The Inequalities of Merris and Foregger for Permanents

Abstract

Conjectures on permanents are well-known unsettled conjectures in linear algebra. Let A be an $n\times n$ matrix and $S_n$ be the symmetric group on n element set. The permanent of A is defined as $\mathrm{per}A={\displaystyle \sum _{\sigma \in S_n}}{\displaystyle \prod _{i=1}^n}{a}_{i\sigma \left(i\right)}.$ The Merris conjectured that for all $n\times n$ doubly stochastic matrices (denoted by ${\mathsf{\Omega }}_n$), $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)$, where $A\left(j\right|i)$ denotes the matrix obtained from A by deleting the jth row and ith column. Foregger raised a question whether $\mathrm{per}(t{J}_n+(1-t)A)\le \mathrm{per}A$ for $0\le t\le \frac{n}{n-1}$ and for all $A\in {\mathsf{\Omega }}_n$, where $J_n$ is a doubly stochastic matrix with each entry $\frac{1}{n}$. The Merris conjecture is one of the well-known conjectures on permanents. This conjecture is still open for $n\ge 4$. In this paper, we prove the Merris inequality for some classes of matrices. We use the sub permanent inequalities to prove our results. Foregger’s inequality is also one of the well-known inequalities on permanents, and it is not yet proved for $n\ge 5$. Using the concepts of elementary symmetric function and subpermanents, we prove the Foregger’s inequality for $n=5$ in [0.25, 0.6248]. Let ${\sigma }_k\left(A\right)$ be the sum of all subpermanents of order k. Holens and Dokovic proposed a conjecture (Holen–Dokovic conjecture), which states that if $A\in {\mathsf{\Omega }}_n,A\ne J_n$ and k is an integer, $1\le k\le n,$ then ${\sigma }_k\left(A\right)\ge \frac{{(n-k+1)}^2}{nk}{\sigma }_{k-1}\left(A\right)$. In this paper, we disprove the conjecture for $n=k=4$.

1. Introduction

Let $S_n$ be the symmetric group on n element set and let A be an $n\times n$ matrix. The permanent of A is defined as

$$\mathrm{per}A={\displaystyle \sum _{\sigma \in S_n}}{\displaystyle \prod _{i=1}^n}{a}_{i\sigma \left(i\right)}.$$

A matrix A is said to be doubly stochastic if it is a real non-negative matrix with each row sum and column sum equal to 1. Let ${\mathsf{\Omega }}_n$ denote the set of all $n\times n$ doubly stochastic matrices. For positive integers n and k with $\left(1\le k\le n\right)$, $Q_{k,n}$ denotes the set $\{\left(i_1,\dots ,i_k\right)/1\le i_1<\dots <i_k\le n\}$. For $\alpha ,\beta \in Q_{k,n}$, let $A\left(\alpha /\beta \right)$ be the submatrix of A obtained by deleting the rows indexed by $\alpha$ and columns indexed by $\beta$ and $A[\alpha /\beta ]$ be the submatrix of A with rows and columns indexed by $\alpha$ and $\beta$, respectively.

For $1\le k\le n$, the kth order subpermanent of A is defined by ${\sigma }_k\left(A\right)={\displaystyle \sum _{\alpha ,\beta \in Q_{k,n}}}\mathrm{per}A[\alpha /\beta ].$ In this paper, we use the following results quoted by Minc [1]: If A and B are two $n\times n$ matrices and $1\le k\le n$, then

$$\mathrm{per}A={\displaystyle \sum _{\beta \in Q_{k,n}}}\mathrm{per}A[\alpha /\beta ]\mathrm{per}A\left(\alpha /\beta \right),\mathrm{for}\alpha \in Q_{k,n},$$

and

$$\mathrm{per}(A+B)={\displaystyle \sum _{k=0}^n}{S}_k\left(A,B\right),$$

(1)

where $S_k\left(A,B\right)={\displaystyle \sum _{\alpha ,\beta \in Q_{k,n}}}\mathrm{per}A[\alpha /\beta ]\mathrm{per}B\left(\alpha /\beta \right)$, per$A[\alpha /\beta ]=1$ when $k=0$ and per $B(\alpha /\beta )=1$ when $k=n$.

Elliott H. Lieb [2] gave proofs of some conjectures on permanents. S G Hwang [3] proved that $J_n={\left(\frac{1}{n}\right)}_{n\times n}$ is the unique $\varphi$-maximizing matrix on $K_n$. Lih and Wang [4] proved the monotonicity conjecture for $n=3$. A survey on conjectures on permanents are given in [5,6].

Merris [7] conjectured that if $A\in {\mathsf{\Omega }}_n$ then $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$ He also suggested a method to prove this conjecture. The conjecture is still open for $n\ge 4.$ Subramanian and Somasundaram [8] have proved that if $A\in {\mathsf{\Omega }}_n$ and the polynomial ${\displaystyle \sum _{r=2}^n}r\frac{(n-r)!}{n^{n-r}}{\sigma }_r(A-J_n)t^{r-2}$ has no root in $(0,1)$ then A satisfies Merris conjecture. Furthermore, they proved some sufficient conditions for matrices in ${\Gamma }_k^n$ to satisfy the Merris conjecture, where ${\Gamma }_k^n$ denote the set of $n\times n$ non-negative matrices with each row sum and column sum equal to k.

In Section 2, we prove the Merris inequality for all $n\times n$ non-negative matrices with minimum entry greater than or equal to $\frac{1}{n}$. We prove that if A is an $n\times n$ non-negative matrix with minimum entry greater than or equal to $\frac{1}{n}$ and maximum entry less than or equal to 1, then $n^{2}perA\ge \sum {\lambda }_i$, where ${\lambda }_i^{\prime }s$ are the eigenvalues of $\left[a_{ij}perA\left(i\right|j)\right]$. Furthermore, we give a sufficient condition for a doubly stochastic matrix A to satisfy the Merris conjecture.

Foregger [9] raised a question whether $\mathrm{per}(t{J}_n+(1-t)S)\le \mathrm{per}S$ for $0\le t\le \frac{n}{n-1}$, and $S\in {\mathsf{\Omega }}_n$. He proved in [9] that for $n=3$, $\mathrm{per}(t{J}_3+(1-t)S\le \mathrm{per}S$ for $0\le t\le \frac{3}{2}$ for $S\in {\mathsf{\Omega }}_3$ with equality iff $S=J_3$ or $t=\frac{3}{2}$ and S is (up to permutations of rows and columns) $\frac{1}{2}(I+P)$, where P is a full-cycle permutation matrix. In addition, he proved in [10] that if $S\in {\mathsf{\Omega }}_4$ has all its off-diagonal entries less than or equal to $\frac{9}{20}$ and $t_0<t\le \frac{4}{3}$, where $t_0$ is the unique real root of $106t^3-418t^2+465t-100$ then $\mathrm{per}(t{J}_4+(1-t)S)\le \mathrm{per}S$ with equality if and only if $S=J_4$.

Subramanian and Somasundaram [8] proved that if $A\in {\mathsf{\Omega }}_n,2\le k\le n,$ and the polynomial ${\displaystyle \sum _{r=2}^k}r{c}_r{\sigma }_r(A-J_n)t^{r-2}$ has no root in $(0,1)$, where $c_r=\frac{(k-r)!}{n^{k-r}}{\left(\genfrac{}{}{0pt}{}{n-r}{k-r}\right)}^2$, then ${\sigma }_k(tA+(1-t)J_n)\le {\sigma }_k\left(A\right)$ for all $t\in [0,1]$. In Section 3, we prove that for all $S\in {\mathsf{\Omega }}_5$ and all t such that $0.25\le t\le 0.6248$, $\mathrm{per}(t{J}_5+(1-t)S)\le \mathrm{per}S$.

Holens [11] and Dokovic [12] proposed a conjecture (Holen–Dokovic conjecture), which states that if $A\in {\mathsf{\Omega }}_n,A\ne J_n$ and k is an integer, $1\le k\le n$, then ${\sigma }_k\left(A\right)\ge \frac{{(n-k+1)}^2}{nk}{\sigma }_{k-1}\left(A\right)$. S G Hwang [13] proved the conjecture for an $n-2$ dimensional face of ${\mathsf{\Omega }}_n$. Wanless [14] disproved this conjecture by providing a counterexample of order 22. The smallest order of a counterexample has not been established. In Section 3, we prove that the Holen–Dokovic conjecture fails for $n=k=4$ and thus established that the smallest order of a counterexample to Holen–Dokovic conjecture is 4.

2. Merris Conjecture

Let ${\Gamma }_k^n$ denote the set of $n\times n$ non-negative matrices with each row sum and column sum equal to k. Merris [7] conjectured that for all $n\times n$ doubly stochastic matrices,

$$n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$$

He also raised a question whether

$$n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{max}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)\mathrm{for}\mathrm{all}A\in {\mathsf{\Omega }}_n.$$

The Merris conjecture is one of the well-known conjectures in linear algebra, in particular on permanent. The conjecture is still open for $n\ge 4.$ There is not much progress in this conjecture. Subramanian and Somasundaram [8] have proved that if $A\in {\mathsf{\Omega }}_n$ and the polynomial ${\displaystyle \sum _{r=2}^n}r\frac{(n-r)!}{n^{n-r}}{\sigma }_r(A-J_n)t^{r-2}$ has no root in $(0,1)$ then A satisfies the Merris conjecture, and they also proved some sufficient conditions for matrices in ${\Gamma }_k^n$ to satisfy the Merris conjecture.

A matrix is said to be a positive matrix if all its entries are non-negative [15]. Let $A_i$ be $k\times k$ matrix, $i=1,2,\dots ,n$. The direct sum of the matrices $A_i$ is defined as follows:

${⨁}_{i=1}^n{A}_i=\mathrm{diag}(A_1,A_2,\dots ,A_n)=\left(\begin{array}{cccc}{A}_1& \mathbf{0}& \cdots & \mathbf{0}\\ \mathbf{0}& A_2& \cdots & \mathbf{0}\\ ⋮& ⋮& ⋮\\ \mathbf{0}& \mathbf{0}& \cdots & A_n\end{array}\right)$, where 0 is the zero matrix.

Lemma 1.

If A is a $n\times n$ positive matrix with minimum entry greater than or equal to $\frac{1}{n}$, then

• $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{max}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$
• $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$
• $n^2\mathrm{per}A\ge {\displaystyle \sum _{i,j=1}^n}\mathrm{per}A\left(i\right|j).$

Proof. 

• We need to find ${\displaystyle \underset{1\le i\le n}{max}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$ Let the maximum sum be attained in the kth column, i.e., ${\displaystyle \underset{1\le i\le n}{max}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)={\sum }_{j=1}^n\mathrm{per}A\left(j\right|k)$, where $1\le k\le n$. Let the entries of the kth column be $k_1,k_2,\dots ,k_n$. This implies that $\frac{1}{n}\le k_l$ for each $l=1,2,\dots ,n$ and hence $1\le n{k}_l$ for each $l=1,2,\dots ,n$. Taking the permanent along the kth column, $\mathrm{per}A={\displaystyle \sum _{i=1}^n}{k}_i\mathrm{per}\left(i\right|k)$. Multiplying by n on both sides, $n\mathrm{per}A={\displaystyle \sum _{i=i}^n}n{k}_i\mathrm{per}A\left(i\right|k)$. Since $n{k}_l\ge 1$ for each $l=1,2,\dots ,n$ and since each of the subpermanents is non-negative, this implies that $n\mathrm{per}A\ge {\displaystyle \sum _{i=1}^n}\mathrm{per}A\left(i\right|k)$. This implies that $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{max}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)$.
• From the inequality 1, $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)$.
• From the proof of the inequality 1, $n\mathrm{per}A\ge {\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)$ for each $i=1,2,\dots ,n$. Taking summation over $i,i$ running from 1 to n, $n^2\mathrm{per}A\ge {\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)$. $\Rightarrow n^2\mathrm{per}A\ge {\displaystyle \sum _{i,j=1}^n}\mathrm{per}A\left(i\right|j)$.

□

Theorem 1.

If A is a $n\times n$ positive matrix with constant columns and maximum entry greater than or equal to $\frac{1}{n}$ then A satisfies the inequality

$$nperA\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}perA\left(j\right|i).$$

Proof. 

Suppose $A=\left(a_{ij}\right)=k_j,$ for all i. Then per$A=n!k_1{k}_2\dots k_n.$ ${\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)=n(n-1)!k_1{k}_2\dots k_{i-1}{k}_{i+1}\dots k_n=n!k_1{k}_2\dots k_{i-1}{k}_{i+1}\dots k_n$ $=\frac{\mathrm{per}A}{k_i}$. ${\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)=\frac{\mathrm{per}A}{k_l},$ where $k_l=\max \{k_1,k_2,\dots k_n\}$. Since, $k_l\ge \frac{1}{n}$, we have $n\mathrm{per}A-{\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}perA\left(j\right|i)\ge \frac{\mathrm{per}A}{k_l}-{\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i)=0.$ Therefore, $n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(j\right|i).$ □

Theorem 2.

If A is a $n\times n$ matrix whose minimum entry is greater than or equal to $\frac{1}{n}$ and maximum entry is less than or equal to 1 then $n^2\mathrm{per}A\ge \sum {\lambda }_i$, where ${\lambda }_i$ is an eigenvalue of $\left[a_{ij}\mathrm{per}A\left(i\right|j)\right].$

Proof. 

If A is an $n\times n$ non-negative matrix whose minimum entry is greater than or equal to $\frac{1}{n}$ then from Lemma 1, $n^2\mathrm{per}A\ge {\displaystyle \sum _{i,j=1}^n}\mathrm{per}A\left(i\right|j)$.

$\Rightarrow \mathrm{per}A\ge \frac{1}{n^2}{\displaystyle \sum _{i,j=1}^n}\mathrm{per}A\left(i\right|j)$.

Let $a_{lm}$ be the maximum entry of A. Multiplying on both sides by $a_{lm}$,

$a_{lm}\mathrm{per}A\ge \frac{1}{n^2}{\displaystyle \sum _{i,j=1}^n}{a}_{lm}\mathrm{per}A\left(i\right|j)\ge \frac{1}{n^2}{\displaystyle \sum _{i,j=1}^n}{a}_{ij}\mathrm{per}A\left(i\right|j)$.

$\Rightarrow \mathrm{per}A\ge \frac{1}{n^2{a}_{lm}}{\displaystyle \sum _{i,j=1}^n}{a}_{ij}\mathrm{per}A\left(i\right|j)$.

By the assumption $a_{lm}\le 1,\Rightarrow \frac{1}{a_{lm}}\ge 1$.

$\Rightarrow \mathrm{per}A\ge \frac{1}{n^2}{\displaystyle \sum _{i,j=1}^n}{a}_{ij}\mathrm{per}A\left(i\right|j)\ge \frac{1}{n^2}{\displaystyle \sum _{i=1}^n}{a}_{ii}\mathrm{per}A\left(i\right|i).$

$\Rightarrow n^2\mathrm{per}A\ge tr\left(\left[a_{ij}\mathrm{per}A\left(i\right|j)\right]\right).$

$\Rightarrow n^2\mathrm{per}A\ge \mathrm{Sum\; of\; eigenvalues\; of}\left[a_{ij}\mathrm{per}A\left(i\right|j)\right].$ □

Theorem 3.

Let $A\in {\mathsf{\Omega }}_n$ and $P=\left(\mathrm{per}A(i/j)\right)=\left(p_{ij}\right)$. If kth row of P gives the ${\displaystyle \underset{i}{max}}{\displaystyle \sum _{j=1}^n}{p}_{ij}$ and $min\left\{a_{kj}\right\}=\frac{1}{n}$ then

$$n\mathrm{per}A\ge {\displaystyle \underset{1\le i\le n}{min}}{\displaystyle \sum _{j=1}^n}\mathrm{per}A\left(i\right|j).$$

Proof. 

${\displaystyle \underset{i}{min}}{\displaystyle \sum _{j=1}^n}{p}_{ij}\le {\displaystyle \underset{i}{max}}{\displaystyle \sum _{j=1}^n}{p}_{ij}={\sum }_{j=1}^n{p}_{kj}$$\le n{\displaystyle \sum _{j=1}^n}{a}_{kj}{p}_{kj}=n\mathrm{per}A,$ since $a_{kj}\ge \frac{1}{n}.$ □

Example 1.

$A=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right)⨁\left(\begin{array}{cc}\frac{6}{7}& \frac{1}{7}\\ \frac{1}{7}& \frac{6}{7}\end{array}\right),$ where $⨁$ is the direct sum.

It is easy to see that $P=\left(\begin{array}{ccc}\frac{74}{441}& \frac{74}{441}& \frac{74}{441}\\ \frac{74}{441}& \frac{74}{441}& \frac{74}{441}\\ \frac{74}{441}& \frac{74}{441}& \frac{74}{441}\end{array}\right)⨁\left(\begin{array}{cc}\frac{4}{21}& \frac{2}{63}\\ \frac{2}{63}& \frac{4}{21}\end{array}\right).$ Maximum row sum of $P=\frac{74}{147}$ and the minimum element of the row corresponding to the maximum row sum of $P=\frac{1}{3}>\frac{1}{5}$. Therefore, $5perA\ge {\displaystyle \underset{1\le i\le 5}{min}}{\displaystyle \sum _{j=1}^5}perA\left(i\right|j).$

3. Foregger’s Inequality

Let $J_n$ denote the $n\times n$ matrix with each entry equal to $\frac{1}{n}$. Several authors have considered the problem of finding an upper bound for the permanent of a convex combination of $J_n$ and S, where $S\in {\mathsf{\Omega }}_n$. Lih and Wang [16] discussed convexity inequality on the permanent of doubly stochastic matrices. For example, Marcus and Minc conjectured [17] that if $S\in {\mathsf{\Omega }}_n,n\ge 2$, then $\mathrm{per}\left(\frac{n{J}_n-S}{n-1}\right)\le \mathrm{per}S$, equality holds when $n=2$. If $n\ge 3$ then inequality holds iff $S=J_n.$ They established in [17] that the conjecture is true for $n=2,$ or if S is positive semi-definite symmetric, or if S is in a sufficiently small neighborhood of $J_n.$

E.T.H.Wang conjectured [18] that $\mathrm{per}\left(\frac{n{J}_n+S}{n+1}\right)\le \mathrm{per}S$ and proved the Marcus and Minc conjecture for $n=3$, with a revised statement of the case of equality.

Foregger [9] raised a question whether $\mathrm{per}(t{J}_n+(1-t)S)\le \mathrm{per}S$ for $0\le t\le \frac{n}{n-1}$, and $S\in {\mathsf{\Omega }}_n$. He proved in [9] that for $n=3$, $\mathrm{per}(t{J}_3+(1-t)S\le \mathrm{per}S$ for $0\le t\le \frac{3}{2}$ for $S\in {\mathsf{\Omega }}_3$ with equality iff $S=J_3$ or $t=\frac{3}{2}$ and S is (up to permutations of rows and columns) $\frac{1}{2}(I+P)$, where P is a full-cycle permutation matrix. In addition, he proved in [10] that if $S\in {\mathsf{\Omega }}_4$ has all its off-diagonal entries less than or equal to $\frac{9}{20}$ and $t_0<t\le \frac{4}{3}$, where $t_0$ is the unique real root of $106t^3-418t^2+465t-100$ then $\mathrm{per}(t{J}_4+(1-t)S)\le \mathrm{per}S$ with equality iff $S=J_4$. This Foregger inequality is not yet proved for $n\ge 5.$

Subramanian and Somasundaram [8] proved that if $A\in {\mathsf{\Omega }}_n,2\le k\le n$ and the polynomial ${\displaystyle \sum _{r=2}^k}r{c}_r{\sigma }_r(A-J_n)t^{r-2}$ has no root in $(0,1)$ where $c_r=\frac{(k-r)!}{n^{k-r}}{\left(\genfrac{}{}{0pt}{}{n-r}{k-r}\right)}^2$ then ${\sigma }_k(tA+(1-t)J_n)\le {\sigma }_k\left(A\right)$ for all $t\in [0,1]$. In this paper, we prove that for all $S\in {\mathsf{\Omega }}_5$ and all t such that $0.25\le t\le 0.6248$, $\mathrm{per}(t{J}_5+(1-t)S)\le \mathrm{per}S$. The following theorem is from Ebelein (Theorem 1, [19]).

Theorem 4.

Let $\varphi (x_1,x_2,\dots ,x_n)$ be a real symmetric polynomial of degree at most one in each variable defined for $0\le x_i\le 1$ and ${\displaystyle \sum _{i=1}^n}{x}_i=\gamma$ (γ is a real constant), then the maximum and minimum of $\varphi \left(x\right)$ on the set C = { $x|{\displaystyle \sum _{i=1}^n}{x}_i=\gamma$ and for $i=1,2,\dots ,n,x_i\in [{\alpha }_i,{\beta }_i]$, where $[{\alpha }_i,{\beta }_i]$ is any closed interval contained in $[0,1]$ } and is assumed at least among the points whose components which are not end points are all equal. Moreover, if the maximum or minimum is attained only in the interior of C then it is assumed uniquely at the point $(\frac{\gamma }{n},\frac{\gamma }{n},\dots \frac{\gamma }{n}).$

Let x be an n-dimensional vector. Then the elementary symmetric function of x denoted by $e_r\left(x\right)$ is the sum of products of coordinates of x taken r at a time. Let $x=(x_1,x_2,\dots ,x_n)$. Then $e_r\left(x\right)=e_r(x_1,x_2,\dots ,x_n),r=1,2,\dots ,n$.

Theorem 5.

Let $S\in {\mathsf{\Omega }}_5$ have all its off-diagonal entries less than or equal to $\frac{9}{20}$ and $0.25\le t\le 0.6248$. Then $\mathit{per}(t{J}_5+(1-t)S)\le \mathit{per}S.$

Proof. 

Let $S\left(t\right)=t{J}_5+(1-t)S$. Then by Eberlein and Mudholkar ([20], p. 393)

$$\mathrm{per}S\left(t\right)=-9+{\displaystyle \sum _{T_1\left(S\left(t\right)\right)}}(-e_2+e_3-e_4+2e_5)\left(x\right)+{\displaystyle \sum _{T_2(S\left(t\right)}}(e_2-e_3+e_4-2e_5)\left(x\right),$$

where $e_r$ is the rth elementary symmetric function and $T_r\left(B\right)$ is the set of sums of columns of B, taken r at a time. If $x\in T_1\left(S\left(t\right)\right)$ then $x=t\frac{e}{5}+(1-t)s$ where $s\in T_1\left(S\right)$ and $e={[1,1,1,1,1]}^T$. Hence,

$e_2\left(x\right)=\frac{2}{5}{t}^2+t(1-t)\frac{4}{5}+{(1-t)}^2{e}_2\left(s\right)$,

$e_3\left(x\right)=\frac{2}{25}{t}^3+\frac{6}{25}{t}^2(1-t)+\frac{1}{5}t{(1-t)}^2{e}_2\left(s\right)+{(1-t)}^3{e}_3\left(s\right)$,

$e_4\left(x\right)=\frac{1}{125}{t}^4+\frac{4}{125}{t}^3(1-t)+\frac{1}{25}{t}^2{(1-t)}^2{e}_2\left(s\right)+\frac{1}{5}t{(1-t)}^3{e}_3\left(S\right)+{(1-t)}^4{e}_4\left(s\right)$,

$$\begin{gathered} e_5\left(x\right)=\frac{1}{3125}{t}^5+\frac{4}{625}{t}^4(1-t)+\frac{1}{125}{t}^3{(1-t)}^2{e}_2\left(s\right)+\frac{1}{25}{t}^2{(1-t)}^3{e}_3\left(s\right)+\frac{1}{5}t{(1-t)}^4{e}_4\left(s\right) \\ +{(1-t)}^5{e}_5\left(s\right). \end{gathered}$$

Similarly if $x\in T_2\left(S\left(t\right)\right)$ then there exists $r\in T_2\left(S\right)$ such that $x=\frac{2}{5}te+(1-t)r.$ Hence

$e_2\left(x\right)=\frac{8}{5}{t}^2+\frac{16}{5}t(1-t)+{(1-t)}^2{e}_2\left(r\right)$,

$e_3\left(x\right)=\frac{16}{25}{t}^3+\frac{48}{25}{t}^2(1-t)+\frac{2}{5}t{(1-t)}^2{e}_2\left(r\right)+{(1-t)}^3{e}_3\left(r\right)$,

$e_4\left(x\right)=\frac{16}{125}{t}^4+\frac{64}{125}{t}^3(1-t)+\frac{4}{25}{t}^2{(1-t)}^2{e}_2\left(r\right)+\frac{2}{5}t{(1-t)}^3{e}_3\left(r\right)+{(1-t)}^4{e}_4\left(r\right)$,

$e_5\left(x\right)=\frac{32}{3125}{t}^5+\frac{128}{625}{t}^4(1-t)+\frac{8}{125}{t}^3{(1-t)}^2{e}_2\left(r\right)+\frac{4}{25}{t}^2{(1-t)}^3{e}_3\left(r\right)+\frac{2}{5}t{(1-t)}^4{e}_4\left(r\right)$

$+{(1-t)}^5{e}_5\left(r\right).$

After substitution and simplification we have

$$\mathrm{per}S\left(t\right)=\mathrm{per}S+p_1\left(t\right)+{\displaystyle \sum _{T_1\left(S\right)}}(p_2\left(t\right)e_2+p_3\left(t\right)e_3+p_4\left(t\right)e_4+p_5\left(t\right)e_5)\left(x\right)+$$

$${\displaystyle \sum _{T_2\left(s\right)}}(p_6\left(t\right)e_2+p_7\left(t\right)e_3+p_8\left(t\right)e_4+p_9\left(t\right)e_5)\left(x\right),$$

where

$p_1\left(t\right)=\frac{7246}{3125}{t}^5-\frac{2901}{625}{t}^4+\frac{1426}{125}{t}^3-\frac{448}{25}{t}^2+\frac{76}{5}t$,

$p_2\left(t\right)=\frac{2}{125}{t}^5-\frac{9}{125}{t}^4+\frac{37}{125}{t}^3-\frac{36}{25}{t}^2+\frac{11}{5}t$,

$p_3\left(t\right)=\frac{-2}{25}{t}^5+\frac{11}{25}{t}^4-\frac{46}{25}{t}^3+\frac{92}{25}{t}^2-\frac{16}{5}t$

$p_4\left(t\right)=\frac{2}{5}{t}^5-\frac{13}{5}{t}^4+\frac{32}{5}{t}^3-\frac{22}{5}{t}^2+\frac{22}{5}t$,

$p_5\left(t\right)=-2t^5+10t^4-20t^3+20t^2-10t$,

$p_6\left(t\right)=\frac{-16}{125}{t}^5+\frac{52}{125}{t}^4-\frac{56}{125}{t}^3+\frac{49}{25}{t}^2-\frac{12}{5}t$,

$p_7\left(t\right)=\frac{8}{25}{t}^5-\frac{34}{25}{t}^4+\frac{79}{25}{t}^3-\frac{113}{25}{t}^2+\frac{17}{5}t$

$p_8\left(t\right)=\frac{-4}{5}{t}^5+\frac{21}{5}{t}^4-\frac{44}{5}{t}^3+\frac{46}{5}{t}^2-\frac{24}{5}t$,

$p_9\left(t\right)=2t^5-10t^4+20t^3-20t^2+10t.$

Now use the identities ([20], p. 391)

$${\displaystyle \sum _{T_2\left(A\right)}}{e}_2\left(x\right)=3{\displaystyle \sum _{T_1\left(A\right)}}{e}_2\left(x\right)+10\mathrm{and}{\displaystyle \sum _{T_2\left(A\right)}}{e}_3\left(x\right)={\displaystyle \sum _{T_1\left(A\right)}}{e}_3\left(x\right)+3{\displaystyle \sum _{T_1\left(A\right)}}{e}_2\left(x\right)$$

to write

$$\mathrm{per}S\left(t\right)=\mathrm{per}S+\frac{10}{3}\alpha +p_1\left(t\right)+{\displaystyle \sum _{T_1\left(S\right)}}(p_2+\alpha +\beta +3\gamma )e_2+(p_3+\gamma +\frac{\beta }{3})e_3+p_4\left(t\right)e_4+p_5\left(t\right)e_5)+$$

$${\displaystyle \sum _{T_2\left(S\right)}}((p_6\left(t\right)-\frac{\alpha }{3})e_2+(p_7\left(t\right)-\frac{\beta }{3}-\gamma )e_3)+p_8\left(t\right)e_4+p_9\left(t\right)e_5\left)\right)$$

for any polynomials $\alpha ,\beta$ and $\gamma .$

$$\mathrm{per}S\left(t\right)=\mathrm{per}S+c\left(t\right)+{\displaystyle \sum _{T_1\left(S\right)}}{f}_t\left(s\right)+{\displaystyle \sum _{T_2\left(S\right)}}{g}_t\left(r\right),$$

where $f_t=p_5\left(t\right)e_5+p_4\left(t\right)e_4+(p_3+\gamma +\beta /3)e_3+(p_2+\alpha +\beta +3\gamma )e_2$, $g_t=(p_6\left(t\right)-\alpha /3)e_2+(p_7\left(t\right)-\beta /3-\gamma )e_3+p_8\left(t\right)e_4+p_9\left(t\right)e_5)$ and $c\left(t\right)=10/3\alpha +p_1\left(t\right).$

We assume that all vectors in $T_1$ satisfy the condition $0\le x_1\le 1,0\le x_i\le \frac{9}{20}$, $i=2,3,4,5.$ The functions $f_t$ and $g_t$ are linear combinations of elementary symmetric functions.

From Theorem 4, possible points of maximum of $f_t$ are $[\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5}],[\frac{11}{20},\frac{9}{20},0,0,0],$ $[\frac{11}{40}.\frac{11}{40},\frac{9}{20},0,0]$, $[\frac{11}{60},\frac{11}{60},\frac{11}{60}\frac{9}{20},0],[\frac{11}{80},\frac{11}{80},\frac{11}{80},\frac{11}{80},\frac{9}{20}],[\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},0],$ $[\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0],[1,0,0,0,0],$ $[\frac{1}{10},\frac{9}{20},\frac{9}{20},0,0,],$ $[\frac{1}{20}.\frac{9}{20},\frac{9}{20},\frac{1}{20},0]$, $[\frac{1}{30},\frac{9}{20},\frac{9}{20},\frac{1}{30},\frac{1}{30}]$ and possible points of maximum of $g_t$ are $[\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5}],[\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},0],[\frac{2}{3},\frac{2}{3},\frac{2}{3},0,0],$ $[1,1,0,0,0],[1,\frac{1}{3},\frac{1}{3},\frac{1}{3},0],[1,\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}],$ $[1,\frac{1}{2},\frac{1}{2},0,0]$, $[\frac{1}{15},\frac{9}{10},\frac{9}{10},\frac{1}{15},\frac{1}{15}].$

For each t, a set of linear inequalities must be satisfied in order for $(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5})$ to be a maximum for $f_t$ and for $(\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5})$ to be a maximum for $g_t$. These inequalities are solved numerically for various values of t and then interpolated to find $\alpha$ and $\beta$ (details are shown in Appendix A). Substituting the values of $\alpha$ and $\beta$ we obtain the values for $f_t\left(s\right)$ and $g_t\left(s\right)$ at different points. We have shown the values of $f_t\left(s\right)$ for different values of s and $g_t\left(r\right)$ for different values of r are given in the next two tables, respectively.

s	$f_t\left(s\right)$
$[\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5}]$	$$\begin{gathered} 8t^4/3125-7t^4/625+29/12500t^3-194/625t^2+75397/187500t- \\ 2609/375000 \end{gathered}$$
$[\frac{11}{20}.\frac{9}{20},0,0,0]$	$0.00396t^5-0.01782t^4+0.0647955t^3-0.3564t^2+0.38741175t-0.0043065$
$[\frac{11}{40},\frac{11}{40},\frac{9}{20},0,0]$	$$\begin{gathered} 979/400000t^5-6633/800000t^4+175813/8000000t^3-68013/20000t^2+ \\ 127003657/320000000t-1798797/320000000 \end{gathered}$$
$[\frac{11}{60},\frac{11}{60},\frac{11}{60},\frac{9}{20},0]$	$$\begin{gathered} 201/78125t^5-6061/625000t^4+717247/50000000t^3-5068/15625t^2+ \\ 58906841/150000000t-1817611/300000000 \end{gathered}$$
$[\frac{11}{80},\frac{11}{80},\frac{11}{80},\frac{11}{80},\frac{9}{20}]$	$0.0026t^5-0.0104t^4+0.0095t^3-0.311744t^2+0.3884t-0.0063$
$[\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},0]$	$$\begin{gathered} 8t^5/3125-241/25000t^4+1627/200000t^3-8179/25000t^2+ \\ 969949/2400000t-3131/480000 \end{gathered}$$
$[\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0]$	$$\begin{gathered} 8/3375t^5-26/3375t^4+2581/135000t^3-232/675t^2+32663/81000t- \\ 4697/810000 \end{gathered}$$
$[\frac{1}{2},\frac{1}{2},0,0,0]$	$1/250t^5-9/500t^4+1309/20000t^3-9/25t^2+15653/40000t-87/20000$
$[1,0,0,0,0]$	0
$[\frac{1}{10},\frac{9}{20},\frac{9}{20},0,0]$	$$\begin{gathered} 191/62500t^5-379/31250t^4+78449/2000000t^3-5414/15625t^2+ \\ 14733359t/37500000-1526647/300000000 \end{gathered}$$
$[\frac{1}{20},\frac{9}{20},\frac{9}{20},\frac{1}{20},0]$	$$\begin{gathered} -197/25000t^5+1953/50000t^4-275569/1000000t^3+2349/5000t^2- \\ 606491/2000000t-1281/250000 \end{gathered}$$
$[\frac{1}{30},\frac{9}{20},\frac{9}{20},\frac{1}{30},\frac{1}{30}]$	$$\begin{gathered} 15669/5000000t^5-63953/5000000t^4+972461/25000000t^3- \\ 171569/500000t^2+117437989/300000000t-1543843/300000000 \end{gathered}$$

r	$g_t\left(r\right)$
$[\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5}]$	$-\frac{256}{3125}{t}^5+\frac{144}{625}{t}^4+\frac{1257}{3125}{t}^3+\frac{152}{125}{t}^2-\frac{86134}{46875}t+\frac{291}{31250}$
$[\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},0]$	$-\frac{41}{500}{t}^5+\frac{413}{2000}{t}^4+\frac{3751}{10000}{t}^3+\frac{251}{200}{t}^2-\frac{112961}{60000}t+\frac{131}{15000}$
$[\frac{2}{3},\frac{2}{3},\frac{2}{3},0,0]$	$$\begin{gathered} -\frac{11851}{156250}{t}^5+\frac{94803}{625000}{t}^4+\frac{17709461}{50000000}{t}^3+\frac{159249}{125000}{t}^2-\frac{191045273}{100000000}t+ \\ \frac{291289}{37500000} \end{gathered}$$
$[1,1,0,0,0]$	$-\frac{16}{125}{t}^5+\frac{52}{125}{t}^4-\frac{2183}{5000}{t}^3+\frac{49}{25}{t}^2-\frac{32827}{15000}t+\frac{7}{1200}$
$[1,\frac{1}{3},\frac{1}{3},\frac{1}{3},0]$	$$\begin{gathered} -\frac{12771}{156250}{t}^5+\frac{128943}{625000}{t}^4+\frac{13137261}{50000000}{t}^3+\frac{63973}{50000}{t}^2-\frac{91805513}{50000000}t+ \\ \frac{2329571}{300000000} \end{gathered}$$
$[1,\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}]$	$-\frac{2033}{25000}{t}^5+\frac{2711}{12500}{t}^4+\frac{55171}{200000}{t}^3+\frac{62519}{50000}{t}^2-\frac{4323233}{2400000}t+\frac{1281}{160000}$
$[1,\frac{1}{2},\frac{1}{2},0,0]$	$-\frac{2}{25}{t}^5+\frac{9}{50}{t}^4+\frac{977}{4000}{t}^3+\frac{33}{25}{t}^2-\frac{75423}{40000}t+\frac{437}{60000}$
$[\frac{1}{15},\frac{9}{10},\frac{9}{10},\frac{1}{15},\frac{1}{15}]$	$$\begin{gathered} -\frac{31377}{312500}{t}^5+\frac{354931}{1250000}{t}^4-\frac{1128039}{50000000}{t}^3+\frac{49261}{31250}{t}^2-\frac{602407919}{300000000}t+ \\ \frac{129307}{18750000} \end{gathered}$$

In calculating the elementary symmetric functions and $f_t\left(s\right)$ and $g_t\left(r\right)$ at different points, MATLAB programs were used.

In the Appendix A, we have shown the curves $f_t\left(s\right)$ and $g_t\left(r\right)$ in Figure A1 and Figure A2, respectively. From the figures, $f_t\left(s\right)\le f_t(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5})$ in $(0.25,0.98)$. Furthermore, $g_t\left(r\right)\le g_t(1,\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})$ in $(0.1,0.65)$ and $g_t\left(r\right)\le g_t(\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5})$ in $(0.65,1)$. Therefore, in the interval (0.25, 0.65),

$$\mathrm{per}S\left(t\right)\le {\displaystyle \sum _{T_2\left(S\right)}}{g}_t(1,\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})+{\displaystyle \sum _{T_1\left(S\right)}}{f}_t(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5})+\mathrm{per}S+c\left(t\right).$$

Similarly, in the interval (0.65, 0.98), we have,

$$perS\left(t\right)\le {\displaystyle \sum _{T_2\left(S\right)}}{g}_t(\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5})+{\displaystyle \sum _{T_1\left(S\right)}}{f}_t(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5})+perS+c\left(t\right).$$

Substituting the values, we obtain $\mathrm{per}S\left(t\right)\le \mathrm{per}S$ in $(0.25,0.6248).$ □

Holens [11] and Dokovic [12] proposed a conjecture (Holen–Dokovic conjecture), which states that if $A\in {\mathsf{\Omega }}_n,A\ne J_n$ and k is an integer, $1\le k\le n,$ then ${\sigma }_k\left(A\right)\ge \frac{{(n-k+1)}^2}{nk}{\sigma }_{k-1}\left(A\right)$. Dokovic proved that the conjecture is true for $k\le 3$. Kopotun [21] proved that the conjecture is true for $k=4$ and $n\ge 5$. Wanless [14] disproved this conjecture by providing a counterexample of order 22. The smallest order of a counterexample has not been established. In Theorem 6, we prove that the Holen–Dokovic conjecture fails for $n=k=4.$ Before that, we recall that Foregger [10] proved that if $A\in {\mathsf{\Omega }}_4$ has all its off-diagonal entries less than or equal to $\frac{9}{20}$ and $t_0<t\le \frac{4}{3}$, where $t_0$ is the unique real root of $106t^3-418t^2+465t-100$, then $\mathrm{per}(t{J}_4+(1-t)A)\le \mathrm{per}S$ with equality if and only if $A=J_4$.

Theorem 6.

The Holen–Dokovic conjecture fails for $n=k=4$.

Proof. 

Let $\Delta \left(t\right)=\mathrm{per}A-\mathrm{per}(t{J}_4+(1-t)A)$.

Foregger [10] proved that $\Delta \left(t\right)\ge 0$ in $[t_0,\frac{4}{3}]$ and $t_0$ is the unique real root of $106t^3-418t^2+465t-100$.

Now, $\Delta \left(t\right)=\mathrm{per}A-{\displaystyle \sum _{r=0}^4}{c}_r{(1-t)}^{4-r}{t}^r{\sigma }_r\left(A\right)$, where $c_r=\frac{(n-r)!}{n^{n-r}}.$

Here, $\Delta \left(1\right)=0$ and

$$\begin{gathered} {\Delta }^{\prime }\left(1\right)=-{\displaystyle \sum _{r=0}^4}{c}_r{\sigma }_r\left(A\right){[{(1-t)}^{4-r}r{t}^{r-1}+t^r{(1-t)}^{3-r}(-1)]}_{t=1} \\ =-4c_4\mathrm{per}A+c_3{\sigma }_3\left(A\right) \\ =-4\mathrm{per}A+\frac{1}{4}{\sigma }_3\left(A\right) \end{gathered}$$

${\Delta }^{\prime }\left(1\right)=-4\mathrm{per}A+\frac{1}{4}{\sigma }_3\left(A\right)$

${\Delta }^{\prime }\left(1\right)={\displaystyle \underset{t\to 1}{lim}}\frac{\Delta \left(t\right)}{t-1}$${\Delta }^{{}^{\prime }}\left(1\right)\le 0$ iff $\Delta \left(t\right)\ge 0$ for all $t\in [1-ϵ,1]$ and $\Delta \left(t\right)\le 0$ for all $t\in [1,1+ϵ]$, which is not the case since $\Delta \left(t\right)\ge 0$ for all $t\in [t_0,\frac{4}{3}]$.

Hence, for some $A\in {\mathsf{\Omega }}_4,$${\sigma }_k\left(A\right)<\frac{{(n-k+1)}^2}{nk}{\sigma }_{k-1}\left(A\right)$. □

4. Conclusions

The Merris conjecture is one of the well-known conjectures in linear algebra and it is still open for $n\ge 4$. We proved the Merris inequality for all $n\times n$ non-negative matrices with minimum entry greater than or equal to $\frac{1}{n}$. Furthermore, we gave a sufficient condition for a doubly stochastic matrix A to satisfy the Merris conjecture. Secondly, we proved the Foregger’s inequality. That is, for all $S\in {\mathsf{\Omega }}_5$ with off-diagonal entries less than or equal to $\frac{9}{20}$ and all t such that $0.25\le t\le 0.6248$, $\mathrm{per}(t{J}_5+(1-t)S)\le \mathrm{per}S$. Finally, we proved that the Holen–Dokovic conjecture fails for $n=k=4$ and thus established that the smallest order of a counterexample to the Holen–Dokovic conjecture is $n=4$.