LINEAR OPERATORS THAT PRESERVE ZERO ... - MathNet Korea

J. Korean Math. Soc. 36 (1999), No. 6, pp. 1181{1190

LINEAR OPERATORS THAT PRESERVE ZERO-TERM RANK OF BOOLEAN MATRICES LeRoy. B. Beasley, Seok-Zun Song, and Sang-Gu Lee

Abstract. Zero-term rank of a matrix is the minimum number of

lines (rows or columns) needed to cover all the zero entries of the given matrix. We characterize the linear operators that preserve zero-term rank of the m n matrices over binary Boolean algebra.

1. Introduction and Preliminaries There is much literature on the study of those linear operators on matrices that leave certain properties or subsets invariant. Boolean matrices also have been the subject of research by many authors. But there are few papers on zero-term rank of the matrices over Boolean algebra. Beasley and Pullman characterized those linear operators that preserve Boolean rank in [1] and term rank of matrices over antinegative semirings in [2] and [3]. In [4], [6] and [7], analogues of their work were obtained on certain column rank for Boolean matrices. In this paper, we consider the zero-term rank of Boolean matrices. We obtain characterizations of those linear operators that preserve zero-term rank of m n matrices over binary Boolean algebra. We also extend the results in [2] to zero-term rank preserver on Boolean matrices. We let M = M m;n (B ) denote the set of all m n matrices with entries in B = f0; 1g, the two element Boolean algebra. Arithmetic in B follows the usual rules except that 1 + 1 = 1. Let Eij be the m n matrix whose (i; j )th entry is 1 and whose other entries are all zero, which is called a cell. Let J denote the m n Received June 15, 1999. 1991 Mathematics Subject Classi cation: Primary 15A03, 15A04. Key words and phrases: Boolean linear operator, term-rank, zero-term rank. The authors wish to acknowledge the nancial support of the Korea Research Foundation made in the program year of 1997-1998.

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L. B. Beasley, S. Z. Song, and S. G. Lee

matrix all of whose entries are 1 and = fEij j1 i m; 1 j ng denote the set of cells. The zero-term rank [5] of a matrix X , z(X ), is the minimum number of lines (rows or columns) needed to cover all the zero entries of X . Of course, the term rank [2] of X , t(X ), is de ned similarly for all the nonzero entries of X . If A and B are in M , we say A dominates B (written A B or B A) if aij = 0 implies bij = 0 for all i; j . Then we obtain the following result. Lemma 1.1. For A; B 2 M , A B implies that z (A) z (B ). Proof. If z(B ) = k, then there are k lines which cover all zero entries in B . Since A B , this k lines can also cover all zero entries in A. Hence z(A) k.

2. Zero-Term Rank Preserver over Boolean Matrices

A function T mapping M = M m;n (B ) into itself is called a Boolean linear operator if T preserves sums and 0. Which Boolean linear operators over M preserve zero-term rank? The operations of (1) permuting rows, (2) permuting columns and (3) (if m = n) transposing the matrices in M are all linear operators that preserve zero-term rank of the matrices on M . That these operations and their compositions are the only zero-term rank preservers is one of the consequences of Theorem 2.5 below. Such operations are described more formally in the following de nition. If P and Q are m m and n n permutation matrices respectively and X is an m n Boolean matrix, then a Boolean linear operator T : M ! M is a (P; Q)-operator if (1) T (X ) = PXQ for all X in M or (2) m = n and T (X ) = PX tQ for all X in M . If z(T (X )) = k whenever z(X ) = k, we say T preserves zero-term rank k. So a Boolean linear operator preserves zero-term rank if it preserves zero-term rank k for every k minfm; ng. For term rank preserver and term rank k preserver [2], they are de ned similarly. From now on we will assume that 2 m n unless speci ed otherwise for all m n matrices. And a mapping T will denote a Boolean

Boolean zero-term rank preservers

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linear operator on M = M m;n (B ). Lemma 2.1. Suppose T preserves zero-term rank 1 and T (J ) = J . Then T maps a cell onto a cell and hence it is a bijection on . Proof. If T (Eij ) = 0 for some Eij 2 , then we can choose mn ; 1 cells E1; E2; ; Emn;1, which are dierent from Eij such that

J = T (J ) = T Eij +

!

mn X;1

Ek k=1 mn X;1

= T Eij ) + T (

(2.1)

mn X;1

=T

k=1

P

!k

Ek

!

=1

Ek :

;1 E ) = 1. Since T preserves zero-term rank But z(J ) = 0 andPz( mn k k =1 mn ; 1 1, we have z(T ( k=1 Ek )) = 1. Then we have a contradiction from (2.1). Thus T (Eij ) covers at least one cell in . For some cell Eij 2 , suppose T (Eij ) covers two cells, say Ekl and Euv in . For each cell Ers except for both Ekl and Euv we can choose one cell Eh such that T (Eh) covers Ers because T (J ) = J . Since the number of cells except for both Ekl and Euv is mn ; 2, there exist at most mn ; 1 cells E1; E2; ; Emn;1 containing Eij such that

T

mn X;1 k=1

!

Eh = J:

This is impossible since T preserves zero-term rank 1, but

z

mn X;1 k=1

!

Eh = 1 6= 0 = z(J ):

Hence T (Eij ) covers only one cell, say Ekl , for all Eij 2 . That is, T maps a cell into a cell. Now, we show that T is a bijection on . If

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T (E ) = T (F ) for some dierent cells E and F , then we have J = T (J ) = T (fJ ; (E + F )g + (E + F )) = T (J ; (E + F )) + T (E + F ) = T (J ; (E + F )) + (T (E ) + T (F )) = T (J ; (E + F )) + (T (E ) + T (E )) = T (J ; (E + F )) + T (E ) = T (fJ ; (E + F )g + E ) = T (J ; F ); which is impossible since T preserves zero-term rank 1. Therefore T is a bijection on . Lemma 2.2. If T is bijective on and T preserves zero-term rank 1, then T preserves term rank 1. Proof. Suppose that T does not preserve term rank 1. Then there exist some cells Eij and Eil on the same row (or column) such that

T (Eij + Eil ) = Epq + Ers with p 6= r and q 6= s. Since the zero-term rank of J ; Eij ; Eil is 1 and the zero-term rank of its image under T is

z(T (J ; Eij ; Eil )) = z(J ; Ers ; Epq ) = 2; which is a contradiction. Hence T preserves term rank 1.

Lemma 2.3. If T is bijective on and T preserves zero-term rank 1, then T maps a row of a matrix onto a row (or column if m = n). Proof. Suppose T does not map a row into a row (or a column if m = n). Then T does not preserve term rank 1. This contradicts Lemma 2.2. Hence T maps a row into a row (or a column if m = n). Then the bijectivity of T implies that T maps a row onto a row (or may be a column if m = n).


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Lemma 2.4. Suppose T is bijective on and T preserves zero-term

rank 1. For the case m = n, if T maps a row onto a row (or a column), then all rows of a matrix must be mapped some rows (or columns, respectively) under T . Proof. Let n m X X Ri = Eij ; Cj = Eij j =1

i=1

for i = 1; ; m and j = 1; ; n. Suppose T maps a row, say R1 , onto an ith row Ri and another row, say R2, onto a j th column Cj . Then R1 + R2 has 2n cells but Ri + Cj has 2n ; 1 cells. This contradicts the bijectivity of T on . Hence all rows must be mapped some rows (or columns) under T . Thus we have the following characterization theorem for zero-term rank preserver over M m;n (B ). Theorem 2.5. Suppose T is a Boolean linear operator on M . Then the following statements are equivalent: (i) T is a (P; Q)-operator; (ii) T preserves zero-term rank; (iii) T preserves zero-term rank 1 and T (J ) = J . Proof. Obviously (i) implies (ii) and (ii) implies (iii). We now show that (iii) implies (i). By Lemma 2.1, T is a bijection on . Lemmas 2.3 and 2.4 imply that T maps all rows of a matrix onto rows and columns onto columns. Thus for all m n matrix X , T (X ) = PXQ with some permutation matrices P and Q of degree m and n, respectively. If m = n, then T may map all rows of a matrix onto columns and columns onto rows. By choosing the permutation matrices P and Q appropriately, we have T (X ) = PX tQ for all matrix X . Hence T is a (P; Q);operator. Lemma 2.6. For A; B in M

AB

implies

T (A) T (B ):

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Proof. By de nition of A B , we have aij bij for all i; j . Using

A=

m X n X i=1 j =1

for Eij 2 , we have

aij Eij and B =

m X n X i=1 j =1

bij Eij

0m n 1 XX T (A) = T @ aij Eij A i=1 j =1 m X n X = aij T (Eij ) i=1 j =1 m X n X bij T (Eij ) i=1 j =1

= T (B ):

We say that a Boolean linear operator T strongly preserves zeroterm rank 1 provided that z(T (A)) = 1 if and only if z(A) = 1. For term rank, it is de ned similarly. Lemma 2.7. If T strongly preserves zero-term rank 1, then we have T (J ) = J . Proof. Suppose that T (J ) 6= J . Since T strongly preserves zeroterm rank 1, we have z(T (J )) 2. Then z(T (A)) 2 for all A 2 M by Lemmas 1.1 and 2.6. For any cell Eij 2 M , z (J ; Eij ) = 1 but z(T (J ; Eij )) 2 by above argument. This contradicts to the assumption. Theorem 2.8. Suppose T is a Boolean linear operator on M . Then T preserves zero-term rank if and only if it strongly preserves zero-term rank 1. Proof. Suppose T strongly preserves zero-term rank 1. Lemma 2.7 implies that T (J ) = J . By Theorem 2.5, T preserves zero-term rank. The converse is immediate. Notice that the linear operator T which preserves zero-term rank of Boolean matrices is invertible. Its inverse is a (P t; Qt)-operator.


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3. Relation with Term Rank Preserver In this section, we compare the zero-term rank preserver with term rank preserver, permanent preserver and rook polynomial preserver. Lemma 3.1. If T is bijective on , and preserves zero-term rank 1, then T preserves term rank 2. Proof. By Lemma 2.3, T maps a row of a matrix onto a row (or a column if m = n). So T does not increase term rank of a matrix. Suppose there exists a matrix X such that t(X ) = 2 and t(T (X )) = 1. Since T is bijective on , we can take a submatrix A of X such that A = Eij + Ekl with i 6= k, j 6= l, and T (A) has term rank 1. Then we consider three cases: Case1) Either T maps i-th and k-th rows into one row (or one column if m = n) or T maps j -th and l-th columns into one column (or one row if m = n). This is impossible by the bijectiveness of T on . Case 2) T maps i-th and k-th row into two rows such that images of Eij and Ekl are the same column. Then T must map the j -th and l-th column into one column, which is impossible by the bijectiveness of T on . Case 3) m = n and T maps i-th and k-th rows into two column such that the images of Eij and Ekl are in the same row. Then T must map the i-th and k-th rows into one row, which is also impossible by the bijectiveness of T on . Therefore we conclude that T preserves term rank 2. In [2], Beasley and Pullman obtained characterizations of term rank preserver over arbitrary semirings. We restate their results for Boolean linear operators. Lemma 3.2 ([2]). Suppose T is a Boolean linear operator on M . Then the following are equivalent: (i) T is a (P; Q)-operator; (ii) T preserves term rank; (iii) T preserves term ranks 1 and 2; (iv) T strongly preserves term rank 1. Using this results, we have the following characterizations.

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Theorem 3.3. Suppose T is a Boolean linear operator on M . Then

the followings are equivalent: (i) T preserves zero-term rank; (ii) T preserves zero-term rank 1 and T (J ) = J ; (iii) T strongly preserves zero-term rank 1; (iv) T is a (P; Q)-operator; (v) T preserves term rank; (vi) T preserves term ranks 1 and 2; (vii) T strongly preserves term rank 1. Proof. The equivalence of (i)(iv) comes from Theorems 2.5 and 2.8. And Lemma 3.2 implies the equivalence of (iv)(vii). Therefore we have the results. In order to compare the zero-term rank preserver with the permanent preserver and rook-polynomial preserver, we give their de nitions. For an m n matrix A = [aij ] over B , we de ne the permanent as

per(A) =

X

a1(1)a2(2) am(m);

where the summation extends over all one-to one functions from f1; ; m ; ng. And the rook polynomial of a matrix A is RA(x) = Pg topf1x;j, where p0 = 1 and pj is the sum of the permanents of the j 0 j j j submatrices of A. Beasley and Pullman [2] obtained characterizations of permanent preservers and rook polynomial preservers on matrices over antinegative semirings. We restate their results for Boolean matrices as follows: Lemma 3.4 ([2]). Suppose T is a Boolean linear operator on M . Then these are equivalent: (i) T preserves the permanent; (ii) T preserves term rank; (iii) T preserves the rook-polynomial; (iv) T is a (P; Q)-operator. Using this Lemma 3.4, we have the followings:


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Theorem 3.5. Suppose T is a Boolean linear operator on M . Then

the followings are equivalent: (i) T preserves zero-term rank; (ii) T preserves term rank; (iii) T preserves the permanent; (iv) T preserves the rook-polynomial. Proof. It comes from Theorem 3.3 and Lemma 3.4.

Thus we had characterizations of the Boolean linear operators that preserve zero-term rank. We also extended the characterizations of term rank preservers of Boolean matrices to zero-term rank preservers, permanent preservers and rook polynomial preservers.

References [1] L. B. Beasley and N. J. Pullman, Boolean rank preserving operators and Boolean rank-1 spaces, Linear Algebra Appl. 59 (1984), 55-77. [2] , Term-rank, permanent and rook-polynomial preservers, Linear Algebra Appl. 90 (1987), 33-46. , Operators that preserve semiring matrix functions, Linear Algebra [3] Appl. 99 (1988), 199-216. [4] S. G. Hwang, S. J. Kim and S. Z. Song, Linear operators that preserve maximal column rank of Boolean matrices, Linear and Multilinear Algebra 36 (1994), 305-313. [5] C. R. Johnson and J. S. Maybee, Vanishing minor conditions for inverse zero patterns, Linear Algebra Appl. 178 (1993), 1-15. [6] S. Z. Song, Linear operators that preserve column rank of Boolean matrices, Proc. Amer. Math. Soc. 119 (1993), 1085-1088. [7] S. Z. Song and S. G. Lee, Column ranks and their preservers of general Boolean matrices, J. Korean Math. Soc. 32, no. 3, 531-540.

LeRoy. B. Beasley Department of Mathematics Utah State University Logan, Utah 84322-4125 USA E-mail : [email protected]

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Seok-Zun Song Department of Mathematics Cheju National University Cheju 690-756, Korea E-mail : [email protected] Sang-Gu Lee Department of Mathematics SungKyunKwan University Suwon 440-746, Korea E-mail : [email protected]