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### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Possible numbers of x’s in an {x, y}-matrix with a given rank

Chao Ma
Published Online: 2017-07-21 | DOI: https://doi.org/10.1515/math-2017-0079

## Abstract

Let x, y be two distinct real numbers. An {x, y}-matrix is a matrix whose entries are either x or y. We determine the possible numbers of x’s in an {x, y}-matrix with a given rank. Our proof is constructive.

Keywords: {x, y}-matrix; Rank; Number of x’s

MSC 2010: 15A03; 05B20

## 1 Introduction

A matrix whose entries are either 0 or 1 is called a {0, 1}-matrix. Such matrices arise frequently in combinatorics and graph theory [1-4]. They also have applications in econometrics, probability and statistics [5-7]. In , Hu, Li and Zhan determined the possible numbers of 1’s in a square {0, 1}-matrix with a given rank.

Let x, y be two distinct real numbers. We call a matrix whose entries are either x or y an {x, y}-matrix. Note that {0, 1}-matrices are special {x, y}-matrices. Naturally we have the following generalized problem.

#### Problem

Determine the possible numbers of x’s in an {x, y}-matrix (not necessarily square) with a given rank.

Since multiplying a matrix by a nonzero number does not change its rank, it suffices to consider three kinds of {x, y}-matrices: {0, 1}-matrices, {−1, 1}-matrices and {x, 1}-matrices (−1 < x < 1 and x ≠ 0). Based on these three cases, we will solve the problem in Sections 2-4 respectively.

Denote by Jp,q the p × q matrix all of whose entries are equal to 1, by Ei,j the m × n matrix with its entry in the i-th row and j-th column being 1 and with all other entries being 0. Let A be an m × n matrix and let i1, i2,..., ip, j1, j2,..., jq be integers with 1 ≤ i1 < i2 < ... < ipm, 1 ≤ j1 < j2 < ... < jqn. Denote by A[i1, i2,..., ip|j1, j2,..., jq] the p × q submatrix of A that lies in the rows i1, i2,..., ip and columns j1, j2,..., jq. We abbreviate A[i1, i2,..., ip|i1, i2,..., ip] to A[i1, i2,..., ip]. The symbol A(i, j) denotes the entry of A in the i-th row and j-th column. If A is an {x, 1}-matrix (x ≠ 1), we denote by f(A) the number of 1’s in A. For simplicity we use 0 to denote the zero matrix whose size will be clear from the context.

## 2 {0, 1}-Matrices

#### Theorem 2.1

Let m, n, r be positive integers with r ≤ min{m, n}. Then there exists an m × n {0, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = pq for some integers p, q with 1 ≤ pm, 1 ≤ qn when r = 1;

2. rkmnr + 1 when r ≥ 2.

#### Proof

1. The sufficiency is clear by considering the m × n {0, 1}-matrix $[Jp,q000].$ Now we prove the necessity. Let A be an m × n {0, 1}-matrix with rank A = 1. Then f(A) > 0. By row or column permutations if necessary, without loss of generality we may suppose the first row of A is [J1,q 0], where 1 ≤ qn. For i = 2, 3,... , m, since rank A = 1, the i-th row of A is either a zero row or equal to the first row. Suppose A has p rows equal to [J1,q 0]. Then 1 ≤ pm and f(A) = pq.

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

Let ${A}_{1}=\left[\begin{array}{ll}{I}_{r}& 0\\ 0& 0\end{array}\right]$ be of size m × n, where Ir denotes the identity matrix of order r. Then rank A1 = r and f(A1) = r. Note that the matrices obtained from A1 by successively replacing the 0’s in the strictly upper triangular part of A1 [1, 2,..., r] with 1’s still have rank r. Finally A1 becomes a {0, 1}-matrix with exactly r(r + 1)/2 entries equal to 1 and we denote this matrix by A2.

Clearly, the matrices obtained from A2 by successively replacing the 0’s in A2[1, 2,..., r|r + 1, r + 2,..., n] with 1’s still have rank r. Finally A2 becomes a {0, 1}-matrix with exactly rnr(r − 1)/2 entries equal to 1 and we denote this matrix by A3.

It is easy to verify that the matrices obtained from A3 by successively replacing the 0’s in the strictly lower triangular part of A3[2, 3,..., r|1, 2,..., r − 1] with 1’s still have rank r. Finally A3 becomes a {0, 1}-matrix with exactly rnr + 1 entries equal to 1 and we denote this matrix by A4. In fact, $A4=[Jr,n0]−∑i=1r−1Ei+1,i.$ Thus for every integer k with rkrnr + 1, there exists an m × n {0, 1}-matrix of rank r with exactly k 1’s.

Let $A5=[Jr+s,n0]−∑i=1r−1Ei+1,i+Er+s+1,1,0<_s<_m−r−1.$ Then rank A5 = r and f(A5) = (r + s)nr + 2.

Let $A6=[Jr+s,n0]−∑i=2r−1Ei+1,i+∑i=1tEr+s+1,i,0<_s<_m−r−1,1<_t<_n−1.$ Then rank A6 = rank(A6[2, 3, . . . , r, r + s + 1|1, 2, . . . , r − 1, n]) = r and f(A6) = (r + s)nr + 2 + t.

Note that {(r + s)nr + 2 : s = 0, 1,..., mr − 1} ∪ {(r + s)nr + 2 + t : s = 0, 1,..., mr − 1; t = 1, 2,..., n − 1} = {rnr+2, rnr+3,..., mnr+1}. Thus for every integer k with rnr+2 ≤ kmnr+1, there exists an m × n {0, 1}-matrix of rank r with exactly k 1’s. This completes the proof of sufficiency. ☐

#### Corollary 2.2

(). Let n, r be positive integers with rn. Then there exists an n × n {0, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = pq for some integers p, q with 1 ≤ p, qn when r = 1;

2. rkn2r + 1 when r ≥ 2.

## 3 {x, 1}-Matrices (-1 < x < 1 and x ≠ 0)

#### Theorem 3.1

Let m, n, r be positive integers with r ≤ min{m, n} and let x be a nonzero real number with −1 < x < 1. Then there exists an m × n {x, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = np or mq for some integers p, q with 0 ≤ pm, 0 ≤ qn when r = 1;

2. r − 1 ≤ kmnr + 1 when r ≥ 2.

#### Proof

1. The sufficiency is clear by considering the {x, 1}-matrices $[Jp,nxJm−p,n]and[Jm,qxJm,n−q].$

Now we prove the necessity. Let A be an m × n {x, 1}-matrix with rank A = 1.

If f(A) = 0, A = xJm,n.

If f(A) > 0, by row or column permutations if necessary, we may suppose the first row of A is [J1,q xJ1,nq], where 1 ≤ qn. We distinguish two cases.

#### Case 1

q = n. Then the first row of A is J1,n. Since rank A = 1, for i = 2, 3,... , m, the i-th row of A is either J1,n or xJ1,n. Suppose A has p rows equal to J1,n. Then 1 ≤ pm and f(A) = np.

#### Case 2

1 ≤ q < n. Assume that A(i, 1) = x for some integer i with 2 ≤ im. Since rank A = 1, the i-th row of A is [xJ1,q x2J1,nq]. Since −1 < x < 1 and x ≠ 0, x2 ∉ {x, 1}. This contradicts the assumption that A is an {x, 1}-matrix. Thus A(i, 1) = 1 and the i-th row of A is equal to the first row for i = 2, 3,... ,m. Then f(A) = mq.

• ii. The necessity is clear. Now we prove the sufficiency.

Let ${A}_{1}=\left[\begin{array}{lll}x{J}_{r-1}+\left(1-x\right){I}_{r-1}& & \phantom{\rule{4mm}{0ex}}x{J}_{r-1,n-r+1}\\ \phantom{\rule{6mm}{0ex}}x{J}_{m-r+1,r-1}& & x{J}_{m-r+1,n-r+1}\end{array}\right],$ where ${J}_{r-1}\stackrel{\mathrm{△}}{=}{J}_{r-1,r-1}.$ Then rank A1 = rank(A1[1, 2,... , r]) = r and f(A1) = r − 1. It is easy to verify that the matrices obtained from A1 by successively replacing the x’s in the strictly upper triangular part of A1[1, 2,... , r − 1] with 1’s still have rank r. Finally A1 becomes an {x, 1}-matrix with exactly r(r − 1)/2 entries equal to 1 and we denote this matrix by A2.

Note that the matrices obtained from A2 by successively replacing the x’s in A2[1, 2,... , r−1|r, r + 1,..., n−1] with 1’s still have rank r. Finally A2 becomes an {x, 1}-matrix with exactly (r − 1)nr(r − 1)/2 entries equal to 1 and we denote this matrix by A3. And it is easy to verify that the matrices obtained from A3 by successively replacing A3(2, n), A3(3, n),..., A(r − 1, n) with 1’s still have rank r. Finally A3 becomes an {x, 1}-matrix with exactly (r − 1)n − (r − 1)(r − 2)/2 − 1 entries equal to 1 and we denote this matrix by A4.

It is also easy to verify that the matrices obtained from A4 by successively replacing the x’s in the strictly lower triangular part of A4[2, 3,..., r − 1|1, 2,... , r − 2] with 1’s still have rank r. Finally A4 becomes an {x, 1}-matrix with exactly (r − 1)nr + 1 entries equal to 1 and we denote this matrix by A5. In fact, $A5=[Jr−1,nxJm−r+1,n]+(x−1)(∑i=1r−2Ei+1,i+E1,n).$ Thus for every integer k with r − 1 ≤ k ≤ (r − 1)nr + 1, there exists an m × n {x, 1}-matrix of rank r with exactly k 1’s.

Let $A6=[Jr+s−1,nxJm−r−s+1,n]+(x−1)(∑i=1r−2Ei+1,i+Er−1,n)+(1−x)Er+s,n,0<_s<_m−r.$ Then rank A6 = rank(A6[1, 2,..., r − 1, r + s|1, 2,..., r − 1, n]) = r and f(A6) = (r + s − 1)nr + 2.

Let $A7=[Jr+s−1,nxJm−r−s+1,n]+(x−1)∑i=1r−2Ei+1,i+(1−x)∑i=1tEr+s,n−i,0<_s<_m−r,1<_t<_n−1.$ Then rank A7 = rank(A7[1, 2,..., r − 1, r + s|1, 2,..., r − 2, n − 1, n]) = r and f(A7) = (r + s − 1)nr + 2 + t.

Note that {(r + s − 1)nr + 2 : s = 0, 1,... , mr} ∪ {(r + s − 1)nr + 2 + t : s = 0, 1,..., mr; t = 1, 2,... , n − 1} = {(r − 1)nr + 2, (r − 1)nr + 3,..., mnr + 1}. Thus for every integer k with (r − 1)nr + 2 ≤ kmnr + 1, there exists an m × n {x, 1}-matrix of rank r with exactly k 1’s. This completes the proof of sufficiency. ☐

#### Corollary 3.2

Let n, r be positive integers with rn and let x be a nonzero real number with −1 < x < 1. Then there exists an n × n {x, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = np for some integer p with 0 ≤ pn when r = 1;

2. r − 1 ≤ kn2r + 1 when r ≥ 2.

## 4 {-1, 1}-Matrices

#### Theorem 4.1

Let m, n, r be positive integers with r ≤ min{m, n}. Then there exists an m × n {−1, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = pq + (mp)(nq) for some integers p, q with 0 ≤ pm, 0 ≤ qn when r = 1;

2. r − 1 ≤ kmnr + 1 when r ≥ 2.

#### Proof

1. The sufficiency is clear by considering the {−1, 1}-matrix $[Jp,q−Jp,n−q−Jm−p,qJm−p,n−q].$

Now we prove the necessity. Let A be an m × n {−1, 1}-matrix with rank A = 1. We distinguish two cases.

#### Case 1

f(A) = 0. Then A = −Jm,n.

#### Case 2

f(A) > 0. By row or column permutations if necessary, we may suppose the first row of A is [J1,qJ1,nq], where 1 ≤ qn. Since rank A = 1, if A(i, 1) = 1 for some integer i with 2 ≤ im, then the i-th row of A is equal to the first row. Otherwise the i-th row of A is [−J1,q J1,nq]. Suppose A has p rows equal to [J1,qJ1,nq]. Then 1 ≤ pm and f(A) = pq + (mp)(nq).

• ii. The proof is the same as Theorem 3.1 (ii). Here we let x = − 1. ☐

#### Corollary 4.2

Let n, r be positive integers with rn. Then there exists an n × n {−1, 1}-matrix of rank r with exactly k 1’s if and only if

1. k = pq + (np)(nq) for some integers p, q with 0 ≤ p, qn when r = 1;

2. r − 1 ≤ kn2r + 1 when r ≥ 2.

## Acknowledgement

The author is grateful to Professor Xingzhi Zhan for helpful suggestions. This research was supported by the National Natural Science Foundation of China (Grant Nos. 11601322, 61573240).

## References

• 

Konvalina J., Liu Y.H., Zero-one matrices without consecutive ones, Appl. Math. Lett., 1991, 4, 35-37

• 

Kündgen A., Leander G., Thomassen C., Switchings, extensions, and reductions in central digraphs, J. Combin. Theory Ser. A, 2011, 118, 2025-2034

• 

McKay B.D., Wang X., Asymptotic enumeration of 0-1 matrices with equal row sums and equal column sums, Linear Algebra Appl., 2003, 373, 273-287

• 

Spinrad J.P., Doubly lexical ordering of dense 0-1 matrices, Inf. Process. Lett., 1993, 45, 229-235

• 

Chen Y., Diaconis P., Holmes S.P., Liu J.S., Sequential Monte Carlo methods for statistical analysis of tables, J. Amer. Statist. Assoc., 2005, 100, 109-120

• 

Louchard G., Prodinger H., Random 0-1 rectangular matrices: A probabilistic analysis, Period. Math. Hungar., 2003, 47, 169-193

• 

Magnus J.R., Neudecker H., Symmetry, 0-1 matrices and Jacobians: A review, Econometric Theory, 1986, 2, 157-190

• 

Hu Q., Li Y., Zhan X., Possible numbers of ones in 0-1 matrices with a given rank, Linear Multilinear Algebra, 2005, 53, 435-443

Accepted: 2017-06-05

Published Online: 2017-07-21

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 974–977, ISSN (Online) 2391-5455,

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