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 ba 1= and ba 0= . 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
i }|{ LIiy , where I is an index set,
(a). yxyx iIiiIi )(=)( and
(b). yxyx iIiiIi )(=)( 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). aa 0= (b). aba (c). baba 0= (d). =)( babab (e). cba )( acab (f). cbbacbba )()(=)()(
(g). 1 iini 1 ini 1 baba ini )()()( . Let L be a distributive lattice with the greatest element 1 and the least element 0 .
Let ,nm LM )( be the set of all nm matrices over L (Lattice Matrices). We shall denote ,nn LM )( by n LM )( . Also, aij denotes the element of L which stands in the th ji ),( entry of ,nm LMA )( .
For (= ij ), (= ij ), ij ,nm LMcCbBaA )()(= , define
ijijij njmibacCBA ),1,2,=,,1,2,=(==
T jiij mjniacCA ),1,2,=,,1,2,=(==
aA cC aaijij njmiLa ),1,2,=,,1,2,=(,for== 1748 Geena Joy and K. V. Thomas
For ij ,km ()(= ), ij ,nk ()(= ), ij ,nm LMcCLMbBLMaA )()(= , define AB == njmibacC ),1,2,=,,1,2,=( . ij 1 kl ljil 1, if = ji I ij =)( 0n ij nji ),1,2,=,(0=)( 0, if ji
0 k1 k For ij n LMaA )()(= , define =,= for 0, kkAAAIA is integer.an
Lemma 2.2 [4] For any ij ,km ()(= ), ij ,rk ()(= ), ij ,nr LMcCLMbBLMaA )()(= , (a). A ABBC )(=)( C (b). TT =)( AA .
Let n LMA )( . Then A is called nilpotent, if there exists some integer k 1 such k that A = 0n ; if aij 0= , > ji , 1,2,=, nji , then A is called upper triangular; if aij 0= , < ji , 1,2,=, nji , then A is called lower triangular. The matrix A is said to be invertible, if there is a n LMB )( such that BAAB == I .
Theorem 2.3 [12, 15] Let LMA )( be nilpotent. Then aaa 0= , for n ii 211 iii mi
21 iiii m n},{1,2,},,,,{ .
Let n LMBA )(, . If there exists an invertible matrix n LMP )( such that = PB 1AP , then B is said to be similar to A .
For n LMA )( , the permanent A|| of A is defined as | |= , (2)2(1)1 aaaA nn )( Sn where Sn denotes the symmetric group of all permutations of the indices n}{1,2, . Let us use the following notations:
n SS n |{= is }even
n SS n |{= is odd}.
Now the semi-permanants of n LMA )( are defined as follows: The Determinant and Rank of a Lattice Matrix 1749
n =)( (2)2(1)1 aaaAp nn )( Sn n =)( (2)2(1)1 aaaAq nn )( . Sn
Lemma 2.4 [6] Let n LMA )( . Then
(a). n n )()( =| AAqAp |
T T (b). n n ApAp )(=)( and n n AqAq )(=)( (c). let B be the matrix obtained from A by interchanging columns A and i Aj . Then n n AqBp )(=)( and n n ApBq )(=)( (d). let B be the matrix obtained from A by replacing Aj by Aj . Then
n n ApBp )(=)( and n n AqBq )(=)( (e). let = cbA , for some ,1,2,= nj , where 1 LMcb )(, . Suppose that j n B is the matrix obtained from A by replacing Aj by b and C is the matrix obtained from A by replacing Aj by c . Then n n n CpBpAp )()(=)( and
n n n CqBqAq )()(=)( . From the proofs of Theorem 25, Theorem 32 and Theorem 34 in [14], we have the following Lemma.
Lemma 2.5 [14] Let n LMBA )(, . Then
(a). if A is upper triangular, then n =)( 1 ni aAp ii and n Aq 0=)(
(b). if A is invertible, then n n AqAp 0=)()( and n n AqAp 1=)()(
(c). if , BA are similar lattice matrices, then n n BpAp )(=)( and n n BqAq )(=)( . Let L be a complete and completely distributive lattice with the greatest element 1 and the least element 0 . Let n LMA )( and L . Then is called a characteristic root of A , if satisfies the characteristic equation of A , which is defined as
n n1 n2 n3 n4 1 2 3 4 ApAqApAq )()()()( b
n1 n2 n3 1 2 3 ApAqAp )()()(= c , where 1750 Geena Joy and K. V. Thomas
Ap =)( ( ( ) ( ) aaa )( ), k 1 1 2 2 rrrrr k rk 1 < << ( , ,),, is even 1 2 rrr k 1 2 rrrSn k
Aq =)( ( ( ) ( ) aaa )( ), k 1 1 2 2 rrrrr k rk 1 < << ( , ,, ), is odd 1 2 rrr k 1 2 rrrSn k
21 rrrS k ),,,( is the symmetric group of all permutations of the set 21 rrr k },,,{ ,
n Apbnk )(=,,1,2,= and n Aqc )(= , when n is even; n Aqb )(= and n Apc )(= , when n is odd.
Theorem 2.6 [10] Let L be a complete and completely distributive lattice with 1 and 0 , and n LMA )( . Then the least characteristic roots of A is
n n ()(( )) n n ApAqAqAp )()(( ).
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 n LV )( be the set of all column T T vectors ( n -vectors) over L . Denote e ,1)(1,1,= and n LV )(,0)(0,0,=0 .
We endow n LV )( with the following operations: For T T 21 n 21 n n LVyyyyxxxx )(),,,(=,),,,(= and La , define
T Addition: 2211 yxyxyxyx nn ),,,(= and
T scalar multiplication: xa 21 axaxax n ),,,(= .
Then n LV )( has the properties of a linear space except for the lack of additive inverse for non-zero elements. We call the elements of n LV )( 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 n LVx )( and La , we have (a). x 0=0 (b). a 0=0 .
The Determinant and Rank of a Lattice Matrix 1751
Let S be a non-empty subset of n LV )( . Then is a subspace of n LV )( if it is closed under addition and scalar multiplication. A vector y is a linear combination of vectors ,,, xxx , when there exists scalars 21 k 21 ,,, aaa k such that = 1 xay iiki . Let xxxSS ),,,(= be the set of all possible linear combinations of the vectors 21 k ,,, xxx . Then S is a subspace of LV )( and is called the linear span of the 21 k n vectors 21 ,,, xxx k . Here 21 xxx k },,,{ 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 ,nm LMA )( . 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 C Arank )( . The dimension of AR )( is called the row rank of A and is denoted by R Arank )( .
It is known that C R ArankArank )()( , in general case. This is demonstrated in the following example.
Example 3.2 Consider the lattice ihgfL ,1},,,{0,= , whose diagrammatical representation is as follows:
It is easy to see that L is a distributive lattice with 1 and 0 . 0 gf = )( Let A hhf 3 LM . hhi Then the column space AC )( is spanned by the vectors T T hhghhfif T }),,(,),,(,),{(0, and this is a minimal spanning subset of AC )( .
Therefore, C Arank 3=)( . 1752 Geena Joy and K. V. Thomas
Now the row space AR )( is spanned by the vectors T T hhihhfgf T }),,(,),,(,),{(0, . But, this is not a minimal spanning subset of AR )( . We can see that T T hhigf }),,(,),{(0, is a minimal spanning subset of AR )( . Therefore, R Arank 2=)( . Let A be the j th column of LMA )( . Then the matrix A can be written as j ,nm [= 1 AAAA nj ].
Theorem 3.3 (a). Let ,nm LMA )( and ,km LMB )( . Then BCAC )(=)( if and only if A = BU and B = AV , for some ,nk ( ), ,kn LMVLMU )( .
(b). Let ,nm LMA )( and ,nk LMB )( . Then BRAR )(=)( if and only if A =UB and B =VA , for some ,km ( ), ,mk LMVLMU )( . Proof. (a) First assume that BCAC )(=)( . Then j ),1( njBCA . Therefore, = 1 klj lj l ,1 njBuA . Let
lj ),1(= ,1 njkluU . Thus we get A = BU , where ,nk LMU )( . Also, l ),1( klACB . Therefore, = 1 njl jl j ,1 klAvB . Let
jl ),1(= ,1 klnjvV . Thus we get B =VA , where ,kn LMV )( .
Conversely assume that A = BU and B = AV , for some ,nk ( ), ,kn LMVLMU )( . Then for 1 nj , = 1 BuA and for 1 kl , = 1 AvB . Therefore, klj lj l njl jl j j ),1( njBCA and l ),1( klACB . Thus BCAC )()( and ACBC )()( . Hence BCAC )(=)( . Similarly we can prove (b).
Definition 3.4 Let ,nm LMA )( . The factor rank (or schein rank) of A is the smallest integer r for which there exists ,rm LMB )( and ,nr LMC )( such that
A = BC . The factor rank of A is denoted by f Arank )( .
In otherwords, the factor rank of A means that each column or row of the matrix A is a linear combination of f Arankr )(= columns or rows (not necessarily belonging to the matrix A ), and this number cannot be decreased.
The Determinant and Rank of a Lattice Matrix 1753
Lemma 3.5 Let n LMA )( and B be a submatrix of A , then
f f ArankBrank )()( .
Theorem 3.6 Let ,nm LMA )( . Then
(a). f C ArankArank )()(
(b). f R ()( ArankArank ). Proof. (a) Let =)( kArank . Then there exists a spanning subset { ,1 klB } of C l AC )( , where ),,,(= T ,1 klbbbB . Therefore, for 1 nj , 21 lll ml = 1 klj lj BuA l , where lj ,1,1 njklLu .
Let il ),1(= ,1 klmibB and lj ),1(= ,1 njkluU . Then A = BU , where ,km LMB )( and ,nk LMU )( .
Hence f C ArankkArank )(=)( . Similarly we can prove (b).
Theorem 3.7 Let ,km LMA )( and ,nk LMB )( . Then
(a). rank f AB f Arank )()(
(b). rank f AB f Brank )()( .
Proof. Let f =)( rArank . Then there exists ,rm LMC )( and ,kr LMD )( such that
A = CD . Therefore, AB CDB == CF , where F = DB ,nr LM )( .
Hence rank f AB f Arankr )(=)( . Similarly we can prove (b).
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 n LMA )( . Then the determinant of A is defined as
n n ()(( )) n n ()(( ApAqAqAp )) and is denoted by AD )( . 1754 Geena Joy and K. V. Thomas
We have
n n ()((=)( )) n n ()(( ApAqAqApAD ))
n n ()((= )) n n ()(( AqApAqAp ))( Lemmaby 2.1(f))
n n AqAp )()( Lemma(by .2.1(b)) Therefore, AAD ||)( .
If L is a Boolean lattice, then n n ()((=)( )) n n ()(( ApAqAqApAD )) .
Now we discuss some properties of the determinant of a lattice matrix.
Theorem 4.2 Let n LMA )( . Then AD 0=)( if and only if n n AqAp )(=)( . Proof. We have
n n ()((0=)( )) n n ()(( ApAqAqApAD )) 0=
n n =)()( 0, n n ApAqAqAp 0=)()(
n ()( ), nn n ApAqAqAp )()( Lemma(by 2.1(c))
n n (=)( AqAp ).
Theorem 4.3 Let n LMA )( and L . Then (a). ADAD T )(=)( (b). if B is the matrix obtained from A by interchanging Ai and Aj , then ADBD )(=)( (c). if B is the matrix obtained from A by replacing Aj by Aj , then ADBD )()( (d). if two columns of A are identical, then AD 0=)( . Proof. (a) We have
T T T T T n n ()((=)( )) n n ()(( ApAqAqApAD ))
n n ()((= )) n n ()(( ApAqAqAp )) Lemma(by 2.4(b)) (= AD ). The Determinant and Rank of a Lattice Matrix 1755
(b) We have
n n ()((=)( )) n n ()(( BpBqBqBpBD ))
n n ()((= )) n n ()(( AqApApAq )) Lemma(by 2.4(c)) (= AD ). (c) We have
n n ()((=)( )) n n ()(( BpBqBqBpBD ))
n n ()((= )) n n ()(( ApAqAqAp )) Lemma(by 2.4(d))
n n ()(( )) n n ()(( ApAqAqAp )) Lemma(by 2.1(e))
(( n n ()( )) n n ()(( ApAqAqAp ))) (AD ).
(d) We have n n AqAp )(=)( . Therefore, AD 0=)( .
Theorem 4.4 Let LMA )( and LVB )( , 1 kl . Suppose that each column n nl of the matrix A is a linear combination of Bl , 1 kl . Then AD 0=)( , provided that < nk . Proof. Suppose that = 1 BuA llklj , for some 1 nj . Then n =)( n ([ 1 AAApAp nj ]) = n ([ 11 ABuAp nllkl ]) = 1 nlkl ([ 1 ABApu nl ]) Lemma(by 2.4(e)) Similarly n =)( 1 nlkl ([ 1 ABAquAq nl ]) . Therfore, n n =)()( 1 nlkl ([ 1 nl ]) 1 nlkl ([ 1 ABAquABApuAqAp nl ]) 1 ( nlkl ([ 1 nlnl ([]) 1 ABAquABApu nl ])) Lemma(by 2.1(g)) 1 ( nlkl ([ 1 nnl ([]) 1 ABAqABApu nl ])) Lemma(by 2.1(e)) 1 lkl ([ 1 ABADu nl ]). Similarly n n )()( 1 lkl ([ 1 ABADuApAq nl ]). 1756 Geena Joy and K. V. Thomas Therfore, )( 1 lkl ([ 1 ABADuAD nl ]) . 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 { l ,1 klB } . Since < nk , the determinants of such matrices contain identical columns and therefore are equal to zero. Hence, AD 0=)( .
Theorem 4.5 Let n LMA )( . If f <)( nArank , then AD 0=)( . Proof. Let <=)( nrArank . Then there exists LVB )( , 1 rl such that each f nl column of the matrix A is a linear combination of l ,1 rlB . Hence from Theorem 4.4, it follows that AD 0=)( .
Corollary 4.6 Let n LMA )( . If AD 0>)( , then
f C R =)(=)(=)( nArankArankArank . Proof. It follows from Theorem 3.6 and Theorem 4.5.
Theorem 4.7 Let n LMBA )(, . Then D AB BDAD )()()( .
Proof. We have D AB pn AB qn ()((=)( AB)) qn AB pn ()(( AB)) .
Consider pn AB qn AB)()( )()()()()()(= (1)1 ABAB (2)2 AB nn )( (1)1 ABAB (2)2 AB nn )( Sn Sn
(= 1 2 (1) (2) bbbaaa )( ) 1 kk 2 nkn 1 kk 2 kn n 1 , ,, kkk n Sn 1 2 n
( 1 2 (1) (2) bbbaaa )( ) 1 kk 2 nkn 1 kk 2 kn n 1 , ,, kkk n Sn 1 2 n
1 2 ( (1) (2) bbbaaa )( (1) (2) bbb )( ) 1 kk 2 nkn 1 kk 2 kn n 1 kk 2 kn n 1 , ,, kkk n 1 2 n Sn Sn Lemma(by 2.1(g)) ( ) (2)2(1)1 nn )( bbbaaa nn )()((2)(2)(1)(1) bbb nn )()((2)(2)(1)(1) S n Sn Sn Lemma(by 2.1(a)) The Determinant and Rank of a Lattice Matrix 1757
( ( )) (2)2(1)1 nn )( bbbaaa nn )()((2)(2)(1)(1) bbb nn )()((2)(2)(1)(1) Sn Sn Sn
( ( )) (2)2(1)1 nn )( bbbaaa nn )()((2)(2)(1)(1) bbb nn )()((2)(2)(1)(1) Sn Sn Sn
( ( )) (2)2(1)1 nn )( (2)2(1)1 bbbaaa nn )( (2)2(1)1 bbb nn )( Sn Sn Sn
( ( )) (2)2(1)1 nn )( (2)2(1)1 bbbaaa nn )( (2)2(1)1 bbb nn )( Sn Sn Sn
n ( )( n n ()( )) ( )( nn n ()( BpBqAqBqBpAp )).
Therefore, pn AB qn AB n ()()( )( n n ()( )) ( )( nn n ()( BpBqAqBqBpAp )) . Similarly, we can prove that pn AB qn AB n ()()( )( n n ()( )) ( )( nn n ()( ApAqBqAqApBp )) qn AB pn AB ()()( )( nn n ()( )) n ( )( n n ()( BqBpAqBpBqAp )) qn AB pn AB ()()( )( nn n ()( )) n ( )( n n ()( AqApBqApAqBp )) . Hence,
D AB pn AB qn ()((=)( AB)) qn AB pn ()(( AB))
n (( )( n n ()( )) ( )( nn n ()( ))) (( )( nn n ()( )) n ( )( n n ()( BqBpAqBpBqApBpBqAqBqBpAp )))
n ( )(( n n ()( )) n n ()(( ))) n ( )(( n n ()( )) n n ()(( BpBqBqBpAqBpBqBqBpAp )))
n n ()(( (()) n n ()( )) n n ()(( ))) =| (| BDABpBqBqBpAqAp ). Also, D AB ADB )(||)( . Therefore, D AB)( (| (| (|)) (| ADBBDA )) (| (| (|)) (| BDBADA )) 1758 Geena Joy and K. V. Thomas
()( BDAD ). Hence, the proof is complete.
Theorem 4.8 Let n LMA )( be nilpotent. Then AD 0=)( .
Proof. Assume that A is nilpotent. Then by Theorem 2.3, we have n n AqAp 0=)(=)( . Hence AD 0=)( .
Theorem 4.9 Let n LMA )( be triangular (upper or lower). Then =)( 1 ni aAD ii . Proof. Assume that A is upper triangular. Then by Lemma 2.5(a), we have
n =)( 1 ni aAp ii and n Aq 0=)( . Hence =)( 1 ni aAD ii . If A is lower triangular, then AT is upper triangular. Hence by Theorem 4.3, we get T that =)(=)( 1 ni aADAD ii .
Theorem 4.10 Let n LMA )( be invertible. Then AD 1=)( . Proof. Assume that A is invertible. Then by Lemma 2.5(b), we have
n n AqAp 0=)()( and n n AqAp 1=)()( . Hence
n n ()((=)( )) n n ()(( ApAqAqApAD ))
n n ()((= )) n n ()(( AqApAqAp )) Lemma(by 2.1(f)) =1.
Theorem 4.11 Let n LMBA )(, be similar lattice matrices. Then BDAD )(=)( .
Proof. Let n LMBA )(, be similar lattice matrices. Then by Lemma 2.5(c), we have
n n BpAp )(=)( and n n BqAq )(=)( . Hence BDAD )(=)( .
Theorem 4.12 Let L be a complete and completely distributive lattice with 1 and
0 and n LMA )( . Then n n ()((=)( )) n n ()(( ApAqAqApAD )) is the least characteristic root of A . Proof. It follows from Theorem 2.6.
The Determinant and Rank of a Lattice Matrix 1759
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 n LMA )( . 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
| B |> 0 . The permanent rank of A is denoted by p Arank )( . 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 BD 0>)( . The rank of A is denoted by Arank )( . The rank of a zero matrix is 0 .
Theorem 5.2 Let n LMBA )(, . Then
(a). p ArankArank )()(
(b). f ArankArank )()( (c). rank AB ({)( ), (BrankArankmin )}.
Proof. (a) We have AAD ||)( . Hence p ArankArank )()( . (b) Let =)( kArank . Then there exists a kk )( -submatrix B of A such that
BD 0>)( . By Corollary 4.6, we get f f ArankBrankBrankkArank )()(=)(==)( .
Hence f ArankArank )()( . (c) Let =)( kArank . Assume that C is a kk ))1()1(( -submatrix of AB . Let the rows and columns of C be iii n},{1,2,},,,{ and jjj n},{1,2,},,,{ , respectively. 21 k1 21 k1 Then for some jjjj },,,{ , = ABC , where 21 k1 1 llnlj ),,,(= T ,1 nlAAAA . 21 ililil k1l Proceeding as in Theorem 4.4, we have )( 1 ([ CACDBCD ]) . Since 1 jljlnl k1 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 kk ))1()1(( -submatrices of A . Since the determinants of 1760 Geena Joy and K. V. Thomas all kk ))1()1(( -submatrices of A are equal to zero, we have CD 0=)( . Therefore, rank AB (=)( Arankk ). Similarly we can prove that rank AB Brank )()( . Hence rank AB ({)( ), (BrankArankmin )}.
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.
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1762 Geena Joy and K. V. Thomas