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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Rank relations between a {0, 1}-matrix and its complement

Chao Ma / Jin Zhong
Published Online: 2018-03-20 | DOI: https://doi.org/10.1515/math-2018-0020

Abstract

Let A be a {0, 1}-matrix and r(A) denotes its rank. The complement matrix of A is defined and denoted by Ac = JA, where J is the matrix with each entry being 1. In particular, when A is a square {0, 1}-matrix with each diagonal entry being 0, another kind of complement matrix of A is defined and denoted by A = JIA, where I is the identity matrix. We determine the possible values of r(A) ± r(Ac) and r(A) ± r(A) in the general case and in the symmetric case. Our proof is constructive.

Keywords: {0, 1}-matrix; Complement matrix; Rank

MSC 2010: 15A03; 15B36; 05B20; 05C50

1 Introduction

A {0, 1}-matrix is an integer matrix with each entry being 0 or 1. {0, 1}-matrices are closely related to graph theory and combinatorial mathematics [1, 2, 3, 4]. They also have a wide range of practical applications in statistics and probability [5, 6, 7, 8].

Denote by Mm, n{0, 1} the set of m × n {0, 1}-matrices, and we abbreviate Mn, n{0, 1} as Mn{0, 1}. Let AMm, n{0, 1}. Then the matrix Ac = Jm, nA is called the complement of A, where Jm, n is the m × n matrix with each entry being 1. It is clear that A and Ac are mutually complementary, i.e., (Ac)c = A.

Recall that the adjacency matrix of a digraph D is the square matrix A = (aij), where aij is the number of arcs (i, j) in D. A digraph is called strict if it has no loops or parallel arcs. Thus the adjacency matrix of a strict digraph is a {0, 1}-matrix with each diagonal entry being 0. The complement of a strict digraph D, denoted by D, is also a strict digraph on the same vertices such that (i, j) is an arc in D if and only if (i, j) is not an arc in D. Let A be the n × n adjacency matrix of a strict digraph D. Then the adjacency matrix of D is JnInA, where Jn = Jn, n and In is the identity matrix of order n. Denote by Ωn{0, 1} the set of n × n {0, 1}-matrices with each diagonal entry being 0. Thus for AΩn{0, 1}, we define another kind of complement matrix of A to be A = JnInA. It is also clear that A and A are mutually complementary, i.e., (A) = A.

In this paper, we mainly investigate the rank relations between a {0, 1}-matrix and its complement. Denote by r(A) the rank of a matrix A. In Section 2, we determine the possible values of r(A) ± r(Ac) for AMm, n{0, 1} in the general case and in the symmetric case. In Section 3, we determine the possible values of r(A) ± r(A) for AΩn{0, 1} in the general case and in the symmetric case.

We use Om, n to denote the m × n zero matrix. On, n will be abbreviated as On. Denote by Eij the matrix with its entry in the i-th row and j-th column being 1 and with all other entries being 0.

2 Rank relations between A and Ac

First we determine the possible values of r(A) ± r(Ac) in the general case.

Theorem 2.1

Let m, n ≥ 2 be positive integers. Then there exists AMm, n{0, 1} with r(A) − r(Ac) = k if and only if −1 ≤ k ≤ 1.

Proof

Since r(Ac) = r(Jm, nA) ≤ r(Jm, n)+r(A) = 1+r(A), r(Ac) − r(A) ≤ 1. Likewise, r(A) − r(Ac) ≤ 1. This proves the necessity.

For the sufficiency, note that r(A) − r(Ac) = ± 1 if A = Om, n or Jm, n, and r(A) = r(Ac) if A=[Jm,1Om,n1]. This completes the proof. □

Theorem 2.2

Let m, n ≥ 2 be positive integers. Then there exists AMm, n{0, 1} with r(A)+r(Ac) = k if and only if 1 ≤ k ≤ 2 min{m, n}.

Proof

The necessity is clear. Now we prove the sufficiency.

Suppose mn. Let A1=IpOp,npOmp,pOmp,npMm,n{0,1} with 0 ≤ pm − 1. Clearly r(A1) = p. It is easy to verify that A1c=JpIpJp,npJmp,pJmp,np has rank p+1. Then r(A1)+r( A1c) = 2p+1, 0 ≤ pm − 1. Thus for every odd k with 1 ≤ k ≤ 2m − 1, there exists AMm, n{0, 1} such that r(A)+r(Ac) = k.

Let A2=IqOq,mqOq,nmOmq,qJmqOmq,nmMm,n{0,1} with 1 ≤ qm − 1. Clearly r(A2) = q+1. It is easy to verify that A2c=JqIqJq,mqJq,nmJmq,qOmqJmq,nm has rank q+1. Then r(A2)+r( A2c) = 2q+2, 1 ≤ qm − 1. Thus for every even k with 4 ≤ k ≤ 2m, there exists AMm, n{0, 1} such that r(A)+r(Ac) = k.

Let A3 = Jm,1Om,n1Mm, n{0, 1}. Then r(A3)+r( A3c) = 1+1 = 2.

Thus for every integer k with 1 ≤ k ≤ 2m, there exists AMm, n{0, 1} such that r(A)+r(Ac) = k.

If m > n, the argument is similar. This completes the proof. □

Next we consider the case when AMn{0, 1} is symmetric.

Theorem 2.3

Let n ≥ 2 be a positive integer. Then there exists symmetric AMn{0, 1} with r(A) − r(Ac) = k if and only if −1 ≤ k ≤ 1.

Proof

For the necessity, the argument is the same as that of Theorem 2.1.

For the sufficiency, note that r(A) − r(Ac) = −1, 0, 1 if A = On, In, Jn, respectively. □

Lemma 2.4

Let AMn{0, 1} be symmetric. Then r(A)+r(Ac) ≠ 2.

Proof

Assume that there exists symmetric AMn{0, 1} such that r(A)+r(Ac) = 2. If r(A) = 0, then A = On and thus Ac = Jn, which contradicts the assumption that r(A)+r(Ac) = 2. If r(A) = 2, then r(Ac) = 0, Ac = On. This implies A = Jn and thus r(A) = 1, a contradiction. If r(A) = 1, by the proof of Theorem 4(i) in [9], it follows that A is permutation similar to JpOp,npOnp,pOnp with 1 ≤ pn. Thus Ac is permutation similar to OpJp,npJnp,pJnp, which implies that r(Ac) = 0 when p = n and r(Ac) = 2 when 1 ≤ pn − 1. Then r(A)+r(Ac) = 1 or 3, a contradiction. Thus r(A)+r(Ac) ≠ 2 for any symmetric AMn{0, 1}. □

Note that the matrices A1, A2Mm, n{0, 1} in the proof of Theorem 2.2 are symmetric when m = n. By Theorem 2.2 and Lemma 2.4, we have the following result.

Theorem 2.5

Let n ≥ 2 be a positive integer. Then there exists symmetric AMn{0, 1} with r(A)+r(Ac) = k if and only if 1 ≤ k ≤ 2n with k ≠ 2.

3 Rank relations between A and A

In this section, we only consider AΩn{0, 1} which corresponds to the adjacency matrix of a strict digraph. Recall that for an n × n matrix A = (aij), the main diagonal of A is the list of entries a11, a22, …, ann, and the secondary diagonal of A is the list of entries a1n, a2, n − 1, …, an1. Let C1 be the square matrix whose entries above the main diagonal are all 1’s, while other entries are all 0’s. The size of C1 will be clear from the context.

First we determine the possible values of r(A) ± r(A) in the general case.

Theorem 3.1

Let n ≥ 2 be a positive integer. Then there exists AΩn{0, 1} with r(A) − r(A) = k if and only if

  1. k = 0, ±2 when n = 2;

  2. nkn when n ≥ 3.

Proof

  1. The case n = 2 is trivial.

  2. The necessity is clear. Now we prove the sufficiency.

    Let B1=C1Jp,npOnp,pOnpΩn{0,1}, where 1 ≤ pn − 1. It is easy to verify that r(B1) = p, r(B1) = n − 1. Thus for every integer k with −n+2 ≤ kn − 2, there exists AΩn{0, 1} such that r(A) − r(A) = k.

    Let B2=C2On1,nΩn{0,1}, where C2 = [0, 1, …, 1, 0]. Then r(B2) = 1, r(B2) = n. Note that r(On) = 0, r(On) = n. Thus for k = ±(n − 1), ± n, there exists AΩn{0, 1} such that r(A) − r(A) = k. This proves the sufficiency. □

Theorem 3.2

Let n ≥ 2 be a positive integer. Then there exists AΩn{0, 1} with r(A)+r(A) = k if and only if

  1. k = 2 when n = 2;

  2. nk ≤ 2n when n ≥ 3.

Proof

  1. Trivial.

  2. r(A)+r(A) ≤ 2n is clear. Since A+A = JnIn, n = r(A+A) ≤ r(A)+r(A). This proves the necessity.

    For the sufficiency, first note that the matrix B1Ωn{0, 1} in the proof of Theorem 3.1, (ii) implies that for every integer k with nk ≤ 2n − 2, there exists AΩn{0, 1} such that r(A)+r(A) = k.

    Let B3 = C1E1nΩn{0, 1}. Then r(B3) = n − 1, r(B3) = n.

    Let B4 = B3+En1Ωn{0, 1}. Then r(B4) = r(B4) = n.

    Thus for k = 2n − 1, 2n, there exists AΩn{0, 1} such that r(A)+r(A) = k. This proves the sufficiency. □

    Next we consider the case when AΩn{0, 1} is symmetric.

Lemma 3.3

Let G=G1Jp,npJnp,pJnpInpΩn{0, 1} with 1 ≤ pn − 1, where G1 is the p × p matrix whose entries on the main diagonal, on the secondary diagonal and above the secondary diagonal are all 0’s, while other entries are all 1’s. Then r(G) = n if p is odd, and r(G) = n − 1 if p is even.

Proof

First consider the case when p is odd. If p = n − 1, then n = p+1 is even and it is clear that det G ≠ 0. If p < n − 1, for i = p+1, p+2, …, n − 1, subtract the last row of G from the i-th row, and then add the i-th column to the last column. Using Laplace expansion formula, we deduce that det G ≠ 0.

When p is even, the p2-th row (column) of G is identical to the ( p2+1)-th row (column). Denote by G2 the submatrix of G obtained by deleting the p2-th row and the p2-th column. Note that G2 = G3Jp1,npJnp,p1JnpInpΩn1{0,1}, where G3 is the (p − 1) ×(p − 1) matrix which has the same form as G1. Since p − 1 is odd, r(G) = r(G2) = n − 1 by what we have just proved. □

Lemma 3.4

Let HΩn{0, 1} be the matrix whose entries on the main diagonal, on the secondary diagonal and above the secondary diagonal are all 0’s, while other entries are all 1’s. Then

  1. r(H+E1,n12+En12,1)=n for odd n ≥ 5, r(H+E1,n12+En12,1En+32,nEn,n+32)=n for odd n ≥ 7; r(H¯E1,n12En12,1+En+32,n+En,n+32)=n

  2. r(H+E1,n2+En2,1)=n for even n ≥ 4, r(H+E1,n2+En2,1+En21,n2+1+En2+1,n21)=n1 for even n ≥ 6.

Proof

  1. When n ≥ 5 is odd, n12>1. Then E1,n12En12,1. Using Laplace expansion formula, it is easy to verify that det(H+E1,n12+En12,1)0 for odd n ≥ 5 and det(H+E1,n12+En12,1En+32,nEn,n+32)0 for odd n ≥ 7.

    When n ≥ 7 is odd, subtract the second row of H¯E1,n12En12,1+En+32,n+En,n+32 from the first row, and then subtract the second column from the first column. Using Laplace expansion formula, we deduce that det(H¯E1,n12En12,1+En+32,n+En,n+32)0.

  2. When n ≥ 4 is even, n2>1. Then E1,n2En2,1. Using Laplace expansion formula, it is easy to verify that det(H+E1,n2+En2,1)0 for even n ≥ 4.

    When n ≥ 6 is even, n2 −1 > 1. Note that the ( n2)-th row (column) of H+E1,n2+En2,1+En21,n2+1+En2+1,n21 is the sum of the first row (column) and the ( n2 − 1)-th row (column). Then r(H+E1,n2+En2,1+En21,n2+1+En2+1,n21)n − 1. Using Laplace expansion formula and the fact that J3I3 is nonsingular, we can deduce that the submatrix of H+E1,n2+En2,1+En21,n2+1+En2+1,n21 obtained by deleting the ( n2)-th row and the (n2+2)-th column is nonsingular. Thus r(H+E1,n2+En2,1+En21,n2+1+En2+1,n21)=n1.

Finally we determine the possible values of r(A) ± r(A) in the symmetric case.

Theorem 3.5

Let n ≥ 2 be a positive integer. Then there exists symmetric AΩn{0, 1} with r(A) − r(A) = k if and only if

  1. k = ±2 when n = 2;

  2. k = 0, ±3 when n = 3;

  3. nkn with k ≠ ±(n − 1) when n ≥ 4.

Proof

(i) and (ii) are easy to verify.

(iii) First we prove the necessity. It is clear that −nr(A) − r(A) ≤ n. If r(A) − r(A) = −(n − 1), then either r(A) = 0, r(A) = n − 1, or r(A) = 1, r(A) = n. For the former case, A = On and thus A = JnIn is nonsingular, a contradiction. For the later case, note that any nonzero symmetric AΩn{0, 1} always has a 2 × 2 submatrix 0110. Then r(A) ≥ 2, a contradiction. Thus r(A) − r(A) ≠ −(n − 1). Likewise, r(A) − r(A) ≠ n − 1.

Next we prove the sufficiency. We will use the symmetric matrices G1Ωp{0, 1} and G, HΩn{0, 1} in Lemmas 3.3 and 3.4.

Note that r(G) = r(G1) = p − 1 if p is odd, and r(G) = r(G1) = p if p ≥ 2 is even. Then by Lemma 3.3, r(G) − r(G) = n+1 − p for odd p, and r(G) − r(G) = n − 1 − p for even p ≥ 2. When n ≥ 5 is odd, for odd p with 1 ≤ pn − 2, r(G) − r(G) can be 3, 5, 7, …, n − 2, n; for even p with 2 ≤ pn − 1, r(G) − r(G) can be 0, 2, 4, …, n − 5, n − 3. Thus k can be 0, ±2, ± 3, …, ± (n − 2), ± n for odd n ≥ 5. When n ≥ 4 is even, for odd p with 1 ≤ pn − 1, r(G) − r(G) can be 2, 4, 6, …, n − 2, n; for even p with 2 ≤ pn − 2, r(G) − r(G) can be 1, 3, 5, …, n − 5, n − 3. Thus k can be ±1, ±2, …, ±(n − 2), ± n for even n ≥ 4.

By Lemma 3.4 (i), for odd n ≥ 5, r(H+E1,n12+En12,1)=n. Since r(H+E1,n12+En12,1¯)=r(H¯E1,n12En12,1) = n − 1 when n is odd, r(H+E1,n12+En12,1)r(H¯E1,n12En12,1)=1. Thus k can be ±1 for odd n ≥ 5. By Lemma 3.4 (ii), for even n ≥ 4, r(H+E1,n2+En2,1)=n. Since r(H+E1,n2+En2,1¯)=r(H¯E1,n2En2,1)=n when n is even, k can be 0 for even n ≥ 4.

Thus for k = 0, ± 1, ± 2, …, ±(n − 2), ± n with n ≥ 4, there exists symmetric AΩn{0, 1} such that r(A) − r(A) = k. This completes the proof. □

Theorem 3.6

Let n ≥ 2 be a positive integer. Then there exists symmetric AΩn{0, 1} with r(A)+r(A) = k if and only if

  1. k = ±2 when n = 2;

  2. k = 3, 4 when n = 3;

  3. k = 4, 5, 6, 8 when n = 4;

  4. nk ≤ 2n when n ≥ 5.

Proof

(i) and (ii) are easy to verify.

(iii) Denote by f(A) the number of 1’s in A. Then for symmetric AΩ4{0, 1}, f(A) and f(A) are even with f(A)+f(A) = 12. Since A and A are mutually complementary, we may suppose f(A) ≤ 6.

If f(A) = 0, then A = O4 and thus r(A)+r(A) = 0 + 4 = 4.

If f(A) = 2, under permutation similarity, it suffices to consider the case A = E12+E21. A direct computation shows that r(A)+r(A) = 2 + 3 = 5.

If f(A) = 4, under permutation similarity, it suffices to consider the cases A = E12+E21+E13+E31 and A = E12+E21+E34+E43. A direct computation shows that r(A)+r(A) = 2+4 = 6 in the first case and r(A)+r(A) = 4 + 2 = 6 in the second case.

If f(A) = 6, under permutation similarity, it suffices to consider the cases A = E12+E21+E13+E31+E23+E32, A = E12+E21+E13+E31+E14+E41 and A = E12+E21+E13+E31+E24+E42. A direct computation shows that r(A)+r(A) = 3+2 = 5 in the first case, r(A)+r(A) = 2+3 = 5 in the second case and r(A)+r(A) = 4 + 4 = 8 in the third case.

Therefore, r(A)+r(A) can only be 4, 5, 6, 8 for symmetric AΩ4{0, 1}.

(iv) The necessity has been proved in Theorem 3.2 (ii). Now we prove the sufficiency.

By Lemma 3.3 and the proof of Theorem 3.5 (iii), k can be n − 1+p for 1 ≤ pn − 1.

By Lemma 3.4 (i), for odd n ≥ 5, r(H+E1,n12+En12,1)+r(H¯E1,n12En12,1)=n+(n1)=2n1. By Lemma 3.4 (ii), for even n ≥ 6, r(H+E1,n2+En2,1+En21,n2+1+En2+1,n21)+r(H¯E1,n2En2,1En21,n2+1En2+1,n21) = (n − 1)+n = 2n − 1. Thus k can be 2n − 1 when n ≥ 5.

By Lemma 3.4 (i), for odd n ≥ 7, r(H+E1,n12+En12,1En+32,nEn,n+32)+r(H¯E1,n12En12,1+En+32,n+En,n+32) = n+n = 2n. By Lemma 3.4 (ii), for even n ≥ 4, r(H+E1,n2+En2,1)+r(H¯E1,n2En2,1)=n+n=2n. Note that r0011000011100011100001100+r0100110100010100010110010=5+5=10.

Thus k can be 2n when n ≥ 4.

Then for symmetric AΩn{0, 1} with n ≥ 5, r(A)+r(A) can be n, n+1, …, 2n. □

4 Conclusion

This paper considers two kinds of complement matrices Ac and A of a {0, 1}-matrix A. If A is a square {0, 1}-matrix with each diagonal entry being 0, then A and its complement A correspond to a strict digraph D and its complement D. We mainly discuss their rank relations. As is shown in the proof, we construct a {0, 1}-matrix A for each possible value of r(A) ± r(Ac) and r(A) ± r(A) in both general and symmetric cases.

Acknowledgement

The authors are grateful to the anonymous referees for their valuable comments and suggestions, which helped improve the original manuscript of this paper. This research was supported by the National Natural Science Foundation of China (Grant Nos. 11601322, 11661041, 61573240, 11661040), the Program of Qingjiang Excellent Young Talents, Jiangxi University of Science and Technology (JXUSTQJYX2017007).

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About the article

Received: 2017-09-04

Accepted: 2017-12-31

Published Online: 2018-03-20


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 190–195, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0020.

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© 2018 Ma and Zhong, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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