Abstract

Nonsingular -matrices and positive stable matrices play an important role in the stability of neural network system. In this paper, some criteria for nonsingular -matrices are obtained by the theory of diagonally dominant matrices and the obtained result is introduced into identifying the stability of neural networks. So the criteria for nonsingular -matrices are expanded and their application on neural network system is given. Finally, the effectiveness of the results is illustrated by numerical examples.

1. Introduction

The research on data mining based on neural networks has a great significance. Recently, as one kind of artificial neural networks, Hopfield neural network is used for association rules mining and remarkable results are obtained. Nonsingular -matrices and positive stable matrices play an important role in the stability of neural network system. However, it is rather difficult in practice to determine whether a matrix is a nonsingular -matrix or not. Therefore, it is of a great theoretical and practical value to study the numerical methods for judging the nonsingular -matrices, to provide the concise and practical criteria. Up to now, within the scope of the field, many researchers have done a lot of in-depth studies and acquired some very valuable results in many respects, such as nonsingular -matrix properties and criteria (see [1–9]). In this paper, some criteria for nonsingular -matrices are obtained by the theory of diagonally dominant matrices and the obtained result is introduced into identifying the stability of neural networks. So the criteria for nonsingular -matrices are expanded and their application on neural network system is given. Effectiveness of the results is illustrated by numerical examples. For convenience, we are dealing with nonsingular -matrices, calling them shortly -matrices.

Next, we will introduce some notations.

Let , and let . denotes the set of all by complex matrices: and (for all ).

If (for all ), then is said to be a (strictly) diagonally dominant matrix and is denoted by (); if (for all ), then is said to be a (strictly) double diagonally dominant matrix and is denoted by (). It is well known that an equivalent definition of -matrices is given by demanding that there exist positive numbers such that (for all ); that is, there exists a positive diagonal matrix such that (see [1]). So, we always assume that (for all ).

2. Definitions and Lemmas

It is learned that the class of -double diagonally dominant matrices play a central role in identifying -matrices. So, we will start with its definition and some background results.

Definition 1 (see [2]). Let ; if there exists some , satisfying then is called a (strictly) -double diagonally dominant matrix and is denoted by ().

Lemma 2 (see [2]). Let ; if , then is an -matrix.

Lemma 3 (see [3]). Let , if there exists some , satisfying and, for every with , there exists a nonzero elements chain or such that or , , where then is an -matrix.
Let denote the set of all circuits of length in (directed graph of the matrix ). Recall that a circuit in is an ordered sequence of vertices (), where are all distinct and () are arcs of . Let denote the set of all arcs.

Lemma 4 (see [4]). Let be an irreducible complex matrix. Suppose there exists some , satisfying
If there exists some arc and such that then is an -matrix.

3. Criteria for -Matrices

In the rest of the paper, we will use the notations:

It is obvious to deduce that .

Let It is obvious to observe

The following are our main results. First, we give an equivalent representation for strictly -double diagonally dominant matrices.

Lemma 5. Let ; then if and only if and for any , , satisfying

Proof. Sufficiency. From inequality (9), for any , , it follows that
Recalling that , for any , we have . So there exists some positive number such that
Let ; it is easy to see and ; that is,
By both ends of inequality (13) multiplied by , we have ; that is,
The inequality above implies that
By inequality (12) again, for any , it is obvious that ; that is,
By both ends of inequality (16) multiplied by , we have ; that is,
The inequality above implies that
Moreover, for any , and any , it is obvious that
Recalling that , for any , there exists some such that
Therefore, we have by Definition 1.
Necessity. Suppose ; then , and, for any , there exists some such that that is,
Then by the notations of and , we have . Furthermore, by both ends of the inequality multiplied by , we get . Therefore, it can be seen that
Following a similar argument for any , we have
Combining inequalities (23) and (24), we obtain inequality (9). The proof is completed.

As its application, some new practical criteria for -matrices are obtained.

Theorem 6. Let , , and, for any , , satisfying then is an -matrix.

Proof. By Lemma 5, we obtain , and further using Lemma 2, we conclude that is an -matrix.

Theorem 7. is an H-matrix if satisfies either of the conditions:(1);(2).

Proof. (1) Suppose ; then, for any , by , there exists some positive number , such that
Let ; then we have , which implies that
For any , and any , it is obvious that
Next, similarly as in the proof of Sufficiency of Lemma 5, we conclude that is an -matrix.
(2) Suppose ; then for any , by , there exists some positive number such that
Let ; then we have , which implies that
Similarly, we conclude that is an -matrix.

Theorem 8. Let , , and, for any , , satisfying
If, for every pair of indices , with there exists two nonzero elements chains or and or with or , and or , , where then is an -matrix.

Proof. Similarly as in the proof of Sufficiency of Lemma 5 combined with inequality (31), we can prove that for any , and , there exists some such that
Moreover, for any , and any , it is obvious that
Recalling that , we conclude that there exists some , such that
By equality (32), we know that, for every pair of indices , with there exists two nonzero elements chains or and or with or , and or , , where By Lemma 3, it follows that is an -matrix.

Similarly as in the proof of Theorem 8, we can obtain the following result.

Theorem 9. Let , , and, for any , , satisfying
If, for every pair of indices , , there exists two nonzero elements chains or and or with or and or such that , where then is an -matrix.

Theorem 10. Let be an irreducible complex matrix, , and, for any , , satisfying
If there exists some arc and such that then is an -matrix.

Proof. With the same argument as in the proof of Theorem 8, we can obtain that, for any , there exists some such that
By inequality (42), we know that there exists some arc and such that
Recalling that is irreducible, it follows that is an -matrix by Lemma 4.

4. Algorithm and Program

Algorithm for Theorem 6.(1)Input matrix ;(2)calculate and (for all ) (denoted in the Introduction of the paper);(3)define indices , , and ;(4)if , then the criterion is invalid;(5)if , then calculate (for all ) and (for all );(6)calculate and verify the condition of Theorem 6. If the condition is satisfied, then output “ is an -matrix.”

We write the related program by the above algorithm using MATLAB Software. All the results are calculated by MATLAB 7.0. The procedures are shown in Procedure 1.

A=input(“please input a matrix”)
M1= ;M2= ;M6= ;F= ;B= ;
n=size(A,1);RA=zeros(n,1);CA=zeros(n,1);
for k=1:n
    A=abs(A);
    RA(k)=[sum(A(k,:))-A(k,k)];
    CA(k)=[sum(A(:,k))-A(k,k)];
end
for i=1:n-1
    for j=i+1:n
    RR=RA(i)*RA(j);
    aa=abs(A(i,i)*A(j,j));
    CC=CA(i)*CA(j);
    if RR<aa&aa<CC
     M1=[M1;i,j];
     alpha=aa/RR;beta=CC/aa;gamma=alpha*beta;
     F=[F,alpha,beta,gamma];
    elseif CC<aa&aa<RR
     M2=[M2;i,j];
     x=aa/CC;y=RR/aa;z=x*y;
     B=[B,x,y,z];
    elseif RR>=aa&CC>=aa
     M6=1;break;
     Show=“the criterion is invalid”;
     end
    end
end
if M6==1
    “the criterion is invalid”;
    elseif size(M1,1)==0|size(M2,1)==0
     “A is an H-matrix”
     else
      k1=size(F,1);k2=size(B,1);
      for i=1:k1
      F2(i)=log(F(i,2))/log(F(i,3));
      end
      for i=1:k2
      B2(i)=log(B(i,2))/log(B(i,3));
      end
      if max(B2)+max(F2)<1
       show=“A is an H-matrix”
     end
end
Procedure 1 

5. Numerical Examples

Example 1. Let
Then we have
But, we notice . The condition does not satisfy either Theorem 2 or Theorem  3 in [5], so we cannot obtain that is an -matrix.
According to the notations of this paper, we have
By calculating, we obtain and then
It satisfies the condition of Theorem 6, and then is an -matrix.
We consider the following Hopfield type continuous neural networks: where,
Notice that is an -matrix, and then is a nonsingular -matrix, which ensures existence, uniqueness, and global exponential stability of the equilibrium point of the above neural networks by [10].

Example 2. Let
By calculating, we have
It satisfies the condition (1) of Theorem 7, and then is an -matrix.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This work is supported by Fund of Inner Mongolia University for the Nationalities (NMD1303).