorem 2.3]).
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix. Then [figure omitted; refer to PDF]
In this paper, we also focus on the estimation problem of the infinity norm of the inverse of Nekrasov matrices and give an improvement of the bound in Theorem 3 (Theorem 2.2 in [9]). Numerical example is given to illustrate the corresponding results.
2. Bounds for the Infinity Norm of the Inverse of Nekrasov Matrices
In order to obtain a new bound, we start with the following lemmas and notations. Given a matrix A = [ a i j ] , by A = D - L - U we denote the standard splitting of A into its diagonal ( D ) , strictly lower ( - L ) , and strictly upper ( - U ) triangular parts. And by [ A ] i j denote the ( i , j ) -entry of A ; that is, [ A ] i j = a i j . Furthermore, we denote | A | = [ | a i j | ] .
Lemma 5 (see [10]).
Let A = [ a i j ] ∈ C n , n be a nonsingular H -matrix. Then [figure omitted; refer to PDF]
Lemma 6 (see [11]).
Given any matrix A = [ a i j ] ∈ C n , n , n ...5; 2 , with a i i ...0; 0 for all i ∈ N , then [figure omitted; refer to PDF] where e ∈ C n , n is the vector with all components equal to 1.
Lemma 7 (see [12]).
A matrix A = [ a i j ] ∈ C n , n , n ...5; 2 , is a Nekrasov matrix if and only if [figure omitted; refer to PDF] that is, if and only if E - ( | D | - | L | ) - 1 | U | is an SDD matrix, where E is the identity matrix.
Let C = E - ( | D | - | L | ) - 1 | U | = [ c i j ] . Then from Lemma 7, C is SDD when A is a Nekrasov matrix. Note that c 11 = 1 , c k 1 = 0 , k = 2,3 , ... , n , and c 1 k = - | a 1 k | / | a 11 | , k = 2,3 , ... , n , which leads to the following lemma.
Lemma 8.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix and [figure omitted; refer to PDF] where D ( μ ) = diag ... ( μ , 1 , ... , 1 ) and μ > r 1 ( A ) / | a 11 | . Then C ( μ ) is SDD.
Proof.
It is not difficult from (12) to see that [ C ( μ ) ] k 1 = μ c k 1 for all k ∈ N and [ C ( μ ) ] k j = c k j for all k ∈ N and j ...0; 1 . Hence [figure omitted; refer to PDF] and for i = 2 , ... , n , [figure omitted; refer to PDF] From the fact that C is SDD and μ > r 1 ( A ) / | a 11 | , we have that C ( μ ) is SDD. The proof is completed.
The main result of this paper is the following theorem.
Theorem 9.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix. Then for μ > r 1 ( A ) / | a 11 | , [figure omitted; refer to PDF]
Proof.
Let C ( μ ) = C D ( μ ) = ( E - ( | D | - | L | ) - 1 | U | ) D ( μ ) , where D ( μ ) = diag ... ( μ , 1 , ... , 1 ) . From (12), we have [figure omitted; refer to PDF] which implies that [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Furthermore, since a Nekrasov matrix is an H -matrix, we have, from Lemma 5, [figure omitted; refer to PDF]
First, we estimate || [ ( | D | - | L | ) Δ ] - 1 || ∞ . Since ( | D | - | L | ) Δ is an M -matrix and there exists a positive diagonal matrix Δ such that ( | D | - | L | ) Δ e = e , see [9], we get [figure omitted; refer to PDF]
Secondly, we estimate || C ( μ ) - 1 Δ || ∞ . From Lemma 8, C ( μ ) is SDD. Obviously, multiplying the left-hand side of C ( μ ) by diagonal matrix Δ - 1 does not change SDD property, so Δ - 1 C ( μ ) is also SDD. Thus, Varah's bound (4) can be applied as follows: [figure omitted; refer to PDF] In addition, since z ( A ) = [ z 1 ( A ) , ... , z n ( A ) ] T = | D | ( | D | - | L | ) - 1 e and ( | D | - | L | ) Δ e = e , see [9, 13], we have [figure omitted; refer to PDF] Substituting (22) into (21), we get that [figure omitted; refer to PDF] Finally, from (20), (23), z 1 ( A ) = 1 , and the fact that || D ( μ ) || ∞ = max ... { μ , 1 } , we have [figure omitted; refer to PDF] The conclusions follow.
Example 10.
Consider the Nekrasov matrix A 1 in [2, 9], where [figure omitted; refer to PDF] By computation, h 1 ( A ) = 3.2000 , h 2 ( A ) = 8.2000 , h 3 ( A ) = 2.9609 , h 4 ( A ) = 0.7359 , z 1 ( A ) = 1 , z 2 ( A ) = 2 , z 3 ( A ) = 1.2971 , and z 4 ( A ) = 1.2394 . By the bound of Theorem 3 (the bound of Theorem 2.2 in [9]), we have [figure omitted; refer to PDF] By Theorem 9, we have [figure omitted; refer to PDF] In fact, || A 1 - 1 || ∞ = 0.1921 .
Remark 11.
Example 10 shows that by choosing the value of μ , the bound in Theorem 9 is better than that in Theorem 3 in some cases. We further observe the bound in Theorem 9 by Figure 1 and find that there is an interval such that for any μ in this interval, the bound in Theorem 9 for the matrix A 1 is always smaller than that in Theorem 3. An interesting problem arises: whether there is an interval of μ such that the bound in Theorem 9 for any Nekrasov matrix is smaller than that in Theorem 3. In the following section, we will study this problem.
Figure 1: The bounds in Theorems 9 and 3.
[figure omitted; refer to PDF]
3. The Choice of μ
In this section, we determine the value of μ such that the bound for || A - 1 || ∞ in Theorem 9 is less than or equal to that in [9]. First, we consider the Nekrasov matrix A = [ a i j ] ∈ C n , n with [figure omitted; refer to PDF] and give the following lemma.
Lemma 12.
Let a , b , and c be positive real numbers, and 0 < a ( b - c ) < 1 . Then [figure omitted; refer to PDF]
Proof.
We only need to prove that ( 1 + a c ) / a b - 1 > 0 and 1 / a ( b - c ) - ( 1 + a c ) / a b > 0 . In fact, [figure omitted; refer to PDF] The proof is completed.
Lemma 13.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix with [figure omitted; refer to PDF] Then [figure omitted; refer to PDF]
Proof.
Let a = max ... i ...0; 1 ... ( z i ( A ) / ( | a i i | - h i ( A ) ) ) , b = | a 11 | , and c = h 1 ( A ) . From (28), we get 0 < a ( b - c ) < 1 . Then from Lemma 12, the first and second inequalities in (32) hold.
We now give an interval of μ such that the bound in Theorem 9 is less than that in Theorem 3.
Lemma 14.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix with [figure omitted; refer to PDF] Then for each μ ∈ ( 1 , ( 1 / ( | a 11 | - h 1 ( A ) ) ) / max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , [figure omitted; refer to PDF]
Proof.
From Lemma 13, we have [figure omitted; refer to PDF] and max ... { μ , 1 } = μ .
( I ) For μ ∈ ( 1 , ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) ] , then [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] Consider the function f ( x ) = x / ( x | a 11 | - h 1 ( A ) ) , x ∈ [ 1 , ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) ] . It is easy to prove that f ( x ) is a monotonically decreasing function of x . Hence, for any μ ∈ ( 1 , ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) ] , [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Hence, [figure omitted; refer to PDF]
( II ) For ∈ [ ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , ( 1 / ( | a 11 | - h 1 ( A ) ) ) / max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , then [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] Consider the function [figure omitted; refer to PDF] Obviously, g ( x ) is a monotonically increasing function of x . Hence, for any μ ∈ [ ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , ( 1 / ( | a 11 | - h 1 ( A ) ) ) / max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Hence, [figure omitted; refer to PDF] The conclusion follows from ( I ) and ( II ) .
Lemma 14 provides an interval of μ such that the bound in Theorem 9 is better than the bound in Theorem 3 (the bound in [9]). Moreover, we can determine the optimal value of μ by the following theorem.
Theorem 15.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix with [figure omitted; refer to PDF] Then [figure omitted; refer to PDF] Furthermore, [figure omitted; refer to PDF]
Proof.
From the proof of Lemma 14, we have that [figure omitted; refer to PDF] is decreasing and that [figure omitted; refer to PDF] is increasing. Therefore, the minimum of f ( x ) and g ( x ) is [figure omitted; refer to PDF] which implies that (50) holds. Again by Lemma 14, (51) follows easily.
Remark 16.
Theorem 15 provides a method to determine the optimal value of μ for a Nekrasov matrix A = [ a i j ] ∈ C n , n with [figure omitted; refer to PDF] Also consider the matrix A 1 in Example 10. By computation, we get [figure omitted; refer to PDF] Hence, by Theorem 15, we can obtain that the bound in Theorem 9 reaches its minimum [figure omitted; refer to PDF] at μ = ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) = 1.0639 (also see Figure 1).
Next, we study the bound in Theorem 9 for the Nekrasov matrix A = [ a i j ] ∈ C n , n with [figure omitted; refer to PDF]
Theorem 17.
Let A = [ a i j ] ∈ C n , n be a Nekrasov matrix with [figure omitted; refer to PDF] Then we can take μ ∈ [ ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , 1 ] such that [figure omitted; refer to PDF]
Proof.
From (59), we get ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) ...4; 1 . Then, for μ > r 1 ( A ) / | a 11 | = h 1 ( A ) / | a 11 | , we have [figure omitted; refer to PDF]
( I ) For μ ∈ ( h 1 ( A ) / | a 11 | , ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) ) , then max ... { μ , 1 } = 1 and [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] ( II ) For μ ∈ [ ( 1 + max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) · h 1 ( A ) ) / ( | a 11 | · max ... i ...0; 1 ( z i ( A ) / ( | a i i | - h i ( A ) ) ) ) , 1 ] , then max ... { μ , 1 } = 1 and [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] ( III ) For μ ∈ ( 1 , + ∞ ) , then max ... { μ , 1 } = μ and [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] The conclusion follows from (I), (II), and (III).
Remark 18.
Theorems 15 and 17 provide the value of μ ; that is, [figure omitted; refer to PDF] such that the bound in Theorem 9 is not worse than that in Theorem 3 for a Nekrasov matrix A = [ a i j ] ∈ C n , n . In particular, for the Nekrasov matrix A with 1 / ( | a 11 | - h 1 ( A ) ) > max ... i ...0; 1 ( z i ( A ) / ( | a i i - h i ( A ) | ) ) , the bound in Theorem 9 is better than that in Theorem 3.
Example 19.
Consider the following five Nekrasov matrices in [2, 9]: [figure omitted; refer to PDF] Obviously, A 2 , A 3 , and A 4 are SDD. And it is not difficult to verify that A 4 satisfies the conditions in Theorem 15 and A 2 , A 3 , A 5 , and A 6 satisfy the conditions in Theorem 17. We compute by Matlab 7.0 the upper bounds for the infinity norm of the inverse of A i , i = 2 , ... , 6 , which are shown in Table 1. It is easy to see from Table 1 that this example illustrates Theorems 15 and 17.
Table 1: The upper bounds for | | A i - 1 | | ∞ , i = 2 , ... , 6 .
Matrix | A 2 | A 3 | A 4 | A 5 | A 6 |
Exact | | A - 1 | | ∞ | 0.2390 | 0.8759 | 0.2707 | 1.1519 | 0.4474 |
Varah (4) | 1 | 1.4286 | 0.5556 | -- | -- |
Cvetkovic et al. (5) | 0.8848 | 1.8076 | 0.6200 | 1.4909 | 1.1557 |
Cvetkovic et al. (6) | 0.6885 | 0.9676 | 0.7937 | 2.4848 | 0.5702 |
Kolotilina (7) | 0.5365 | 0.9676 | 0.5556 | 1.4138 | 0.4928 |
Theorem 9 | 0.5365 | 0.9676 | 0.5038 | 1.4138 | 0.4928 |
4. Conclusions
In this paper, we give an improvement on the infinity norm bound for the inverse of a Nekrasov matrix in [9]. In particular, for the Nekrasov matrix A = [ a i j ] ∈ C n , n with [figure omitted; refer to PDF] we prove that new bound is better than that in [9]. However, for the Nekrasov matrix A with [figure omitted; refer to PDF] we only obtain that new bound is equal to that in [9]. For this case, we try to found some better bounds in future. On the other hand, our bound only considers one parameter μ , that is, D ( μ ) = diag ... ( μ , 1 , ... , 1 ) , which poses an interesting problem: whether we further improve this bound by introducing more parameters. In future, we will research this problem.
Acknowledgments
This research is supported by the National Natural Science Foundation of China (11361074 and 11326242), the Science Foundation of the Education Department of Yunnan Province of China (2013FD002), and Science and Technology Innovation Fund projects of Yunnan University (ynuy201366).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Copyright © 2014 Lei Gao et al. Lei Gao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
A new upper bound which involves a parameter for the infinity norm of the inverse of Nekrasov matrices is given. And we determine the optimal value of the parameter such that the bound improves the results of Kolotilina, 2013. Numerical examples are given to illustrate the corresponding results.
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