## Abstract

We prove that a positive matrix with all permutation products equal is diagonally equivalent to *J*, the all-1s matrix. Then we give a simple proof of the rank inequality for diagonally magic matrices (J. Inequal. Appl. 2015:318, 2015).

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# A note on the rank inequality for diagonally magic matrices

## Abstract

## Introduction

## Main result

### Theorem 2.1

### Theorem 2.2

### Proof

### Proof of Theorem 2.1

## References

## Acknowledgements

## Author information

### Authors and Affiliations

### Corresponding author

## Additional information

### Competing interests

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### Keywords

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*Journal of Inequalities and Applications*
**volume 2016**, Article number: 66 (2016)

We prove that a positive matrix with all permutation products equal is diagonally equivalent to *J*, the all-1s matrix. Then we give a simple proof of the rank inequality for diagonally magic matrices (J. Inequal. Appl. 2015:318, 2015).

We denote by \(\mathbb{C}^{n\times n}\) and \(\mathbb{R}^{n\times n}\) the sets of \(n \times n\) complex matrices and \(n \times n\) real matrices, respectively. For a positive integer *n*, let \(S_{n}\) be the set of all *n*! permutations of \(\{1, 2, \ldots, n\}\). If \(A=(a_{i,j}) \in\mathbb{C}^{n\times n}\) and \(\sigma\in S_{n}\), then the sequence \(a_{1,\sigma(1)}, a_{2,\sigma(2)}, \ldots, a_{n,\sigma (n)}\) is called the *transversal* of *A* [2]. Let \(A\in \mathbb{C}^{n\times n}\), \(1\leq i_{1} \leq i_{2}\leq\cdots\leq i_{k}\leq n\), and \(1\leq j_{1} \leq j_{2}\leq\cdots\leq j_{s}\leq n\). We denote by \(A[i_{1}, i_{2}, \ldots, i_{k}|j_{1}, j_{2}, \ldots, j_{s}]\) the \(k\times s\) submatrix of *A* that lies in the rows \(i_{1}, i_{2}, \ldots, i_{k}\) and columns \(j_{1}, j_{2}, \ldots, j_{s}\). Denote by \(A(i_{1}, i_{2}, \ldots, i_{k}|j_{1}, j_{2}, \ldots, j_{s})\) the \((n-k)\times(n-s)\) submatrix of *A* obtained by deleting the rows \(i_{1}, i_{2}, \ldots, i_{k}\) and columns \(j_{1}, j_{2}, \ldots, j_{s}\). A matrix \(A=(a_{i,j})\in\mathbb{C}^{n\times n}\) is called *diagonally magic* if

$$ \sum_{i=1}^{n} a_{i,\sigma(i)}=\sum_{i=1}^{n} a_{i,\pi(i)} $$

for all \(\sigma, \pi\in S_{n}\).

Obviously, the zero matrix \(0_{n\times n}\) and \(J=[1]_{n\times n}\), the matrix of all ones, are diagonally magic matrices. In [1], we prove that

$$ B_{n}= \begin{pmatrix} 1&2 & \cdots& n\\ n+1&n+2 & \cdots& 2n\\ \vdots&\vdots& \ddots& \vdots\\ (n-1)n+1&(n-1)n+2 & \cdots& n^{2} \end{pmatrix} $$

and the Henkel matrix

$$ C_{n}= \begin{pmatrix} 1&2 & \cdots& n\\ 2&3 & \cdots& n+1\\ \vdots&\vdots& \ddots& \vdots\\ n&n+1 & \cdots& 2n-1 \end{pmatrix} $$

are diagonally magic matrices. So, there are a lot of diagonally magic matrices. The nonnegative matrices \(B_{n}\) and \(C_{n}\) have been a hot research area [3, 4].

The rank inequality for diagonally magic matrices can be stated as follows.

([1], Theorem 2.1)

*Let*
\(A \in\mathbb{C}^{n\times n}\)
*be a diagonally magic matrix*. *Then*
\(\operatorname {rank}(A)\leq2\).

There are diagonally magic matrices of ranks 0, 1, 2. Indeed, \(\operatorname {rank}(0_{n\times n} )=0\), \(\operatorname {rank}([1]_{n\times n} )=1\), and \(\operatorname {rank}(B_{n})=\operatorname {rank}(C_{n})=2\).

The purpose of this note is to give a simple proof of Theorem 2.1. Our proof depends only on the following fact.

*Let*
\(C =(c_{i,j})\in\mathbb{R}^{n\times n}\)
*be a positive matrix with*

$$\prod_{i=1}^{n}c_{i,\gamma(i)}=\prod _{j=1}^{n}c_{j,\tau(j)} $$

*for all*
\(\gamma, \tau\in S_{n}\). *Then there exist positive diagonal matrices*
\(X=\operatorname {diag}(x_{1},x_{2},\ldots,x_{n})\)
*and*
\(Y= \operatorname {diag}(y_{1},y_{2},\ldots,y_{n})\)
*such that*

$$C=XJY. $$

Let *B* be a \(k\times k\) submatrix of *C*. Then there are row and column indices \(\alpha=(i_{1},i_{2},\ldots,i_{k})\) and \(\beta=(j_{1},j_{2},\ldots,j_{k})\) such that \(B=C[\alpha|\beta]\). Note that the union of a transversal of *B* and a transversal of \(C(\alpha|\beta)\) is a transversal of *C*. Choose an arbitrary but fixed transversal *T* of the square matrix \(C(\alpha|\beta)\). For any \(\sigma,\pi\in S_{k}\), \(c_{i_{1},j_{\sigma(1)}},\ldots ,c_{i_{k},j_{\sigma(k)}}\) and the entries in *T* constitute a transversal of *C*, whereas \(c_{i_{1},j_{\pi(1)}},\ldots,c_{i_{k},j_{\pi(k)}}\) and the entries in *T* also constitute a transversal of *C*. Let *b* be the product of the entries in *T*. Obviously, \(b>0\). Since

$$\prod_{i=1}^{n}c_{i,\gamma(i)}=\prod _{j=1}^{n}c_{j,\tau(j)} $$

for all \(\gamma, \tau\in S_{n}\), we have

$$b\prod_{t=1}^{k} c_{i_{t},j_{\sigma(t)}}=b\prod _{t=1}^{k} c_{i_{t},j_{\pi(t)}}, $$

which yields

$$\prod_{t=1}^{k} c_{i_{t},j_{\sigma(t)}}=\prod _{t=1}^{k} c_{i_{t},j_{\pi(t)}}. $$

Particularly, this shows that any \(2\times2\) submatrix

$$B= \begin{pmatrix} c_{i_{1},j_{1}} &c_{i_{1},j_{2}}\\ c_{i_{2},j_{1}}&c_{i_{2},j_{2}} \end{pmatrix} $$

of *C* satisfies

$$ c_{i_{1},j_{1}}c_{i_{2},j_{2}}=c_{i_{1},j_{2}}c_{i_{2},j_{1}}. $$

(1)

For any \(x_{1}>0\), let

$$ y_{j}=\frac{c_{1,j}}{x_{1}} $$

(2)

for \(j=1,2,\ldots,n\) and

$$ x_{i}=\frac{c_{i,1}}{c_{1,1}} x_{1} $$

(3)

for \(i=2,3,\ldots,n\). According to (1), (2), and (3), we have

$$c_{i,j}=\frac{c_{i,1}c_{1,j}}{c_{1,1}}=x_{i}y_{j} $$

for all \(i,j=1,2,\ldots,n\). Let \(X=\operatorname {diag}(x_{1},x_{2},\ldots,x_{n})\) and \(Y= \operatorname {diag}(y_{1},y_{2},\ldots,y_{n})\). Obviously, *X* and *Y* are positive diagonal matrices, and we have

$$C=XJY. $$

This completes the proof. □

We are now ready to present our proof of Theorem 2.1.

First, let *A* be real. Let *A* be a diagonally magic matrix. Then the elementwise exponential \(C =\exp (A):=(c_{i,j})\in\mathbb{R}^{n\times n}\) is a positive matrix with all permutation products equal. Hence, by Theorem 2.2 it is diagonally equivalent to *J*, the all-1s matrix, that is,

$$c_{i,j} = x_{i}y_{j},\quad i,j=1, 2, \ldots, n, $$

for suitable positive vectors \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) and \(y=(y_{1}, y_{2}, \ldots, y_{n})^{T}\). Hence, if \(q = (\log (x_{1}), \log (x_{2}), \ldots, \log (x_{n}) )^{T}\) and \(r= (\log (y_{1}), \log(y_{2}), \ldots, \log (y_{n}) )^{T}\), then

$$a_{i,j} = \log (x_{i}) + \log (y_{j}),\quad i,j=1, 2, \ldots, n. $$

Hence,

$$A = e_{n}\cdot q^{T} + r\cdot e_{n}^{T}, $$

where \(e_{n}=(\underbrace{1, 1,\ldots, 1}_{n})^{T}\). Thus, *A* is the sum of two matrices of rank 1 and, hence, at most of rank 2.

Now let *A* be complex, so that

$$A = B + iC \quad (B, C \text{ real}). $$

Since *A* is a diagonally magic matrix, so are *B* and *C*. Hence, both *B* and *C* are of the form

$$\begin{aligned}& B=e_{n}\cdot q_{1}^{T} + r_{1}\cdot e_{n}^{T} \quad \text{with } q_{1}, r_{1} \text{ real}, \\& C=e_{n}\cdot q_{2}^{T} + r_{2}\cdot e_{n}^{T} \quad \text{with } q_{2}, r_{2} \text{ real,} \end{aligned}$$

and hence

$$ A=e_{n}\cdot(q_{1}+iq_{2})^{T} + (r_{1}+ir_{2})\cdot e^{T}_{n}. $$

(4)

The matrix *A* has rank at most 2. This completes the proof. □

According to (4), we obtain that a diagonally magic matrix *A* can be presented in the form

$$ A=e_{n}\cdot x^{T} + y\cdot e^{T}_{n}= \begin{pmatrix} x_{1}+y_{1}&x_{1}+y_{2} & \cdots& x_{1}+y_{n}\\ x_{2}+y_{1}&x_{2}+y_{2} & \cdots& x_{2}+y_{n}\\ \vdots&\vdots& \ddots& \vdots\\ x_{n}+y_{1}&x_{n}+y_{2} & \cdots& x_{n}+y_{n} \end{pmatrix} . $$

(5)

If \(A=(a_{i,j}) \in\mathbb{C}^{n\times n}\) is a diagonally magic matrix, for any \(x_{1}\in\mathbb{C}\), let

$$y_{j}=a_{1,j}-x_{1} $$

for \(j=1,2,\ldots,n\) and

$$x_{i}=a_{i,1}-a_{1,1}+x_{1} $$

for \(i=2,3,\ldots,n\). By (5) we have

$$a_{i,j}=x_{i}+y_{j} $$

for all \(i,j=1,2,\ldots,n\). For example,

$$\begin{aligned} B_{n} =& \begin{pmatrix} 1&2 & \cdots& n\\ n+1&n+2 & \cdots& 2n\\ \vdots&\vdots& \ddots& \vdots\\ (n-1)n+1&(n-1)n+2 & \cdots& n^{2} \end{pmatrix} \\ =& \begin{pmatrix} 0+1&0+2 & \cdots& 0+n\\ n+1&n+2 & \cdots& n+n\\ \vdots&\vdots& \ddots& \vdots\\ (n-1)n+1&(n-1)n+2 & \cdots& (n-1)n+n \end{pmatrix} \end{aligned}$$

and

$$ C_{n}= \begin{pmatrix} 1&2 & \cdots& n\\ 2&3 & \cdots& n+1\\ \vdots&\vdots& \ddots& \vdots\\ n&n+1 & \cdots& 2n-1 \end{pmatrix} = \begin{pmatrix} 1+0&1+1 & \cdots& 1+(n-1)\\ 2+0&2+1 & \cdots& 2+(n-1)\\ \vdots&\vdots& \ddots& \vdots\\ n+0&n+1 & \cdots& n+(n-1) \end{pmatrix} . $$

By (5) we can get the characteristic polynomial, the eigenvalues, and the eigenvectors of *A*. In fact, the characteristic polynomial of *A* is

$$ p_{A}(\lambda)=\lambda^{n-2} \Biggl( \lambda^{2}-\lambda \Biggl(\sum_{i=1}^{n}(x_{i}+y_{i}) \Biggr)+\sum_{i=1}^{n}x_{i}\sum _{j=1}^{n}y_{j}-n\sum _{i=1}^{n}(x_{i}y_{i}) \Biggr). $$

(6)

From (6) we can see that *the algebraic multiplicity of the eigenvalue* 0 of the diagonally magic matrix *A* is *at least*
\(n-2\).

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This work is supported by the National Natural Science Foundation of China (No. 11501126, No. 11471122), the Youth Natural Science Foundation of Jiangxi Province (No. 20151BAB211011), the Science Foundation of Jiangxi Provincial Department of Education (No. GJJ150979), the Supporting the Development for Local Colleges and Universities Foundation of China - Applied Mathematics Innovative team building, and the Education Department of Jiangxi Province (No. 15YB106).

The authors declare that they have no competing interests.

All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

**Open Access** This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Zhou, D., Cai, Q. & Chen, X. A note on the rank inequality for diagonally magic matrices.
*J Inequal Appl* **2016**, 66 (2016). https://doi.org/10.1186/s13660-016-1013-4

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DOI: https://doi.org/10.1186/s13660-016-1013-4

- 15A03
- 15A06

- rank
- diagonally magic matrices
- positive matrix