A note on the rank inequality for diagonally magic matrices

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).

1 Introduction

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].

2 Main result

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

Theorem 2.1

([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.

Theorem 2.2

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

$$C=XJY. $$

Proof

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.

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 $n-2$.