%0 Journal Article
%T Bound on the Minimum Eigenvalue of -Matrices Involving Hadamard Products
%A Kun Du
%A Guiding Gu
%A Guo Liu
%J Algebra
%D 2013
%I Hindawi Publishing Corporation
%R 10.1155/2013/102438
%X 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¨C6]. 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 H£¿lder 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
%U http://www.hindawi.com/journals/algebra/2013/102438/