## 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.

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

#### Open Access

## Abstract

## 1 Introduction

#### Problem

## 2 {0, 1}-Matrices

#### Theorem 2.1

#### Proof

#### Corollary 2.2

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

#### Theorem 3.1

#### Proof

#### Case 1

#### Case 2

#### Corollary 3.2

## 4 {-1, 1}-Matrices

#### Theorem 4.1

#### Proof

#### Case 1

#### Case 2

#### Corollary 4.2

## Acknowledgement

## References

## About the article

More options …# Open Mathematics

### formerly Central European Journal of Mathematics

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

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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

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 [8], 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.

*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 *J*_{p,q} the *p* × *q* matrix all of whose entries are equal to 1, by *E*_{i,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 *i*_{1}, *i*_{2},..., *i _{p}*,

*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*

*k*=*pq for some integers p*,*q with*1 ≤*p*≤*m*, 1 ≤*q*≤*n when r*= 1;*r*≤*k*≤*mn*−*r*+ 1*when r*≥ 2.

The sufficiency is clear by considering the

*m*×*n*{0, 1}-matrix $$\left[\begin{array}{ll}{J}_{p,q}& 0\\ 0& 0\end{array}\right].$$ 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 [*J*_{1,q}0], where 1 ≤*q*≤*n*. 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 [*J*_{1,q}0]. Then 1 ≤*p*≤*m*and*f*(*A*) =*pq*.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 *I _{r}* denotes the identity matrix of order

Clearly, the matrices obtained from *A*_{2} by successively replacing the 0’s in *A*_{2}[1, 2,..., *r*|*r* + 1, *r* + 2,..., *n*] with 1’s still have rank *r*. Finally *A*_{2} becomes a {0, 1}-matrix with exactly *rn* − *r*(*r* − 1)/2 entries equal to 1 and we denote this matrix by *A*_{3}.

It is easy to verify that the matrices obtained from *A*_{3} by successively replacing the 0’s in the strictly lower triangular part of *A*_{3}[2, 3,..., *r*|1, 2,..., *r* − 1] with 1’s still have rank *r*. Finally *A*_{3} becomes a {0, 1}-matrix with exactly *rn* − *r* + 1 entries equal to 1 and we denote this matrix by *A*_{4}. In fact,
$${A}_{4}=\left[\begin{array}{l}{J}_{r,n}\\ \phantom{\rule{2.3mm}{0ex}}0\end{array}\right]-\sum _{i=1}^{r-1}{E}_{i+1,i}.$$
Thus for every integer *k* with *r* ≤ *k* ≤ *rn* − *r* + 1, there exists an *m* × *n* {0, 1}-matrix of rank *r* with exactly *k* 1’s.

Let
$${A}_{5}=\left[\begin{array}{l}{J}_{r+s,n}\\ \phantom{\rule{5mm}{0ex}}0\end{array}\right]-\sum _{i=1}^{r-1}{E}_{i+1,i}+{E}_{r+s+1,1},0\underset{\_}{<}s\underset{\_}{<}m-r-1.$$
Then rank *A*_{5} = *r* and *f*(*A*_{5}) = (*r* + *s*)*n* − *r* + 2.

Let
$${A}_{6}=\left[\begin{array}{l}{J}_{r+s,n}\\ \phantom{\rule{5mm}{0ex}}0\end{array}\right]-\sum _{i=2}^{r-1}{E}_{i+1,i}+\sum _{i=1}^{t}{E}_{r+s+1,i},0\underset{\_}{<}s\underset{\_}{<}m-r-1,1\underset{\_}{<}t\underset{\_}{<}n-1.$$
Then rank *A*_{6} = rank(*A*_{6}[2, 3, . . . , *r*, *r* + *s* + 1|1, 2, . . . , *r* − 1, *n*]) = *r* and *f*(*A*_{6}) = (*r* + *s*)*n* − *r* + 2 + *t*.

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

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

*k*=*pq for some integers p*,*q with*1 ≤*p*,*q*≤*n when r*= 1;*r*≤*k*≤*n*^{2}−*r*+ 1*when r*≥ 2.

*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*

*k*=*np or mq for some integers p*,*q with*0 ≤*p*≤*m*, 0 ≤*q*≤*n when r*= 1;*r*− 1 ≤*k*≤*mn*−*r*+ 1*when r*≥ 2.

The sufficiency is clear by considering the {

*x*, 1}-matrices $$\left[\begin{array}{l}\phantom{\rule{4.2mm}{0ex}}{J}_{p,n}\\ x{J}_{m-p,n}\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}[{J}_{m,q}\phantom{\rule{2mm}{0ex}}x{J}_{m,n-q}].$$

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

If *f*(*A*) = 0, *A* = *xJ*_{m,n}.

If *f*(*A*) > 0, by row or column permutations if necessary, we may suppose the first row of *A* is [*J*_{1,q} *xJ*_{1,n−q}], where 1 ≤ *q* ≤ *n*. We distinguish two cases.

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

1 ≤ *q* < *n*. Assume that *A*(*i*, 1) = *x* for some integer *i* with 2 ≤ *i* ≤ *m*. Since rank *A* = 1, the *i*-th row of *A* is [*xJ*_{1,q} *x*^{2}*J*_{1,n−q}]. Since −1 < *x* < 1 and *x* ≠ 0, *x*^{2} ∉ {*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 ${A}_{1}=\left[\begin{array}{lll}x{J}_{r-1}+(1-x){I}_{r-1}& & \phantom{\rule{4mm}{0ex}}x{J}_{r-1,n-r+1}\\ \phantom{\rule{6mm}{0ex}}x{J}_{m-r+1,r-1}& & x{J}_{m-r+1,n-r+1}\end{array}\right],$ where ${J}_{r-1}\stackrel{\mathrm{\u25b3}}{=}{J}_{r-1,r-1}.$ Then rank *A*_{1} = rank(*A*_{1}[1, 2,... , *r*]) = *r* and *f*(*A*_{1}) = *r* − 1. It is easy to verify that the matrices obtained from *A*_{1} by successively replacing the *x*’s in the strictly upper triangular part of *A*_{1}[1, 2,... , *r* − 1] with 1’s still have rank *r*. Finally *A*_{1} becomes an {*x*, 1}-matrix with exactly *r*(*r* − 1)/2 entries equal to 1 and we denote this matrix by *A*_{2}.

Note that the matrices obtained from *A*_{2} by successively replacing the *x*’s in *A*_{2}[1, 2,... , *r*−1|*r*, *r* + 1,..., *n*−1] with 1’s still have rank *r*. Finally *A*_{2} becomes an {*x*, 1}-matrix with exactly (*r* − 1)*n* − *r*(*r* − 1)/2 entries equal to 1 and we denote this matrix by *A*_{3}. And it is easy to verify that the matrices obtained from *A*_{3} by successively replacing *A*_{3}(2, *n*), *A*_{3}(3, *n*),..., *A*(*r* − 1, *n*) with 1’s still have rank *r*. Finally *A*_{3} 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 *A*_{4}.

It is also easy to verify that the matrices obtained from *A*_{4} by successively replacing the *x*’s in the strictly lower triangular part of *A*_{4}[2, 3,..., *r* − 1|1, 2,... , *r* − 2] with 1’s still have rank *r*. Finally *A*_{4} becomes an {*x*, 1}-matrix with exactly (*r* − 1)*n* − *r* + 1 entries equal to 1 and we denote this matrix by *A*_{5}. In fact,
$${A}_{5}=\left[\begin{array}{l}\phantom{\rule{4.5mm}{0ex}}{J}_{r-1,n}\\ x{J}_{m-r+1,n}\end{array}\right]+(x-1)(\sum _{i=1}^{r-2}{E}_{i+1,i}+{E}_{1,n}).$$
Thus for every integer *k* with *r* − 1 ≤ *k* ≤ (*r* − 1)*n* − *r* + 1, there exists an *m* × *n* {*x*, 1}-matrix of rank *r* with exactly *k* 1’s.

Let
$${A}_{6}=\left[\begin{array}{l}\phantom{\rule{4.5mm}{0ex}}{J}_{r+s-1,n}\\ x{J}_{m-r-s+1,n}\end{array}\right]+(x-1)(\sum _{i=1}^{r-2}{E}_{i+1,i}+{E}_{r-1,n})+(1-x){E}_{r+s,n},0\underset{\_}{<}s\underset{\_}{<}m-r.$$
Then rank *A*_{6} = rank(*A*_{6}[1, 2,..., *r* − 1, *r* + *s*|1, 2,..., *r* − 1, *n*]) = *r* and *f*(*A*_{6}) = (*r* + *s* − 1)*n* − *r* + 2.

Let
$${A}_{7}={\displaystyle \left[\begin{array}{l}\phantom{\rule{4.5mm}{0ex}}{J}_{r+s-1,n}\\ x{J}_{m-r-s+1,n}\end{array}\right]+(x-1)\sum _{i=1}^{r-2}{E}_{i+1,i}+(1-x)\sum _{i=1}^{t}{E}_{r+s,n-i},0\underset{\_}{<}s\underset{\_}{<}m-r,1\underset{\_}{<}t\underset{\_}{<}n-1.}$$
Then rank *A*_{7} = rank(*A*_{7}[1, 2,..., *r* − 1, *r* + *s*|1, 2,..., *r* − 2, *n* − 1, *n*]) = *r* and *f*(*A*_{7}) = (*r* + *s* − 1)*n* − *r* + 2 + *t*.

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

*Let n*, *r be positive integers with r* ≤ *n 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*

*k*=*np for some integer p with*0 ≤*p*≤*n when r*= 1;*r*− 1 ≤*k*≤*n*^{2}−*r*+ 1*when r*≥ 2.

*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*

*k*=*pq*+ (*m*−*p*)(*n*−*q*)*for some integers p*,*q with*0 ≤*p*≤*m*, 0 ≤*q*≤*n when r*= 1;*r*− 1 ≤*k*≤*mn*−*r*+ 1*when r*≥ 2.

The sufficiency is clear by considering the {−1, 1}-matrix $$\left[\begin{array}{ll}\phantom{\rule{6mm}{0ex}}{J}_{p,q}& -{J}_{p,n-q}\\ -{J}_{m-p,q}& {J}_{m-p,n-q}\end{array}\right].$$

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

*f*(*A*) = 0. Then *A* = −*J*_{m,n}.

*f*(*A*) > 0. By row or column permutations if necessary, we may suppose the first row of *A* is [*J*_{1,q} − *J*_{1,n−q}], where 1 ≤ *q* ≤ *n*. Since rank *A* = 1, if *A*(*i*, 1) = 1 for some integer *i* with 2 ≤ *i* ≤ *m*, then the *i*-th row of *A* is equal to the first row. Otherwise the *i*-th row of *A* is [−*J*_{1,q} *J*_{1,n−q}]. Suppose *A* has *p* rows equal to [*J*_{1,q} − *J*_{1,n−q}]. Then 1 ≤ *p* ≤ *m* and *f*(*A*) = *pq* + (*m* − *p*)(*n* − *q*).

ii. The proof is the same as Theorem 3.1 (ii). Here we let

*x*= − 1. ☐

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

*k*=*pq*+ (*n*−*p*)(*n*−*q*)*for some integers p*,*q with*0 ≤*p*,*q*≤*n when r*= 1;*r*− 1 ≤*k*≤*n*^{2}−*r*+ 1*when r*≥ 2.

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).

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**Received**: 2017-02-21

**Accepted**: 2017-06-05

**Published Online**: 2017-07-21

**Citation Information: **Open Mathematics, Volume 15, Issue 1, Pages 974–977, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0079.

© 2017 Ma. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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