J. Korean Math. Soc. 41 (2004), No. 2, pp. 397–406

RANK AND PERIMETER PRESERVERS

OF BOOLEAN RANK-1 MATRICES

Seok-Zun Song, LeRoy B. Beasley,

Gi-Sang Cheon and Young-Bae Jun

Abstract. For a Boolean rank-1 matrix A = abt, we define the

perimeter of A as the number of nonzero entries in both a and b.We characterize the Boolean linear operators that preserve rankand perimeter of Boolean rank-1 matrices.

1. Introduction and preliminaries

The Boolean algebra consists of the set B = {0, 1} equipped withtwo binary operations, addition and multiplication. The operations aredefined as usual except that 1 + 1 = 1.

There are many papers on linear operators that preserve the rankof matrices over several semirings. Boolean matrices also have beenthe subject of research by many authors. Beasley and Pullman [1] ob-tained characterizations of rank-preserving operators of Boolean matri-ces. They were unable to find necessary and sufficient conditions for aBoolean operator to preserve the rank of Boolean rank-1 matrices. Thisremains open until now. We consider this problem by adding conditionson perimeter of the Boolean rank-1 matrices.

Let Mm,n(B) denote the set of all m× n matrices with entries in theBoolean algebra B. The usual definitions for adding and multiplyingmatrices apply to Boolean matrices as well. Throughout this paper, weshall adopt the convention that m ≤ n unless otherwise specified.

If an m×n Boolean matrix A is not zero, then its Boolean rank, b(A),is the least k for which there exist m× k and k × n Boolean matrices B

and C with A = BC. The Boolean rank of the zero matrix is 0. It is

Received February 27, 2003.2000 Mathematics Subject Classification: Primary 15A03, 15A04.Key words and phrases: Boolean linear operator, perimeter, (U,V)-operator.This work was supported by Korea Research Foundation Grant KRF-2002-042-

C00001.

398 Seok-Zun Song, LeRoy B. Beasley, Gi-Sang Cheon and Young-Bae Jun

well known that b(A) is the least k such that A is the sum of k matricesof Boolean rank 1 ([3], [4]).

Let ∆m,n = {(i, j)|1 ≤ i ≤ m, 1 ≤ j ≤ n}, and Eij be the m × n

matrix whose (i, j)th entry is 1 and whose other entries are all 0, andEm,n = {Eij |(i, j) ∈ ∆m,n}. We call Eij a cell.

An n × n Boolean matrix A is said to be invertible if for some X,AX = XA = In, where In is the n×n identity matrix. This matrix X isnecessarily unique when it exists. It is well known that the permutationmatrices are the only invertible Boolean matrices (see [2]).

If A and B are in Mm,n(B), we say A dominates B (written B ≤ A

or A ≥ B) if aij = 0 implies bij = 0 for all i, j. Equivalently, B ≤ A ifand only if A + B = A.

Also lowercase, boldface letters will represent vectors, all vectors u

are column vectors (ut is a row vector) for u ∈ Bm [= Mm,1(B)].

It is easy to verify that the Boolean rank of A is 1 if and only if thereexist nonzero (Boolean) vectors a ∈ Mm,1(B) and b ∈ Mn,1(B) such thatA = abt. And these vectors a and b are uniquely determined by A. Wecall a the left factor, and b the right factor of A. Therefore there areexactly (2m − 1)(2n − 1) rank-1 m × n Boolean matrices.

For any vector u ∈ Mm,1(B), let |u| be the number of nonzero entriesin u, and when A = abt is not zero, define the perimeter of A, P (A), as|a|+ |b|. Since the factorization of A as abt is unique, the perimeter ofA is also unique.

Proposition 1.1. If A, B and A+B are rank-1 matrices in Mm,n(B),then P (A + B) < P (A) + P (B).

Proof. Let A = axt, B = byt and A + B = czt be the factorizationsof A, B and A + B. We have for all i, j

(1.1) aix + biy = ciz

and

(1.2) xja + yjb = zjc.

Now, consider three cases :

Case (1) B ≤ A. Then A + B = A. Thus P (A + B) = P (A) <

P (A) + P (B) since P (B) 6= 0, as required.

Case (2) A ≤ B. It is similar to the case (1).

Case (3) A 6≤ B and B 6≤ A.

Rank and perimeter preservers 399

Subcase (3.1) a 6≤ b and b 6≤ a. From (1.1) we have for some i, j,ai = 1 and bi = 0, and aj = 0 and bj = 1 so that x = ciz and y = cjz.But x 6= 0 and y 6= 0, so x = y = z. Thus

P (A + B) = P(

(a + b)zt)

= |a + b| + |z|

< (|a| + |z|) + (|b| + |z|) = P (A) + P (B),

as required.Subcase (3.2) a ≤ b. Then x 6≤ y. Thus a = zjc = c for some j from

(1.2). But c ≥ b from (1.2). Therefore a = b = c and we have

P (A + B) = P(

c(x + y)t)

= |c| + |x + y|

< (|c| + |x|) + (|c| + |y|) = P (A) + P (B),

as required.Subcase (3.3) b ≤ a. It is similar to the subcase (3.2).

A mapping T : Mm,n(B) → Mm,n(B) is called a Boolean linear op-

erator if T preserves sums and 0. A Boolean linear operator T :Mm,n(B) → Mm,n(B) is invertible if T is injective and surjective. Aswith vector spaces over fields, the inverse, T−1, of a Boolean linear op-erator T is also linear.

Beasley and Pullman obtained the following:

Lemma 1.2.([1]) If T is a Boolean linear operator on Mm,n(B), then

the following statements are equivalent:

(a) T is invertible;(b) T is surjective;(c) T permutes Em,n.

In this paper, we obtain characterizations of the linear operators thatpreserve the rank and the perimeter of every Boolean rank-1 matrix.These are motivated by analogous results for the preserver of Booleanrank. However, we obtain results and proofs in the view of the perimeteranalog.

In the following, “linear operator” on Mm,n(B) means “Boolean linearoperator”.

2. Linear operators preserving the rank and perimeter

of Boolean rank-1 matrices.

In this section, we will characterize the linear operators that preservethe rank and the perimeter of every Boolean rank-1 matrix.

Suppose T is a linear operator on Mm,n(B). Then

400 Seok-Zun Song, LeRoy B. Beasley, Gi-Sang Cheon and Young-Bae Jun

(i) T is a (U, V )-operator if there exist invertible matrices U and V

such that T (A) = UAV for all A in Mm,n(B), or m = n and T (A) =UAtV for all A in Mm,n(B).

(ii) T preserve Boolean rank 1 if b(T (A)) = 1 whenever b(A) = 1 forall A ∈ Mm,n(B).

(iii) T preserve perimeter k of Boolean rank-1 matrices if P (T (A)) =k whenever P (A) = k for all A ∈ Mm,n(B) with b(A) = 1.

The following example shows that not all rank-1 preserving operatorsT are of the form T (X) = UXV for some invertible matrices U, V .

Example 2.1. Let T : M2,3(B) → M2,3(B) be defined by

T

([

a b c

d e f

])

=

[

a 0 b

0 0 0

]

+ (c + d + e + f)

[

1 1 11 1 1

]

.

Then T preserves Boolean rank 1, but T

([

0 0 00 1 0

])

=

[

1 1 11 1 1

]

. So

we can not find invertible matrices U and V such that T (X) = UXV

for X =

[

0 0 00 1 0

]

∈ M2,3(B). This shows that T is not a (U, V )-

operator.

Proposition 2.2. If T is a (U, V )-operator on Mm,n(B), then T

preserves both rank and perimeter of Boolean rank-1 matrices.

Proof. Let A be a matrix in Mm,n(B) with b(A) = 1 and A = aut

be the factorization of A with P (A) = |a| + |u|. Since T (A) = UAV =(Ua)(utV ) = (Ua)(V tu)t and U, V are invertible matrices(and hencepermutations), we have

b (T (A)) = b(

(Ua)(V tu)t)

= 1,

andP (T (A)) = |Ua| + |V tu| = |a| + |u| = P (A).

If m = n and T (A) = UAtV , then we can show that b (T (A)) = 1 andP (T (A)) = |a| + |u| by a similar method as above.

Hence (U, V )-operators preserve the rank and perimeter of everyBoolean rank-1 matrix.

Since a Boolean matrix has perimeter 2 if and only if it is a cell, wehave the following Lemma:

Lemma 2.3. Let T be a linear operator on Mm,n(B). If T preserves

Boolean rank 1 and perimeter 2 of Boolean rank-1 matrices, then (1) T

maps a cell into a cell and (2) T maps a row (or a column) of a Boolean

matrix into a row or a column if m=n.

Rank and perimeter preservers 401

Proof. (1) follows from the property that T preserves perimeter 2.Suppose (2) is not true. Then there exists two distinct cells Ei,j , Ei,h insome ith row such that T (Ei,j) and T (Ei,h) lie in two different rows anddifferent columns. Then the Boolean rank of Ei,j + Ei,h is 1 but that ofT (Ei,j + Ei,h) = T (Ei,j) + T (Ei,h) is 2. Therefore T does not preserveBoolean rank 1, a contradiction.

Now, we give an example of a linear operator that preserves Booleanrank 1 and perimeter 2.

Example 2.4. Let T : M2,2(B) → M2,2(B) be defined by

T

([

a b

c d

])

= (a + b + c + d)

[

1 00 0

]

.

Then T is a linear operator and preserves Boolean rank 1. And T

preserves perimeter 2 of each Boolean rank-1 matrix. But T does not

preserve perimeter 3; for, A =

[

1 10 0

]

=

[

10

]

[1, 1] has Boolean rank 1

and perimeter 3, but T (A) = (1 + 1)

[

1 00 0

]

=

[

10

]

[1, 0] has Boolean

rank 1 and perimeter 2.Moreover, T is not a (U, V )-operator: if there existed invertible ma-

trices U and V such that T (X) = UXV for all X ∈ M2,2(B), then for

j = 1, 2, we have T (E1,j) = Ue1ejtV = u1vj

t where e1 =

[

10

]

, e2 =

[

01

]

and u1 is the first column of U and vj is the jth column of V .

But T (E1,1) =

[

1 00 0

]

=

[

10

]

[1, 0] and T (E1,2) =

[

1 00 0

]

=

[

10

]

[1, 0]

and hence V =

[

1 10 0

]

, which is not invertible. This contraction implies

that T is not a (U, V )-operator.

Let Ri = {Ei,j |1 ≤ j ≤ n}, Cj = {Ei,j |1 ≤ i ≤ m}, R = {Ri|1 ≤ i ≤m} and C = {Cj |1 ≤ j ≤ n}. Define T ∗(Ri) = {T (Ei,j)|1 ≤ j ≤ n} forall i and T ∗(Cj) = {T (Ei,j)|1 ≤ i ≤ m} for all j.

Lemma 2.5. Let T be a linear operator on Mm,n(B). Suppose that

T preserves Boolean rank 1 and perimeters 2 and p (≥ 3) of Boolean

rank-1 matrices. Then,

(1) T maps two distinct cells in a row (or a column) into two distinct

cells in a row or in a column,

402 Seok-Zun Song, LeRoy B. Beasley, Gi-Sang Cheon and Young-Bae Jun

(2) if T maps a row of a Boolean matrix A into a row, then T maps

each row of A into a row of T (A). Similarly, if T maps a column

of a Boolean matrix A into a column, then T maps each column

of A into a column of T (A).

Proof. (1) Suppose T (Ei,j) = T (Ei,h) = Ei′,j′ for some distinct pairs(i, j) 6= (i, h). Then T maps the ith row into the i′th row and boththe jth and hth column into the j′th column by Lemma 2.3. Thus forany Boolean rank-1 matrix A with perimeter p (≥ 3) which dominatesEi,j and Ei,h, we can show that T (A) has perimeter at most p-1, acontradiction. Thus T maps distinct cells in a row into distinct cells ineither a row or a column by Lemma 2.3(2).

(2) If not, then there exist rows Ri and Rj such that T (Ri) ⊆ Rr andT (Rj) ⊆ Cs for some r, s. Consider a matrix D = Ei,p+Ei,q +Ej,p+Ej,q

with Boolean rank 1. Then

T (D) = T (Ei,p +Ei,q)+T (Ej,p +Ej,q) = (Er,p′ +Er,q′)+ (Ep′′,s +Eq′′,s)

for some p′ 6= q′ and p′′ 6= q′′ by (1). Therefore b (T (D)) 6= 1 and T doesnot preserve Boolean rank 1, a contradiction. Hence T maps each rowinto a row. Similarly, T maps each column into a column.

Now we have interesting examples:

Example 2.6. (1) Consider a linear map T on M2,3(B) such that

T (

[

a1,1 a1,2 a1,3

a2,1 a2,2 a2,3

]

) =

[

a1,1 + a2,2 a1,2 + a2,3 a1,3 + a2,1

0 0 0

]

.

Then T maps each row and each column into the first row. Moreover,T preserves both Boolean rank and perimeters 2, 3 and 4 of Booleanrank-1 matrices. But T does not preserve perimeter 5 of Boolean rank-1matrix and this map T is not a (U, V)-operator.

(2) Consider a linear map L on Mm,n(B) with m ≥ 2 and n ≥ 4 suchthat

L(A) = B = (bi,j)

where A = (ai,j), bi,j = 0 with i ≥ 2 and b1,j =∑m

i=1ai,r with r ≡

i + (j − 1) (mod n) and 1 ≤ r ≤ n. Then L maps each row and eachcolumn into the first row. And L preserves both Boolean rank andperimeters 2, 3 and n + 1 of Boolean rank-1 matrices. But L does notpreserve perimeters k (k ≥ 4 and k 6= n+1) of Boolean rank-1 matrices:For, if 4 ≤ k ≤ n, then we can choose a 2 × (k − 2) submatrix withperimeter k which is mapped to distinct k position in the first row of B

Rank and perimeter preservers 403

under L. Then this 1 × k submatrix has perimeter k + 1. Therefore L

does not preserve perimeter k of Boolean rank-1 matrices.

Theorem 2.7. Let T be a linear operator on Mm,n(B) with m ≥ 2and n ≥ 4. Then the following are equivalent:

(1) T is a (U, V )-operator;(2) T preserves both rank and perimeter of Boolean rank-1 matrices;(3) T preserves both rank and perimeters 2 and k (k ≥ 4, k 6= n + 1)

of Boolean rank-1 matrices.

Proof. (1) implies (2) by Proposition 2.2. It is obvious that (2)implies (3). We now show that (3) implies (1). We will first show thatT (Ei,j) = Er,l and the corresponding mapping f : ∆m,n → ∆m,n definedby f(i, j) = (r, l) is a bijection.

By Lemma 2.3, T (Ei,j) = Er,l for some (r, l) ∈ ∆m,n. Without lossof generality, we may assume that T maps the ith row into the rth row.Then T maps each row into a row. Suppose T (Ei,j) = T (Ep,q) = Er,l forsome pairs (i, j) 6= (p, q). If i = p or j = q, then we have a contradictionby Lemma 2.5. So let i 6= p and j 6= q.

If k = n + k′ ≥ n + 2, consider the matrix

D =n

∑

j′=1

Ei,j′ +n

∑

q′=1

Ep,q′ +k′−2

∑

g=1

n∑

h=1

Eg,h

with Boolean rank 1 and perimeter n + k′ = k. Then T maps the ithand pth row of D into the rth row by Lemma 2.5. Then the perimeterof T (D) is less than n + k′ = k, a contradiction.

If 4 ≤ k ≤ n, we will show that we can choose a 2× (k−2) submatrixfrom the ith and pth row whose image under T has 1 × k submatrixin the rth row as follows: Since T (Ei,j) = T (Ep,q) = Er,l, T maps theith row and the pth row into the rth row. But T maps distinct cells ineach row (or column) to distinct cells by Lemma 2.5. Now, choose Ei,j ,Ep,j but do not choose Ei,q, Ep,q. Since there is a cell Ep,h (h 6= j, q)in the pth row such that T (Ep,h) = T (Ei,q) but T (Ei,h) 6= T (Ep,j), wechoose 2 × 2 submatrix Ei,j + Ei,h + Ep,j + Ep,h whose image underT is 1 × 4 submatrix in the rth row. And we can choose a cell Ep,s

(s 6= q, j, h) such that T (Ei,s) 6= T (Ep,j), T (Ep,q), T (Ep,h). Then wehave 2× 3 submatrix Ei,j + Ei,h + Ei,s + Ep,j + Ep,h + Ep,s whose imageunder T is 1 × 5 submatrix in the rth row. Similarly, we can choosea 2 × (k − 2) submatrix whose image under T is a 1 × k submatrix inthe rth row. This shows that T does not preserve the perimeter k of aBoolean rank-1 matrix, a contradiction.

404 Seok-Zun Song, LeRoy B. Beasley, Gi-Sang Cheon and Young-Bae Jun

Therefore we have shown that T permutes Em,n and the correspond-ing mapping f is a bijection on ∆m,n.

Now, there are two cases : (a) T maps R onto R and maps C onto C

or (b) T maps R onto C and C onto R.Case (a). Let α(i) = r if and only if T ∗(Ri) = Rr. Then α permutes

{1, 2, . . . , m}. Let β(j) = h if and only if T ∗(Cj) = Ch. Then β permutes{1, 2, . . . , n}. Let U be the m×m permutation matrix corresponding toα, and V be the n × n permutation matrix corresponding to β. Thenfor all X, T (X) = UXV .

Case (b). In this case m = n. Let α(i) = r if and only if T ∗(Ri) = Cr

and β(j) = h if and only if T ∗(Cj) = Rh. Then α and β permute{1, 2, . . . , m}. By choosing the permutation matrices U and V corre-sponding to α and β respectively, we have T (X) = UXtV for all X.

We say that a linear operator T strongly preserves perimeter k ofBoolean rank-1 matrices if P (T (A)) = k if and only if P (A) = k.

Consider the linear operator on M2,2(B) defined by

T

([

a b

c d

])

= (a + b + c + d)

[

0 10 0

]

.

Then T preserves both rank and perimeter 2 of Boolean rank-1 matrices

but does not strongly preserve perimeter 2, since T

([

1 11 1

])

=

[

0 10 0

]

with P

([

1 11 1

])

= 4 but P

([

0 10 0

])

= 2.

Theorem 2.8. Let T be a linear operator on Mm,n(B). Then T

preserves both rank and perimeter of Boolean rank-1 matrices if and

only if it preserves perimeter 3 and strongly preserves perimeter 2 of

Boolean rank-1 matrices.

Proof. Suppose T preserves perimeter 3 and strongly preserves per-imeter 2 of Boolean rank-1 matrices. Then T maps each row into a row(or a column if m = n). Since T strongly preserves perimeter 2, T mapseach cell onto a cell. That is, T permutes Em,n by Lemma 1.2. ThusT preserves both rank and perimeter of Boolean rank-1 matrices by thesimilar method in the proof of Theorem 2.7.

The converse is immediate.

Theorem 2.9. Let T be a linear operator on Mm,n(B) that preserves

the rank of Boolean rank-1 matrices. Then T preserves perimeter of

Boolean rank-1 matrices if and only if it strongly preserves perimeter 2of Boolean rank-1 matrices.

Rank and perimeter preservers 405

Proof. Suppose T strongly preserves perimeter 2 of Boolean rank-1matrices. Then T maps each cell onto a cell and hence permutes Em,n byLemma 1.2. Since T preserves Boolean rank 1, it maps a row into a row(or a column if m = n). Thus T preserves both rank and perimeter ofBoolean rank-1 matrices by the similar method in the proof of Theorem2.7.

The converse is immediate.

Consider the linear operator on M2,2(B) defined by

T

([

a b

c d

])

=

[

a b

d c

]

.

Then T is invertible but does not preserve Boolean rank 1, since

T

([

1 01 0

])

=

[

1 00 1

]

.

However any (U, V )-operator is invertible. Its inverse is(

U t, V t)

-oper-ator.

Corollary 2.10. Let T be a linear operator on Mm,n(B) that pre-

serves both rank and perimeter of Boolean rank-1 matrices. Then T is

invertible.

Thus we obtain characterizations of linear operators that preserverank and perimeter of Boolean rank-1 matrices.

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] L. B. Beasley, S. Z. Song and S. G. Lee, Zero term rank preservers, Linear Mul-

tilinear Algebra 48 (2001), 313–318.[3] D. A. Gregory and N. J. Pullman, Semiring rank : Boolean rank and nonnegative

rank factorizations, J. Combin. Inform. System Sci. 8 (1983), 223–233.[4] K. H. Kim, Boolean Matrix Theory and Applications, Pure Appl. Math. Vol. 70,

Marcel Dekker, New York, 1982.

Seok-Zun SongDepartment of MathematicsCheju National UniversityJeju 690-756, KoreaE-mail : szsong@cheju.cheju.ac.kr

406 Seok-Zun Song, LeRoy B. Beasley, Gi-Sang Cheon and Young-Bae Jun

LeRoy B. BeasleyDepartment of Mathematics and StatisticsUtah State UniversityLogan, Utah 84322-3900, USAE-mail : lbeasley@math.usu.edu

Gi-Sang CheonDepartment of MathematicsDaejin UniversityPocheon 487-711, KoreaE-mail : gscheon@road.daejin.ac.kr

Young-Bae JunDepartment of Mathematics EducationGyeongsang National UniversityChinju 660-701, KoreaE-mail : ybjun@nongae.gsnu.ac.kr