Applied Mathematical Sciences, Vol. 5, 2011, no. 6, 273 - 278
A New Criteria for a Matrix is not Generalized Strictly Diagonally Dominant Matrix
Zhijun Guo 1 and Jianguang Yang
Department of Mathematics, Hunan City University Yiyang, Hunan 413000, P.R. China
Abstract In this paper, we obtained some new results about a matrix is not generalized strictly diagonally dominant matrix by using gauss elimina- tion and some exists theories, which improve the corresponding results of [4]. Advantages are illustrated by numerical examples.
Mathematics Subject Classification: 15A47
Keywords: Gauss elimination, diagonally dominant matrix, diagonally dominant matrix, bi-diagonally dominant matrix
1 Introduction
Generalized strictly diagonally dominant matrix play an important role in numerical algebra, control theory, electric system, economic mathematics and elastic dynamics and so on. But it isn’t easy to determine whether a matrix is or not a generalized strictly diagonally dominant matrix in reality. In this paper, we obtained some new results about a matrix is not generalized strictly diagonally dominant matrix.
2 Preliminary Notes
n×n n×n n×n Let C (R ) be the set of complex (real) n×n matrices, A =(aij) ∈ C , N = {1, 2, ···,n}, for any i, j ∈ N, we denote: αi = |aij|,βi = |aij|, α˜i = |aji|, β˜i = |aji|, j∈N1 j∈N2 j∈N1 j∈N2 j =i j =i j =i j =i Λi = |aij| = αi + βi,Si = |aji| =˜αi + β˜i, j =i j =i [email protected] 274 Z. Guo and J. Yang
where N = N1 ⊕ N2,N1 = {i||aii| > Λi,i∈ N},N2 = {i|0 < |aii|≤Λi(A),i∈ (k) (k) N}. In this paper, we denote A =(aij ) is a matrix that A after kth steps gauss elimination. n×n Definition 2.1 Let A =(aij) ∈ C . (i) If |aii|≥Λi, ∀i ∈ N, then A is a diagonally dominant matrix; (ii) If |aii| > Λi, ∀i ∈ N, then A is a strictly diagonally dominant matrix, we denote it by A ∈ D0; (iii) If there exists a positive diagonal matrix D = diag{d1,d2,...,dn}, such that AD = B ∈ D0, then A is a generalized strictly diagonally dominant matrix, we denote it by A ∈ D . It is obvious that if N1 = Φ be an empty set, then A ∈ D ,ifN2 = Φ, then A ∈ D . n×n Definition 2.2 Let A =(aij) ∈ C . (i) If there exists an α ∈ (0, 1] and
|aii|≥αΛi +(1− α)Si (1.1) for any i ∈ N, then A is an α diagonally dominant matrix. We denote it by A ∈ D0(α); (ii) If all strictly inequality holds in (1.1), then A is a strictly α diagonally dominant matrix. We denote it by A ∈ D(α); (iii) If there exists a positive diagonal matrix D, such that AD = B ∈ D(α), then A is a generalized strictly α diagonally dominant matrix. We denote it by A ∈ D (α). n×n Definition 2.3 Let A =(aij) ∈ C . (i) If there exists an α ∈ (0, 1] and α 1−α |aiiajj|≥(ΛiΛj) (SiSj) (1.2) for any i = j, (i, j ∈ N), then A is an α bi-diagonally dominant matrix. We α denote it by A ∈ D0 ; (ii) If all strictly inequality holds in (1.2), then A is a strictly α bi-diagonally dominant matrix. We denote it by A ∈ Dα; (iii) If there exists a positive diagonal matrix D, such that AD = B ∈ Dα, then A is a generalized strictly α bi-diagonally dominant matrix. We denote it by A ∈ D α. n×n n×n Definition 2.4 Let A =(aij) ∈ C ,ifμ(A)=(mij) ∈ R , |aii|,i= j, mij = −|aij|,i= j. then we say μ(A) is a comparison matrix of A. n×n A matrix A =(aij) ∈ C is called nonsingular H-matrix if its comparison matrix μ(A) is a nonsingular M-matrix. Criteria for a matrix is not generalized strictly diagonally dominant matrix 275
3 Main Results
In this section, we will give some criteria for a matrix is not generalized strictly diagonally dominant matrix.
Lemma 3.1 ([1]) If A ∈ D(α),then μ(A) is a nonsingular M-matrix and A ∈ D .
Lemma 3.2 ([2]) If A ∈ Dα,then A ∈ D .
n×n Lemma 3.3 ([3]) Let A =(aij) ∈ C ,ifA is a nonsingular H-matrix, then there exists at least one strict diagonally dominant row, namely N1 =Φ .
n×n Lemma 3.4 ([4]) Let A =(aij) ∈ C , if there exist N1, N2 such that N1 ∪ N2 = N, and (|aii|−αi)(|ajj|−βj) ≤ βiαj for any i ∈ N1,j ∈ N2, then A is not a generalized strictly diagonally dominant matrix and μ(A) is not a nonsingular M-matrix.
n×n (n−1) Theorem 3.5 Let A =(aij) ∈ C ,ifA ∈ D0, then A ∈ D0.
Proof. As A ∈ D0, we have a11 = 0 and |aii| > |aij| = |ai1| + |aij| j =i j =1,i Λ1 > |ai1| + |aij| |a11| j =1,i |ai1| |ai1| = |a1j| + |a1i| + |aij| |a11| j =1,i |a11| j =1,i it follows that