AlgebraVolume 2013 (2013), Article ID 102438, 4 pageshttp://dx.doi.org/10.1155/2013/102438

Research Article

Bound on the Minimum Eigenvalue of -Matrices Involving Hadamard Products

Kun Du,¹ Guiding Gu,² and Guo Liu¹

¹School of Finance, Shanghai University of Finance and Economics, Shanghai 200433, China²Department of Applied Mathematics, Shanghai University of Finance and Economics, Shanghai 200433, China

Received 5 March 2013; Accepted 4 June 2013

Academic Editor: Sorin Dascalescu

Copyright © 2013 Kun Du 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.

We present a new lower bound on the minimum eigenvalue of -matrices involving Hadamard products , and we show that our lower bound is larger than the lower bound . Three examples verify our result.

1. Introduction

In [1], it is shown by Theorem  5.7.15 that if are   -matrices for all , and satisfy , then where is defined as entrywise and any scalar definition of such that is allowed. This theorem provided a beautiful result about the minimum eigenvalue of -matrices involving Hadamard products, but sometimes this inequality could be very weak. For example, are -matrices, and see the details in Section 3.

A lot of works have been done on the minimum eigenvalue of -matrices and -matrices involving Hadamard products, see the results in [1–6].

In this paper, we present a new lower bound by including diagonal entries and prove that our bound is larger than the bound in (1).

We now introduce some notations, see [1]. The Hadamard product of and is defined by . We define and . Let and denote by for every . We denote , where , and denote by the class of all real matrices all of whose off-diagonal entries are nonpositive. Let , then the minimum eigenvalue of is defined by ; is a eigenvalue of .

For two real matrices , the Fan product of and , denoted by , is defined by and then . The comparison matrix of a given matrix is defined by A matrix is an -matrix if its comparison matrix is an -matrix. For , we define .

Last, for , we introduce a new definition , where

2. Main Results

To prove the main theorem, we need several lemmas.

Lemma 1. Let for all . If and , then

Proof. When , we can prove this inequality by the Hölder inequality and by induction. When , apparently, , similarly, we can prove this inequality.

If , then for some and some . Let . Then . If there is a such that , then , but this is a contradiction. So, . If is the right Perron eigenvector of , then . If is the left Perron eigenvector of , then . So, similar to the Perron-Frobenius theorem, we have the following: if became irreducible, then there exist positive vectors and such that , and , and being called right and left Perron eigenvectors of , respectively.

Lemma 2. If is irreducible, and for a nonnegative nonzero vector , then .

Proof. , where is irreducible and . By the Perron-Frobenius theorem, has a positive left Perron vector , that is, . Note that . Hence, . Since , we have .

Lemma 3. Let for all . If and , then

Proof. It is quite evident that the conclusion holds with equality for . For , we have two cases.Case  1. is irreducible. Then are irreducible for all . Thus, are irreducible for all . Let be the right Perron eigenvectors of for all . If for all , then for every , we have Let . Let . Then for all . Then, for all , we have By Lemma 1, the “≥” hold. By Lemma 2, we have Case  2. is reducible. We denote by the permutation matrix () with the remaining zero, then both are irreducible for any chosen positive real number . Now we substitute for in the previous case, and then by letting , the result follows by continuity.

Lemma 4 (see [7]). Let for all , and satisfying . Then

The following is our main theorem.

Theorem 5. Let and . If is   -matrices for all , then

Proof. Without loss of generality, we can assume that for all . Let , then . Now we have: The first “≥” hold by Lemma 3. Because is -matrix, we have , which means that . So the second “≥” hold by and Lemma 4.

3. Examples

In this section, we present three examples to illustrate our improved bound.

Example 6. We take the matrices , , and in Section 1. It is easy to get , so the low bound in (1) is ; on the other side, we can get , which is greatly larger than the lower bound , that is, However, our lower bound in Theorem 5 is which is the exact value of .

Example 7. Let It is easy to see that and are -matrices and . Thus, we have but our bound is which is close to the exact value of 9799.

Example 8. Take the -matrices It is easy to see that but our bound is Also, our bound is much closer to the exact value than the low bound in (1).

Acknowledgments

The authors would like to thank the anonymous reviewers for their helpful comments and valuable suggestions which improved the quality of this paper. This work is supported by the Cultivation Fund of the Key Scientific and Technical Innovation Project, Ministry of Education of China (no. 708040), the Leading Academic Discipline Program, 211 Project for Shanghai University of Finance and Economics (the third phase), and Graduate Research and Innovation Fund of Shanghai University of Finance and Economics (CXJJ-2012-385).

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