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

Chao Ma; Jin Zhong

doi:10.1515/math-2018-0020

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

Chao Ma and Jin Zhong

Published/Copyright: March 20, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

Abstract

Let A be a {0, 1}-matrix and r(A) denotes its rank. The complement matrix of A is defined and denoted by A^c = J − A, 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 = J − I − A, where I is the identity matrix. We determine the possible values of r(A) ± r(A^c) 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 M_{m, n}{0, 1} the set of m × n {0, 1}-matrices, and we abbreviate M_{n, n}{0, 1} as M_n{0, 1}. Let A ∈ M_{m, n}{0, 1}. Then the matrix A^c = J_{m, n} − A is called the complement of A, where J_{m, n} is the m × n matrix with each entry being 1. It is clear that A and A^c are mutually complementary, i.e., (A^c)^c = A.

Recall that the adjacency matrix of a digraph D is the square matrix A = (a_ij), where a_ij 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 J_n − I_n − A, where J_n = J_{n, n} and I_n is the identity matrix of order n. Denote by Ω_n{0, 1} the set of n × n {0, 1}-matrices with each diagonal entry being 0. Thus for A ∈Ω_n{0, 1}, we define another kind of complement matrix of A to be A = J_n − I_n − A. 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(A^c) for A ∈ M_{m, 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 A ∈Ω_n{0, 1} in the general case and in the symmetric case.

We use O_{m, n} to denote the m × n zero matrix. O_{n, n} will be abbreviated as O_n. Denote by E_ij 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 A^c

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

Theorem 2.1

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

Proof

Since r(A^c) = r(J_{m, n} − A) ≤ r(J_{m, n})+r(A) = 1+r(A), r(A^c) − r(A) ≤ 1. Likewise, r(A) − r(A^c) ≤ 1. This proves the necessity.

For the sufficiency, note that r(A) − r(A^c) = ± 1 if A = O_{m, n} or J_{m, n}, and r(A) = r(A^c) if A=[Jm,1Om,n−1]. This completes the proof. □

Theorem 2.2

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

Proof

The necessity is clear. Now we prove the sufficiency.

Suppose m ≤ n. Let A1=IpOp,n−pOm−p,pOm−p,n−p∈Mm,n{0,1} with 0 ≤ p ≤ m − 1. Clearly r(A₁) = p. It is easy to verify that A1c=Jp−IpJp,n−pJm−p,pJm−p,n−p has rank p+1. Then r(A₁)+r( A1c) = 2p+1, 0 ≤ p ≤ m − 1. Thus for every odd k with 1 ≤ k ≤ 2m − 1, there exists A ∈ M_{m, n}{0, 1} such that r(A)+r(A^c) = k.

Let A2=IqOq,m−qOq,n−mOm−q,qJm−qOm−q,n−m∈Mm,n{0,1} with 1 ≤ q ≤ m − 1. Clearly r(A₂) = q+1. It is easy to verify that A2c=Jq−IqJq,m−qJq,n−mJm−q,qOm−qJm−q,n−m has rank q+1. Then r(A₂)+r( A2c) = 2q+2, 1 ≤ q ≤ m − 1. Thus for every even k with 4 ≤ k ≤ 2m, there exists A ∈ M_{m, n}{0, 1} such that r(A)+r(A^c) = k.

Let A₃ = Jm,1Om,n−1 ∈ M_{m, n}{0, 1}. Then r(A₃)+r( A3c) = 1+1 = 2.

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

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

Next we consider the case when A ∈ M_n{0, 1} is symmetric.

Theorem 2.3

Let n ≥ 2 be a positive integer. Then there exists symmetric A ∈ M_n{0, 1} with r(A) − r(A^c) = 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(A^c) = −1, 0, 1 if A = O_n, I_n, J_n, respectively. □

Lemma 2.4

Let A ∈ M_n{0, 1} be symmetric. Then r(A)+r(A^c) ≠ 2.

Proof

Assume that there exists symmetric A ∈ M_n{0, 1} such that r(A)+r(A^c) = 2. If r(A) = 0, then A = O_n and thus A^c = J_n, which contradicts the assumption that r(A)+r(A^c) = 2. If r(A) = 2, then r(A^c) = 0, A^c = O_n. This implies A = J_n 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,n−pOn−p,pOn−p with 1 ≤ p ≤ n. Thus A^c is permutation similar to OpJp,n−pJn−p,pJn−p, which implies that r(A^c) = 0 when p = n and r(A^c) = 2 when 1 ≤ p ≤ n − 1. Then r(A)+r(A^c) = 1 or 3, a contradiction. Thus r(A)+r(A^c) ≠ 2 for any symmetric A ∈ M_n{0, 1}. □

Note that the matrices A₁, A₂ ∈ M_{m, 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 A ∈ M_n{0, 1} with r(A)+r(A^c) = k if and only if 1 ≤ k ≤ 2n with k ≠ 2.

3 Rank relations between A and A

In this section, we only consider A ∈Ω_n{0, 1} which corresponds to the adjacency matrix of a strict digraph. Recall that for an n × n matrix A = (a_ij), the main diagonal of A is the list of entries a₁₁, a₂₂, …, a_nn, and the secondary diagonal of A is the list of entries a_1n, a_{2, n − 1}, …, a_n1. 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 A ∈Ω_n{0, 1} with r(A) − r(A) = k if and only if

k = 0, ±2 when n = 2;
−n ≤ k ≤ n when n ≥ 3.

Proof

The case n = 2 is trivial.
The necessity is clear. Now we prove the sufficiency.
Let B1=C1Jp,n−pOn−p,pOn−p∈Ωn{0,1}, where 1 ≤ p ≤ n − 1. It is easy to verify that r(B₁) = p, r(B₁) = n − 1. Thus for every integer k with −n+2 ≤ k ≤ n − 2, there exists A ∈ Ω_n{0, 1} such that r(A) − r(A) = k.
Let B2=C2On−1,n∈Ωn{0,1}, where C₂ = [0, 1, …, 1, 0]. Then r(B₂) = 1, r(B₂) = n. Note that r(O_n) = 0, r(O_n) = n. Thus for k = ±(n − 1), ± n, there exists A ∈ Ω_n{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 A ∈ Ω_n{0, 1} with r(A)+r(A) = k if and only if

k = 2 when n = 2;
n ≤ k ≤ 2n when n ≥ 3.

Proof

Trivial.
r(A)+r(A) ≤ 2n is clear. Since A+A = J_n − I_n, n = r(A+A) ≤ r(A)+r(A). This proves the necessity.
For the sufficiency, first note that the matrix B₁ ∈ Ω_n{0, 1} in the proof of Theorem 3.1, (ii) implies that for every integer k with n ≤ k ≤ 2n − 2, there exists A ∈ Ω_n{0, 1} such that r(A)+r(A) = k.
Let B₃ = C₁ − E_1n ∈ Ω_n{0, 1}. Then r(B₃) = n − 1, r(B₃) = n.
Let B₄ = B₃+E_n1 ∈ Ω_n{0, 1}. Then r(B₄) = r(B₄) = n.
Thus for k = 2n − 1, 2n, there exists A ∈ Ω_n{0, 1} such that r(A)+r(A) = k. This proves the sufficiency. □
Next we consider the case when A ∈ Ω_n{0, 1} is symmetric.

Lemma 3.3

LetG=G1Jp,n−pJn−p,pJn−p−In−p ∈ Ω_n{0, 1} with 1 ≤ p ≤ n − 1, where G₁is 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 = G3Jp−1,n−pJn−p,p−1Jn−p−In−p∈Ωn−1{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 H ∈ Ω_n{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

r(H+E1,n−12+En−12,1)=nfor odd n ≥ 5, r(H+E1,n−12+En−12,1−En+32,n−En,n+32)=nfor odd n ≥ 7; r(H¯−E1,n−12−En−12,1+En+32,n+En,n+32)=n
r(H+E1,n2+En2,1)=nfor even n ≥ 4, r(H+E1,n2+En2,1+En2−1,n2+1+En2+1,n2−1)=n−1for even n ≥ 6.

Proof

When n ≥ 5 is odd, n−12>1. Then E1,n−12≠En−12,1. Using Laplace expansion formula, it is easy to verify that det(H+E1,n−12+En−12,1)≠0 for odd n ≥ 5 and det(H+E1,n−12+En−12,1−En+32,n−En,n+32)≠0 for odd n ≥ 7.
When n ≥ 7 is odd, subtract the second row of H¯−E1,n−12−En−12,1+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,n−12−En−12,1+En+32,n+En,n+32)≠0.
When n ≥ 4 is even, n2>1. Then E1,n2≠En2,1. 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+En2−1,n2+1+En2+1,n2−1 is the sum of the first row (column) and the ( n2 − 1)-th row (column). Then r(H+E1,n2+En2,1+En2−1,n2+1+En2+1,n2−1) ≤ n − 1. Using Laplace expansion formula and the fact that J₃ − I₃ is nonsingular, we can deduce that the submatrix of H+E1,n2+En2,1+En2−1,n2+1+En2+1,n2−1 obtained by deleting the ( n2)-th row and the (n2+2)-th column is nonsingular. Thus r(H+E1,n2+En2,1+En2−1,n2+1+En2+1,n2−1)=n−1.

□

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 A ∈ Ω_n{0, 1} with r(A) − r(A) = k if and only if

k = ±2 when n = 2;
k = 0, ±3 when n = 3;
−n ≤ k ≤ n 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 −n ≤ r(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 = O_n and thus A = J_n − I_n is nonsingular, a contradiction. For the later case, note that any nonzero symmetric A ∈ Ω_n{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 G₁ ∈ Ω_p{0, 1} and G, H ∈ Ω_n{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 ≤ p ≤ n − 2, r(G) − r(G) can be 3, 5, 7, …, n − 2, n; for even p with 2 ≤ p ≤ n − 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 ≤ p ≤ n − 1, r(G) − r(G) can be 2, 4, 6, …, n − 2, n; for even p with 2 ≤ p ≤ n − 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,n−12+En−12,1)=n. Since r(H+E1,n−12+En−12,1¯)=r(H¯−E1,n−12−En−12,1) = n − 1 when n is odd, r(H+E1,n−12+En−12,1)−r(H¯−E1,n−12−En−12,1)=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,n2−En2,1)=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 A ∈ Ω_n{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 A ∈ Ω_n{0, 1} with r(A)+r(A) = k if and only if

k = ±2 when n = 2;
k = 3, 4 when n = 3;
k = 4, 5, 6, 8 when n = 4;
n ≤ k ≤ 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 A ∈ Ω₄{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 A ∈ Ω₄{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 ≤ p ≤ n − 1.

By Lemma 3.4 (i), for odd n ≥ 5, r(H+E1,n−12+En−12,1)+r(H¯−E1,n−12−En−12,1)=n+(n−1)=2n−1. By Lemma 3.4 (ii), for even n ≥ 6, r(H+E1,n2+En2,1+En2−1,n2+1+En2+1,n2−1)+r(H¯−E1,n2−En2,1−En2−1,n2+1−En2+1,n2−1) = (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,n−12+En−12,1−En+32,n−En,n+32)+r(H¯−E1,n−12−En−12,1+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,n2−En2,1)=n+n=2n. Note that

r0011000011100011100001100+r0100110100010100010110010=5+5=10.

Thus k can be 2n when n ≥ 4.

Then for symmetric A ∈ Ω_n{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 A^c 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(A^c) 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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Received: 2017-09-04

Accepted: 2017-12-31

Published Online: 2018-03-20

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https://doi.org/10.1515/math-2018-0020

Keywords for this article

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

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

Article

Abstract

1 Introduction

2 Rank relations between A and Ac

Theorem 2.1

Proof

Theorem 2.2

Proof

Theorem 2.3

Proof

Lemma 2.4

Proof

Theorem 2.5

3 Rank relations between A and A

Theorem 3.1

Proof

Theorem 3.2

Proof

Lemma 3.3

Proof

Lemma 3.4

Proof

Theorem 3.5

Proof

Theorem 3.6

Proof

4 Conclusion

Acknowledgement

References

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2 Rank relations between A and A^c