the determinant and rank of a lattice matrixare studied in [3, 10, 14]. in [10], y. j. tan discussed...

Global Journal of Pure and Applied Mathematics.

ISSN 0973-1768 Volume 13, Number 6 (2017), pp. 1745-1761

© Research India Publications

http://www.ripublication.com

The Determinant and Rank of a Lattice Matrix

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 deals with the determinant of a matrix over a dually Browerian,

distributive lattice L with the greatest element 1 and the least element 0 , and

proves that determinant of product of matrices over L is less than equal to the

product of determinants of the matrices. Also, this paper introduces the concept

of rank of a matrix over L . We demonstrate that the rank of a marix over L

does not exceed the factorisation rank of the matrix and prove that the rank of

product of matrices over L does not exceed the ranks of the factors.

Keywords: Distributive lattice, Dually Browerian lattice, Lattice matrix,

Determinant of a lattice matrix, Rank of a lattice matrix

2010 MSC: 15B99

1 INTRODUCTION

The notion of lattice matrices appeared firstly in the work, ‘Lattice matrices’ [4] 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 [4].

1746 Geena Joy and K. V. Thomas

The theory of determinant of Boolean matrices appeared firstly in the work of O. B.

Sokolov [9]. Since then, a number of researchers have studied the determinant theory

for Boolean matrices and lattice matrices (see [2, 6, 7, 8]). In [8], V. B. Poplavskii

introduced the notion of minor rank of a Boolean matrix and discussed some of its

properties. Further E. E. Marenich [6] discussed the determinant rank for matrices over

a Browerian, distributive lattice with 1 and 0 .

The eigenproblems and characteristic roots of matrices over a complete and completely

distributive lattice with the greatest element 1 and the least element 0 are studied in

[3, 10, 14]. In [10], Y. J. Tan discussed the eigenproblems of lattice matrices and

provided the least element for the set of all characteristic roots of a lattice matrix.

Further, G. Joy and K. V. Thomas [3] discussed the eigenproblems of nilpotent lattice

matrices and introduced the concept of non-singular lattice matrices. Also, K. V.

Thomas and G. Joy [14] studied the characteristic roots of different types of lattice

matrices and introduced the concept of similar lattice matrices. The least element for

the set of all characteristic roots of a lattice matrix is taken as the determinant of a lattice

matrix.

In the present work, we define the determinant of a matrix over a dually Browerian,

distributive lattice L with the greatest element 1 and the least element 0 , and prove

that determinant of product of matrices over L is less than equal to the product of

determinants of the matrices. Also, this paper introduces the concept of rank of a matrix

over L . We demonstrate that the rank of a marix over L does not exceed the

factorisation rank of the matrix and prove that the rank of product of matrices over L

does not exceed the ranks of the factors.

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, 4, 5, 10, 11, 12, 13, 14, 15].

A partially ordered set ),( L is a lattice if for all Lba , , the least upper bound of

},{ ba and the greatest lower bound of },{ ba exist in L . For any Lba , , the least

upper bound and the greatest lower bound is denoted by ba and ba (or ab ),

respectively.

A lattice L is called a complete lattice if for any LH , both the least upper bound

}|{ Hyy and the greatest lower bound }|{ Hyy of H exist in L . A lattice

),,,( L is a distributive lattice if the operations and are distributive with

respect to each other.

An element La is called greatest element of L if ax , Lx . An element

Lb is called least element of L if xb , Lx . We use 1 and 0 to denote the

greatest element and the least element of L , respectively.

A complement of an element La is an element Lb for which 1=ba and

0=ba . The complement of a is denoted by a . If for any La , a exists, then

The Determinant and Rank of a Lattice Matrix 1747

L is said to be a complemented lattice. A complemented distributive lattice is called a

Boolean lattice.

For any Lba , , the least element Lx satisfying the inequality axb is called

the relative lower pseudocomplement of b in a and is denoted by ba . If for any

Lba , , ba exists, then L is said to be a dually Brouwerian lattice. If L is a

Boolean lattice, then baba = .

A lattice L is said to be completely distributive, if for any Lx and any

LIiyi }|{ , where I is an index set,

(a). )(=)( iIiiIi yxyx and

(b). )(=)( iIiiIi yxyx holds.

It is known [1] that a complete lattice L is dually Brouwerian if and only if (b) is

satisfied in L . Therefore, a complete and completely distributive lattice L is dually

Brouwerian.

Lemma 2.1 [10, 12, 13] Let L be a distributive and dually Brouwerian lattice with 1 and 0. Then for any Lcba ,, , we have

(a). 0=aa

(b). aba

(c). 0=baba

(d). babab =)(

(e). acabcba )(

(f). )()(=)()( cbbacbba

(g). )()()( 111 iniiniiini baba .

Let L be a distributive lattice with the greatest element 1 and the least element 0 .

Let )(, LM nm be the set of all nm matrices over L (Lattice Matrices). We shall

denote )(, LM nn by )(LM n . Also, ija denotes the element of L which stands in the

th),( ji entry of )(, LMA nm .

For )()(=),(=),(= , LMcCbBaA nmijijij , define

),1,2,=,,1,2,=(== njmibacCBA ijijij

),1,2,=,,1,2,=(== mjniacCA jiijT

),1,2,=,,1,2,=(,for== njmiLaaacCaA ijij


For )()(=),()(=),()(= ,,, LMcCLMbBLMaA nmijnkijkmij , define

),1,2,=,,1,2,=(==1

njmibacCAB ljilklij .

),1,2,=,(0=)(if0,

=if1,=)( nji

jiji

I ijij n0

For )()(= LMaA nij , define integer.an is 0,for =,= 10 kkAAAIA kk

Lemma 2.2 [4] For any )()(=),()(=),()(= ,,, LMcCLMbBLMaA nrijrkijkmij ,

(a). CABBCA )(=)(

(b). AA TT =)( .

Let )(LMA n . Then A is called nilpotent, if there exists some integer 1k such

that n0=kA ; if 0=ija , ji > , nji 1,2,=, , then A is called upper triangular; if

0=ija , ji < , nji 1,2,=, , then A is called lower triangular. The matrix A is

said to be invertible, if there is a )(LMB n such that IBAAB == .

Theorem 2.3 [12, 15] Let )(LMA n be nilpotent. Then 0=211

imiiiii aaa , for

},{1,2,},,,,{ 21 niiii m .

Let )(, LMBA n . If there exists an invertible matrix )(LMP n such that

APPB 1= , then B is said to be similar to A .

For )(LMA n , the permanent || A of A is defined as

,|=| )((2)2(1)1 nnnS

aaaA

where nS denotes the symmetric group of all permutations of the indices }{1,2, n .

Let us use the following notations:

}even is |{= nn SS

}odd is |{= nn SS .

Now the semi-permanants of )(LMA n are defined as follows:


)((2)2(1)1=)( nnnSn aaaAp

)((2)2(1)1=)( nnnSn aaaAq

.

Lemma 2.4 [6] Let )(LMA n . Then

(a). |=|)()( AAqAp nn

(b). )(=)( ApAp nT

n and )(=)( AqAq nT

n

(c). let B be the matrix obtained from A by interchanging columns iA

and

jA

. Then )(=)( AqBp nn and )(=)( ApBq nn

(d). let B be the matrix obtained from A by replacing jA

by jA

. Then )(=)( ApBp nn and )(=)( AqBq nn

(e). let cbAj

= , for some nj ,1,2,= , where )(, 1 LMcb n . Suppose that

B is the matrix obtained from A by replacing jA

by b

and C is the matrix

obtained from A by replacing jA

by c . Then )()(=)( CpBpAp nnn and )()(=)( CqBqAq nnn .

From the proofs of Theorem 25, Theorem 32 and Theorem 34 in [14], we have the

following Lemma.

Lemma 2.5 [14] Let )(, LMBA n . Then

(a). if A is upper triangular, then iinin aAp 1=)( and 0=)(Aqn

(b). if A is invertible, then 0=)()( AqAp nn and 1=)()( AqAp nn

(c). if BA, are similar lattice matrices, then )(=)( BpAp nn and )(=)( BqAq nn .

Let L be a complete and completely distributive lattice with the greatest element 1

and the least element 0 . Let )(LMA n and L . Then is called a

characteristic root of A , if satisfies the characteristic equation of A , which is

defined as

bApAqApAq nnnnn 4

4

3

3

2

2

1

1 )()()()(

,)()()(= 3

3

2

2

1

1 cApAqAp nnn

where


),(=)( )()2

(2

)1

(1

even is ,),,2

,1

(<<2

<1

1krkrrrrr

krrrSnkrrrk aaaAp

),(=)( )()2

(2

)1

(1

odd is ),,,2

,1

(<<2

<1

1krkrrrrr

krrrSnkrrrk aaaAq

),,,( 21 krrrS is the symmetric group of all permutations of the set },,,{ 21 krrr ,

)(=,,1,2,= Apbnk n and )(= Aqc n , when n is even; )(= Aqb n and )(= Apc n ,

when n is odd.

Theorem 2.6 [10] Let L be a complete and completely distributive lattice with 1 and 0 , and )(LMA n . Then the least characteristic roots of A is

).)()(())()(( ApAqAqAp nnnn

3 THE COLUMN RANK, ROW RANK AND FACTOR RANK OF A

LATTICE MATRIX

In this section, we generalize the concepts and propositions for Boolean matrices in [8]

to the case of lattice matrices.

Let L be a distributive lattice with 1 and 0 , and )(LVn be the set of all column

vectors ( n -vectors) over L . Denote Te ,1)(1,1,=

and )(,0)(0,0,=0 LVnT

.

We endow )(LVn with the following operations: For

)(),,,(=,),,,(= 2121 LVyyyyxxxx nT

nT

n

and La , define

Addition: Tnn yxyxyxyx ),,,(= 2211

and

scalar multiplication: Tnaxaxaxxa ),,,(= 21

.

Then )(LVn has the properties of a linear space except for the lack of additive inverse

for non-zero elements. We call the elements of )(LVn as lattice vectors(or simply L-vectors) and the elements of L as scalars.

In this paper, a space means an L -space and a vector means an L -vector.

Lemma 3.1 For )(LVx n and La , we have

(a). 0=0x

(b). 0=0

a .


Let S be a non-empty subset of )(LVn . Then S is a subspace of )(LVn if it is

closed under addition and scalar multiplication.

A vector y is a linear combination of vectors kxxx

,,, 21 , when there exists scalars

kaaa ,,, 21 such that iiki xay 1= .

Let ),,,(= 21 kxxxSS

be the set of all possible linear combinations of the vectors

kxxx

,,, 21 . Then S is a subspace of )(LVn and is called the linear span of the

vectors kxxx

,,, 21 . Here },,,{ 21 kxxx

is a spanning subset of S . Among the

spanning subsets of S , the one with the smallest cardinality, d , are bases of S ; The

dimension of S is that number d . The dimension of the zero space is zero.

Let )(, LMA nm . Then the column space of A is the linear span of the set of all

columns of A and is denoted by )(AC . The row space of A is the linear span of the

set of all rows of A and is denoted by )(AR . The dimension of )(AC is called the

column rank of A and is denoted by )(ArankC . The dimension of )(AR is called the

row rank of A and is denoted by )(ArankR .

It is known that )()( ArankArank RC , in general case. This is demonstrated in the

following example.

Example 3.2 Consider the lattice ,1},,,{0,= ihgfL , whose diagrammatical representation is as follows:

It is easy to see that L is a distributive lattice with 1 and 0 .

Let )(

0

= 3 LMhhihhfgf

A

.

Then the column space )(AC is spanned by the vectors }),,(,),,(,),{(0, TTT hhghhfif and this is a minimal spanning subset of )(AC .

Therefore, 3=)(ArankC .


Now the row space )(AR is spanned by the vectors }),,(,),,(,),{(0, TTT hhihhfgf . But, this is not a minimal spanning subset of )(AR . We can see that

}),,(,),{(0, TT hhigf is a minimal spanning subset of )(AR . Therefore, 2=)(ArankR .

Let jA

be the thj column of )(, LMA nm . Then the matrix A can be written as

][= 1 nj AAAA

.

Theorem 3.3 (a). Let )(, LMA nm and )(, LMB km . Then )(=)( BCAC if and only if BUA = and AVB = , for some )(),( ,, LMVLMU knnk .

(b). Let )(, LMA nm and )(, LMB nk . Then )(=)( BRAR if and only if UBA = and VAB = , for some )(),( ,, LMVLMU mkkm .

Proof. (a) First assume that )(=)( BCAC .

Then njBCAj ),1(

. Therefore, njBuA lljklj ,1= 1

. Let

njkluU lj ,1),1(= . Thus we get BUA = , where )(, LMU nk .

Also, klACBl ),1(

. Therefore, klAvB jjlnjl ,1= 1

. Let

klnjvV jl ,1),1(= . Thus we get VAB = , where )(, LMV kn .

Conversely assume that BUA = and AVB = , for some )(),( ,, LMVLMU knnk .

Then for nj 1 , lljklj BuA

1= and for kl 1 , jjlnjl AvB

1= . Therefore,

njBCAj ),1(

and klACBl ),1(

. Thus )()( BCAC and )()( ACBC .

Hence )(=)( BCAC .

Similarly we can prove (b).

Definition 3.4 Let )(, LMA nm . The factor rank (or schein rank) of A is the smallest integer r for which there exists )(, LMB rm and )(, LMC nr such that

BCA = . The factor rank of A is denoted by )(Arank f .

In otherwords, the factor rank of A means that each column or row of the matrix A is

a linear combination of )(= Arankr f columns or rows (not necessarily belonging to

the matrix A ), and this number cannot be decreased.


Lemma 3.5 Let )(LMA n and B be a submatrix of A , then )()( ArankBrank ff .

Theorem 3.6 Let )(, LMA nm . Then

(a). )()( ArankArank Cf

(b). ).()( ArankArank Rf

Proof. (a) Let kArankC =)( . Then there exists a spanning subset },1{ klBl

of

)(AC , where klbbbB Tmllll ,1),,,(= 21

. Therefore, for nj 1 ,

lljklj BuA

1= , where njklLulj ,1,1 .

Let klmibB il ,1),1(= and njkluU lj ,1),1(= . Then BUA = ,

where )(, LMB km and )(, LMU nk .

Hence )(=)( ArankkArank Cf .


Theorem 3.7 Let )(, LMA km and )(, LMB nk . Then

(a). )()( ArankABrank ff

(b). )()( BrankABrank ff .

Proof. Let rArank f =)( . Then there exists )(, LMC rm and )(, LMD kr such that

CDA = . Therefore, CFCDBAB == , where )(= , LMDBF nr .

Hence )(=)( ArankrABrank ff .


4 THE DETERMINANT OF A LATTICE MATRIX

Let L be a dually Brouwerian, distributive lattice with the greatest element 1 and the

least element 0 , respectively. In this section, we introduce the determinant of a matrix

over L , using the semi-permanants.

Definition 4.1 Let )(LMA n . Then the determinant of A is defined as ))()(())()(( ApAqAqAp nnnn and is denoted by )(AD .


We have

))()(())()((=)( ApAqAqApAD nnnn

2.1(f)) Lemmaby ())()(())()((= AqApAqAp nnnn

.2.1(b)) Lemma(by )()( AqAp nn

Therefore, ||)( AAD .

If L is a Boolean lattice, then ))()(())()((=)( ApAqAqApAD nnnn .

Now we discuss some properties of the determinant of a lattice matrix.

Theorem 4.2 Let )(LMA n . Then 0=)(AD if and only if )(=)( AqAp nn .

Proof. We have

0=))()(())()((0=)( ApAqAqApAD nnnn

0=)()(0,=)()( ApAqAqAp nnnn

2.1(c)) Lemma(by )()(),()( ApAqAqAp nnnn

).(=)( AqAp nn

Theorem 4.3 Let )(LMA n and L . Then

(a). )(=)( TADAD

(b). if B is the matrix obtained from A by interchanging iA

and jA

, then

)(=)( ADBD

(c). if B is the matrix obtained from A by replacing jA

by jA

, then

)()( ADBD

(d). if two columns of A are identical, then 0=)(AD .

Proof. (a) We have

))()(())()((=)( Tn

Tn

Tn

Tn

T ApAqAqApAD

2.4(b)) Lemma(by ))()(())()((= ApAqAqAp nnnn

).(= AD


(b) We have

))()(())()((=)( BpBqBqBpBD nnnn

2.4(c)) Lemma(by ))()(())()((= AqApApAq nnnn

).(= AD

(c) We have

))()(())()((=)( BpBqBqBpBD nnnn

2.4(d)) Lemma(by ))()(())()((= ApAqAqAp nnnn

2.1(e)) Lemma(by ))()(())()(( ApAqAqAp nnnn

)))()(())()((( ApAqAqAp nnnn

).(AD

(d) We have )(=)( AqAp nn . Therefore, 0=)(AD .

Theorem 4.4 Let )(LMA n and )(LVB nl

, kl 1 . Suppose that each column

of the matrix A is a linear combination of lB

, kl 1 . Then 0=)(AD , provided that nk < .

Proof. Suppose that llklj BuA

1= , for some nj 1 . Then

])([=)( 1 njnn AAApAp

])([= 11 nllkln ABuAp

2.4(e)) Lemma(by ])([= 11 nlnlkl ABApu

Similarly ])([=)( 11 nlnlkln ABAquAq

.

Therfore,

])([])([=)()( 1111 nlnlklnlnlklnn ABAquABApuAqAp

2.1(g)) Lemma(by ]))([])([( 111 nlnlnlnlkl ABAquABApu

2.1(e)) Lemma(by ]))([])([( 111 nlnnlnlkl ABAqABApu

]).([ 11 nllkl ABADu

Similarly ])([)()( 11 nllklnn ABADuApAq

.


Therfore, ])([)( 11 nllkl ABADuAD

.

Since this chain can be continued with the use of analogous reasoning applied to each

column of the matrix A , we find that )(AD is less than or equal to some linear

combination of the determinants of the square matrices constructed from },1{ klBl

. Since nk < , the determinants of such matrices contain identical columns and

therefore are equal to zero. Hence, 0=)(AD .

Theorem 4.5 Let )(LMA n . If nArank f <)( , then 0=)(AD .

Proof. Let nrArank f <=)( . Then there exists )(LVB nl

, rl 1 such that each

column of the matrix A is a linear combination of rlBl ,1

. Hence from Theorem

4.4, it follows that 0=)(AD .

Corollary 4.6 Let )(LMA n . If 0>)(AD , then nArankArankArank RCf =)(=)(=)( .

Proof. It follows from Theorem 3.6 and Theorem 4.5.

Theorem 4.7 Let )(, LMBA n . Then )()()( BDADABD .

Proof. We have ))()(())()((=)( ABpABqABqABpABD nnnn .

Consider )()( ABqABp nn

)((2)2(1)1)((2)2(1)1 )()()()()()(= nn

nSnn

nS

ABABABABABAB

)(= )((2)2

(1)12

21

1,,

2,

11

nnkkknnkkknnkkk

nS

bbbaaa

)( )((2)2

(1)12

21

1,,

2,

11

nnkkknnkkknnkkk

nS

bbbaaa

)( )((2)2

(1)1

)((2)2

(1)12

21

1,,

2,

11

nnkkk

nSnnkkk

nSnnkkk

nnkkkbbbbbbaaa

2.1(g)) Lemma(by

)( )()((2)(2)(1)(1))()((2)(2)(1)(1))((2)2(1)1 nn

nSnn

nSnn

nSbbbbbbaaa

2.1(a)) Lemma(by


))(( )()((2)(2)(1)(1))()((2)(2)(1)(1))((2)2(1)1 nn

nSnn

nSnn

nS

bbbbbbaaa

))(( )()((2)(2)(1)(1))()((2)(2)(1)(1))((2)2(1)1 nn

nSnn

nSnn

nS

bbbbbbaaa

))(( )((2)2(1)1)((2)2(1)1)((2)2(1)1 nn

nSnn

nSnn

nS

bbbbbbaaa

))(( )((2)2(1)1)((2)2(1)1)((2)2(1)1 nn

nSnn

nSnn

nS

bbbbbbaaa

)).()()(())()()(( BpBqAqBqBpAp nnnnnn

Therefore,

))()()(())()()(()()( BpBqAqBqBpApABqABp nnnnnnnn .

Similarly, we can prove that

))()()(())()()(()()( ApAqBqAqApBpABqABp nnnnnnnn

))()()(())()()(()()( BqBpAqBpBqApABpABq nnnnnnnn

))()()(())()()(()()( AqApBqApAqBpABpABq nnnnnnnn .

Hence,

))()(())()((=)( ABpABqABqABpABD nnnn

)))()()(())()()((()))()()(())()()((( BqBpAqBpBqApBpBqAqBqBpAp nnnnnnnnnnnn

)))()(())()()((()))()(())()()((( BpBqBqBpAqBpBqBqBpAp nnnnnnnnnn

).(|=|)))()(())()((())()(( BDABpBqBqBpAqAp nnnnnn

Also, )(||)( ADBABD .

Therefore,

))(|(|))(|(|)( ADBBDAABD

))(|(|))(|(| BDBADA


).()( BDAD

Hence, the proof is complete.

Theorem 4.8 Let )(LMA n be nilpotent. Then 0=)(AD .

Proof. Assume that A is nilpotent. Then by Theorem 2.3, we have 0=)(=)( AqAp nn

. Hence 0=)(AD .

Theorem 4.9 Let )(LMA n be triangular (upper or lower). Then iini aAD 1=)( .

Proof. Assume that A is upper triangular. Then by Lemma 2.5(a), we have

iinin aAp 1=)( and 0=)(Aqn . Hence iini aAD 1=)( .

If A is lower triangular, then TA is upper triangular. Hence by Theorem 4.3, we get

that iiniT aADAD 1=)(=)( .

Theorem 4.10 Let )(LMA n be invertible. Then 1=)(AD .

Proof. Assume that A is invertible. Then by Lemma 2.5(b), we have

0=)()( AqAp nn and 1=)()( AqAp nn . Hence

))()(())()((=)( ApAqAqApAD nnnn

2.1(f)) Lemma(by ))()(())()((= AqApAqAp nnnn

1.=

Theorem 4.11 Let )(, LMBA n be similar lattice matrices. Then )(=)( BDAD .

Proof. Let )(, LMBA n be similar lattice matrices. Then by Lemma 2.5(c), we have

)(=)( BpAp nn and )(=)( BqAq nn . Hence )(=)( BDAD .

Theorem 4.12 Let L be a complete and completely distributive lattice with 1 and 0 and )(LMA n . Then ))()(())()((=)( ApAqAqApAD nnnn is the least characteristic root of A . Proof. It follows from Theorem 2.6.


5 THE RANK OF A LATTICE MATRIX

In this section, we introduce the concept of rank of a lattice matrix and discuss some of

its properties. For any ,, Lba we use the notation ba to denote ab and

ab .

Definition 5.1 Let )(LMA n . The permanent rank of a nonzero matrix A is the greatest integer k for which there exists a )( kk -submatrix B of A such that

0|>| B . The permanent rank of A is denoted by )(Arankp . The permanent rank of a zero matrix is 0 . The determinant rank (or rank) of a nonzero matrix A is the greatest integer k for

which there exists a )( kk -submatrix B of A such that 0>)(BD . The rank of A

is denoted by )(Arank . The rank of a zero matrix is 0 .

Theorem 5.2 Let )(, LMBA n . Then

(a). )()( ArankArank p

(b). )()( ArankArank f

(c). )}(),({)( BrankArankminABrank .

Proof. (a) We have ||)( AAD . Hence )()( ArankArank p .

(b) Let kArank =)( . Then there exists a )( kk -submatrix B of A such that

0>)(BD . By Corollary 4.6, we get )()(=)(==)( ArankBrankBrankkArank ff .

Hence )()( ArankArank f .

(c) Let kArank =)( .

Assume that C is a ))1()1(( kk -submatrix of AB . Let the rows and columns of

C be },{1,2,},,,{ 121 niii k and },{1,2,},,,{ 121 njjj k , respectively.

Then for some },,,{ 121 kjjjj , llnlj ABC

1= , where

nlAAAA Tlkililil

,1),,,(=

121

.

Proceeding as in Theorem 4.4, we have ])([)(11

1

kjljlnl CACDBCD

. Since

this chain can be continued with the use of analogous reasoning applied to each column

of the matrix C , we find that )(CD is less than or equal to linear combination of

determinants of some ))1()1(( kk -submatrices of A . Since the determinants of


all ))1()1(( kk -submatrices of A are equal to zero, we have 0=)(CD .

Therefore, ).(=)( ArankkABrank

Similarly we can prove that )()( BrankABrank .

Hence )}(),({)( BrankArankminABrank .

ACKNOWLEDGEMENT

The first author is thankful to the University Grants Commission, New Delhi, India for

the award of Teacher Fellowship under the XII Plan period.

REFERENCES

[1] G. Birkhoff, Lattice Theory, 3 rd ed., Amer. Math. Soc. Providence, RI, 1967.

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the determinant and rank of a lattice matrixare studied in [3, 10, 14]. in [10], y. j. tan discussed...

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