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On the Eigenproblems of Nilpotent Lattice Matrices

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On the Eigenproblems of Nilpotent Lattice Matrices Advances in Fuzzy Mathematics. ISSN 0973-533X Volume 12, Number 4 (2017), pp. 855-866 © Research India Publications http://www.ripublication.com On the Eigenproblems of Nilpotent Lattice Matrices Geena Joy Department of Mathematics, Union Christian College, Aluva, Kochi, Kerala, India – 683102 K. V. Thomas Department of Mathematics, Bharata Mata College, Thrikkakara, Kochi, Kerala, India – 682021 Abstract This paper discusses the eigenproblems of nilpotent lattice matrices and describes the eigenspace of a nilpotent lattice matrix. Also, this paper introduces the concept of non-singular lattice matrices. We prove that every invertible lattice matrix is non-singular and every nilpotent lattice matix is singular. Keywords: Distributive lattice, Nilpotent lattice matrix, eigenspace, Non-singular lattice matrix, Invertible lattice matrix 2010 MSC: 15B99 1. INTRODUCTION The notion of lattice matrices appeared firstly in the work, ‘Lattice matrices’ [2] by G. Give’on in 1964. A matrix is called a lattice matrix if its entries belong to a distributive lattice. All Boolean matrices and fuzzy matrices are lattice matrices. Lattice matrices in various special cases become useful tools in various domains like the theory of switching nets, automata theory and the theory of finite graphs [2]. The concept of nilpotent lattice matrix was introduced firstly by G. Give’on [2]. A m square lattice matrix A is called nilpotent if A = 0n , for some positive integer m , where 0n is the zero matrix. In [2], Give’on proved that an nn lattice matrix A is n nilpotent if and only if A = 0n . Since then, a number of researchers have carried out their research in the area of nilpotent lattice matrices (see [6, 8, 9, 10]). The eigenvector-eigenvalue problems (for short, eigenproblems) of matrices over a 856 Geena Joy and K. V. Thomas complete and completely distributive lattice with the greatest element 1 and the least element 0 (lattice matrices) were discussed firstly by Y. J. Tan [5]. In [7], Y. J. Tan raised an open problem ’How to describe the eigenspace for a given lattice matrix?’. In this paper, we describe the eigenspace of a given nilpotent lattice matrix. Further, we introduce the concept of non-singular lattice matrices. In classical theory of matrices over a field, the theory of non-singular matrices coincides with the theory of invertible matrices. In the case of lattice matrices, every non-singular lattice matrix need not be invertible. Also, we prove that every nilpotent lattice matrix is singular. 2. PRELIMINARIES We recall some basic definitions and results on lattice theory and lattice matrices which will be used in the sequel. For details see [1, 2, 3, 4, 7, 9, 10]. A partially ordered set (,)L is a lattice if for all a, b L , the least upper bound of {,}a b and the greatest lower bound of {,}a b exist in L . For any a, b L , the least upper bound and the greatest lower bound is denoted by a b and a b (or ab ), respectively. An element a L is called greatest element of L if x a , x L . An element b L is called least element of L if b x , x L . We use 1 and 0 to denote the greatest element and the least element of L , respectively. A nonvoid subset K of a lattice L is a sublattice of L , if for any a, b K , a b , a b K . A lattice L is called a complete lattice if for any HL , both the least upper bound {y | y H } and the greatest lower bound {y | y H } of H exist in L . A lattice (,,,)L is a distributive lattice if the operations and are distributive with respect to each other. For any a, b L , the greatest element x L satisfying the inequality a x b is called the relative pseudocomplement of a in b and is denoted by a b . If for any a, b L , a b exists, then L is said to be a Brouwerian lattice. For a L , a 0 is denoted by ac . A lattice L is said to be completely distributive if for any x L and any {yi | i I } L , where I is an index set, (a). x( i I y i ) = i I ( x y i ) and (b). x( i I y i ) = i I ( x y i ) holds. It is known [1] that a complete lattice L is Brouwerian if and only if (a) is satisfied in L . Therefore, a complete and completely distributive lattice L is Brouwerian. Unless otherwise specified, we shall assume throughout the paper that L is a complete and completely distributive lattice with the greatest element 1 and the least element 0 , respectively. Lemma 2.1 [4, 7] Let {ai ,1 i n} L . Then we have c c (a). (1i na i ) = 1i na i c c (b). 1i na i () 1i n a i . On the Eigenproblems of Nilpotent Lattice Matrices 857 Lemma 2.2 Let a, b L such that ac= b c = 0.Then (ab )c = 0. Proof. If possible, assume that (ab )c > 0 . Then there exists y > 0 such that (ab )y = 0. That is, there exists y > 0 such that (ay)( by) = 0 . c c Let y1 = ay and y2 = by . Then y1, y 2 > 0 . Otherwise, a, b > 0 . Now by1 =b (ay ) = ( ab )y = 0 and ay2 =a (by ) = ( ab )y = 0 . Thus there exist y1, y 2 > 0 such that by1= ay 2 = 0, which is a contradiction. Hence (ab )c = 0. Let MLn () be the set of all n n matrices over L (Lattice Matrices). We th shall denote by Aij the element of L which stands in the (,)i j entry of AML n () . For AABBCCML= ( ij ), = ( ij ), = (ij ) n ( ) , define A B= C Cij = A ij B ij (,=1,2,,) i j n aA =CC ij =aA ij ,for a L (i , j =1,2, , n ) n AB =C C = A B( i , j =1,2, , n ) (Multiplica tion) ij ik kj k=1 T A= C Cij = A ji (,=1,2,,)( i j n Transposition) 1, ifi= j (I )ij = (0n )ij =0 (,i j =1,2, ,) n 0, ifi j A0 = I , Ak1 = A k A for k 0, k is an integer. It is clear that MLn () is a semigroup with the identity I and the zero 0n with respect to the multiplication. Also, the transposition in MLn () has the following properties: (a). (AB )TTT = BA (b). (AATT ) = . Let AML n () . Then A is called nilpotent, if there exists some integer k 1 k such that A = 0n . n Theorem 2.3 [2] Let AML n () . Then A is nilpotent if and only if A = 0n . Theorem 2.4 [10] Let AML () be nilpotent. Then a a a = 0 , for n ii1i 1 i 2 imi {i , i1 , i 2 , , im } {1,2, ,n }. Let AML n () . Then A is said to be invertible if there is a BML n () such that AB= BA = I . Here B is called the inverse of A and is denoted by A1 . 858 Geena Joy and K. V. Thomas A set S={ a1 , a 2 , , am } of elements of L is a decomposition of 1 in L if and only if i i ma i =1. The set S is called orthogonal if and only if ai a j = 0 , for all i, j =1,2, , m , i j . Hence S is called an orthogonal decomposition of 1 in L if and only if it is orthogonal and a decomposition of 1 in L . Theorem 2.5 [2] Let AML n () . Then A is invertible if and only if each row and each column of A is an orthogonal decomposition of 1 in L . Let ABML,() n . If there exists an invertible matrix PML n () such that BP= 1AP , then B is said to be similar to A . For AML n () , the permanent of A is defined as | A|= a a a , 1 (1) 2 (2) n () n Sn where Sn denotes the symmetric group of all permutations of the indices {1,2,n }. Lemma 2.6 [9] Let AML n () be invertible. Then | A|= 1. Let VLn () be the set of all column vectors or n -vectors over L . For T T x=(,,,),=(,,,) x1 x 2 xn y y1 y 2 yn Vn () L and a L , define T Addition: x y= ( x1 y 1 , x 2 y 2 , , xn y n ) , scalar multiplication: ax = (ax , ax , , ax )T and 1 2 n T 0 = (0,0, ,0)VLn ( ) . Then VL() is a lattice vector space over L . The elements of VL() are called n n as vectors and the elements of L are called as scalars. Also, 0 is called the zero vector. Denote by e = (1,1, ,1)T . For x=(,,,),=(,,,) x x xT y y y yT V () L , define 1 2 n 1 2 n n x y xi y i ( i =1,2, , n ) T x y= ( x1 y 1 , x 2 y 2 , , xn y n ) T x y= ( x1 y 1 , x 2 y 2 , , xn y n ) xc= ( x c , x c , , xc ) T 1 2 n y= Ax y= A x ,for A M (L) (i =1,2, , n). i ij j n 1j n c c c Lemma 2.7 [7] For x,() y Vn L , we have (x y ) = x y . T A vector x= ( x1 , x 2 , , xn ) is an ortho vector when xi x j = 0 , for T i, j =1,2, , n and i j .
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