The following two problems were posed by de Caen [4] (see also [6]):

BINARY RANKS AND BINARY FACTORIZATIONS OF NONNEGATIVE INTEGER MATRICES JIN ZHONG Abstract A matrix is binary if each of its entries is either or The binary rank of a nonnegative integer matrix A is the smallest integer b such that A = BC, where B and C are binary matrices, and B has b columns In this paper, bounds for the binary rank are given, and nonnegative integer matrices that attain the lower bound are characterized Moreover, binary ranks of nonnegative integer matrices with low ranks are determined, and binary ranks of nonnegative integer Jacobi matrices are estimated Key words Nonnegative integer matrix, Rank, Binary rank, Binary factorization AMS subject classifications 5A23, 5A36 Introduction A matrix is binary if each of its entries is either or Let B m n be the set of all m n binary matrices We denote by Z m n + the set of all m n nonnegative integer matrices For A Z m n +, if there exist two matrices B B m k and C B k n such that A = BC, then we say that A = BC is a binary factorization of A, where B and C are called a left and a right binary factor of A, respectively The smallest k that makes the factorization possible is called the binary rank of A and denoted by b(a) It is easily seen from the definition of b(a) that rank(a) b(a) The problem of obtaining such decompositions of nonnegative integer matrices arises frequently in a variety of scientific contexts such as symmetric designs, combinatorial optimization, probability and statistics De Caen [4] made several possible interpretations of binary factorizations of nonnegative integer matrices, which touched on symmetric designs, bipartite graphs, directed graphs, and combinatorial designs For example, a binary factorization of a nonnegative integer matrix A corresponds to a partition of the edge-set of G(A) into bicliques Pullman and Stanford [8] studied the biclique partitions of a special class of graphs, the regular bipartite graphs The following two problems were posed by de Caen [4] (see also [6]): Problem Given a nonnegative integer matrix A, determine b(a) Problem 2 Find some restrictions on the structure of factors B and C, Received by the editors on September 23, 2 Accepted for publication on June 9, 22 Handling Editor: Bryan L Shader Faculty of Science, Jiangxi University of Science and Technology, Ganzhou, 34, China (zhongjin984@26com) This research was supported by the NSFC grant 977 54

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 54 especially in the extreme case where the number of columns of B is equal to b(a) In a special symmetric version of Problem, Berman and Xu [3] defined a matrix to be {,} completely positive ({,}-cp, for short) matrix if A Z+ n n can be expressed as A = BB T with B B n k They call the smallest possible number of columns of B in such a factorization the {,}-rank of A In this paper, we mainly deal with Problem Determining the exact binary rank and computing the corresponding factorization of a given nonnegative integer matrix, however, are known to be NP-hard (see [7]) The aim of this paper is to give bounds for binary rank, characterize the nonnegative integer matrices that attain the lower bound, establish some necessary conditions for a nonnegative integer matrix to achieve the upper bound and compute the binary ranks of some special classes of nonnegative integer matrices Let A Z m n + We refer to the binary column rank of A, denoted by bc(a), as the smallest nonnegative integer q for which there exist vectors v,v 2,,v q in B n such that eachcolumn ofacan be representedas a linear combinationof v,v 2,,v q with coefficients from the set {,} The binary row rank of A, denoted by br(a), is defined as the binary column rank of A T, the transpose of A The following result gives equivalent characterizations of binary rank; its proof is almost the same as that of [5] and we omit it Lemma 3 Let A Z m n + and let q be a nonnegative integer Then the following statements are equivalent: (i) q = bc(a); (ii) q = br(a); (iii) q = b(a); (iv) q is the smallest integer for which there exist vectors b,b 2,,b q and c,c 2,,c q in B n such that A = q b i c T i A We refer to the representation of A in (iv) as a binary rank-one decomposition of 2 Bounds for the binary ranks of nonnegative integer matrices In this section, we study upper and lower bounds for the binary ranks of nonnegative integer matrices For A = (a ij ) Z m n +, denote A = maxa ij, A r = maxa ij and i,j i j A c = maxa ij It is clear that A r = A T c Let j n B n denote the vector j i of all s

542 Jin Zhong Lemma 2 Let A Z m n + Then, A b(a) min{ A r, A c } Proof The first inequality is clear To prove the second, we will show that any nonnegative integer matrix A can be expressed as a product A = BC for some binary matrices B and C, where B has min{ A r, A c } columns For A = (a ij ) Z m n +, without loss of generality, assume A r A c Let r i be the largest component of the i-th row of A, i =,,m Let B = m jt r i and let C = [C T,C2 T,,Cm] T T, where C i B ri n and the j-th column of C i has exactly a ij s, j =,,n Now, it follows that A = BC and B has A r columns Thus, b(a) A r Corollary 22 Let A B m n If rank(a) = min{m,n}, then b(a) = rank(a) Proof Since A is binary, by Lemma 2, b(a) min{ A r, A c } min{m,n} Ontheotherhand, itisobviousthatrank(a) b(a) Hence, ifrank(a) = min{m,n}, then b(a) = rank(a) = min{m,n} Notice that the upper and lower bounds given in Lemma 2 are sharp Corollary 22 implies that any full-rank binary matrix achieves the upper bound The following theorem characterizes the nonnegative integer matrices that attain the lower bound Theorem 23 Let A = (a ij ) Z m n + and let a ij be a maximum entry of A If min{m,n} = 2, then b(a) = A if and only if (i) a kj = max j n a kj for any k, k m; (ii) a ih = max i m a ih for any h, h n; (iii) a ij +a st a it +a sj for any s,t, s m, t n, s i, t j If min{m,n} > 2, let A = [A T,AT 2,,AT m ]T be row partition of A Then b(a) = A if and only if A satisfies (i), (ii), (iii) and there exists a positive integer i i Ai such that = BC for some matrices B B m A and C B A n and A i for every j {,,m}\{i,i }, A j = x j C for some x j B A Proof First, we consider the case min{m,n} = 2 Since b(a) = b(a T ), we may assume m = 2 If b(a) = A and A = BC is a binary factorization of A, where B B 2 A, C B A n B Let us partition B and C as B = and C = [C,C 2,,C n ] Since a ij = A, it follows that Bi T = C j = j A For any positive integer k, k 2, a kj = B k C j for any j, j n Since C j = j A, it follows that a kj = B k C j = max a kj Similarly, for any positive integer h, h n, a ih = B i C h for j n B 2

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 543 any i, i m Since B i = j T A, it follows that a ih = B i C h = max i 2 a ih Observe that a ij = B i C j and for any nonnegative integers s and t, s i, t j, a st = B s C t, a it = B i C t, a sj = B s C j It is not difficult to check that (B i B s )(C j C t ), which is equivalent to B i C j + B s C t B i C t + B s C j, ie, a ij +a st a it +a sj [ B B 2 Conversely, we may assume a = A Let us partition B and C as B = ] and C = [C,C 2,,C n ], respectively Let B T = C = j A and let B 2 = (ja T 2,) Now, B is fixed Since a 2 +a j a a 2j min{a 2,a j } for every Cj j, 2 j n, let C j =, where C j B a2 and C j2 B (a a2), such C j2 that C j has a 2j s and C j2 has a j a 2j s, j = 2,,n Now, it is readily seen that A = BC If min{m,n} > 2, let us partition A as A = [A T,A T 2,,A T m] T Then a ij is in A i If A satisfies conditions (i), (ii) and (iii), then for a positive integer i i, it follows from the case min{m,n} = 2 that there exist two binary matricesb B 2 A and C B A n Ai such that = B C For every j {,,m}\{i,i }, A i if A j = x j C for some x j B A, let B be the matrix whose i -th row and i -th row are the first row and the second row of B, respectively, and j-th row, j {,,m}\{i,i }, is x j Then A = BC, and thus, b(a) = A Conversely, if b(a) = A, then by the above argument, A satisfies (i), (ii) and (iii) Moreover, it Ai can be seen from A = BC that C is a right binary factor of the submatrix and for every j {,,m}\{i,i }, A j can be expressed as A j = x j C for some x j B A Next, we will give some necessary conditions for b(a) = min{ A r, A c } Since b(a) = b(a T ), we assume A r A c in the sequel Lemma 24 Let A = (a ij ) Z n n + be a triangular matrix If a ii = max j n a ij for every i n, then b(a) = A r Proof Since b(a) = b(a T ), we assume that A is upper triangular Suppose that A = BC is a binary factorization of A, where B B n k and C B k n Notice that the first row of B and the first column of C have at least a s and without loss of generalityweassumethatthe firstrowofb is oftheform(ja T, )andthe firstcolumn of C is of the form (ja T, ) T Exchanging some rows of B and the corresponding columns of C if necessary, it follows from a 2 = and a 22 = max a 2j that the j n second row of B is of the form ( T a,ja T 22, ) Similarly, exchanging some rows of B A i

544 Jin Zhong and the corresponding columns of C if necessary, it follows from a 3 = a 32 = and a 33 = max a 3j that the third row of B is of the form ( T a j n +a 22,ja T 33, ) Continuing in this way we conclude that the last row of B is of the form ( T n,j a T nn, ) Thus, a ii b(a) A r Ontheotherhand, bylemma2,b(a) A r Hence, b(a) = A r A matrix A is defined to be nondegenerate if A has no zero row or zero column Let φ(a) be the number of zeros of A Theorem 25 Let A Z n n + be nondegenerate If b(a) = A r, then n φ(a) n(n ) Moreover, let k and n be positive integers such that n k n(n ) Then there exists a matrix A Z n n + such that φ(a) = k and b(a) = A r Proof If b(a) = A r, we claim that there is at most one row of A all of whose components are nonzero Suppose that all components of row i and row j of A are nonzero, i,j n, i j Assume that a it and a js are the maximum components of row i and row j of A, respectively Let A ai a in = and a j a jn [ ] ja T B = it ja T Then, it is not difficult to see that there exists a binary js matrix C B (ait+ajs ) n such that A = BC, ie, b(a ) a it + a js Thus, b(a) A r, contradicting the assumption that b(a) = A r Hence, there is at most one row of A all of whose components are nonzero, ie, φ(a) n Since A is nondegenerate, it follows that φ(a) n(n ) Let k and n be positive integers such that n k n(n ) For an upper triangular matrix A Z+ n n, assigning the values of the diagonal entries of A such that a ii = max j n a ij for every i n Then, no matter what the entries in the strictly upper triangular part of A are, we have b(a) = A r Hence, if n(n ) 2 k n(n ), then we can find an upper triangular matrix A Z n n + with k n(n ) 2 zeros in its strictly upper triangular part such that b(a) = A r Next, we will show that if n k n(n ) 2, then there exists a matrix A Z+ n n such that φ(a) = k and b(a) = A r By Corollary 22, it suffices to show that for every k, n k n(n ) 2, there exists a nonsingular binary matrix A B n n such that φ(a) = k Let A = B n n

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 545 That is, a i+,i = for every i n, and the remaining entries of A are A direct computation shows that det(a) =, ie, A is nonsingular Moreover, observe that replacing any a ij, 3 i n, j i 2, by will not change the determinant of A Hence, if k [n, n(n ) 2 ], then there exists a nonsingular binary matrix A B n n such that φ(a) = k This completes the proof A natural question is: For a nondegenerate nonnegative integer matrix A, if b(a) = A r, then how small can rank(a) be? The following theorem shows that for a nondegenerate nonnegative integer matrix A of order n 4, if b(a) = A r, then rank(a) 3 For a matrix A Z+ n n, let µ, ν be nonempty ordered subsets of {,2,,n} Denote by A[µ ν] the submatrix of A with rows indexed by µ and columns indexed by ν If µ = ν, then A[µ µ] is abbreviated to A[µ] Theorem 26 Let A Z n + be nondegenerate, n 4 If b(a) = A r, then rank(a) 3 Proof We will show in Theorem 3 that if rank(a) =, then b(a) = A Hence, for a nondegenerate matrix A of order n 2, if rank(a) =, then b(a) < A r Next, we will show that for a nondegenerate nonnegative integer matrix A of order n 4, if rank(a) = 2, then b(a) < A r We have shown in the proof of Theorem 25 that if A Z+ n n with b(a) = A r, then there is at most one row of A all of whose components are nonzero Hence, there exists a submatrix of A, say Ã, which is permutation equivalent to one of the following three forms (I), (II), (III) If à is of type (I) or type (II), then we only consider the first three rows of à Without loss of generality, we assume à = a a 2 a 3 a 4 a n a 22 a 23 a 24 a 2n a 32 a 33 a 34 a 3n Since A is nondegenerate, the first column of à has at least one nonzero component and we may assume a We consider the following three cases:

546 Jin Zhong Case : a 22, a 32 In this case, for any 3 j n, if rank(a) = 2, then Ã[{,2,3},{,2,j}] is singular, which implies a 22 a 3j = a 2j a 32 Hence, either a 2j and a 3j are positive integers or a 2j = a 3j = It follows from the second and the third row of à that b(ã) < à r, which implies b(a) < A r, a contradiction Case 2: a 22 =, a 32 or a 22, a 32 = Since A is nondegenerate, the second row of à has at least one nonzero component, say, a 2j If a 22 = and a 32, it follows that Ã[{,2,3},{,2,j }] is nonsingular, contradicting the assumption that rank(a) = 2 Similarly, if a 22 and a 32 =, then we can find a 3 3 nonsingular sbumatrix of Ã, also a contradiction Case 3: a 22 = a 32 = If at least one of a 23 and a 33 is nonzero, then applying the argument of Case and Case 2 to a 23 and a 33 we conclude that if rank(a) = 2, then b(a) < A r If a 23 = a 33 =, then we consider a 24 and a 34 Continuing in this way we conclude that if rank(a) = 2, then b(a) < A r If à is of type (III), then without loss of generality, we assume a a 2 a 3 a 4 a n à = a 22 a 23 a 24 a 2n a 3 a 33 a 34 a 3n a 4 a 42 a 44 a 4n If rank(a) = 2, then det(ã[{2,3,4},{,2,3}])= a 22a 33 a 4 +a 23 a 3 a 42 = Since A is nonnegative, one must have that at least one of a 22, a 33 and a 4 is zero and at least one of a 23, a 3 and a 42 is zero Thus, it can be reduced to type (I) or type (II) By the above argument, we conclude that if rank(a) = 2, then b(a) < A r This completes the proof We do not know if 3 is the best possible lower bound in Theorem 26 We will show in the next section that if n = 4, then there is a nondegenerate binary matrix A of order 4 satisfies rank(a) = 3 and b(a) = 4 Now, we pose the following question: For any n 5, there is a nondegenerate matrix A Z+ n n such that rank(a) = 3 and b(a) = A r? Or there is a nondegenerate binary matrix A B n n such that rank(a) = 3 and b(a) = n? 3 Computing binary ranks of nonnegative integer matrices with low ranks Let M m,n (S) denote the set of all m n matrices whose entries belong to the set S We refer to the factor rank of A M m,n (S), denoted by r S (A), as the minimum integer k such that A = BC for some matrices B M m,k (S) and C M k,n (S) For example, if S = R +, the set of nonnegative real numbers, then r R +(A) is the nonnegative rank of A (see [5]) Recently, comparing real rank with various ranks over various semirings has been object of interest For example, in [5] and

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 547 independently in [], it was shown that for any A M m,n (R + ), if rank(a) 2, then r R +(A) = rank(a) In [2], it was proved that the nonnegative rank of a product of two nonnegative matrices is not greater than the product of ranks of these two matrices In this section, we will compute the binary ranks of nonnegative integer matrices with low ranks Theorem 3 Let A Z m n + If rank(a) =, then b(a) = A Proof Exchanging some rows and some columns of A if necessary, we assume a = A and a a 2 a n, a a 2 a m For convenience, we assume a n and a m Then a ij for any i,j, i m, j n Since rank(a) =, any two rows of A are linearly dependent It follows from a /a i = a j /a ij that a j a ij for any i, j, where 2 i m and 2 j n Similarly, any two columns of A are linearly dependent and it follows from a /a k = a h /a hk that a h a hk for any h, k, where 2 h m and 2 k n Moreover, for any i,j, i m, 2 j n, let a i /a i+, = a ij /a i+,j = k Then k and (a i a i+, ) (a ij a i+,j ) = a i+, (k ) a i+,j (k ) = (k )(a i+, a i+,j ) Let t i = a i a i+,, i m and t m = a m Partition A as A = [A,A 2,,A n ] Then, A can be expressed as A = + + } {{ } t m + + + } {{ } t m, + + + + ; } {{ } t that is, A can be expressed as a linear combination of the above a vectors in B n with coefficients all equal Moreover, for every j, 2 j n, A j has a similar representation as that of A, we need only replace t i by t ij, where i m, t ij = a ij a i+,j and t mj = a mj Noticethata i a i+, max 2 j n (a ij a i+,j )forany i m,anda m a mj for any 2 j n, ie, t i t ij for any i m Thus, for every j n, A j can be expressed as a linear combination of a vectors in B n with coefficients from {,} Hence, by Lemma 3, b(a) = A It seems that it is not easy to determine the binary rank of a given nonnegative integer matrix of rank 2 However, for binary matrices with rank 2, we are able to determine their binary ranks Theorem 32 Let A B m n If rank(a) = 2, then b(a) = 2

548 Jin Zhong Proof Let A = [A T,AT 2,,AT m] T B m n If rank(a) = 2 and min{m,n} = 2, then it follows from Corollary 22 that b(a) = 2 If min{m,n} > 2, then there exist two rows of A such that each row of A is a linear combination of these two rows Exchanging some rows of A if necessary, assume that these two rows are A and A 2 For j = 3,,m, let A j has the representation A j = α j A +β j A 2 Since rank(a) = 2, there exists a positive integer i such that (A ) i =, (A 2 ) i = or (A ) i =, (A 2 ) i =, where (A j ) i is the i-th component of A j, j =,2 We only consider the case (A ) i =, (A 2 ) i = since the other case can be proved in a similar way Without loss of generality, we assume (A ) = and (A 2 ) = It follows that for any 3 j m, α j = α j (A ) + β j (A 2 ) {,} Notice that for any positive integer k, 2 k n, ((A ) k,(a 2 ) k ) {(,),(,),(,),(,)} We consider the following two cases Case : There exists at least one k, 2 k n, such that (A ) k = and (A 2 ) k = In this case, for any 3 j m, it follows from α j (A ) k +β j (A 2 ) k {,} that β j {,} Thus, any row of A can be expressed as a linear combination of A and A 2 with coefficients from the set {,} Hence, A has the following binary factorization: which implies b(a) = 2 A = α 3 β 3 α m β m [ A Case 2: Foranypositiveintegerl,((A ) l,(a 2 ) l ) {(,),(,),(,)},2 l n We exclude the case ((A ) l,(a 2 ) l ) = (,) since otherwise A will have a zero column Let à = A Then à is of the form A 2 à = (3) or à = A 2 [ ], ] (32) Observe that there exists a positive integer l such that (A ) l = (A 2 ) l = in both forms above It follows from α j (A ) l +β j (A 2 ) l {,} that α j +β j {,} Thus, if α j =, then β j {,} If α j =, then β j {,} Observe that if α j = and β j =, then A j = [ ] (33)

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 549 or A j = [ ] (34) Thus, if à is of the form (3), then any row of A is equal to one of the four types: () A ; (2) A 2 ; (3) ; (4) the vector in (33) If à is of the form (32), then any row of A is equal to one of the four types: () A ; (2) A 2 ; (3) ; (4) the vector in (34) First, we consider the case that à is of the form (32) Without loss of generality, suppose that A h s, 3 h t m, are of the form (33) and the rest rows of A equal A, A 2 or Then A = γ t+ δ t+ γ m δ m [ ], (35) where (,), if (α j,β j ) = (,), (γ j,δ j ) = (,), if (α j,β j ) = (,), (,), if (α j,β j ) = (,) Now, it is readily seen from (35) that b(a) = 2 [ If à is of the form (32), replace the right factor of (35) by continuing in a similar way to above, we have b(a) = 2 This completes the proof ] For a binary matrix A, if rank(a) 3, one may wonder if rank(a) = b(a) is still true? The answer is negative A simple example is provided by A = It is easyto seethat rank(a) = 3 Cohenand Rothblum [5] showthat rank R +(A) = 4 Hence, 4 = b(a) = rank R +(A) > rank(a) = 3

55 Jin Zhong 4 Binary ranks of nonnegative integer Jacobi matrices A matrix A is called a Jacobi matrix if it is of the form a b b a 2 b 2 b 2 a 3 b 3 A =, (4) an b n b n a n where a i are real and b i are positive Theorem 4 Let A = (a ij ) be a nonnegative integer Jacobi matrix of the form (4) If a ii = max j n a ij for every i n, then n b(a) a + max{a i b i,b i }+a n b n i=2 Proof SupposethatA = BC isabinaryfactorizationofa, whereb B n k, C B k n Let us partitionb and C asb = [B T,BT 2,,BT n ]T and C = [C,C 2,,C n ], respectively Without loss of generality, assume that B is of the form (ja T, ) Then C is of the form (ja T, ) T, C i s, i = 3,,n, are of the form ( T a, ) T and B i s, i = 3,,n, are of the form ( T a, ) Hence, to guarantee B 2 C 3 = b 2, B 2 should have at least a + b 2 components and without loss of generality, assume that B 2 is of the form (,,,j }{{} b T 2, ) Then C 3 is of the form ( T a,jb T 2, ) T and C i s, i = a 4,,n, are of the form ( T a, T a +b 2, ) T If a 2 b b 2, then it is possible to assign the values of the first a components of B 2 and the first a + b 2 components of C 2 such that B C 2 = B 2 C = b and B 2 C 2 = a 2 If a 2 b b 2, since the first a components of C 2 have at most b s, to guarantee B 2 C 2 = a 2, B 2 should have at least a + a 2 b components Similarly, it follows from a 24 = and a 34 = b 3 that B 3 should have at least a +max{a 2 b,b 2 }+b 3 components By comparing a 3 b 2 with b 3 and using a similar argument as above we conclude that B 3 should have at least a + max{a 2 b,b 2 } + max{a 3 b 2,b 3 } components Continuing in this way we conclude that B n should have at least a + n max{a i b i,b i } components Moreover, let t = a + n i=2 i=2 max{a i b i,b i } Then it follows from the last row of A that B n is of the form ( T t,jt b n, ) Hence, to guarantee B n C n = a n,

Binary Ranks and Binary Factorizations of Nonnegative Integer Matrices 55 B n should have at least a + n b(a) a + n i=2 i=2 max{a i b i,b i }+a n b n max{a i b i,b i } + a n b n components, ie, Corollary 42 [3] Let A be a nonnegative integer Jacobi matrix of the form (4) If A is diagonally dominant, then b(a) = n n a i b i Proof Let jb T j T a b jb T jb T 2 ja T 2 b b 2 jb T B = 2 ja T 3 b 2 jb T n ja T n b n 2 jb T n ja T n b n Then A = BB T and B has n a i n theotherhand, bytheorem4, b(a) n b i columns Thus, b(a) n a i n a i n b i On b i Hence, b(a) = n a i n b i Acknowledgment The author thanks Prof Xingzhi Zhan for suggesting him to study this topic The author would also like to thank the referee for his valuable comments and suggestions toward improving the paper REFERENCES [] LB Beasley, SJ Kirkland, and BL Shader Rank comparisons Linear Algebra and its Applications, 22:7 88, 995 [2] LB Beasley and TJ Laffey Real rank versus nonnegative rank Linear Algebra and its Applications, 43:233 2335, 29 [3] A Berman and CQ Xu {, } Completely positive matrices Linear Algebra and its Applications, 399:35 5, 25 [4] D de Caen A survey of binary factorizations of nonnegative integer matrices Journal of Combinatorial Math and Combinatorial Computing, 2:5, 987 [5] JE Cohen and UG Rothblum Nonnegative ranks, decompositions, and factorizations of nonnegative matrices Linear Algebra and its Applications, 9:49 68, 993 [6] ER Van Dam The combinatorics of Dom de Caen Designs, Codes and Cryptography, 34:37 48, 25

552 Jin Zhong [7] J Orlin Contentment in graph theory: Covering graphs with cliques Indigationes Mathematicae, 8(5):46 424, 977 [8] NJ Pullman and M Stanford The biclique numbers of regular bigraphs Congressus Numerantium, 56:237 249, 987