# On vector matrix game and symmetric dual vector optimization problem

## Abstract

A vector matrix game with more than two skew symmetric matrices, which is an extension of the matrix game, is defined and the symmetric dual problem for a nonlinear vector optimization problem is considered. Using the Kakutani fixed point theorem, we prove an existence theorem for a vector matrix game. We establish equivalent relations between the symmetric dual problem and its related vector matrix game. Moreover, we give an example illustrating the equivalent relations.

## 1 Introduction

A matrix game is defined by B of a real $mÃ—n$ matrix together with the Cartesian product ${S}_{n}Ã—{S}_{m}$ of all n-dimensional probability vectors ${S}_{n}$ and all m-dimensional probability vectors ${S}_{m}$; that is, ${S}_{n}:=\left\{x={\left({x}_{1},â€¦,{x}_{n}\right)}^{T}âˆˆ{\mathbb{R}}^{n}:{x}_{i}â‰§0,{âˆ‘}_{i=1}^{n}{x}_{i}=1\right\}$, where the symbol T denotes the transpose. A point $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆˆ{S}_{n}Ã—{S}_{n}$ is called an equilibrium point of a matrix game B if ${x}^{T}B\stackrel{Â¯}{y}â‰¦{\stackrel{Â¯}{x}}^{T}B\stackrel{Â¯}{y}â‰¦{\stackrel{Â¯}{x}}^{T}By$ for all $x,yâˆˆ{S}_{n}$ and $\stackrel{Â¯}{x}B\stackrel{Â¯}{y}=v$, where v is value of the game. If $n=m$ and B is skew symmetric, then we can check that $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆˆ{S}_{n}Ã—{S}_{n}$ is an equilibrium point of the game B if and only if $B\stackrel{Â¯}{x}â‰¦0$ and $B\stackrel{Â¯}{y}â‰¦0$. When B is an $nÃ—n$ skew symmetric matrix, $\stackrel{Â¯}{x}âˆˆ{S}_{n}$ is called a solution of the matrix game B if $B\stackrel{Â¯}{x}â‰¦0$ [1].

Consider the linear programming problem (LP) and its dual (LD) as follows:

(LP) â€ƒMinimize ${c}^{T}x$ â€ƒsubject to $Axâ‰§b$, $xâ‰§0$,

(LD) â€ƒMaximize ${b}^{T}y$ â€ƒsubject to ${A}^{T}yâ‰¦c$, $yâ‰§0$,

where $câˆˆ{\mathbb{R}}^{n}$, $xâˆˆ{\mathbb{R}}^{n}$, $bâˆˆ{\mathbb{R}}^{m}$, $yâˆˆ{\mathbb{R}}^{m}$, $A=\left[{a}_{ij}\right]$ is an $mÃ—n$ real matrix.

Now consider the matrix game associated with the following $\left(n+m+1\right)Ã—\left(n+m+1\right)$ skew symmetric matrix B:

$B=\left[\begin{array}{ccc}0& {A}^{T}& âˆ’c\\ âˆ’A& 0& b\\ {c}^{T}& âˆ’{b}^{T}& 0\end{array}\right].$

Dantzig [1] gave the complete equivalence between the linear programming duality and the matrix game B. Many authors [2â€“5] have extended the equivalence results of Dantzig [1] to several kinds of scalar optimization problems. Very recently, Hong and Kim [6] defined a vector matrix game and generalized the equivalence results of Dantzig [1] to a vector optimization problem by using the vector matrix game.

Recently, Kim and Noh [4] established equivalent relations between a certain matrix game and symmetric dual problems. Symmetric duality in nonlinear programming, in which the dual of the dual is the primal, was first introduced by Dorn [7]. Dantzig, Eisenberg and Cottle [8] formulated a pair of symmetric dual nonlinear problems and established duality results for convex and concave functions with non-negative orthant as the cone. Mond and Weir [9] presented two pairs of symmetric dual vector optimization problems and obtained symmetric duality results concerning pseudoconvex and pseudoconcave functions.

In this paper, a vector matrix game with more than two skew symmetric matrices, which is an extension of the matrix game, is defined and a nonlinear vector optimization problem is considered. We formulate a symmetric dual problem for the nonlinear vector optimization problem and establish equivalent relations between the symmetric dual problem and the corresponding vector matrix game. Moreover, we give a numerical example for showing such equivalent relations.

## 2 Vector matrix game and existence theorem

Throughout this paper, we will denote the relative interior of ${S}_{p}$ by ${\stackrel{o}{S}}_{p}$, and we will use the following conventions for vectors in the Euclidean space ${\mathbb{R}}^{n}$ for vectors $x:=\left({x}_{1},â€¦,{x}_{n}\right)$ and $y:=\left({y}_{1},â€¦,{y}_{n}\right)$:

Consider the nonlinear programming problem (VOP):

where $X=\left\{xâˆˆ{\mathbb{R}}^{n}:g\left(x\right)â‰§b,xâ‰§0\right\}$, $f:{\mathbb{R}}^{n}â†’{\mathbb{R}}^{p}$, $g:{\mathbb{R}}^{n}â†’{\mathbb{R}}^{m}$ are continuously differentiable. The gradient $\mathrm{âˆ‡}f\left(x\right)$ is an $nÃ—p$ matrix, and $\mathrm{âˆ‡}g\left(x\right)$ is an $nÃ—m$ matrix.

Definition 2.1 [10]

A point $\stackrel{Â¯}{x}âˆˆX$ is said to be an efficient solution for (VOP) if there exists no other feasible point $xâˆˆX$ such that $\left({f}_{1}\left(x\right),â€¦,{f}_{p}\left(x\right)\right)â‰¤\left({f}_{1}\left(\stackrel{Â¯}{x}\right),â€¦,{f}_{p}\left(\stackrel{Â¯}{x}\right)\right)$.

Now, we define solutions for a vector matrix game as follows.

Definition 2.2 [6]

Let ${B}_{i}$, $i=1,â€¦,p$, be real $nÃ—n$ skew-symmetric matrices. A point $\stackrel{Â¯}{x}âˆˆ{S}_{n}$ is said to be a vector solution of the vector matrix game ${B}_{i}$, $i=1,â€¦,p$ if $\left({\stackrel{Â¯}{x}}^{T}{B}_{1}x,â€¦,{\stackrel{Â¯}{x}}^{T}{B}_{p}x\right)â‰°\left({\stackrel{Â¯}{x}}^{T}{B}_{1}\stackrel{Â¯}{x},â€¦,{\stackrel{Â¯}{x}}^{T}{B}_{p}\stackrel{Â¯}{x}\right)â‰°\left({x}^{T}{B}_{1}\stackrel{Â¯}{x},â€¦,{x}^{T}{B}_{p}\stackrel{Â¯}{x}\right)$ for any $xâˆˆ{S}_{n}$.

We proved the characterization of a vector solution of the vector matrix game in [6].

Lemma 2.1 [6]

Let ${B}_{i}$, $i=1,â€¦,p$, be an $nÃ—n$ skew symmetric matrix. Then $\stackrel{Â¯}{y}âˆˆ{S}_{n}$ is a vector solution of the vector matrix game ${B}_{i}$, $i=1,â€¦,p$, if and only if there exists $\mathrm{Î¾}âˆˆ{\stackrel{o}{S}}_{p}$ such that $\left({âˆ‘}_{i=1}^{p}{\mathrm{Î¾}}_{i}{B}_{i}\right)\stackrel{Â¯}{y}â‰¦0$.

Remark 2.1 Let ${B}_{i}$, $i=1,â€¦,p$, be an $nÃ—n$ skew symmetric matrix. From Lemma 2.1, we can obtain the following remark saying that the vector matrix game can be solved by fixed point problems; $\stackrel{Â¯}{y}âˆˆ{S}_{n}$ is a vector solution of the vector matrix game ${B}_{i}$, $i=1,â€¦,p$, if and only if there exists $\mathrm{Î¾}âˆˆ{\stackrel{o}{S}}_{p}$ such that $\stackrel{Â¯}{y}âˆˆ{F}_{\mathrm{Î¾}}\left(\stackrel{Â¯}{y}\right)$, where ${F}_{\mathrm{Î¾}}\left(x\right)=\left\{yâˆˆ{S}_{n}âˆ£yâˆˆxâˆ’\left({âˆ‘}_{i=1}^{p}{\mathrm{Î¾}}_{i}{B}_{i}\right)xâˆ’{\mathbb{R}}_{+}^{n}\right\}$.

Noticing Remark 2.1, we can obtain an existence theorem for the vector matrix game.

Theorem 2.1 Let ${B}_{i}$, $i=1,â€¦,p$, be an $nÃ—n$ skew symmetric matrix. Then there exists a vector solution of the vector matrix game ${B}_{i}$, $i=1,â€¦,p$.

Proof Let $\mathrm{Î¾}âˆˆ{\stackrel{o}{S}}_{p}$. Define a multifunction ${F}_{\mathrm{Î¾}}:{S}_{n}â†’{S}_{n}$ by, for any $xâˆˆ{S}_{n}$,

${F}_{\mathrm{Î¾}}\left(x\right)=\left\{yâˆˆ{S}_{n}|yâˆˆxâˆ’\left(\underset{i=1}{\overset{p}{âˆ‘}}{\mathrm{Î¾}}_{i}{B}_{i}\right)xâˆ’{\mathbb{R}}_{+}^{n}\right\}.$

Then the multifunction ${F}_{\mathrm{Î¾}}$ is closed and hence upper semi-continuous, and so it follows from the well-known Kakutani fixed point theorem [11] that the multifunction ${F}_{\mathrm{Î¾}}$ has a fixed point. So, by Remark 2.1, there exists a vector solution of the vector matrix game ${B}_{i}$, $i=1,â€¦,p$.â€ƒâ–¡

## 3 Equivalence relations

Now, we consider the nonlinear symmetric programming problem (SP) together with its dual (SD) as follows:

where $f:=\left({f}_{1},â€¦,{f}_{p}\right):{\mathbb{R}}^{n}Ã—{\mathbb{R}}^{m}â†’{\mathbb{R}}^{p}$ are continuously differentiable.

Consider the vector matrix game defined by the following $\left(n+m+1\right)Ã—\left(n+m+1\right)$ skew symmetric matrix ${B}_{i}\left(x,y\right)$, $i=1,â€¦,p$, related to (SP) and (SD):

${B}_{i}\left(x,y\right)=\left[\begin{array}{ccc}0& âˆ’x{\mathrm{âˆ‡}}_{y}{f}_{i}{\left(x,y\right)}^{T}& âˆ’{\mathrm{âˆ‡}}_{x}{f}_{i}\left(x,y\right)\\ {\mathrm{âˆ‡}}_{y}{f}_{i}\left(x,y\right){x}^{T}& 0& {\mathrm{âˆ‡}}_{y}{f}_{i}\left(x,y\right)\\ {\mathrm{âˆ‡}}_{x}{f}_{i}{\left(x,y\right)}^{T}& âˆ’{\mathrm{âˆ‡}}_{y}{f}_{i}{\left(x,y\right)}^{T}& 0\end{array}\right].$

Now, we give equivalent relations between (SD) and the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,â€¦,p$.

Theorem 3.1 Let $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ be feasible for (SP) and (SD), with ${\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)={\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)Ã—\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)=0$. Let ${z}^{âˆ—}=1/\left(1+{âˆ‘}_{i}{\stackrel{Â¯}{x}}_{i}+{âˆ‘}_{j}{\stackrel{Â¯}{y}}_{j}\right)$, ${x}^{âˆ—}={z}^{âˆ—}\stackrel{Â¯}{x}$ and ${y}^{âˆ—}={z}^{âˆ—}\stackrel{Â¯}{y}$. Then $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)$ is a vector solution of the vector matrix game ${B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$, $i=1,â€¦,p$.

Proof Let $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ be feasible for (SP) and (SD). Then the following holds:

(3.1)
(3.2)
(3.3)
(3.4)

Multiplying (3.3) by $\stackrel{Â¯}{x}â‰§0$ gives $âˆ’\stackrel{Â¯}{x}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{y}=0$ and from (3.2),

$âˆ’\stackrel{Â¯}{x}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{y}âˆ’{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0.$
(3.5)

Multiplying (3.1) by ${\stackrel{Â¯}{x}}^{T}\stackrel{Â¯}{x}â‰§0$, ${\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right){\stackrel{Â¯}{x}}^{T}\stackrel{Â¯}{x}â‰¦0$. It implies that since ${\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0$,

${\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right){\stackrel{Â¯}{x}}^{T}\stackrel{Â¯}{x}+{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0.$
(3.6)

From (3.3) we have

${\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{x}âˆ’{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{y}=0.$
(3.7)

But ${z}^{âˆ—}>0$ by (3.4), from (3.5), (3.6) and (3.7), we get

(3.8)
(3.9)
(3.10)

From (3.8), (3.9) and (3.10), we have the following inequality:

$\left(\underset{i=1}{\overset{p}{âˆ‘}}{\stackrel{Â¯}{\mathrm{Î¾}}}_{i}{B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)\right)\left(\begin{array}{c}{x}^{âˆ—}\\ {y}^{âˆ—}\\ {z}^{âˆ—}\end{array}\right)â‰¦0.$

By Lemma 2.1, $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)$ is a vector solution of the vector matrix game ${B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$, $i=1,â€¦,p$.â€ƒâ–¡

Theorem 3.2 Let $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)$ with ${z}^{âˆ—}>0$ be a vector solution of the vector matrix game ${B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$, $i=1,â€¦,p$, where $\stackrel{Â¯}{x}={x}^{âˆ—}/{z}^{âˆ—}$ and $\stackrel{Â¯}{y}={y}^{âˆ—}/{z}^{âˆ—}$. Then there exists $\stackrel{Â¯}{\mathrm{Î¾}}âˆˆ{\stackrel{o}{S}}_{p}$ such that $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ is feasible for (SP) and (SD), and ${\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)={\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)=0$. Moreover, if ${f}_{i}\left(â‹\dots ,y\right)$, $i=1,â€¦,p$, are convex for fixed y and ${f}_{i}\left(x,â‹\dots \right)$, $i=1,â€¦,p$, are concave for fixed x, then $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$ is efficient for (SP) with fixed $\stackrel{Â¯}{\mathrm{Î¾}}$ and $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$ is efficient for (SD) with fixed $\stackrel{Â¯}{\mathrm{Î¾}}$.

Proof Let $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)$ with ${z}^{âˆ—}>0$ be a vector solution of the vector matrix game ${B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$, $i=1,â€¦,p$. Then by Lemma 2.1, there exists $\stackrel{Â¯}{\mathrm{Î¾}}âˆˆ{\stackrel{o}{S}}_{p}$ such that

$\left(\underset{i=1}{\overset{p}{âˆ‘}}{\stackrel{Â¯}{\mathrm{Î¾}}}_{i}{B}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)\right)\left(\begin{array}{c}{x}^{âˆ—}\\ {y}^{âˆ—}\\ {z}^{âˆ—}\end{array}\right)â‰¦0.$

Thus, we get

(3.11)
(3.12)
(3.13)
(3.14)

Dividing (3.11), (3.12) and (3.13) by ${z}^{âˆ—}>0$, we have

(3.15)
(3.16)
(3.17)

From (3.14),

$\stackrel{Â¯}{x}â‰§0,\phantom{\rule{2em}{0ex}}\stackrel{Â¯}{y}â‰§0.$
(3.18)

By (3.16), ${\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)\left({\stackrel{Â¯}{x}}^{T}\stackrel{Â¯}{x}+1\right)â‰¦0$. It implies that since ${\stackrel{Â¯}{x}}^{T}\stackrel{Â¯}{x}+1>0$,

$âˆ’{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰§0.$
(3.19)

From (3.15), $âˆ’\stackrel{Â¯}{x}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{y}â‰¦{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$. Using (3.18) and (3.19), we obtain $0â‰¦âˆ’\stackrel{Â¯}{x}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right){\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)}^{T}\stackrel{Â¯}{y}â‰¦{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$. It implies that $âˆ’{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0$. From (3.17), ${\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦{\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$. But since $\stackrel{Â¯}{x}â‰§0$ and ${\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰§0$, ${\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰§0$ and since $\stackrel{Â¯}{y}â‰§0$ and ${\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0$, ${\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0$. Then we have

$0â‰¦{\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦{\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)â‰¦0.$

Hence, ${\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)={\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$. Thus, $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ is feasible for (SP) and (SD) with ${f}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)={f}_{i}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$, $i=1,â€¦,p$. Since $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ is feasible for (SD), by weak duality in [9], $\left({f}_{1}\left(x,y\right)âˆ’{y}^{T}{\mathrm{âˆ‡}}_{y}\left({\mathrm{Î¾}}^{T}f\right)\left(x,y\right),â€¦,{f}_{p}\left(x,y\right)âˆ’{y}^{T}{\mathrm{âˆ‡}}_{y}\left({\mathrm{Î¾}}^{T}f\right)\left(x,y\right)\right)â‰°\left({f}_{1}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right),â€¦,{f}_{p}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{y}}^{T}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)\right)$ and $\left({f}_{1}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right),â€¦,{f}_{p}\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)âˆ’{\stackrel{Â¯}{x}}^{T}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)\right)â‰°\left({f}_{1}\left(u,v\right)âˆ’{u}^{T}{\mathrm{âˆ‡}}_{u}\left({\mathrm{Î¾}}^{T}f\right)\left(u,v\right),â€¦,{f}_{p}\left(u,v\right)âˆ’{u}^{T}{\mathrm{âˆ‡}}_{u}\left({\mathrm{Î¾}}^{T}f\right)\left(u,v\right)\right)$ for any feasible $\left(u,v,\mathrm{Î¾}\right)$ of (SP) and (SD). Therefore, $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$ is efficient for (SP) with fixed $\stackrel{Â¯}{\mathrm{Î¾}}$ and $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)$ is efficient for (SD) with fixed $\stackrel{Â¯}{\mathrm{Î¾}}$.â€ƒâ–¡

Now, we give an example illustrating Theorems 3.1 and 3.2.

Example 3.1 Let ${f}_{1}\left(x,y\right)={x}^{2}âˆ’{y}^{2}$ and ${f}_{2}\left(x,y\right)=yâˆ’x$. Consider the following vector optimization problem (SP) together with its dual (SD) as follows:

Now, we determine the set of all vector solutions of the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,2$. Let

${B}_{i}\left(x,y\right)=\left(\begin{array}{ccc}0& âˆ’x{\mathrm{âˆ‡}}_{y}{f}_{i}{\left(x,y\right)}^{T}& âˆ’{\mathrm{âˆ‡}}_{x}{f}_{i}\left(x,y\right)\\ âˆ’{\mathrm{âˆ‡}}_{y}{f}_{i}\left(x,y\right){x}^{T}& 0& {\mathrm{âˆ‡}}_{y}{f}_{i}\left(x,y\right)\\ {\mathrm{âˆ‡}}_{x}{f}_{i}{\left(x,y\right)}^{T}& âˆ’{\mathrm{âˆ‡}}_{y}{f}_{i}{\left(x,y\right)}^{T}& 0\end{array}\right).$

Then

${B}_{1}\left(x,y\right)=\left(\begin{array}{ccc}0& 2xy& âˆ’2x\\ âˆ’2xy& 0& âˆ’2y\\ 2x& 2y& 0\end{array}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{B}_{2}\left(x,y\right)=\left(\begin{array}{ccc}0& âˆ’x& 1\\ x& 0& 1\\ âˆ’1& âˆ’1& 0\end{array}\right).$

Let $\left(x,y\right)âˆˆ{\mathbb{R}}^{2}$ and $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)âˆˆ{S}_{3}$ be a vector solution of the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,2$, if and only if there exist ${\mathrm{Î¾}}_{1}>0$, ${\mathrm{Î¾}}_{2}>0$, ${\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1$ such that

$\left({\mathrm{Î¾}}_{1}\left(\begin{array}{ccc}0& 2xy& âˆ’2x\\ âˆ’2xy& 0& âˆ’2y\\ 2x& 2y& 0\end{array}\right)+{\mathrm{Î¾}}_{2}\left(\begin{array}{ccc}0& âˆ’x& 1\\ x& 0& 1\\ âˆ’1& âˆ’1& 0\end{array}\right)\right)\left(\begin{array}{c}{x}^{âˆ—}\\ {y}^{âˆ—}\\ {z}^{âˆ—}\end{array}\right)â‰¦\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$

â‡” there exist ${\mathrm{Î¾}}_{1}>0$, ${\mathrm{Î¾}}_{2}>0$, ${\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1$ such that

$\left(\begin{array}{c}x\left(2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){y}^{âˆ—}âˆ’\left(2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){z}^{âˆ—}\\ âˆ’x\left(2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){x}^{âˆ—}âˆ’\left(2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){z}^{âˆ—}\\ \left(2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){x}^{âˆ—}+\left(2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}\right){y}^{âˆ—}\end{array}\right)â‰¦\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$

Thus, we determine the set of all the vector solutions of the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,2$.

1. (I)

the case that $x>0$:

2. (a)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(0,0,1\right)$.

3. (b)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(0,\mathrm{Î±},1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

4. (c)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(0,1,0\right)$.

5. (d)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\mathrm{Î±},0,1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

6. (e)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left({x}_{1},{x}_{2},{x}_{3}\right)âˆ£{x}_{1}â‰§0,{x}_{2}â‰§0,{x}_{3}â‰§0,{x}_{1}+{x}_{2}+{x}_{3}=1\right\}$.

7. (f)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(0,1,0\right)$.

8. (g)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(1,0,0\right)$.

9. (h)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

10. (i)

$2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$, $2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(0,1,0\right)$.

11. (II)

the case that $x=0$:

12. (a)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(1âˆ’\mathrm{Î±},\mathrm{Î±},0\right)âˆ£\mathrm{Î±}â‰¦\frac{{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}},y>0,{\mathrm{Î¾}}_{1}>0,{\mathrm{Î¾}}_{2}>0,{\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1\right\}$.

13. (b)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

14. (c)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

15. (III)

the case that $x<0$:

16. (a)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\frac{2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},âˆ’\frac{2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},âˆ’\frac{2xy{\mathrm{Î¾}}_{1}âˆ’x{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}}\right)$: $2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}>0,2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0,2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0,{\mathrm{Î¾}}_{1}>0,{\mathrm{Î¾}}_{2}>0,{\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1\right\}$.

17. (b)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}=0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right):\left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}$.

18. (c)

$2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0$: $\left({x}^{âˆ—},{y}^{âˆ—},{z}^{âˆ—}\right)=\left(1,0,0\right)$.

Let $\left(x,y\right)âˆˆ{\mathbb{R}}^{2}$ and ${S}_{\left(x,y\right)}$ be the set of vector solutions of the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,2$. From (I), (II) and (III),

$\begin{array}{rcl}\underset{\left(x,y\right)âˆˆ{\mathbb{R}}^{2}}{â‹ƒ}S\left(x,y\right)& =& \left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}âˆª\left\{\left(0,\mathrm{Î±},1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}\\ âˆª\left\{\left(\mathrm{Î±},0,1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}\\ âˆª\left\{\left(\mathrm{Î±},\mathrm{Î²},\mathrm{Î³}\right)âˆ£\mathrm{Î±}â‰§0,\mathrm{Î²}â‰§0,\mathrm{Î³}â‰§0,\mathrm{Î±}+\mathrm{Î²}+\mathrm{Î³}=1\right\}\\ âˆª\left\{\left(\frac{2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},âˆ’\frac{2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},\\ âˆ’\frac{2xy{\mathrm{Î¾}}_{1}âˆ’x{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}}\right)|x<0,2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}>0,\\ 2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0,\phantom{\rule{0.25em}{0ex}}2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0,{\mathrm{Î¾}}_{1}>0,{\mathrm{Î¾}}_{2}>0,{\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1\right\}.\end{array}$

Let $\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y},\stackrel{Â¯}{\mathrm{Î¾}}\right)$ be feasible for (SP) and (SD) with $\stackrel{Â¯}{y}{\mathrm{âˆ‡}}_{y}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)=\stackrel{Â¯}{x}{\mathrm{âˆ‡}}_{x}\left({\stackrel{Â¯}{\mathrm{Î¾}}}^{T}f\right)\left(\stackrel{Â¯}{x},\stackrel{Â¯}{y}\right)=0$. We can easily check that

Thus,

Therefore, Theorem 3.1 holds.

Let $\left(x,y\right)âˆˆ{\mathbb{R}}^{2}$ and ${S}_{\left(x,y\right)}$ be the set of vector solutions of the vector matrix game ${B}_{i}\left(x,y\right)$, $i=1,2$. Then

$\begin{array}{rcl}\underset{\left(x,y\right)âˆˆ{\mathbb{R}}^{2}}{â‹ƒ}{S}_{\left(x,y\right)}& =& \left\{\left(\mathrm{Î±},1âˆ’\mathrm{Î±},0\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}âˆª\left\{\left(0,\mathrm{Î±},1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}\\ âˆª\left\{\left(\mathrm{Î±},0,1âˆ’\mathrm{Î±}\right)âˆ£0â‰¦\mathrm{Î±}â‰¦1\right\}\\ âˆª\left\{\left(\mathrm{Î±},\mathrm{Î²},\mathrm{Î³}\right)âˆ£\mathrm{Î±}â‰§0,\mathrm{Î²}â‰§0,\mathrm{Î³}â‰§0,\mathrm{Î±}+\mathrm{Î²}+\mathrm{Î³}=1\right\}\\ âˆª\left\{\left(\frac{2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},âˆ’\frac{2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}},\\ âˆ’\frac{2xy{\mathrm{Î¾}}_{1}âˆ’x{\mathrm{Î¾}}_{2}}{2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}}\right)|x<0,2y{\mathrm{Î¾}}_{1}âˆ’2x{\mathrm{Î¾}}_{1}âˆ’2xy{\mathrm{Î¾}}_{1}+x{\mathrm{Î¾}}_{2}>0,\\ 2y{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}>0,2x{\mathrm{Î¾}}_{1}âˆ’{\mathrm{Î¾}}_{2}<0,{\mathrm{Î¾}}_{1}>0,{\mathrm{Î¾}}_{2}>0,{\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1\right\}.\end{array}$

So,

Let F be the set of all feasible solutions of (SP) and let G be the set of all feasible solutions of (SD). Then we can check that $\left\{\left(\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}},\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}},{\mathrm{Î¾}}_{1},{\mathrm{Î¾}}_{2}\right)âˆ£{\mathrm{Î¾}}_{1}>0,{\mathrm{Î¾}}_{2}>0,{\mathrm{Î¾}}_{1}+{\mathrm{Î¾}}_{2}=1\right\}âŠ‚Fâˆ©G$ and $\left(\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}}\right){\mathrm{âˆ‡}}_{y}\left({\mathrm{Î¾}}^{T}f\right)\left(\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}},\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}}\right)=\left(\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}}\right){\mathrm{âˆ‡}}_{x}\left({\mathrm{Î¾}}^{T}f\right)\left(\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}},\frac{{\mathrm{Î¾}}_{2}}{2{\mathrm{Î¾}}_{1}}\right)=0$. Therefore, Theorem 3.2 holds.

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## Acknowledgements

The authors would like to thank the referees for giving valuable comments for the revision of the paper.

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Correspondence to Gue Myung Lee.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

The authors, together discussed and solved the problems in the manuscript. All authors read and approved the final manuscript.

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Hong, J.M., Kim, M.H. & Lee, G.M. On vector matrix game and symmetric dual vector optimization problem. Fixed Point Theory Appl 2012, 233 (2012). https://doi.org/10.1186/1687-1812-2012-233